happpyyyyy new years !!!!!!!
unfortunatly schools about to start up again but oh well the holidays were nice while they lasted so here we go again...
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve.
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve.
limits:
Rule #1 - When the degree (exponent) of the bottom is GREATER than the degree of the top, the limit is Zero.
Rule #2 - When the degree (exponent) of the bottom is SMALLER than the degree of the top, the limit is infinity. (positive or negative)
Rule #3 - When the degrees are equal, the limit is the coeffecients.
linierazation:
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Second Derivative:
first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
once you have everything plugged in and ready to go you then solve for d^2y/d^2x
Intermediate Value Theorem:
1. if f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
* basically you cannot skip any y value
HOW TO FIND THE EQUATION OF A TANGENT LINE:
1. take f1(x)
2. plug x in to find your slope m
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
Thursday, December 31, 2009
Post #20
Well hello everyone! I hope yall Christmas went well and wish yall a Happy New Year starting tomorrow! Don't go out and party too much tonight .. have fun and stay safe, I'll see yall Monday.
So since we're doing blogs with nothing new to restate I'm gonna go over the first things we learned in Calculus this year.
Derivatives - There were thirty-six formulas that we copied down in order to help us. These formulas help a lot once you understand the concept. For instance, if you want to take the derivative of ANY number, it will ALWAYS be zero. However, anything with an exponet will have a derivative by multipling the exponet with the number in front of the variable. Once you multiply that, your answer will take the place of the number in front of the variable. But, you're not done yet! Take the exponet and subtract one. Your answer will take the place of the exponet. Then you solve. Remember, though, that if you have a negative exponet you have to put it at the bottom of a fraction in order for it to become positive. [You can't have a negitive exponet!]
Product Rule - For the product rule, you must have multiplication. You will keep the first part of the problem and multiply it by the derivative of the second part, and then add the second part of the formula, which is to keep the second part of the problem and multiply it by the derivative of the first. You'll solve and that's your answer.
Quotient Rule - For the quotient rule you have a fraction. You take the bottom and then multiply it by the derivative of the top then you'll sbtract the top multiplied by the derivative of the bottom. This will all be over the bottom squared. Then you solve, and that's your answer.
Average Velocity - Average velocity is when you take the derivative of your problem and plug in the two numbers from the problem. Then you'll take the answer of t1 and t2 in order to subtract t1 from t2. After, you will divide it by the numbers they gave you for t1 and t2, so you'll do...t2 minus t1 from the problem not the answer you get when you plug them into the problem then solve for your answer.
Instantaneous Velocity - Instantaneous velocity is when you take the derivative of the problem and put in the one number they give you. What you get is your answer!
DON'T EVER FORGET THE TRIG CHART EITHER!
Well yall stay safe and have a good time for the rest of our break. See yall soon!
~ElliE~
So since we're doing blogs with nothing new to restate I'm gonna go over the first things we learned in Calculus this year.
Derivatives - There were thirty-six formulas that we copied down in order to help us. These formulas help a lot once you understand the concept. For instance, if you want to take the derivative of ANY number, it will ALWAYS be zero. However, anything with an exponet will have a derivative by multipling the exponet with the number in front of the variable. Once you multiply that, your answer will take the place of the number in front of the variable. But, you're not done yet! Take the exponet and subtract one. Your answer will take the place of the exponet. Then you solve. Remember, though, that if you have a negative exponet you have to put it at the bottom of a fraction in order for it to become positive. [You can't have a negitive exponet!]
Product Rule - For the product rule, you must have multiplication. You will keep the first part of the problem and multiply it by the derivative of the second part, and then add the second part of the formula, which is to keep the second part of the problem and multiply it by the derivative of the first. You'll solve and that's your answer.
Quotient Rule - For the quotient rule you have a fraction. You take the bottom and then multiply it by the derivative of the top then you'll sbtract the top multiplied by the derivative of the bottom. This will all be over the bottom squared. Then you solve, and that's your answer.
Average Velocity - Average velocity is when you take the derivative of your problem and plug in the two numbers from the problem. Then you'll take the answer of t1 and t2 in order to subtract t1 from t2. After, you will divide it by the numbers they gave you for t1 and t2, so you'll do...t2 minus t1 from the problem not the answer you get when you plug them into the problem then solve for your answer.
Instantaneous Velocity - Instantaneous velocity is when you take the derivative of the problem and put in the one number they give you. What you get is your answer!
DON'T EVER FORGET THE TRIG CHART EITHER!
Well yall stay safe and have a good time for the rest of our break. See yall soon!
~ElliE~
Wednesday, December 30, 2009
First post of holidays
Well I am finally getting to these because I have finally been home where I have internet and have time since I have been home. One thing I understand is how to take derivatives of some different things. The reason is that the formulas on the derivatives are not hard to understand. One of the formulas is the product rule. This rule is one of the easiest to me. It is because you just follow the formula and its easy algebra. All you do is write the first take the derivative of the second and add that to where you write the second and take the derivative of the first. Also the derivatives any number. This is easy because the derivative of any number is equal to 0. Another thing is to take a derivative of anything with an x raised to a number. All that has to be done is multiply the coefficient by the exponent and then subtract the exponent by one. An example of this is 3x^3+4x^2-5x-5. The derivative is 9x^2+8x-5.Also I understand how to do the average speed and instantaeous speed even though I did not really understand this at all at the beginning. The reason I started to understand this is because i realize that average has to do with over a time period and slope and instantaneous is has to do with now. This helps me do these problems. Also I am very comfortable with tangent lines. All you do is:
1. Take the derivative of the equation
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into point slope form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
1. Take the derivative of the equation
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into point slope form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
Help for second semester
Several files were posted on edline to help with problematic areas that you guys have mentioned on the blog. Please visit edline to take advantage of these files.
Tuesday, December 29, 2009
second post of the three we needed.
ok i just decided to do this because i forgot earlier in the week. and i went to the wayne concert and it was AWESOMEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE. ok on a serious note. MATH. its been such a good couple of days w/out it so here we go.
LIMIT RULES:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree.
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
ok i think thats good enough for this one. remember people. lil wayne is awesomeeeee :) and i kinda dont know related rates and angles of elevation. lol
LIMIT RULES:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree.
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
ok i think thats good enough for this one. remember people. lil wayne is awesomeeeee :) and i kinda dont know related rates and angles of elevation. lol
Monday, December 28, 2009
Post 19
Ok, so looking back at my notes from the beginning of the year, I realized how simple limits and derivitaves and limits were/are. When we first covered them, they seemed to be so hard, but looking back on it now I really understand them.
Ok, first of all, limits. Limits are used to find where an x value is going on a graph. There are two different kinds of limits, limits that appraoch infinity or negative infinity or limits that approach a number. If you are solving a limit approaching infinity, you do these things:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degrees
Limits approaching numbers can be solved/found in a few different ways. First, you can plug the number x is going to into the x values of the limit and see what it comes out to be. Sometimes you can do this and come out with an actual number, but other times, the bottom comes out to zero. Many people think this means the limit is undefined, but this is not always correct. If the limit comes out to be zero, you have to use other methods to solve it. Another easy way to solve a limit is to try to break up the limit and cancle what you can, then plug in your x value. If this does not work, then you would have to plug in the limit into your calculator. If you are working with the definition of a derivative, all you have to do is take the derivative of the last term, and that is your limit
For derivatives, I will give an example
x^4 + 3x^3 + 6x
Ok for derivatives, you multiply the coefficient in front of your variable by it's exponent, then subtract 1 from your exponent.
4x^3 + 9x^2 + 6
There is also the first derivative test:
1. take first derivative
2. set equal to zero
3. solve for critical values
4. set up intervals
5. test intervals
The first derivative test can be used to find critical values on a graph, makimums, and minimums.
There is also a second derivative test. It has the same steps as the first derivative test except you take two derivatives instead of one. You can use the second derivative test to find points of inflection on a graph and where the graph concaves up or down.
I'm still a little shaky on linerization and I still absolutly do not understand angle of elevation
Ok, first of all, limits. Limits are used to find where an x value is going on a graph. There are two different kinds of limits, limits that appraoch infinity or negative infinity or limits that approach a number. If you are solving a limit approaching infinity, you do these things:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degrees
Limits approaching numbers can be solved/found in a few different ways. First, you can plug the number x is going to into the x values of the limit and see what it comes out to be. Sometimes you can do this and come out with an actual number, but other times, the bottom comes out to zero. Many people think this means the limit is undefined, but this is not always correct. If the limit comes out to be zero, you have to use other methods to solve it. Another easy way to solve a limit is to try to break up the limit and cancle what you can, then plug in your x value. If this does not work, then you would have to plug in the limit into your calculator. If you are working with the definition of a derivative, all you have to do is take the derivative of the last term, and that is your limit
For derivatives, I will give an example
x^4 + 3x^3 + 6x
Ok for derivatives, you multiply the coefficient in front of your variable by it's exponent, then subtract 1 from your exponent.
4x^3 + 9x^2 + 6
There is also the first derivative test:
1. take first derivative
2. set equal to zero
3. solve for critical values
4. set up intervals
5. test intervals
The first derivative test can be used to find critical values on a graph, makimums, and minimums.
There is also a second derivative test. It has the same steps as the first derivative test except you take two derivatives instead of one. You can use the second derivative test to find points of inflection on a graph and where the graph concaves up or down.
I'm still a little shaky on linerization and I still absolutly do not understand angle of elevation
Sunday, December 27, 2009
Post #I completely forgot the number
I almost forgot about the blog again! The holidays are throwing me off. I never know what day it is haha. Alright well here we go again..
MAX or MIN
First we need the steps:
1. First Derivative Test
2. Plug critical values into ORIGINAL function to get Y-values
3. Plug endpoints into ORIGINAL function to get Y-values
4. Highest Y-value is absolute max and lowest Y-value is absolute min
Next we can do an example problem:
f(x) = 2(x)^4 - 4(x)^3 on [0,5]
8(x)^3 - 8(x)^2
8x^2[(x)^2-1]
Set equal to zero to get x = 0,-1,1
Using 0, -1, 1 and the point in the problem, plug ALL values in for x.
smallest is the min and largest is the max
MIN [1,-2] and MAX [5,750]
*Remember absolute max and mins are written as a point or just the y-value.
I also want to point out that taking the tangent line is so easy. I don't know why I always forget it. It is simply take derivative, plug in x-value into derivative, and then put into point slope form. Simple? Yeah I know..now.
Now I'm putting the steps for optimization because I STILL don't understand it! I just don't know what my problem is with optimization..I can know the steps or formulas but still fail miserably at the problem.
Steps for optimization:
1.Identify primary and secondary equations. The primary will be the one you are maximizing or minimizing, and the secondary will be the other one.
2.Solve secondary equation for one variable, and plug into the primary. (if the primary only has one variable, this step is not necessary)
3.Take the derivative of the primary, and set it equal to zero; solve for x.
4.Plug into secondary equation to find the other value, check end points if necessary.
Any advice for this? Thank youuuuu :)
MAX or MIN
First we need the steps:
1. First Derivative Test
2. Plug critical values into ORIGINAL function to get Y-values
3. Plug endpoints into ORIGINAL function to get Y-values
4. Highest Y-value is absolute max and lowest Y-value is absolute min
Next we can do an example problem:
f(x) = 2(x)^4 - 4(x)^3 on [0,5]
8(x)^3 - 8(x)^2
8x^2[(x)^2-1]
Set equal to zero to get x = 0,-1,1
Using 0, -1, 1 and the point in the problem, plug ALL values in for x.
smallest is the min and largest is the max
MIN [1,-2] and MAX [5,750]
*Remember absolute max and mins are written as a point or just the y-value.
I also want to point out that taking the tangent line is so easy. I don't know why I always forget it. It is simply take derivative, plug in x-value into derivative, and then put into point slope form. Simple? Yeah I know..now.
Now I'm putting the steps for optimization because I STILL don't understand it! I just don't know what my problem is with optimization..I can know the steps or formulas but still fail miserably at the problem.
Steps for optimization:
1.Identify primary and secondary equations. The primary will be the one you are maximizing or minimizing, and the secondary will be the other one.
2.Solve secondary equation for one variable, and plug into the primary. (if the primary only has one variable, this step is not necessary)
3.Take the derivative of the primary, and set it equal to zero; solve for x.
4.Plug into secondary equation to find the other value, check end points if necessary.
Any advice for this? Thank youuuuu :)
2nd Post for vacation
Here is my 2nd post for the holidays and im loving the holidays but only have a week left so im a little bummed.
1st: Related RAtes
1. Identify all variables and equations2. identify what you are looking for3. make a sketch and label4.write an equation involving your variables5. take the derivative with respect to time6. substitute in derivative and solveEx: The variable x and y are differentiable functions of t and are related by the equation
y=2x^3-x+4 when x=2 (dy/dt)=-1
Find (dy/dt) when x=2
(dy/dt)=6x^2
(dy/dt)-(dy/dt)=6(2)^2(-1)-(-1)
(dy/dt)=-23
Another problem i have and still have is Angle of elevation i don't know where my problem is i think its just starting the problem thats really hard for me so if someone can help with that thanks.
1st: Related RAtes
1. Identify all variables and equations2. identify what you are looking for3. make a sketch and label4.write an equation involving your variables5. take the derivative with respect to time6. substitute in derivative and solveEx: The variable x and y are differentiable functions of t and are related by the equation
y=2x^3-x+4 when x=2 (dy/dt)=-1
Find (dy/dt) when x=2
(dy/dt)=6x^2
(dy/dt)-(dy/dt)=6(2)^2(-1)-(-1)
(dy/dt)=-23
Another problem i have and still have is Angle of elevation i don't know where my problem is i think its just starting the problem thats really hard for me so if someone can help with that thanks.
Post #19
Happy late Christmas =)
Indefinite Integration:
- All the same properties of derivatives apply
Polynomials: instead of subtracting one from the exponent, add one. Take the reciprocal of the exponent and put it in front of the variable instead of bringing exponent to front.
Don't forget +c.
Examples: integrate x^3 dx
3+1 = 4 so 1/4 x ^4 + c
You can check your answers by taking the derivative of the function
4(1/4) = 1 4-1=3 = x^3
2x^2 + 6x + 5 integrated is:
2/3 x^3 + 3x^2 + 5x + c
Other Functions:
-Work backwards from taking a derivative
integrate sin x dx:
-cos x + c
Integrate sec^2 dx:
tan x + c
Integrate 2 csc x cot x dx:
-2 csc x + c
Definite Integrals:
b S a f(x) dx = f(b) - f(a)
Definite integrals will always equal a number.
Examples:
3 S 0 x^2 dx
First step is to integrate the function: 1/3 x ^3 on [0,3]
Then plug into the formula: 1/3(3)^3 - [1/3(0)^3] = 9
2 S 0 2x^2- 3x + 2 dx
Integrate: 2/3 x^3 - 3/2 x^2 + 2x on [0,2]
Plug in: 2/3(2)^3 - 3/2(2)^2 + 2(2) - [2/3(0)^3 - 3/2(0)^2 + 2(0)] = 10/3
Average value is similar to definite integrals except the formula is 1/b-a b S a f(a)
Example: Find the average value of f(x) = x^2 on [0,5]
5 S 0 x^2 dx
1/5-0 5 S 0 x^2 dx
Integrate: 1/5 [1/3 x^3] on [0,5]
Plug in: 1/5 [1/3 (5)^3 - 1/3 (0) ^3]
1/5 [ 125/3 - 0 ] = 125/15 = 25/3
19th post
so here we go..............
related rates:
Steps:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation(s) involving your variables (only have 1 unknown)
5. Take the derivative with respect to TIME!
6. Substitute in the Derivative and solve
limits:
Rule #1 - When the degree (exponent) of the bottom is GREATER than the degree of the top, the limit is Zero.
Rule #2 - When the degree (exponent) of the bottom is SMALLER than the degree of the top, the limit is infinity. (positive or negative)
Rule #3 - When the degrees are equal, the limit is the coeffecients.
linierazation:
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Second Derivative:
first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
once you have everything plugged in and ready to go you then solve for d^2y/d^2x
Intermediate Value Theorem:
1. if f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
* basically you cannot skip any y value
HOW TO FIND THE EQUATION OF A TANGENT LINE:
1. take f1(x)
2. plug x in to find your slope m
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
related rates:
Steps:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation(s) involving your variables (only have 1 unknown)
5. Take the derivative with respect to TIME!
6. Substitute in the Derivative and solve
limits:
Rule #1 - When the degree (exponent) of the bottom is GREATER than the degree of the top, the limit is Zero.
Rule #2 - When the degree (exponent) of the bottom is SMALLER than the degree of the top, the limit is infinity. (positive or negative)
Rule #3 - When the degrees are equal, the limit is the coeffecients.
linierazation:
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Second Derivative:
first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
once you have everything plugged in and ready to go you then solve for d^2y/d^2x
Intermediate Value Theorem:
1. if f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
* basically you cannot skip any y value
HOW TO FIND THE EQUATION OF A TANGENT LINE:
1. take f1(x)
2. plug x in to find your slope m
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
First Holiday Post
Hope everyone had a great Christmas and has a excellent New Year.
So we started off the year going over pre-calculus from last year and learning the basics of derivatives.
So, taking limits:
If the equation is an infinite limit, and it is a fraction, then you use the degree to find the limit.
- If the largets degrees of the top and bottom are equal to each other, then use the the coefficients of the largest degrees in fraction form... simplify if necessary.
- If the degree of the top is larger than the degree of the bottom, then the limit is positive or negative infinity, depending on your coefficients.
- If the degree of the top is smaller than the degree of the bottom, then the limit is 0.
And when I was stressing when we first were learning derivatives, now they are almost natural to me...
So we started off the year going over pre-calculus from last year and learning the basics of derivatives.
So, taking limits:
If the equation is an infinite limit, and it is a fraction, then you use the degree to find the limit.
- If the largets degrees of the top and bottom are equal to each other, then use the the coefficients of the largest degrees in fraction form... simplify if necessary.
- If the degree of the top is larger than the degree of the bottom, then the limit is positive or negative infinity, depending on your coefficients.
- If the degree of the top is smaller than the degree of the bottom, then the limit is 0.
And when I was stressing when we first were learning derivatives, now they are almost natural to me...
Example of taking derivatives:
3x^2 + 5x + 212309483290781075843157834097175173847534
You take the derivates using the exponents and get:
(3)(2)x + 5
= 6x + 5
Example 2:
2/x^2
you use the quotient rule { [copybottom*deriv.top - copytop*deriv.bottom] / bottom^2}
[(x^2)(0) - (2)(2x)] / (x^2)^2
Simplify:
= -4x / x^4
Simplify:
-4/x^3
Example 3:
tan(x/3)
Can also be written as: tan(1/3*x)
take derivative: sec^2(1/3*x) * (1/3)
Simplify: [sec^2(x/3)]/[3]
Example 4:
5x^8 + 9x^7 - x^6 + 78x^5 + 8x^4 - 10x^3 + 56x^2 - 23x + 9*23*1993
Pretty much felt like busting my calculator out :)...
Anyway:
40x^7 + 63x^6 - 6x^5 + 390x^4 + 32x^3 - 30x^2 + 112x - 23
Well I guess that's it for this post and once again:
Merry Christmas and Happy New Year!!!
Post Number Nineteen
Hope everyone had a good Christmas!
Well i'm just gonna go through different things in this blog. I may be able to tell you this stuff but trust me i remember nothing :(
well i guess i'm in for an awakening soon.
here it goessssssss
Tangent Lines:
First take the derivative of f(x)
Then plug in your x to find the slope
Plug x into the original to get your y
After finding your y and your slope, plug into point slope formula:
y – y1 = m(x – x1)
Integration:
There are two types of integration, definite and indefinite.
In definite integration the answer is always a number.
It uses the integral [a, b].
The formula is bSa f(x) dx = f(b) – f(a) = #
Indefinite integration gives you an equation.
All derivative rules apply S x^n dx = x^(n+1)/(n+1) + C
Riemann Sums:
LRAM RRAM MRAM and TRAM are used for approximations
Uses the interval [a, b], deltax = (b – a)/n
LRAM:
Estimated from the left side, rectangles are drawn from the x-axis and then up to the graph, they then go over.
Deltax [f(a) + f(a + deltax) + … + f(b – deltax)]
RRAM:
Estimated from the right side, rectangles are drawn from the end of the interval of the graph itself.
Deltax [f(a + deltax) + f(b)]
MRAM:
Midpoints of LRAM and RRAM
deltax [f(midpoint1) + f(midpoint2) + f(midpoint n + 1)]
TRAM:
Most accurate of all Riemann sums
deltax/2 [f(a) + 2f(a + deltax) + 2f(a + 2deltax) + … + f(b)]
Well i'm just gonna go through different things in this blog. I may be able to tell you this stuff but trust me i remember nothing :(
well i guess i'm in for an awakening soon.
here it goessssssss
Tangent Lines:
First take the derivative of f(x)
Then plug in your x to find the slope
Plug x into the original to get your y
After finding your y and your slope, plug into point slope formula:
y – y1 = m(x – x1)
Integration:
There are two types of integration, definite and indefinite.
In definite integration the answer is always a number.
It uses the integral [a, b].
The formula is bSa f(x) dx = f(b) – f(a) = #
Indefinite integration gives you an equation.
All derivative rules apply S x^n dx = x^(n+1)/(n+1) + C
Riemann Sums:
LRAM RRAM MRAM and TRAM are used for approximations
Uses the interval [a, b], deltax = (b – a)/n
LRAM:
Estimated from the left side, rectangles are drawn from the x-axis and then up to the graph, they then go over.
Deltax [f(a) + f(a + deltax) + … + f(b – deltax)]
RRAM:
Estimated from the right side, rectangles are drawn from the end of the interval of the graph itself.
Deltax [f(a + deltax) + f(b)]
MRAM:
Midpoints of LRAM and RRAM
deltax [f(midpoint1) + f(midpoint2) + f(midpoint n + 1)]
TRAM:
Most accurate of all Riemann sums
deltax/2 [f(a) + 2f(a + deltax) + 2f(a + 2deltax) + … + f(b)]
Ash's 19th Post
So, first off, I almost forgot about this!
YAY FACEBOOK! (when I see some people's names, I automatically think "Calculus"..don't ask)
Anywho, Christmas was fun, New Years is going to be great! On with the math!!
My brother needs help in Pre-Algebra and he asked me. Okay, no big deal, Pre-Alg, righT? I saw his problems and tried to take the derivative. Then I tried to use some of the formulas I learned. That's when it hit me....IT'S PRE-ALGEBRA! -.- Apparently, I fail completely at not using Calculus now.... -.-
So, because I know a lot of us have problems with algebra (and I forgot my Calc notebook in my locker), how about I go over some algebra rules?
When multiplying, JUST MULTIPLY! NO PRODUCT RULE! (=])
2x(3)+4 does NOT equal 2(3)+0+4
When dividing, JUST DIVIDE! NO QUOTIENT RULE! (=])
2x/3 does NOT equal (2)(3)-(0)
Now that that's cleared up, let's move on.
What to explain, what to explain...or to clarify? Hmmm
Simplifying!
Lots of us mess up while simplifying an insanely beast problem, right? RIGHT!
So, here are some tips that I've decided to share.
1. Is the problem REALLY supposed to be that beast? I mean seriously, yes, lots are huge, but if it takes up 3 lines in your notebook for one equation, chances are that you mess up somewhere. Always take it slow and double check as you go. (hehe, the rhymes)
2. Once all actual solving is done, (now this is just what I do) find the largest exponential term(s) and underline them. Now, simplify those and write it on your next line, or wherever you write the final. Keep going with this process until all terms are done. Double check.
Now, those are my methods, but I'm not sure if it'll work.
Now, for my questions.
How do you work on the AP exam? Okay, so there's a specific way to work more efficiently on the ACT and whatnot, right? So, is there a method that anyone uses that they'd like to share with me? It'd be greatly appreciated!!
Happy New Year to all!!
YAY FACEBOOK! (when I see some people's names, I automatically think "Calculus"..don't ask)
Anywho, Christmas was fun, New Years is going to be great! On with the math!!
My brother needs help in Pre-Algebra and he asked me. Okay, no big deal, Pre-Alg, righT? I saw his problems and tried to take the derivative. Then I tried to use some of the formulas I learned. That's when it hit me....IT'S PRE-ALGEBRA! -.- Apparently, I fail completely at not using Calculus now.... -.-
So, because I know a lot of us have problems with algebra (and I forgot my Calc notebook in my locker), how about I go over some algebra rules?
When multiplying, JUST MULTIPLY! NO PRODUCT RULE! (=])
2x(3)+4 does NOT equal 2(3)+0+4
When dividing, JUST DIVIDE! NO QUOTIENT RULE! (=])
2x/3 does NOT equal (2)(3)-(0)
Now that that's cleared up, let's move on.
What to explain, what to explain...or to clarify? Hmmm
Simplifying!
Lots of us mess up while simplifying an insanely beast problem, right? RIGHT!
So, here are some tips that I've decided to share.
1. Is the problem REALLY supposed to be that beast? I mean seriously, yes, lots are huge, but if it takes up 3 lines in your notebook for one equation, chances are that you mess up somewhere. Always take it slow and double check as you go. (hehe, the rhymes)
2. Once all actual solving is done, (now this is just what I do) find the largest exponential term(s) and underline them. Now, simplify those and write it on your next line, or wherever you write the final. Keep going with this process until all terms are done. Double check.
Now, those are my methods, but I'm not sure if it'll work.
Now, for my questions.
How do you work on the AP exam? Okay, so there's a specific way to work more efficiently on the ACT and whatnot, right? So, is there a method that anyone uses that they'd like to share with me? It'd be greatly appreciated!!
Happy New Year to all!!
post 19
ok so christmas was good :) can't wait for new years!
three formulas for trig inverse functions:
1) 1/du arcsin u/a + C
2) 1/dua arctan u/a + C
3) 1/dua arcsec absolute value of u/a + C
Now on to volume by disks. First off, what is a disk? It's a solid object!The formula is
pi S[r(x)]^2dx
First you have to sketch the graph. Just plug into your y equals in your calculator and sketch! Depending on what axis it tells you to do the problem on is how you reflect the graph. Once you sketch it, shade the graph and then start working the problem!
Ex: Find the volume of the solid formed by revolving the region bounded by the graph of f(x) = square root of x and on y=0 x=3 about the x-axis.
First you would sketch the graph, which looks like a curved line coming out of (0,0) heading to the right to x=3. You then reflect that about the x axis and shade, which alternatively gives you a shape that looks like half of a football. Now to start the problem. Pi S(0 - 3) (square root of x)^2 dxPi S(0 - 3) x dxPi[1/2x^2] 0 to 3=9pi/2
Volume by washers is the same as volume by disks except it has a hole in it. That means it is bounded by two graphs. The formula for volume by washers is
Pi S top^2 - bottom^2 dx.
also area for both disks and washers is the same as their volume formula except everything that usually is squared, isn't squared.
what i don't understand is lram, rram, mram, and tram and how to use them. it just doesn't click in my head for some reason.
everyone be safe on new years :)
three formulas for trig inverse functions:
1) 1/du arcsin u/a + C
2) 1/dua arctan u/a + C
3) 1/dua arcsec absolute value of u/a + C
Now on to volume by disks. First off, what is a disk? It's a solid object!The formula is
pi S[r(x)]^2dx
First you have to sketch the graph. Just plug into your y equals in your calculator and sketch! Depending on what axis it tells you to do the problem on is how you reflect the graph. Once you sketch it, shade the graph and then start working the problem!
Ex: Find the volume of the solid formed by revolving the region bounded by the graph of f(x) = square root of x and on y=0 x=3 about the x-axis.
First you would sketch the graph, which looks like a curved line coming out of (0,0) heading to the right to x=3. You then reflect that about the x axis and shade, which alternatively gives you a shape that looks like half of a football. Now to start the problem. Pi S(0 - 3) (square root of x)^2 dxPi S(0 - 3) x dxPi[1/2x^2] 0 to 3=9pi/2
Volume by washers is the same as volume by disks except it has a hole in it. That means it is bounded by two graphs. The formula for volume by washers is
Pi S top^2 - bottom^2 dx.
also area for both disks and washers is the same as their volume formula except everything that usually is squared, isn't squared.
what i don't understand is lram, rram, mram, and tram and how to use them. it just doesn't click in my head for some reason.
everyone be safe on new years :)
19th post
Happy Holidays to everyone and i hope everyone had a great Christmas. Let's start off by reviewing first derivative test.
You take the derivative of the function and then you solve for x. Your x values will be considered your critical values. You then set these values into intervals between infinity and negative infinity. You then plug in numbers into the derivative to see if the function is increasing or decreasing on that interval. If the number is positive it is increasing, if the number is negative it is decreasing. These numbers are also considered max's and min's. To find the absolute max and mins, you take the point given to you and plug it into the orginal function to create another point. Then you take the derivative and find your critical values and plug those in as well. Whichever point is the biggest is your absolute max and whichever point is the lowest is your absolute min.
For the second derivative test, you take the derivative of the original function and take the derivative again. You solve for your critical values again like in the first derivative test, and you set them up into intervals again. This time when plugging in numbers between the intervals, if the number is positive it is concave up, if the number is negative it is concave down. There is a point of inflection wherever there is a change in concavity in the function.
Lets also go over the rules of related rates:
First define all of the given in the problem.
Second, figure out which formula they are asking you to use
Plug in all of your given.
Take the derivative and solve for the unknown.
I hope everyone has a Happy New Year!!
Bye for now :)
You take the derivative of the function and then you solve for x. Your x values will be considered your critical values. You then set these values into intervals between infinity and negative infinity. You then plug in numbers into the derivative to see if the function is increasing or decreasing on that interval. If the number is positive it is increasing, if the number is negative it is decreasing. These numbers are also considered max's and min's. To find the absolute max and mins, you take the point given to you and plug it into the orginal function to create another point. Then you take the derivative and find your critical values and plug those in as well. Whichever point is the biggest is your absolute max and whichever point is the lowest is your absolute min.
For the second derivative test, you take the derivative of the original function and take the derivative again. You solve for your critical values again like in the first derivative test, and you set them up into intervals again. This time when plugging in numbers between the intervals, if the number is positive it is concave up, if the number is negative it is concave down. There is a point of inflection wherever there is a change in concavity in the function.
Lets also go over the rules of related rates:
First define all of the given in the problem.
Second, figure out which formula they are asking you to use
Plug in all of your given.
Take the derivative and solve for the unknown.
I hope everyone has a Happy New Year!!
Bye for now :)
19th post
Okay, so christmas was great because we did not have school and because i got an i phone, but now it is time to do this blog, and then i could get on with my break. so i am going to do my blog on the first derivative test. the first derivative test is used to find absolute maximums and minimums, to do these, first yhou take the derivative of the equation given to you, then you set it equal to zero and solve for x, then you set up intervals using the x values, then you plug in numbers within the intervals into the derivative to find where your relative maximums and minimums are, then you plug the x values into the original equation to find the absolute maximums and minimums.
So here is an example problam i had:
Find the absolute maximums and minimums of the function f(x)=x^2-8x+4
First, you take the derivative of it:
f'(x)= 2x-8
Then you set it equal to zero and solve for x.
2x-8=0 2x=8 x=4
Then you set up the intervals
(-infinity, 4) U (4, infinity)
then you plug in a number found within the interval into the first derivative to find relative max's and min's.
(-infinity, 4)= negative number
(4, infinity)= positive number
so at x=4 there is a minimum
after that, you plug in that x value into the original equation to find the absolute max's and min's. (in this case it is only a min)
x=4 is the absolute minimum (4,-12)
I still don't know how to do MRAM, i know you have to use LRAM and RRAM to do it, i just don't know how it all comes together, but w/e. happy holidays!
So here is an example problam i had:
Find the absolute maximums and minimums of the function f(x)=x^2-8x+4
First, you take the derivative of it:
f'(x)= 2x-8
Then you set it equal to zero and solve for x.
2x-8=0 2x=8 x=4
Then you set up the intervals
(-infinity, 4) U (4, infinity)
then you plug in a number found within the interval into the first derivative to find relative max's and min's.
(-infinity, 4)= negative number
(4, infinity)= positive number
so at x=4 there is a minimum
after that, you plug in that x value into the original equation to find the absolute max's and min's. (in this case it is only a min)
x=4 is the absolute minimum (4,-12)
I still don't know how to do MRAM, i know you have to use LRAM and RRAM to do it, i just don't know how it all comes together, but w/e. happy holidays!
Saturday, December 26, 2009
Post #19
Average Speed
First of all, remember that a slope is the y value, or dy, of a derivative.
Example:
A ball is flung from a little child. It's path is projected as y=4.9t2m in "t" seconds. What is the average speed of the ball from 0 to 3 seconds?
1. Set up equations and intervals: (f(b)-f(a))/(b-a) 4.9t^2 [0,3]
2. Plug in a and b values for t: f(b)=4.9(3)2=44.1 f(a)=4.9(0)2=0
3. Plug into main equation and solve: (44.1-0)/(3-0)=14.7m/s
Average speed is used for many different things, from finding the speed at which a cannonball was launched out of a cannon from how fast a cheetah runs in a straight line trying to catch it's prey. The concept behind average speed is a fairly simple concept that many people understand right away. You're basically finding the slope of the equation using calculus and algebra. If I ask someone what the average speed of a ball from [3,4] if it's path was graphed as y=x.
y=(4) y=(3) (4-3)/(4-3)=1
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve.
First of all, remember that a slope is the y value, or dy, of a derivative.
Example:
A ball is flung from a little child. It's path is projected as y=4.9t2m in "t" seconds. What is the average speed of the ball from 0 to 3 seconds?
1. Set up equations and intervals: (f(b)-f(a))/(b-a) 4.9t^2 [0,3]
2. Plug in a and b values for t: f(b)=4.9(3)2=44.1 f(a)=4.9(0)2=0
3. Plug into main equation and solve: (44.1-0)/(3-0)=14.7m/s
Average speed is used for many different things, from finding the speed at which a cannonball was launched out of a cannon from how fast a cheetah runs in a straight line trying to catch it's prey. The concept behind average speed is a fairly simple concept that many people understand right away. You're basically finding the slope of the equation using calculus and algebra. If I ask someone what the average speed of a ball from [3,4] if it's path was graphed as y=x.
y=(4) y=(3) (4-3)/(4-3)=1
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve.
Post #19
Week 19
Well so my Christmas was pretty good, hope everyone else is as satisfied as me :-)
For my blog I guess I will explain the things I know really well..
Maxs and mins have to be one of the easiest things to do in Calculus. You are given a function, such as 2x^2 + 4x + 5. To find the max or min of this function, you take the derivative. The derivative is 4x + 4. Set this equal to 0, so 4x+4=0 and solve for x. When you solve, you get x=-1. So the x value of the vertex is -1. To find the actual point, you plug this in to the original to get the y-value of the vertex.
Point of inflection, which is the point where concavity changes, is also really easy and is like the same thing. The only difference is you take the second derivative. So for x^3 + 2x^2 + x + 4 ... it would be 3x^2 + 4x + 1 for first, 6x + 4 for second derivative. Set 6x+4=0 and solve for x to get that x=-2/3.
For maxs and mins, to find out if it is a max or a min, you have to use the derivative test. You have to set up intervals and then test them by plugging in points. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing. If it goes increasing, point, decreasing then it is a max. If it goes decreasing, point, increasing, then it is a min.
Anyway, I don't know what else to talk about for now. Enjoy the rest of your holidays!
AND BTW, according to http://www.wordcounttool.com/, my post is 267 words :-)
Well so my Christmas was pretty good, hope everyone else is as satisfied as me :-)
For my blog I guess I will explain the things I know really well..
Maxs and mins have to be one of the easiest things to do in Calculus. You are given a function, such as 2x^2 + 4x + 5. To find the max or min of this function, you take the derivative. The derivative is 4x + 4. Set this equal to 0, so 4x+4=0 and solve for x. When you solve, you get x=-1. So the x value of the vertex is -1. To find the actual point, you plug this in to the original to get the y-value of the vertex.
Point of inflection, which is the point where concavity changes, is also really easy and is like the same thing. The only difference is you take the second derivative. So for x^3 + 2x^2 + x + 4 ... it would be 3x^2 + 4x + 1 for first, 6x + 4 for second derivative. Set 6x+4=0 and solve for x to get that x=-2/3.
For maxs and mins, to find out if it is a max or a min, you have to use the derivative test. You have to set up intervals and then test them by plugging in points. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing. If it goes increasing, point, decreasing then it is a max. If it goes decreasing, point, increasing, then it is a min.
Anyway, I don't know what else to talk about for now. Enjoy the rest of your holidays!
AND BTW, according to http://www.wordcounttool.com/, my post is 267 words :-)
Post #19
Alright ladies and gents, here goes nothing. Today i'll explain how to do integrals. Because everyone is going back and expaining derivatives..so, of course, let me do the exact opposite.
Now, the first thing you have to do is locate the problem.
Then, read the problem.
You may ask, is this an integral or a derivative. Well, before you ask that terrible question, you know it's an integral if it has a fancy looking S infront of it.
Next, predict a method of solving; since it's a integral...
Perhaps, you should remember what you learned in class.
Then, perform the steps which include: adding one to the exponent and dividing it by the coefficient.
Finally, simplify fully.
So, lets do it with a problem.
Locate, Read, Predict, Think about the Problem
S 2x^3 + 7x + 6
Solvinggg....
S 2/4x^4 + 7/2x^2 +6x
Simplifying fully...
S 1/2x^4 + 7/2x^2 +6x
And tada. There you have it. You have successfully completed an integral problem. Now, you might thing this is all fun and games, and really easy. But don't fool yourself, it isnt always going to be that easy..such as when substitution comes along. [which i STILL do not understand]. So, if anyone wants to take a wackin at trying to explain it to me, that'd be greattt...or probably not, because we don't have to do comments..so nevermind. hahha, i made a funny on accident!
I hope everyone had a good christmas and got everything they wanteddddd!!!
Now, the first thing you have to do is locate the problem.
Then, read the problem.
You may ask, is this an integral or a derivative. Well, before you ask that terrible question, you know it's an integral if it has a fancy looking S infront of it.
Next, predict a method of solving; since it's a integral...
Perhaps, you should remember what you learned in class.
Then, perform the steps which include: adding one to the exponent and dividing it by the coefficient.
Finally, simplify fully.
So, lets do it with a problem.
Locate, Read, Predict, Think about the Problem
S 2x^3 + 7x + 6
Solvinggg....
S 2/4x^4 + 7/2x^2 +6x
Simplifying fully...
S 1/2x^4 + 7/2x^2 +6x
And tada. There you have it. You have successfully completed an integral problem. Now, you might thing this is all fun and games, and really easy. But don't fool yourself, it isnt always going to be that easy..such as when substitution comes along. [which i STILL do not understand]. So, if anyone wants to take a wackin at trying to explain it to me, that'd be greattt...or probably not, because we don't have to do comments..so nevermind. hahha, i made a funny on accident!
I hope everyone had a good christmas and got everything they wanteddddd!!!
Thursday, December 24, 2009
Post #19
Heyy...hope everyone's christmas holidays are going good!
let's go over Rolle's Theorem:
In order to be able to do Rolle's Theorem you have to have a continuous equation and it has to be differentual. When you plug in the points that are given to you, it has to equal to the same number that way f(a)=f(b)=0 [or any other number]
If all of this is passed, then you can start using the theorem:
find the first derivative, set it equal to zero and solve. The number[s] you get for x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be your answer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2-2x [0,2]
it is continuous and differentiable.
plug in: f(0)=0
f(2)=0
take the first derivative: 2x-2=0
now solve: x= -1
since -1 isn't between 0 and 2 it is not an answer. Therefore there is no answers!
&&.. Mean Value Theorem:
In order to be able to do Mean Value Theorem you have to have a continuous equation and it has to be differentual. When you have the points in the problem [a,b] you plug those to the equation. When you solve and find numbers you take [the "b" answer] - [the "a" answer] at the top of the equation, and from the original points given you take [b - a] for the bottom of the equation. When you solve and get a number that's the number that you set the first derivative to. [Take the derivative of the equation given and then set equal to that number] The number[s] you get for the x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be you anwer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2 [-2,1]
it is continuous and differentiable.
plug in: f(-2)=4
f(1) =1
so you plug into the equation...1-4 = -3
----- ----
1+2 = 1
since that equals to -3 when you take the first derivative [2x] you will set it equal to -3.
giving you x= -3/2 [which is between [-2,1] so c = -3/2
SANTA CLAUSE COMES TONIGHT!!...
MERRY CHRISTMAS,
FROM MY FAMILY TO YOURS
let's go over Rolle's Theorem:
In order to be able to do Rolle's Theorem you have to have a continuous equation and it has to be differentual. When you plug in the points that are given to you, it has to equal to the same number that way f(a)=f(b)=0 [or any other number]
If all of this is passed, then you can start using the theorem:
find the first derivative, set it equal to zero and solve. The number[s] you get for x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be your answer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2-2x [0,2]
it is continuous and differentiable.
plug in: f(0)=0
f(2)=0
take the first derivative: 2x-2=0
now solve: x= -1
since -1 isn't between 0 and 2 it is not an answer. Therefore there is no answers!
&&.. Mean Value Theorem:
In order to be able to do Mean Value Theorem you have to have a continuous equation and it has to be differentual. When you have the points in the problem [a,b] you plug those to the equation. When you solve and find numbers you take [the "b" answer] - [the "a" answer] at the top of the equation, and from the original points given you take [b - a] for the bottom of the equation. When you solve and get a number that's the number that you set the first derivative to. [Take the derivative of the equation given and then set equal to that number] The number[s] you get for the x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be you anwer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2 [-2,1]
it is continuous and differentiable.
plug in: f(-2)=4
f(1) =1
so you plug into the equation...1-4 = -3
----- ----
1+2 = 1
since that equals to -3 when you take the first derivative [2x] you will set it equal to -3.
giving you x= -3/2 [which is between [-2,1] so c = -3/2
SANTA CLAUSE COMES TONIGHT!!...
MERRY CHRISTMAS,
FROM MY FAMILY TO YOURS
Monday, December 21, 2009
Post 18
So we're off for two weeks. It's really nice considering we don't have any homework over the holidays except for the blog, which I am extremly greatful for. With the blog I will make an a for the nine weeks. Last week was exam week and it was a lot of reviewing. For review, we took two practice ap tests. We also had a take home portion of the exam. Although this was a part of the exam, it also helped with review. To review, I got together with Chelsea, John, and Mabile. We asked questions, worked problems, and taught each other what we did not know.
In my last blog, I had problems deciphoring between average value, average rate of change, and the mean value theorem. After last week of exams and review, I think I have finally gotten them figured out.
First there is average value. It seems like it would be difficult, but it's really easy to do. It's just taking a definite integral and putting a fraction in front of the integral, just as it would be done if something were missing from the integral. In the front, the fraction is 1/a-b. After the fraction is figured out, it is placed in front of the integral and the integral is solved accordingly.
For average rate of change, you're only finding slope. For average rate of change, slope is found by the formula f(b)-f(a)/b-a
For the mean value theorem, the same formula as average rate of change is to be used except this number is to be set equal to the derivative of the function.
Something I still really don't understand is angle of elevation. I used to really understand linerization, but sometimes I get confused on which number in the problem is what and what formulas to plug it into. Back to angle of elevation, I'm always really lost on how to do the problems. I really don't understand how to work the ones where there is a person and a docking station and a plane flying over it or something? I know I would have to make a right triangle to figure it out and use most of the same rules for linerization, but I get lost after that.
In my last blog, I had problems deciphoring between average value, average rate of change, and the mean value theorem. After last week of exams and review, I think I have finally gotten them figured out.
First there is average value. It seems like it would be difficult, but it's really easy to do. It's just taking a definite integral and putting a fraction in front of the integral, just as it would be done if something were missing from the integral. In the front, the fraction is 1/a-b. After the fraction is figured out, it is placed in front of the integral and the integral is solved accordingly.
For average rate of change, you're only finding slope. For average rate of change, slope is found by the formula f(b)-f(a)/b-a
For the mean value theorem, the same formula as average rate of change is to be used except this number is to be set equal to the derivative of the function.
Something I still really don't understand is angle of elevation. I used to really understand linerization, but sometimes I get confused on which number in the problem is what and what formulas to plug it into. Back to angle of elevation, I'm always really lost on how to do the problems. I really don't understand how to work the ones where there is a person and a docking station and a plane flying over it or something? I know I would have to make a right triangle to figure it out and use most of the same rules for linerization, but I get lost after that.
jessie's 18th post
i was in mississippi this weekend so i hope that mrs robinson sees mine...but i thought i should put the stuff im shaky with on here just to review a little..sooo
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
The steps are:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
The steps are:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
post 18
Ok so things ive learned in calculus involve taking derivatives.
take a derivative
1) mulitply exponent by the coefficient
2) subtract one from the exponent.
its fairly simple
Quotient rule
U/v = (v(u)' - u(v)')/ v^2
1) copy the bottom
2) take derivative of the top
3) - the copy the top
4) derivative of bottom
5) put all over the bottom squared
Product rule
UV = u(v)' + v(u)'
1) copy the first
2) derivative the second
3) copy the second
4) derivative of the first
Some things that i still dont fully get are tangent lines. i need help with these
take a derivative
1) mulitply exponent by the coefficient
2) subtract one from the exponent.
its fairly simple
Quotient rule
U/v = (v(u)' - u(v)')/ v^2
1) copy the bottom
2) take derivative of the top
3) - the copy the top
4) derivative of bottom
5) put all over the bottom squared
Product rule
UV = u(v)' + v(u)'
1) copy the first
2) derivative the second
3) copy the second
4) derivative of the first
Some things that i still dont fully get are tangent lines. i need help with these
Post #18
In Calculus 18 weeks ago, I learned about derivatives, average speed, and instantaneous speed. For each of these math terms, there are different steps to follow. Many formulas are used and it is necessary to remember all of them. We learned around ten derivative formulas which include quotient rule, product rule, sum and difference rules, trig function rules, and rules for numbers with and without variables, and taking derivates of quadratics.I am very comfortable with taking derivatives by using the product rule.
The product rule is used when you are multiplying two terms such as this problem x² (x+4). First, you must know the formula, which is uv^1 + vu^1. In simpler terms, this formula means (copy the first term)(derivative of the second term) + (copy the second term)(derivative of the first term). Following along with the problem so far, you would have (x²)(1) + (x+4)(2x). You get x² because as the formula states, the first part of the problem is copying the first term. Next, you have to take the derivative of (x+4).
Knowing the derivative rule for variables without an exponent is equal to 1 and the derivative of any number is equal to 0, you obtain (1+0) for the derivative of (x+4). So far your problem should look like (x²)(1). The next step is to place the addition sign next in the equation. Then you should copy the second term (x+4), and take the derivative of the first. The derivative of x² is obtained by multiplying the exponent by the coefficient and then subtracting one from the exponent. (x²) = (2)(1) x²-1 = 2x.
Therefore your problem will then become (x²)(1) +(x+4)(2x). Simple algebra is used from then on, you should distribute the x² to the one and distribute the 2x to (x+4). Then add like terms to obtain your final answer of 3x² +8x.
One thing I am not comfortable with is finding instantaneous speed. Problems such as finding the instantaneous speed at t=2 where y=16t² confuses me. At first I though you just plug it in, but then I realized there was a special formula that confuses me. I know the formula, but I don’t understand how you decipher what is your “h” and what is the f(x). After the formula is plugged in and you do the algebra, what number are you supposed to plug in?
word..
The product rule is used when you are multiplying two terms such as this problem x² (x+4). First, you must know the formula, which is uv^1 + vu^1. In simpler terms, this formula means (copy the first term)(derivative of the second term) + (copy the second term)(derivative of the first term). Following along with the problem so far, you would have (x²)(1) + (x+4)(2x). You get x² because as the formula states, the first part of the problem is copying the first term. Next, you have to take the derivative of (x+4).
Knowing the derivative rule for variables without an exponent is equal to 1 and the derivative of any number is equal to 0, you obtain (1+0) for the derivative of (x+4). So far your problem should look like (x²)(1). The next step is to place the addition sign next in the equation. Then you should copy the second term (x+4), and take the derivative of the first. The derivative of x² is obtained by multiplying the exponent by the coefficient and then subtracting one from the exponent. (x²) = (2)(1) x²-1 = 2x.
Therefore your problem will then become (x²)(1) +(x+4)(2x). Simple algebra is used from then on, you should distribute the x² to the one and distribute the 2x to (x+4). Then add like terms to obtain your final answer of 3x² +8x.
One thing I am not comfortable with is finding instantaneous speed. Problems such as finding the instantaneous speed at t=2 where y=16t² confuses me. At first I though you just plug it in, but then I realized there was a special formula that confuses me. I know the formula, but I don’t understand how you decipher what is your “h” and what is the f(x). After the formula is plugged in and you do the algebra, what number are you supposed to plug in?
word..
Post #18
Sooooo, Calculus we meet again. Like Stephanie did last week, I'm also kicking it old school.
I'm fully confident in finding the derivatives of -4x^3+2x^2-3x+1. For each term, you take the exponent and multiply it by the constant and subtract one from the exponent. Also, the derivative of any number is 0. for -4x^3: 3(-4)= -12 3-1=2 for 2x^2: 2(2)=4 2-1=1 for -3x: -3(1) = -1 1-1=0 So, then you have -12x^2+4x-3.
Also, I think I'm getting average speed, but can anyone tell me if I'm doing this right: Given the position equation s(t)=3t^2-3t-4, find the anverage velocity from t=2 to t=4. So, I ended up getting (2,4). I then plugged in 2 and then 4 to the equation 3t^2-3t-4. When I plugged in I got 2 and 32. After that I plugged (2,4) and 2, 32 in to the slope formula. For my final answer I got 15.
I think I'm having trouble more with the algebra kind of stuff. for example: [(-3x^2+5x-5)(sinx)] I understand copy the first, derivative of second, minus, copy the second, derivative of first (-3x^2+5x-5)(cosx)-(sinx)(-6x+5) but then what's the next step to solving?
I'm fully confident in finding the derivatives of -4x^3+2x^2-3x+1. For each term, you take the exponent and multiply it by the constant and subtract one from the exponent. Also, the derivative of any number is 0. for -4x^3: 3(-4)= -12 3-1=2 for 2x^2: 2(2)=4 2-1=1 for -3x: -3(1) = -1 1-1=0 So, then you have -12x^2+4x-3.
Also, I think I'm getting average speed, but can anyone tell me if I'm doing this right: Given the position equation s(t)=3t^2-3t-4, find the anverage velocity from t=2 to t=4. So, I ended up getting (2,4). I then plugged in 2 and then 4 to the equation 3t^2-3t-4. When I plugged in I got 2 and 32. After that I plugged (2,4) and 2, 32 in to the slope formula. For my final answer I got 15.
I think I'm having trouble more with the algebra kind of stuff. for example: [(-3x^2+5x-5)(sinx)] I understand copy the first, derivative of second, minus, copy the second, derivative of first (-3x^2+5x-5)(cosx)-(sinx)(-6x+5) but then what's the next step to solving?
Post #18 or something like it
I completely forgot how to do this so..
Related Rates:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation involving your variables
5. Take the derivative with respect to time
6. Substitute in derivative and solve.
Example:
y= x^1/2 find dy/dt when x=-4 and dx/dt = 3
1. x=4; dx/dt=3
2. dy/dt = ?
3. equation: y=x^1/2
4. dy/dt = 1/2 x ^-1/2 dx/dt
5. 1/2 (-4)^-1/2 (3)
dy/dt = -3/4
The variables x and y are differentiable functions of t and are related by the equation y=2x^3-x+4 when x=2. dx/dt=-1 Find dy/dt when x=2.
1. x=2; dx/dt = -1
2. dy/dt = ?
3. you are already given the equation: y = 2x^3 - x +4
4. Derivative with respect to time: dy/dt= 6x^2 dx/dt - dx/dt
5. Plug in: 6(2)^2 (-1) - (-1)
6. Solve: 6(4)(-1)+1 = -23
A little harder example:
Air is being pumped into a spherical balloon at a rate of 4.5 cubic ft/min. Find the rate of change of the radius when the radius is 2 ft.
1. r=2ft; dv/dt = 4.5 ft^3/min
2. dr/dt = ?
3. Volume of sphere: v=4/3 pi r^3
4. Take derivative: dv/dt = 4 pi r^2 dr/dt
5. Plug in: 4.5 = 4pi(2)^2 dr/dt
6. 4.5=16 pi dr/dt
dr/dt = 9/32 pi ft/min
I know we don't comment this week but for what I am having trouble with:
1. Angle of elevation ( I think I'm just out of practice)
2. Linearization
3. I can always use more help with graphs even though I know we reviewed them plenty of times.
Sunday, December 20, 2009
Post #18
Calculus Week #18
So exams are finally over..thank god, and the holidays are here :-)
As to what I'm going to explain... I really don't know what to explain anymore. I've really explained a LOT of different things..
limits as x goes to infinity with fractions...
1. If the degree on top is larger than degree on bottom, it’s infinity.
2. If the degree on top is equal to the degree on bottom, divide the coefficients.
3. If the degree on top is smaller than degree on bottom, it’s 0.
Let's see...implicits are easy
1. take the derivative of both sides
2. every time you take the derivative of y write it with dy/dx
3. solve for dy/dx
Related Rates
Identify all variable and equations.
Identify what you are looking for.
Make a sketch and label.
Write an equation involving your variables. Know that you can only have one unknown variable, so a secondary equation may be given.
Take the derivative with respect to time. aka: (dy/dt) or (dx/dt).
Substitute in derivative and solve for the rate.
Limit as x goes to infinity of
sin(nx)/a
is n/a.
e^x integration is easy... Whenever working with e^x integration, your u is always set to the exponent of the e. The derivative of e^x is simply e^x times the derivative of x. So when working with these, really all you have to worry about it balancing it out with the derivative of x.
For example, e^(2x). This would become (1/2)e^(2x) + c, The reason behind the (1/2) is because if you take the derivative of e^(2x) it is 2e^(2x). However, we don't want that 2. So we take it out by putting a (1/2) in front, just like we did for other integrals.
good enough for now.
So exams are finally over..thank god, and the holidays are here :-)
As to what I'm going to explain... I really don't know what to explain anymore. I've really explained a LOT of different things..
limits as x goes to infinity with fractions...
1. If the degree on top is larger than degree on bottom, it’s infinity.
2. If the degree on top is equal to the degree on bottom, divide the coefficients.
3. If the degree on top is smaller than degree on bottom, it’s 0.
Let's see...implicits are easy
1. take the derivative of both sides
2. every time you take the derivative of y write it with dy/dx
3. solve for dy/dx
Related Rates
Identify all variable and equations.
Identify what you are looking for.
Make a sketch and label.
Write an equation involving your variables. Know that you can only have one unknown variable, so a secondary equation may be given.
Take the derivative with respect to time. aka: (dy/dt) or (dx/dt).
Substitute in derivative and solve for the rate.
Limit as x goes to infinity of
sin(nx)/a
is n/a.
e^x integration is easy... Whenever working with e^x integration, your u is always set to the exponent of the e. The derivative of e^x is simply e^x times the derivative of x. So when working with these, really all you have to worry about it balancing it out with the derivative of x.
For example, e^(2x). This would become (1/2)e^(2x) + c, The reason behind the (1/2) is because if you take the derivative of e^(2x) it is 2e^(2x). However, we don't want that 2. So we take it out by putting a (1/2) in front, just like we did for other integrals.
good enough for now.
post 18
so for christmas break the only work we have to do is the blog, which is eeaaassyyy!!!! so i am just going to go over some simple stuff that we have learned in this worderful class we call calculus.
substitution:
Substitution takes the place of the derivative rules for problems with a quotient rule and product rule, substitution has a few steps.
1. find u by looking inside the parentheses inside the problem
2. take the derivative of u to find du
3. go into the origional problem and switch out (substitute) the stuff
4. integrate
5. plug in
ok, so ln i kinda understand, ln integration happens when the bottom of a fraction is the u and the top is the derivative of u. so since the derivative of ln(x) is 1/x, the answer is just ln(u)+c like everytime.
e^x integration is pretty decent too, all you have to do for this kind of integration is set your u equal to the exponent of the e. So the derivative of e^x is just e^x times the derivative of x.
i did pretty good on my exam so i do not really have any questions, but since these things require that you post a question i guess i will. umm, for riemann sums, i know how to do LRAM, RRAM, and Trapezoidal, but i suck at MRAM, i was told you have to do LRAM and RRAM to do it, but i just have never done it so it confuses me. if someone could teach me that i'd be good. :)
substitution:
Substitution takes the place of the derivative rules for problems with a quotient rule and product rule, substitution has a few steps.
1. find u by looking inside the parentheses inside the problem
2. take the derivative of u to find du
3. go into the origional problem and switch out (substitute) the stuff
4. integrate
5. plug in
ok, so ln i kinda understand, ln integration happens when the bottom of a fraction is the u and the top is the derivative of u. so since the derivative of ln(x) is 1/x, the answer is just ln(u)+c like everytime.
e^x integration is pretty decent too, all you have to do for this kind of integration is set your u equal to the exponent of the e. So the derivative of e^x is just e^x times the derivative of x.
i did pretty good on my exam so i do not really have any questions, but since these things require that you post a question i guess i will. umm, for riemann sums, i know how to do LRAM, RRAM, and Trapezoidal, but i suck at MRAM, i was told you have to do LRAM and RRAM to do it, but i just have never done it so it confuses me. if someone could teach me that i'd be good. :)
Ash's 18th Blog
Exams are over
Holidays are here
At
Last
Okay, so.... ahh, my battery's dying!!
Since I forgot my notebook to refer back to and my laptop's failing, I think I'm going to type up some tips that I learned last week and this week... =]
1. When taking the left hand, right hand, and regular limits of a graph, they've got to match up in order for it to exist.
2. When they are asking for a limit of a point on a graph, usually they're looking for the open circle... I think? I'm still shaky...can someone clarify AGAIN...I've been told a thousand times...
3. Never. Ever. Ever. Ever. Take the derivative of a graph. The steps that she went through with us...yea...those were amazing, I actually half-way understood that! I think I just need more practice with that way instead of taking the derivative of the graph :)
4. Multiple choice is my best friend.
5. This blog helped a lot more than I ever thought possible. All of those steps written out in English, not math terms like x and y and example problems and whatnot, were so much easier to understand! I sat here for a while going back through most of these blogs, writing down which explanation worked best and was the simplest and easiest to understand.
Oh, I want to figure out the probability that the Saints will lose the Super Bowl...which formula do I use for that? nCr or nPr? And how do I work those..i don't remember...that was also on the ACT, but I think I got it wrong because no formula = no answer. -.-
Holidays are here
At
Last
Okay, so.... ahh, my battery's dying!!
Since I forgot my notebook to refer back to and my laptop's failing, I think I'm going to type up some tips that I learned last week and this week... =]
1. When taking the left hand, right hand, and regular limits of a graph, they've got to match up in order for it to exist.
2. When they are asking for a limit of a point on a graph, usually they're looking for the open circle... I think? I'm still shaky...can someone clarify AGAIN...I've been told a thousand times...
3. Never. Ever. Ever. Ever. Take the derivative of a graph. The steps that she went through with us...yea...those were amazing, I actually half-way understood that! I think I just need more practice with that way instead of taking the derivative of the graph :)
4. Multiple choice is my best friend.
5. This blog helped a lot more than I ever thought possible. All of those steps written out in English, not math terms like x and y and example problems and whatnot, were so much easier to understand! I sat here for a while going back through most of these blogs, writing down which explanation worked best and was the simplest and easiest to understand.
Oh, I want to figure out the probability that the Saints will lose the Super Bowl...which formula do I use for that? nCr or nPr? And how do I work those..i don't remember...that was also on the ACT, but I think I got it wrong because no formula = no answer. -.-
18th post
This past week in calculus we reviewed for our mid- term exam. Mrs. Robinson explained to us the concepts of finding max and mins when looking at a graph. When looking for max and mins, put a filled in circle where the graphs change from increasing to decreasing, for points of inflection, put open circles where the graphs change concavity.
Another thing we reviewed was the equation of a tangent line. When you are given a function and an x point. You take the derivative of the function and plug in the x value to find the slope. To find the y value, plug the x value into the original function and solve.
Another thing was solving for dy/dx. All you do for these is taking the derivative like normal but everytime you take the derivative of a y, you put dy/dx behind it. After the derivative has been taken, you solve for dy/dx.
One thing i have become very good at is finding the relative extrema of a function. If they give you points, you plug the points into the original function. You will get another number and you put the numbers together like points. Then to find more points, you take the derivative of the function and solve for x to find the critical points. You plug the critical points in to the original function and get another point. When everything has been plugged in, you figure out which point is the highest and which one is the lowest. Those points will be your relative max and relative min.
To find out if the function is increasing or decreasing, you take the derivative of the original function and solve for x to find the critical values. Then you plug those points into intervals. You take a number between each interval and plug that number into the derivative, if the number is positve it is increasing, if the number is negative it is decreasing. This is the way you find regular max and mins as well.
Another thing we reviewed was the equation of a tangent line. When you are given a function and an x point. You take the derivative of the function and plug in the x value to find the slope. To find the y value, plug the x value into the original function and solve.
Another thing was solving for dy/dx. All you do for these is taking the derivative like normal but everytime you take the derivative of a y, you put dy/dx behind it. After the derivative has been taken, you solve for dy/dx.
One thing i have become very good at is finding the relative extrema of a function. If they give you points, you plug the points into the original function. You will get another number and you put the numbers together like points. Then to find more points, you take the derivative of the function and solve for x to find the critical points. You plug the critical points in to the original function and get another point. When everything has been plugged in, you figure out which point is the highest and which one is the lowest. Those points will be your relative max and relative min.
To find out if the function is increasing or decreasing, you take the derivative of the original function and solve for x to find the critical values. Then you plug those points into intervals. You take a number between each interval and plug that number into the derivative, if the number is positve it is increasing, if the number is negative it is decreasing. This is the way you find regular max and mins as well.
Yes were finally off of school but we still have to do our blogs... O.o
But thats not that bad :)
Average Speed:
An anvil falls off the roof onto Ryan's head. What is its average speed during the first two seconds if y=16t^2
0:y1=16(0)=0
2: y2= 16(2)^2=64
then you plug into the formula y2-y1/(0ver) x2-x1
after plugging in you would get the awnser of 32
Instantanesous speed:
lim f(x+h)-f(x)/(over)
h
find the speed at t=2
y=16t^2 lim 16(2+h)^2-16(2)^2
f(t)=16t^2
which whould give you 64
And im still having trouble with points of inflection and don't have notes and i know we don't have to comment this week whcih is awesome but i still asked a question.
But thats not that bad :)
Average Speed:
An anvil falls off the roof onto Ryan's head. What is its average speed during the first two seconds if y=16t^2
0:y1=16(0)=0
2: y2= 16(2)^2=64
then you plug into the formula y2-y1/(0ver) x2-x1
after plugging in you would get the awnser of 32
Instantanesous speed:
lim f(x+h)-f(x)/(over)
h
find the speed at t=2
y=16t^2 lim 16(2+h)^2-16(2)^2
f(t)=16t^2
which whould give you 64
And im still having trouble with points of inflection and don't have notes and i know we don't have to comment this week whcih is awesome but i still asked a question.
Post Number Eighteen
For my first post of the holidays, I’m gonna talk about the early weeks of calculus. Hopefully these blogs keep my grade up. The first thing we learned when we first started calculus were derivatives. Although confusing at first, repetition made them sooo easy! We learned many different derivative formulas, some including product rule, quotient rule, chain rule, sum and difference rules, trig function rules, inverse rules, and also quadratics. An example of taking a derivative of a quadratic is 3x^4 + 5x^3 – 2x^2 + x + 1
Taking a derivative of quadratics is easy, simply multiply the exponent to the coefficient and the exponent becomes one less. Therefore, the derivative of the equation given earlier would be 12x^3 + 15x^2 – 4x + 1.
Steps for Product Rule: UV = u(v’) + v(u’)
Copy the first.
Times the derivative of the second
+
Copy the second
Times the derivative of the first
Steps for Quotient Rule: U/V = (v(u’) – u(v’))/ v^2
Copy the bottom
Times the derivative of the top
-
Copy the top
Times the derivative of the bottom
All over the bottom squared
First Derivative Test:
Take the derivative of the original problem
Set the first derivative equal to 0
Solve for x and create intervals for x
Pick a number in the intervals then plug that number in the first derivative for x
Solve the equation
I understand the steps for the first derivative test, but I have trouble applying it for some reason. It used to be so easy, but now I can’t seem to solve problems like these. I think I get lost at the interval part.
Other things I still don’t understand are optimization and angle of elevation.
Hope everyone has a great Christmas :)
Taking a derivative of quadratics is easy, simply multiply the exponent to the coefficient and the exponent becomes one less. Therefore, the derivative of the equation given earlier would be 12x^3 + 15x^2 – 4x + 1.
Steps for Product Rule: UV = u(v’) + v(u’)
Copy the first.
Times the derivative of the second
+
Copy the second
Times the derivative of the first
Steps for Quotient Rule: U/V = (v(u’) – u(v’))/ v^2
Copy the bottom
Times the derivative of the top
-
Copy the top
Times the derivative of the bottom
All over the bottom squared
First Derivative Test:
Take the derivative of the original problem
Set the first derivative equal to 0
Solve for x and create intervals for x
Pick a number in the intervals then plug that number in the first derivative for x
Solve the equation
I understand the steps for the first derivative test, but I have trouble applying it for some reason. It used to be so easy, but now I can’t seem to solve problems like these. I think I get lost at the interval part.
Other things I still don’t understand are optimization and angle of elevation.
Hope everyone has a great Christmas :)
pos #1 holidays
i can't tell you how glad i am to be on vacation, i don't think i could take another week of school right now! thank youuu christmasss!
lets go over related rates because i still need a review.
1. Identify all your variables and equations. Some may be given
2. Identify what you need to solve for.
3. Make a sketch or draw a picture of what you are given
4. Take the derivative of the equations with respect to time.
5. Plug in given numbers to the derivative and solve.
EXAMPLE 1: This problem has the equation and given information.
x^2+y^2=25 dy/dt when x=3, y=4 dx/dt= 8
Derivative: 2x dx/dt + 2y dy/dt= 0
When you plug in you get: 2(3)(8) + 2(4) (dy/dt) = 0
48+ 8 dy/dt=0
dy/dt= -6
mmm.. now how about implicit derivatives.
STEPS:
1. Take the derivative.
2. For each y-term, you put y' or dy/dx behind it.
3. Solve for dy/dy or y'.
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
HELLLP? i still don't unederstand optimiazation from way back in the gap so if anyone wants to show me go for it, thanks. MERRYY CHRISTMAS!
lets go over related rates because i still need a review.
1. Identify all your variables and equations. Some may be given
2. Identify what you need to solve for.
3. Make a sketch or draw a picture of what you are given
4. Take the derivative of the equations with respect to time.
5. Plug in given numbers to the derivative and solve.
EXAMPLE 1: This problem has the equation and given information.
x^2+y^2=25 dy/dt when x=3, y=4 dx/dt= 8
Derivative: 2x dx/dt + 2y dy/dt= 0
When you plug in you get: 2(3)(8) + 2(4) (dy/dt) = 0
48+ 8 dy/dt=0
dy/dt= -6
mmm.. now how about implicit derivatives.
STEPS:
1. Take the derivative.
2. For each y-term, you put y' or dy/dx behind it.
3. Solve for dy/dy or y'.
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
HELLLP? i still don't unederstand optimiazation from way back in the gap so if anyone wants to show me go for it, thanks. MERRYY CHRISTMAS!
Post # 18
This week in Calculus we reviewed for the exam and took the exam that was hard and easy at the same time. So, instead of reviewing things that i already did..i'm going to go back and explain first derivative test.
Before i explain this i'll say something about the concept that if the origional function graph was decreasing, that its first derivative would be starting below the x axis and if the orgional function graph was increasing, that its first derivative would be starting above the x axis. Also when there is a zero on an origional function graph, the zero will become the maximum or minimum on the first derivative graph, and vice versa.
So, for the first derivative test you must take the derivative and set equal to zero then solve for x. After, you must set up intervals, but you must make note of your endpoints.
First derivative test is very easy and should be a give a way problem on the exams and tests. Remember to take your time and do the problem correctly so you can get the give-a-way problem right instead of wasting your points on something that takes little time and is super easy to do!
Before i explain this i'll say something about the concept that if the origional function graph was decreasing, that its first derivative would be starting below the x axis and if the orgional function graph was increasing, that its first derivative would be starting above the x axis. Also when there is a zero on an origional function graph, the zero will become the maximum or minimum on the first derivative graph, and vice versa.
So, for the first derivative test you must take the derivative and set equal to zero then solve for x. After, you must set up intervals, but you must make note of your endpoints.
First derivative test is very easy and should be a give a way problem on the exams and tests. Remember to take your time and do the problem correctly so you can get the give-a-way problem right instead of wasting your points on something that takes little time and is super easy to do!
Friday, December 18, 2009
Post #18
Well, this past week in calculus was .. just two days. Monday we reviewed some more things that we did not understand throughout the year. I can honestly say that I am BEGINNING to understand the stuff that Mrs. Robinson taught us about the picture of a graph, so I think that’s what I’m going to explain.
• If you have f(x):
o Use a solid line to divide where it’s increasing or decreasing: if it’s increasing then the first derivative is positive, and if it’s decreasing then the first derivative is negative.
o Use a dotted line to separate concave up and concave down: if it’s concave up then the second derivative is positive, and if it’s concave down then the second derivative is negative.
o Put a large dot on MAX and MIN: MAXs and MINs mean that it’s a horizontal tangent and the first derivative is zero.
o Put an open circle on points of inflection: points of inflection is where the second derivative is zero.
• If you have f’(x):
o Use a solid line where it’s increasing or decreasing: if it’s increasing then the second derivative is positive, and if it’s decreasing then the second derivative is negative.
o Use a dotted line where it’s above and below the axis: when it’s above then the original is increasing and when it’s below then the original is decreasing.
o Use solid dots on MAX’s and MIN’s then the second derivative is equal to zero and it’s a point of inflection on the derivative.
o Then put open circles on the x intercepts: when the MAX is above the axis then it’s below for the second derivative, and when the MIN is below the axis then it’s above for the second derivative.
So I know that I understand the concept perfectly fine and it’s really simple, it’s just that looking at a graph confuses me so I would need more practice to look at the graph and figure out what is what. I didn’t do so well on my exam so it looks like I’ll be reviewing a lot over the break so I can catch up. THANK GOD IT’S THE HOLIDAYS.
~ElliE~
• If you have f(x):
o Use a solid line to divide where it’s increasing or decreasing: if it’s increasing then the first derivative is positive, and if it’s decreasing then the first derivative is negative.
o Use a dotted line to separate concave up and concave down: if it’s concave up then the second derivative is positive, and if it’s concave down then the second derivative is negative.
o Put a large dot on MAX and MIN: MAXs and MINs mean that it’s a horizontal tangent and the first derivative is zero.
o Put an open circle on points of inflection: points of inflection is where the second derivative is zero.
• If you have f’(x):
o Use a solid line where it’s increasing or decreasing: if it’s increasing then the second derivative is positive, and if it’s decreasing then the second derivative is negative.
o Use a dotted line where it’s above and below the axis: when it’s above then the original is increasing and when it’s below then the original is decreasing.
o Use solid dots on MAX’s and MIN’s then the second derivative is equal to zero and it’s a point of inflection on the derivative.
o Then put open circles on the x intercepts: when the MAX is above the axis then it’s below for the second derivative, and when the MIN is below the axis then it’s above for the second derivative.
So I know that I understand the concept perfectly fine and it’s really simple, it’s just that looking at a graph confuses me so I would need more practice to look at the graph and figure out what is what. I didn’t do so well on my exam so it looks like I’ll be reviewing a lot over the break so I can catch up. THANK GOD IT’S THE HOLIDAYS.
~ElliE~
Tuesday, December 15, 2009
post seventeen
Ok so this week we were preparing for log one and for our exam.
Some things that i understand in calculus include:
Tangent Line:
1. If you are only given a x value, plug into your equation to find the y value.
2. take the derivative of the equation
3. Plug in your x value into the derivative to find the slope
4. Put in point slope
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
STEPS:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
These are fairly simple processes and equations to work with. For what i dont know is linierization. I dont even have notes on this
Some things that i understand in calculus include:
Tangent Line:
1. If you are only given a x value, plug into your equation to find the y value.
2. take the derivative of the equation
3. Plug in your x value into the derivative to find the slope
4. Put in point slope
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
STEPS:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
These are fairly simple processes and equations to work with. For what i dont know is linierization. I dont even have notes on this
Sunday, December 13, 2009
Post #17
Last week in calculus was review week. We didn't learn anything new so I'm going to review some of the old stuff because it will be on our exam.
Tangent Line:
1. If you are only given a x value, plug into your equation to find the y value.
2. take the derivative of the equation
3. Plug in your x value into the derivative to find the slope
4. Put in point slope
EXAMPLE: y=3x^3 + 4x^2 +5 at x=1
1. 3(1)^3+4(1)^2+5 = 12
2. 9x^2 + 8x
3. 9(1)^2 + 8(1) = 17
4. y-12 = 17 (x-1)
Normal line is the same except you take the negative reciprocal of the slope to plug in.
y-12 = -1/17 (x-1)
Rolle's Theorem:
In order to use Rolle's theorem, the fuction must be continuous and differentiable on the interval given and f(a) = f(b)
Example:
-x^2+14x on [0,14]
The function is continuous and differentiable
f(0) = -(0)^2 + 14(0) = 0
f(14) = -(14)^2 + 14(14) = 0
Therefore Rolle's applies
Now take the derivative: -2x+14
and set equal to zero: -2x+14 = 0
solve for x: x=7 or c=7
Mean value theorem:
The function again has to be continuous and differentiable at the interval given and you plug into f(b) - f(a) / b-a
Example: x^2 on [1,7]
The function is continuous and differentiable
First take the derivative: 2x
Then plug into the formula: f(7) - f(1) / 7-1
14-2/ 7-1 = 12/6 which equals 2.
I am still having troubles with graphs and going from one graph to another. I don't understand how to find increasing/decreasing, concave up/down, maxs, mins, and points of inflection. Help please!
17th post
alright, we just reviewed this week, so i am going to go over integration. Integration is pretty much the opposite of a derivative.
there are two types of integration, definite integration and indefinite integration.
indefinite integration is easy, you just take the equation, and do this to it, lets say the equation is x^3+2x^2
to take the integral of something, first you have to raise the exponent by one, so lets take the first part of the problem, x^3, and do that to it. it becomes x^4. Now, you just multiple the whole thing by the inverse of the new exponent. so it would become (1/4)x^4.
then you do the same thing to the second part of the problem. 2x^2 becomes 2x^3, then you have to multiply, and it becomes (2/3)x^3
then you put the two parts back together like normal, and it becomes (1/4)x^4+(2/3)x^3.
but we are not done yet!! for indefinite integrals, you always have to add C at the end of the equation, so your final answer becomes: (1/4)x^4+(2/3)x^3+C
and for definite integrals, you just do the same thing, but for definite integrals, they give you a point, and you just plug in the x value from the point, and you do not add the C at the end!!! so lets just use the example problem from indefinite integration, but they'll also give you a point, let's say (2,2) for example. You would integrate it normally, and get (1/4)x^4+(2/3)x^3, then you plug in your x value, which is 2. so then you get (1/4)(2)^4+(2/3)(2)^3, and after that you can just put it into your calculator, which I do not have with me, and you'll just get some number as your answer. So for definite integration, you'll get a a number answer, and for indefinite integration, you'll get an equation.
there are two types of integration, definite integration and indefinite integration.
indefinite integration is easy, you just take the equation, and do this to it, lets say the equation is x^3+2x^2
to take the integral of something, first you have to raise the exponent by one, so lets take the first part of the problem, x^3, and do that to it. it becomes x^4. Now, you just multiple the whole thing by the inverse of the new exponent. so it would become (1/4)x^4.
then you do the same thing to the second part of the problem. 2x^2 becomes 2x^3, then you have to multiply, and it becomes (2/3)x^3
then you put the two parts back together like normal, and it becomes (1/4)x^4+(2/3)x^3.
but we are not done yet!! for indefinite integrals, you always have to add C at the end of the equation, so your final answer becomes: (1/4)x^4+(2/3)x^3+C
and for definite integrals, you just do the same thing, but for definite integrals, they give you a point, and you just plug in the x value from the point, and you do not add the C at the end!!! so lets just use the example problem from indefinite integration, but they'll also give you a point, let's say (2,2) for example. You would integrate it normally, and get (1/4)x^4+(2/3)x^3, then you plug in your x value, which is 2. so then you get (1/4)(2)^4+(2/3)(2)^3, and after that you can just put it into your calculator, which I do not have with me, and you'll just get some number as your answer. So for definite integration, you'll get a a number answer, and for indefinite integration, you'll get an equation.
post #17
Okay so almost everything we did this week was basically in preparation for the exam on tuesdayyy, so i'm just going to go over some things from the past as a review.
Some things that we just learned that are extremely easy form me are finding the volume and area of disks and washers. atleast i know i'll pass some portion of the exams, assuming ofcourse, that they're even on there.
Area between curves:
formula: bSa top equation-bottom equation
example: Find the area of the region enclosed by y=-4x^2+41x+94 and y=x-2 between x=1 and x=7.
1. draw a picture
2. subtract the two equations: (-4x^2+41x+94-(x-2)-don't square, it's not volumee!
(-4x^2+41x+94-(-x+2)
3.combine like terms and integrate: S(-4x^2+41x+94-(-x+2))
(-4/3)x^3+20x^2+96x
4. plug in 1 and 6 and subtract: (3584/3)-(344/3)= 1080
FORMULA FOR DISKS: pi S[r(x)]^2dx, this formula is almost exactly the same as when you are asked to find the area of a disk, however, while finding the area you do not square the radius (the given equation).
FORMULA FOR WASHERS: pi S[(top)^2-(bottom)^2]dx
TANGENT LINES!
Steps:
Take the first der...f ‘(x).
Plug in x to find your slope: m.
Plug x into original functiong..f(x) to find y (if not already given).
Using your slope..m and (x,y), plug it into slope-intercept form: (y-y1) = m(x-x1)
okayy something i don't understandd! umm i really don't get how to integrate a fraction, which would be integration by substitution i'm assumine. also some things from the past that would be nice to go over is ofcourse optimization, related rates type stuff. other than that, i think i'm okayy, maybe trapezoidal rule but that's about it! help would be WONDERFULL!
Some things that we just learned that are extremely easy form me are finding the volume and area of disks and washers. atleast i know i'll pass some portion of the exams, assuming ofcourse, that they're even on there.
Area between curves:
formula: bSa top equation-bottom equation
example: Find the area of the region enclosed by y=-4x^2+41x+94 and y=x-2 between x=1 and x=7.
1. draw a picture
2. subtract the two equations: (-4x^2+41x+94-(x-2)-don't square, it's not volumee!
(-4x^2+41x+94-(-x+2)
3.combine like terms and integrate: S(-4x^2+41x+94-(-x+2))
(-4/3)x^3+20x^2+96x
4. plug in 1 and 6 and subtract: (3584/3)-(344/3)= 1080
FORMULA FOR DISKS: pi S[r(x)]^2dx, this formula is almost exactly the same as when you are asked to find the area of a disk, however, while finding the area you do not square the radius (the given equation).
FORMULA FOR WASHERS: pi S[(top)^2-(bottom)^2]dx
TANGENT LINES!
Steps:
Take the first der...f ‘(x).
Plug in x to find your slope: m.
Plug x into original functiong..f(x) to find y (if not already given).
Using your slope..m and (x,y), plug it into slope-intercept form: (y-y1) = m(x-x1)
okayy something i don't understandd! umm i really don't get how to integrate a fraction, which would be integration by substitution i'm assumine. also some things from the past that would be nice to go over is ofcourse optimization, related rates type stuff. other than that, i think i'm okayy, maybe trapezoidal rule but that's about it! help would be WONDERFULL!
Ash's 17th Blog
Wow
Only one more week until we can finally be freed from school for two entire weeks! I'm so excited!!
This week we spent...doesn't stuff? I don't actually remember...so let's skip to Friday
Friday: Beast packet :D
Saturday: ACT + Subway (omg $5 6inch..YUM!) + PJs + Walmart (1.5 hours, with Mal's mom to get EIGHT things...) + New Orleans Hamburger (YUM! SEAFOOD!) + Movies (standing outside trying to decide what movie to watch, New Moon or the Blindside...for...thirty...minutes) + NEW MOON! FINALLY! + Calculus packet again
Sunday: Calculus packet + English packet + Tic Tac Toe + Calculus packet..againn
So..yea...that's all I have to say :)
Hmm...what do I know that I can explain?
Umm...Riemann Sums?
I'm pretty sure that's way over-done, but I have no idea what else to explain! =/
Delta = b-a/n (top number above S minus the bottom number below S divided by the number of intervals)
LRAM: delta(f(a)+f(a+delta)+f(a+2delta)...+f(b-delta))
RRAM: delta(f(a+delta)+f(a+2delta)+...f(b))
MRAM: delta(f(mid)+f(mid)...)
Trapezoidal: delta(x)/2 [ f(a) + 2f(a + delta x) + 2f(a + 2deltax)
+ f(b))
Just don't forget to not go over the amount of intervals you have (n).
Now, I've got major questions
1. Trapezoidal graphs...where can I get the software?
2. What's deltax? is that the same thing as delta..???
3. GRAPHING! AHHH!!
4. Also, for limits, are they asking for the open circle or the closed circle? I always get confused! And each time I change it, I get it wrong..Help!!!
5. Another thing for limits, what do you do if it's over zero? I remember something about it...did she or Alex go over that? Anyway, I can't find those notes and I didn't get it before...=/
Okay, this is my extreme problem:
I've never been able to remember the steps for math
I can do it if I know what I'm doing (makes total sense right?)
But I can't remember the steps
Does ANYONE have a way to do remember?? =/
Only one more week until we can finally be freed from school for two entire weeks! I'm so excited!!
This week we spent...doesn't stuff? I don't actually remember...so let's skip to Friday
Friday: Beast packet :D
Saturday: ACT + Subway (omg $5 6inch..YUM!) + PJs + Walmart (1.5 hours, with Mal's mom to get EIGHT things...) + New Orleans Hamburger (YUM! SEAFOOD!) + Movies (standing outside trying to decide what movie to watch, New Moon or the Blindside...for...thirty...minutes) + NEW MOON! FINALLY! + Calculus packet again
Sunday: Calculus packet + English packet + Tic Tac Toe + Calculus packet..againn
So..yea...that's all I have to say :)
Hmm...what do I know that I can explain?
Umm...Riemann Sums?
I'm pretty sure that's way over-done, but I have no idea what else to explain! =/
Delta = b-a/n (top number above S minus the bottom number below S divided by the number of intervals)
LRAM: delta(f(a)+f(a+delta)+f(a+2delta)...+f(b-delta))
RRAM: delta(f(a+delta)+f(a+2delta)+...f(b))
MRAM: delta(f(mid)+f(mid)...)
Trapezoidal: delta(x)/2 [ f(a) + 2f(a + delta x) + 2f(a + 2deltax)
+ f(b))
Just don't forget to not go over the amount of intervals you have (n).
Now, I've got major questions
1. Trapezoidal graphs...where can I get the software?
2. What's deltax? is that the same thing as delta..???
3. GRAPHING! AHHH!!
4. Also, for limits, are they asking for the open circle or the closed circle? I always get confused! And each time I change it, I get it wrong..Help!!!
5. Another thing for limits, what do you do if it's over zero? I remember something about it...did she or Alex go over that? Anyway, I can't find those notes and I didn't get it before...=/
Okay, this is my extreme problem:
I've never been able to remember the steps for math
I can do it if I know what I'm doing (makes total sense right?)
But I can't remember the steps
Does ANYONE have a way to do remember?? =/
Posting...#17
Last week in calculus we did not learn anything new. We took a practic ap monday and wensday then we went over it for tuesday and thursday.
Rolle's Theorem: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b) then there exsists a number c in (a,b) such that
f(b)-f(a)
f ' (c)=----------
b-a
Exampmle: F(x)=x^4 -2x^2 on [-2,2] find c f '(c)=0
Check if its continuous : YES
Check if its differentiable : YES
f(-2)=(-2)^4-2(-2)^2=8
F(2)=(2)^4-2(2)^2=8
f(-2)=f(2)
4x^3-4x=0
4x(x^2-1)=0
X=0,1,-1
Awnser C= -1,0,1
And thats it for roles theroem its pretty easy but for what im having trouble with is even easier so it would be great to get help.
I am confused on Critical Points so can some one explain how to do that for me.
Rolle's Theorem: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b) then there exsists a number c in (a,b) such that
f(b)-f(a)
f ' (c)=----------
b-a
Exampmle: F(x)=x^4 -2x^2 on [-2,2] find c f '(c)=0
Check if its continuous : YES
Check if its differentiable : YES
f(-2)=(-2)^4-2(-2)^2=8
F(2)=(2)^4-2(2)^2=8
f(-2)=f(2)
4x^3-4x=0
4x(x^2-1)=0
X=0,1,-1
Awnser C= -1,0,1
And thats it for roles theroem its pretty easy but for what im having trouble with is even easier so it would be great to get help.
I am confused on Critical Points so can some one explain how to do that for me.
Post Number Seventeen
this week we reviewed for our exam, so i'll just go over stuff that i know and do not know as my blog.
Let's start this off on a good note and explain something i DO understand. Average value :)
The formula for average value is 1/b-a(definite integral)f(a)
For example, Find the average value of f(x) = x^2 on [0,5]
1/5-0 S from 0 to 5 x^2 dx
= 1/5 [1/3x^3]
=1/5[125/3 - 0] = 25/3
Now for something i still don't understand and never did: SUBSTITUTION.
help please?
I also understand rolles and mean value theorems, you just follow them and plug in :)
This may be sad, but i am having trouble finding critical numbers and absolute extrema, those kinds of problems. I need a slight review on this if anyone cares.
I know how to do RRAM and LRAM, but am still having trouble with MRAM and mostly TRAM..
One more thing i understand is volume by disks, so i'll explain that quickly.
Volume by disks is used only with solid objects. The formula used is pi definite integral [R(x)]^2 dx
This is pretty easy, everything is basically given to you, the only place i usually mess up is in the algebra itself.
Well i'm stressing about the exam but hey, what can i do?
Goodnight Everyone :)
Let's start this off on a good note and explain something i DO understand. Average value :)
The formula for average value is 1/b-a(definite integral)f(a)
For example, Find the average value of f(x) = x^2 on [0,5]
1/5-0 S from 0 to 5 x^2 dx
= 1/5 [1/3x^3]
=1/5[125/3 - 0] = 25/3
Now for something i still don't understand and never did: SUBSTITUTION.
help please?
I also understand rolles and mean value theorems, you just follow them and plug in :)
This may be sad, but i am having trouble finding critical numbers and absolute extrema, those kinds of problems. I need a slight review on this if anyone cares.
I know how to do RRAM and LRAM, but am still having trouble with MRAM and mostly TRAM..
One more thing i understand is volume by disks, so i'll explain that quickly.
Volume by disks is used only with solid objects. The formula used is pi definite integral [R(x)]^2 dx
This is pretty easy, everything is basically given to you, the only place i usually mess up is in the algebra itself.
Well i'm stressing about the exam but hey, what can i do?
Goodnight Everyone :)
Post #17
Calculus Week #17
Anyway so we are supposed to explain a concept and ask a question or two about another. Well to be quite honest, this week I actually learned something I should have known a long time ago.
Trapezoidal!
The formula is really simple and easy to remember...anyway, it's
Let x = (b-a)/n
x/2 [f(a) + 2f(a+x) + 2f(a+2x) ... f(b)]
This beats trying to find the actual area under the curve and wondering which one of the answer choices might be close. Lol...
Anyway, let's see...what else can I explain.
Okay a crash course to help you all on the take-home test:
If it asks for a tangent line, take the derivative, plug in x, get your slope, use that in point-slope form with the point you had with that slope.
If it asks for critical values...set derivative equal to 0 and solve for x.
Make sure to test end points.
Relative maximums or minimums are when you have multiple maxs and mins. The relative ones are the ones that are not absolute. The ones that are absolute are the ones that have either the highest y value or lowest y value.
Remember, if you are allowed in calculator, you can check your integration.
fnInt((equation),x,a,b) for f(x) = (equation) on the interval [a,b].
Or just plug it into y= on the plot then go to SECOND, CALC and go to the last option. Then type in a press ENTER then press b and press ENTER. Voila, a graphed picture of the area under your curve.
For the area between two curves, you do the top equation minus bottom equation. If you have no a,b to plug in, set the two equations equal and solve. Use that as your lower and upper bounds.
For volume by discs of a solid rotated about an axis, do the definite integral from a to b of the equation squared...times pi. Pretty simple.
For volume by washers, it's the same thing as area between two curves...except this time you square both equations as well, as in the above example.
Average value is 1/(b-a) times the definite integral from a to b of the function.
Average rate of change is a slope which is a derivative.
Derivative of position is velocity, derivative of velocity is acceleration.
So first derivative is velocity, second derivative is acceleration.
Let's see what else I can cover...
If they give you acceleration and you take the integral to find the velocity equation, don't forget you need to solve for +c. Usually they will give you some information such as v(1) = 1 would be used to plug in and solve for c.
Anyway, I think I gave a pretty decent crash course for the take home test. Don't forget to use your calculator...you can check almost all of the problems in your calculator so you might as well do it!
Anyway so we are supposed to explain a concept and ask a question or two about another. Well to be quite honest, this week I actually learned something I should have known a long time ago.
Trapezoidal!
The formula is really simple and easy to remember...anyway, it's
Let x = (b-a)/n
x/2 [f(a) + 2f(a+x) + 2f(a+2x) ... f(b)]
This beats trying to find the actual area under the curve and wondering which one of the answer choices might be close. Lol...
Anyway, let's see...what else can I explain.
Okay a crash course to help you all on the take-home test:
If it asks for a tangent line, take the derivative, plug in x, get your slope, use that in point-slope form with the point you had with that slope.
If it asks for critical values...set derivative equal to 0 and solve for x.
Make sure to test end points.
Relative maximums or minimums are when you have multiple maxs and mins. The relative ones are the ones that are not absolute. The ones that are absolute are the ones that have either the highest y value or lowest y value.
Remember, if you are allowed in calculator, you can check your integration.
fnInt((equation),x,a,b) for f(x) = (equation) on the interval [a,b].
Or just plug it into y= on the plot then go to SECOND, CALC and go to the last option. Then type in a press ENTER then press b and press ENTER. Voila, a graphed picture of the area under your curve.
For the area between two curves, you do the top equation minus bottom equation. If you have no a,b to plug in, set the two equations equal and solve. Use that as your lower and upper bounds.
For volume by discs of a solid rotated about an axis, do the definite integral from a to b of the equation squared...times pi. Pretty simple.
For volume by washers, it's the same thing as area between two curves...except this time you square both equations as well, as in the above example.
Average value is 1/(b-a) times the definite integral from a to b of the function.
Average rate of change is a slope which is a derivative.
Derivative of position is velocity, derivative of velocity is acceleration.
So first derivative is velocity, second derivative is acceleration.
Let's see what else I can cover...
If they give you acceleration and you take the integral to find the velocity equation, don't forget you need to solve for +c. Usually they will give you some information such as v(1) = 1 would be used to plug in and solve for c.
Anyway, I think I gave a pretty decent crash course for the take home test. Don't forget to use your calculator...you can check almost all of the problems in your calculator so you might as well do it!
Post 17
Ok, so last week we pretty much just reviewed for the upcoming exams. Mrs. Robinson gave us two practice APs, which I failed. They mostly went over things we learned recently like integrals. The first one was much much better than the second one, which is pretty pathetic considering the second one was calculator allowed. We were also given the take home portion of the exam. This portion is only twenty multiple choice questions and is also calculator allowed. The one we will take on exam did will include multiple choice and essay, both of which calculators will not be allowed. Today I was going through the take home portion and I realized I forgot a lot of things we did in the beginning of the year. I thought I knew how to do some problems, but my answers were coming out wrong.
Anyway, two weeks ago we learned how to find volume of a curve. This is done by rotating the curve about an axis or line to make it a solid object. We did this by using both disks and washers. Disks are used when there is only one equation and the object rotated turns out to be a solid. When there are two equations, washers are used since there will be a hole in the figure.
The formula for disks is:
pi S a-b [R(x)]^2 dx
This is basically a definite integral where you plug in the function you are given and square it. Many people forget to square it. This completley changes everything, so don't forget.
The formula for washers is:
pi S a-b top^2-bottom^2 dx
This is also a definite integral where you have to square the functions. I like these because they're very similar to area under a curve, and I also liked those. The only difference is to square both functions given and put a pi on the outside of the integral.
Most of the things we learned recently I'm fine with, but I came to realize I'm confused with a lot of old things. I get average value, average rate of change, and other various things with those words arranged in different ways. I never really caught onto average rate of change, and I still don't know what it is. I also get confused with linerizaton. When I look at the questions, I never know what to do with them. Can someone explain?
Anyway, two weeks ago we learned how to find volume of a curve. This is done by rotating the curve about an axis or line to make it a solid object. We did this by using both disks and washers. Disks are used when there is only one equation and the object rotated turns out to be a solid. When there are two equations, washers are used since there will be a hole in the figure.
The formula for disks is:
pi S a-b [R(x)]^2 dx
This is basically a definite integral where you plug in the function you are given and square it. Many people forget to square it. This completley changes everything, so don't forget.
The formula for washers is:
pi S a-b top^2-bottom^2 dx
This is also a definite integral where you have to square the functions. I like these because they're very similar to area under a curve, and I also liked those. The only difference is to square both functions given and put a pi on the outside of the integral.
Most of the things we learned recently I'm fine with, but I came to realize I'm confused with a lot of old things. I get average value, average rate of change, and other various things with those words arranged in different ways. I never really caught onto average rate of change, and I still don't know what it is. I also get confused with linerizaton. When I look at the questions, I never know what to do with them. Can someone explain?
post 17
This week was mainly a review week in order to get ready for the upcoming midterm exam. We reviewed everything that we learned throughout this year.
Another thing I am going to talk about since I understand is linearization. The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
Also I will talk about integration. Integration is used to find the area under a curve. The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals
Related rates are one thing I understood fromt the start so I am going to talk about them next. The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
Also I am going to talk about taking implicit derivatives since I understand how to take them. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
For things I am not good with is the trapezoidal rule. I get LRAM, RRAM, and MRAM but I just can not get trapezoidal. So if someone can help me before the test that would be great.
Another thing I am going to talk about since I understand is linearization. The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
Also I will talk about integration. Integration is used to find the area under a curve. The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals
Related rates are one thing I understood fromt the start so I am going to talk about them next. The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
Also I am going to talk about taking implicit derivatives since I understand how to take them. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
For things I am not good with is the trapezoidal rule. I get LRAM, RRAM, and MRAM but I just can not get trapezoidal. So if someone can help me before the test that would be great.
post 17
we reviewed mostly this past week so ima show you the formulas andd steps for washers and disks.
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
STEPS:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
im not too good at related rates, optimization, and angles of elevevation still.
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
STEPS:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
im not too good at related rates, optimization, and angles of elevevation still.
post 17
alright so this week in calc we reviewed, reviewed, reviewed, and took two practice ap tests, which i actually did pretty good on ! :-)
calc exam is tuesday, and i just found out that the packet b rob gave us on friday is due the day of the exam! i thought it was our packet for over the christmas holidays! ahhhhh, i need to get started on that. haha. okay well, i'll explain average rate of change.
you are given a problem y=3t+2 and a time iterval from [0,2] seconds. well, your formula is:
f(b)-f(a)/b-a. 0=a,2=b. so, first you must plug in 0 and 2 to find out what your plug ins will be. so you would get [2,8]. then plug it into your formula, 6/2 = 3. so your answer would be three. don't forget to put your units behind it though, whatever that may be.
i'll also, tell you limit rules for finding infinity because i will never ever ever forget these.
if a limit is approaching infinity:
1. top degree < bottom degree it's equal to 0
2. top degree > bottom degree it's equal to +/- infinity. ****
3. top degree = bottom degree divide leading coefficients.
****find out if it's positive or negative infinity by graphing it in your calculator.
also, i'll throw in that volume for disks and washers is squared, while area is not. although it is the exact same formula. figured i would just say that since a lot of people seem to forget those formulas, when really you only have to remember one.
what i dont get still, is LRAM, MRAM, RRAM, TRAM.
also i kinda forgot how to find the equation of a tangent line. don't make funny of me, but i can't remember to save my life and it's driving me crazy cuz i know how simple it is. also, if anyone wants to go over optimization just to refresh my memory, and all the related rates stuff would be nice too :-)
calc exam is tuesday, and i just found out that the packet b rob gave us on friday is due the day of the exam! i thought it was our packet for over the christmas holidays! ahhhhh, i need to get started on that. haha. okay well, i'll explain average rate of change.
you are given a problem y=3t+2 and a time iterval from [0,2] seconds. well, your formula is:
f(b)-f(a)/b-a. 0=a,2=b. so, first you must plug in 0 and 2 to find out what your plug ins will be. so you would get [2,8]. then plug it into your formula, 6/2 = 3. so your answer would be three. don't forget to put your units behind it though, whatever that may be.
i'll also, tell you limit rules for finding infinity because i will never ever ever forget these.
if a limit is approaching infinity:
1. top degree < bottom degree it's equal to 0
2. top degree > bottom degree it's equal to +/- infinity. ****
3. top degree = bottom degree divide leading coefficients.
****find out if it's positive or negative infinity by graphing it in your calculator.
also, i'll throw in that volume for disks and washers is squared, while area is not. although it is the exact same formula. figured i would just say that since a lot of people seem to forget those formulas, when really you only have to remember one.
what i dont get still, is LRAM, MRAM, RRAM, TRAM.
also i kinda forgot how to find the equation of a tangent line. don't make funny of me, but i can't remember to save my life and it's driving me crazy cuz i know how simple it is. also, if anyone wants to go over optimization just to refresh my memory, and all the related rates stuff would be nice too :-)
WEEK 17
Since we took the two practice AP exams this week, I think I'll just explain some problems off of them.
Number 5: If y = e^2x + tan(2x), then y'(pi) =
first take the derivative: y' = e^2x (2) + sec^2(2x)(2)
Then you plug in pi: 2e^2pi + 2sec(2pi)^2 = 2e^2pi + 2
Number 16: If y = sin^2(5x), dy/dx =
dy/dx = 2 (sin 5x)(cos 5x)(5) <--- take care of exponent, take derivative of sin, take derivative of the inside (5x).
dy/dx = 5 (sin(2 * 5x)) <--- 2sinxcosx = sin2x
dy/dx = 5sin(10x)
Number 18:
remember that f(x) = x
f'(x) = v(x)
f''(x) = a(x)
Number 20: If f(1) = 2 and f'(1) = 5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).
Given: x = 1
y = 2
m = 5
find: x = 1.2
y - 2 = 5 (x - 1)
y - 2 = 5 (1.2 - 1)
y = 3
Number 27:
Always remember that AVERAGE VALUE = INTEGRATION
1/ b-a
Number 28: y = x^2 - 3x. Find y'(1).
y'(x) = 2x - 3
y'(1) = 2(1) - 3
y'(1) = -1
y'(1) = 1
Number 11:
remember that average rate of change = f(b) - f(a) / b - a
Number 5: If y = e^2x + tan(2x), then y'(pi) =
first take the derivative: y' = e^2x (2) + sec^2(2x)(2)
Then you plug in pi: 2e^2pi + 2sec(2pi)^2 = 2e^2pi + 2
Number 16: If y = sin^2(5x), dy/dx =
dy/dx = 2 (sin 5x)(cos 5x)(5) <--- take care of exponent, take derivative of sin, take derivative of the inside (5x).
dy/dx = 5 (sin(2 * 5x)) <--- 2sinxcosx = sin2x
dy/dx = 5sin(10x)
Number 18:
remember that f(x) = x
f'(x) = v(x)
f''(x) = a(x)
Number 20: If f(1) = 2 and f'(1) = 5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).
Given: x = 1
y = 2
m = 5
find: x = 1.2
y - 2 = 5 (x - 1)
y - 2 = 5 (1.2 - 1)
y = 3
Number 27:
Always remember that AVERAGE VALUE = INTEGRATION
1/ b-a
Number 28: y = x^2 - 3x. Find y'(1).
y'(x) = 2x - 3
y'(1) = 2(1) - 3
y'(1) = -1
y'(1) = 1
Number 11:
remember that average rate of change = f(b) - f(a) / b - a
Post #17
So after 11 make-up posts I ran out of stuff to say, so i'm going back to the beginning of the year and talking about some stuff.
Complex Derivatives y=ln(e^x) (Chain Rule)
First off one should should identify the steps of your problem. In this case they would be:
1. Natural Log
2. e^x
you problem should be (1/(e^x)).(e^x)'
then you find the derivative of e^x which is e^x . x' (x'=1)
so your final problem should be (1/(e^x)).(e^x)
After this you have to simplify algebraically, giving you (e^x)/(e^x) ,which equals 1.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Complex Derivatives y=ln(e^x) (Chain Rule)
First off one should should identify the steps of your problem. In this case they would be:
1. Natural Log
2. e^x
you problem should be (1/(e^x)).(e^x)'
then you find the derivative of e^x which is e^x . x' (x'=1)
so your final problem should be (1/(e^x)).(e^x)
After this you have to simplify algebraically, giving you (e^x)/(e^x) ,which equals 1.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Post #17
Alrighty then, lets get started.
Today during our study group, I learned average value thanks to Kaitlyn! So I guess I will explain an example of that since I finally get it.
EX: Find the average value of the function f(x)=12x-6x^2 over the interval -5 less than or equal to x less than or greater than 5.
use the formula: 1/b-a S equation given, then solve the definite integral (*S is integral)
1/5-(-5)= 1/10
1/10 (integral form -5 to 5) 12x-6x^2
1/10(12x-2x^3)
1/10(12(5)-2(5)^3)-1/10(12(-5)-2(-5)^3)
1/10(-190)-1/10(190)
-19-(-19)= -38
I do have quite a few questions though..
1. If I get (-1/64)sin(1/8) and I'm suppose to set it equal to 0 then solve..How do I solve that?
2. (like #11 on the packet) For S(3t/t^2+2)^3 I used substitution to solve. For my final answer I got -3/2(t^2+2)^2+c but all the answers have -3/4(t^2+2)^2+c
3. (like #3 on the packet) It says use a graphing utility to graph the function f(x)=12/6-x and locate the absolute extrema of the function on the interval [0,6). If you plug that into your calculator the graph is an asymptote, and I tried taking the derivative and plugging that in too but it is also an asymptote. So how do you get a max and min?
Today during our study group, I learned average value thanks to Kaitlyn! So I guess I will explain an example of that since I finally get it.
EX: Find the average value of the function f(x)=12x-6x^2 over the interval -5 less than or equal to x less than or greater than 5.
use the formula: 1/b-a S equation given, then solve the definite integral (*S is integral)
1/5-(-5)= 1/10
1/10 (integral form -5 to 5) 12x-6x^2
1/10(12x-2x^3)
1/10(12(5)-2(5)^3)-1/10(12(-5)-2(-5)^3)
1/10(-190)-1/10(190)
-19-(-19)= -38
I do have quite a few questions though..
1. If I get (-1/64)sin(1/8) and I'm suppose to set it equal to 0 then solve..How do I solve that?
2. (like #11 on the packet) For S(3t/t^2+2)^3 I used substitution to solve. For my final answer I got -3/2(t^2+2)^2+c but all the answers have -3/4(t^2+2)^2+c
3. (like #3 on the packet) It says use a graphing utility to graph the function f(x)=12/6-x and locate the absolute extrema of the function on the interval [0,6). If you plug that into your calculator the graph is an asymptote, and I tried taking the derivative and plugging that in too but it is also an asymptote. So how do you get a max and min?
Post #17
so..i've been semi-lost this week..so i'm kicking it back, old school.
So, product rule and quotient rule.
First, product rule is done when you need to take the derivative of things being mulitplied. So, it would be something like x(x^2)
So, the first thing you do, is copy the first term times the derivative of the second term plus the copied second term times the derivative of the first term. Product rule is very simple, you just have to remember the rules of simplificiation. Remember to simplify correctly and if you don't remember what to do next on the simplification process, it probably means you're done simplifying.
Now, lets talk about quotient rule..you know you must use quotient rule because you have a fraction.
The rules for quotient rule is copy the bottom times derivative of the top, minus copy the top times the derivative of the bottom. Like product rule, you need to simplify jsut the same. You need to distribute the things necessary and solve it to the simplest terms possible.
The thing i don't understand is the substitution stuff..i always get the same answer after i do the substitution and don't understand the integral.
So, product rule and quotient rule.
First, product rule is done when you need to take the derivative of things being mulitplied. So, it would be something like x(x^2)
So, the first thing you do, is copy the first term times the derivative of the second term plus the copied second term times the derivative of the first term. Product rule is very simple, you just have to remember the rules of simplificiation. Remember to simplify correctly and if you don't remember what to do next on the simplification process, it probably means you're done simplifying.
Now, lets talk about quotient rule..you know you must use quotient rule because you have a fraction.
The rules for quotient rule is copy the bottom times derivative of the top, minus copy the top times the derivative of the bottom. Like product rule, you need to simplify jsut the same. You need to distribute the things necessary and solve it to the simplest terms possible.
The thing i don't understand is the substitution stuff..i always get the same answer after i do the substitution and don't understand the integral.
17th post
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
The steps are:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
related rates:
The steps for related rates are….
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
limits:
Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
To find area between curves:
The formula you use is b(int)a (top eq.) - (bottom eq.).
If a and b is not already given to you, then you much set the equations equal to each other and solve.
You find which equation is top/bottom by graphing both and simply looking to see which one is on top.
If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
The formula for the area of washers is S (top) - (bottom)
The steps are:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
related rates:
The steps for related rates are….
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
limits:
Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
To find area between curves:
The formula you use is b(int)a (top eq.) - (bottom eq.).
If a and b is not already given to you, then you much set the equations equal to each other and solve.
You find which equation is top/bottom by graphing both and simply looking to see which one is on top.
If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
Post....whatever!
So, Let's go with implicit differentiation...just solve for dy / dx.
Example 1: Use implicit differentiation to find the derivative dy / dx where y x + sin y = 1
Solution to Example 1:
Use the sum rule of differentiation to the whole term on the left of the given equation.
d [xy] / dx + d [siny] / dx = d[1]/dx .
Differentiate each term above using product rule to d [xy] / dx and the cain rule to d [siny] / dx.
x dy / dx + y + (dy / dx) cos(y) = 0 .
Note that in calculating d [siny] / dx, we used the chain rule since y is itself a function of x and sin (y) is a function of a function.
Solve for dy/dx to obtain.
dy / dx = -y / (x + cos y)
Example 2: Use implicit differentiation to find the derivative dy / dx where y 4 + x y 2 + x = 3
Solution to Example 2:
Use the differentiation of a sum formula to left side of the given equation.
d[y 4] / dx + d[x y 2] / dx + d[x] / dx = d[3] / dx
Differentiate each term above using power rule, product rule and chain rule.
4y 3 dy / dx + (1) y 2 + x 2y dy / dx + 1 = 0
Solve for dy/dx.
dy/dx = (-1 - y 2) / (4y 3 + 2xy)
Example 3: Find all points on the graph of the equation
x 2 + y 2 = 4
where the tangent lines are parallel to the line x + y = 2
Solution to Example 3:
Rewrite the given line x + y = 2 in slope intercept form: y = -x + 2 and identify the slope as m = -1. The tangent lines are parallel to this line and therefore their slope are equal to -1. The slope of tangent lines at a point can be found by implicity differentiation of x 2 + y 2 = 4
2x + 2y dy/dx = 0
Let P(a , b) be the point of tangency. At point P the slope is -1. Substituting x by a, y by b and dy/dx by -1 in the above equation, we obtain
2a + 2b (-1) = 0
Point P(a , b) is on the graph of x 2 + y 2 = 4, hence
a 2 + b 2 = 4
Solve the system of equations: 2a - 2b = 0 and a 2 + b 2 = 4 to obtain two points
(-sqrt(2) , -sqrt(2)) and (sqrt(2) , sqrt(2))
I DO NOT UNDERSTAND LINEARIZATION????????!!!?!!!!!!! HELLLLPPPPP...i need somebody...>HELLLPPPP....not just anybody.....HELLLLPPPP....i need someone.......anddddd sommmeetthinnng...idk
BYE BYE ADIOS AMIGOS!!! HASTA MANANA!!!
Example 1: Use implicit differentiation to find the derivative dy / dx where y x + sin y = 1
Solution to Example 1:
Use the sum rule of differentiation to the whole term on the left of the given equation.
d [xy] / dx + d [siny] / dx = d[1]/dx .
Differentiate each term above using product rule to d [xy] / dx and the cain rule to d [siny] / dx.
x dy / dx + y + (dy / dx) cos(y) = 0 .
Note that in calculating d [siny] / dx, we used the chain rule since y is itself a function of x and sin (y) is a function of a function.
Solve for dy/dx to obtain.
dy / dx = -y / (x + cos y)
Example 2: Use implicit differentiation to find the derivative dy / dx where y 4 + x y 2 + x = 3
Solution to Example 2:
Use the differentiation of a sum formula to left side of the given equation.
d[y 4] / dx + d[x y 2] / dx + d[x] / dx = d[3] / dx
Differentiate each term above using power rule, product rule and chain rule.
4y 3 dy / dx + (1) y 2 + x 2y dy / dx + 1 = 0
Solve for dy/dx.
dy/dx = (-1 - y 2) / (4y 3 + 2xy)
Example 3: Find all points on the graph of the equation
x 2 + y 2 = 4
where the tangent lines are parallel to the line x + y = 2
Solution to Example 3:
Rewrite the given line x + y = 2 in slope intercept form: y = -x + 2 and identify the slope as m = -1. The tangent lines are parallel to this line and therefore their slope are equal to -1. The slope of tangent lines at a point can be found by implicity differentiation of x 2 + y 2 = 4
2x + 2y dy/dx = 0
Let P(a , b) be the point of tangency. At point P the slope is -1. Substituting x by a, y by b and dy/dx by -1 in the above equation, we obtain
2a + 2b (-1) = 0
Point P(a , b) is on the graph of x 2 + y 2 = 4, hence
a 2 + b 2 = 4
Solve the system of equations: 2a - 2b = 0 and a 2 + b 2 = 4 to obtain two points
(-sqrt(2) , -sqrt(2)) and (sqrt(2) , sqrt(2))
I DO NOT UNDERSTAND LINEARIZATION????????!!!?!!!!!!! HELLLLPPPPP...i need somebody...>HELLLPPPP....not just anybody.....HELLLLPPPP....i need someone.......anddddd sommmeetthinnng...idk
BYE BYE ADIOS AMIGOS!!! HASTA MANANA!!!
Saturday, December 12, 2009
17th post
This week in calculus we took two practice AP exams. One was calculator allowed and the other one was not. I found them to be hard but i didn't do to bad on them lol. Anyways, i'll start off by telling some of the things i know pretty well:
Average Value:
This is something i didn't always understand but today i learned how to do it and it seems pretty easy to me now. You start by understand you will have a number in front of the integral and you find it by 1/b-a... then you have to integrate the function and plug in the numbers. Once you have done that you solve.
Another thing i understand well is Rolle's Theorum: If the function is continuous and differentiable when you plug in the two points they give you.. then Rolle's Theorum can be used. After that, you take the derivative of the function and set it equal to zero and solve for x. Your x value will be your answer.
For mean value theorum, you also have to figure out whether the function is continuous and differentiable by plugging in the points given in the original function. This time, instead of setting the derivative equal to zero, you set it equal to the slope of b and a. To find the slope, you plug into f(b)-f(a)/b-a. Then you set the derivative equal to that number and solve for x.
Another thing is finding critical values. To find critical values, you take the derivative of the function and set it equal to zero and solve for x. What ever number or numbers you get for x is your critical values. Absolute extrema is pretty simple as well. If they give you a point in the problem, then you plug those numbers into the original function to get another number. You also solve for your critical values and plug those into the original function as well. Once you get your second numbers, you set each pair into new sets of points. Your highest point will be your absolute max and you smallest point will be your absolute min.
For some things i do not understand:
I am still having trouble with the trapezoidal rule. I am also having problems with trig inverse functions and solving for x with them and setting them up into intervals. If anyone can help me i would appreciate it. Thanks :)
Average Value:
This is something i didn't always understand but today i learned how to do it and it seems pretty easy to me now. You start by understand you will have a number in front of the integral and you find it by 1/b-a... then you have to integrate the function and plug in the numbers. Once you have done that you solve.
Another thing i understand well is Rolle's Theorum: If the function is continuous and differentiable when you plug in the two points they give you.. then Rolle's Theorum can be used. After that, you take the derivative of the function and set it equal to zero and solve for x. Your x value will be your answer.
For mean value theorum, you also have to figure out whether the function is continuous and differentiable by plugging in the points given in the original function. This time, instead of setting the derivative equal to zero, you set it equal to the slope of b and a. To find the slope, you plug into f(b)-f(a)/b-a. Then you set the derivative equal to that number and solve for x.
Another thing is finding critical values. To find critical values, you take the derivative of the function and set it equal to zero and solve for x. What ever number or numbers you get for x is your critical values. Absolute extrema is pretty simple as well. If they give you a point in the problem, then you plug those numbers into the original function to get another number. You also solve for your critical values and plug those into the original function as well. Once you get your second numbers, you set each pair into new sets of points. Your highest point will be your absolute max and you smallest point will be your absolute min.
For some things i do not understand:
I am still having trouble with the trapezoidal rule. I am also having problems with trig inverse functions and solving for x with them and setting them up into intervals. If anyone can help me i would appreciate it. Thanks :)
Post #17
Alright everyone, we have Calculus Exam Packets this weekend to start on our exam which is comming up on monday! I say we all get together!..ha
but anyways, this past week we took a practice AP Exam.
The first portion was without a calculator so we need to know and remember our formulas and how to read graphs.
~remember that an odd function is symetric about the x axis and an even function is symetric about the y axis.
~remember how to solve in order to get an equation of a tangent line:
take the derivative and find the slope then since you need a point you'll plug in the number they give you to the original equation
~in order to find vertical asymptotes set the denominator equal to zero.
~know that when you’re looking at the derivative of a graph and you want to find a graph that is the original, if it’s increasing then you know that it will have to have a positive slope.
~Average value is not the same thing as average speed!...remember is one over b-a times the integral of the equation.
Tuesday we were unable to get to school due to flooding, so we were off!
Wednesday when we came back, we took the second portion of the AP Exam which allowed calculators!
~if you need to find values that the slope of a tangent line is equal to a certain number:
Take the derivative and set it equal to that number when you solving with a calculator go to y = and plug in the equation look at the graph and go to calc zero and guess what they are and you’ll have your answer!
~When you want to find the smallest positive value, plug the equation into y = and then adjust window, go to second trace and maximum guess the left and right then guess what the max should be and they’ll tell you the answer!
~remember if you need to find the integral of something then go to fnInt( and plug in the equation}, X, the number, the other number [that’s at the top and bottom of the integral] and that will give you the answer!
~know if you’re looking at the derivative of the graph, in order to find the maximums, look at the zeros. [and maximums must go from positive to negative]
~remember the equation for average rate of change: f[b]-f[a] all over b-a then plug in using the points they give you and solve!
Thursday and Friday we went over all of the things we had wrong on the practice AP Exam! B-Rob said that we should've got at least 10 questions right.
DO NOT FORGET:
We have the packet for our take home portion of our exam to do! Good Luck to everyone!
~ElliE~
but anyways, this past week we took a practice AP Exam.
The first portion was without a calculator so we need to know and remember our formulas and how to read graphs.
~remember that an odd function is symetric about the x axis and an even function is symetric about the y axis.
~remember how to solve in order to get an equation of a tangent line:
take the derivative and find the slope then since you need a point you'll plug in the number they give you to the original equation
~in order to find vertical asymptotes set the denominator equal to zero.
~know that when you’re looking at the derivative of a graph and you want to find a graph that is the original, if it’s increasing then you know that it will have to have a positive slope.
~Average value is not the same thing as average speed!...remember is one over b-a times the integral of the equation.
Tuesday we were unable to get to school due to flooding, so we were off!
Wednesday when we came back, we took the second portion of the AP Exam which allowed calculators!
~if you need to find values that the slope of a tangent line is equal to a certain number:
Take the derivative and set it equal to that number when you solving with a calculator go to y = and plug in the equation look at the graph and go to calc zero and guess what they are and you’ll have your answer!
~When you want to find the smallest positive value, plug the equation into y = and then adjust window, go to second trace and maximum guess the left and right then guess what the max should be and they’ll tell you the answer!
~remember if you need to find the integral of something then go to fnInt( and plug in the equation}, X, the number, the other number [that’s at the top and bottom of the integral] and that will give you the answer!
~know if you’re looking at the derivative of the graph, in order to find the maximums, look at the zeros. [and maximums must go from positive to negative]
~remember the equation for average rate of change: f[b]-f[a] all over b-a then plug in using the points they give you and solve!
Thursday and Friday we went over all of the things we had wrong on the practice AP Exam! B-Rob said that we should've got at least 10 questions right.
DO NOT FORGET:
We have the packet for our take home portion of our exam to do! Good Luck to everyone!
~ElliE~
Friday, December 11, 2009
fifth reflection
hmm, well i was gonna do my fifth reflection on area of disks, but it is almost the exact same thing as volume, so i decided against it. and i looked through some stuff and decided i am going to do this one on Riemann Sums!!! (particularly the right Riemann Sum because that is my favorite)
The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. You will need to know this because each of the next formulas require that you know what x is.
LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
alright, they are all basically worked the same, the only difference is the formulas, so I am just going to work an MRAM problem because that is just a good one i came across.
Calculate the right Riemann Sum for q(x)=-x^2-3x on the interval [-2,1] divided into 4 subintervals
so first, you have to find your delta x which you get from using the formula b-a/n and in this specific example, would be 1+2/4, which is 3/4.
Next you just plug into the formula
3/4[f(-1.25)+f(-.5)+f(.25)+f(1)
so then you plug in those values into the formula they gave you to get:
3/4[-(35/16)-1.25-(11/16)-2]
which is then simplified to -147/32
and that is it. for the other ones you just plug into the other formulas to get the answer, but it is worked basically the same way.
yay, i'm finally done all my reflections
The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. You will need to know this because each of the next formulas require that you know what x is.
LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
alright, they are all basically worked the same, the only difference is the formulas, so I am just going to work an MRAM problem because that is just a good one i came across.
Calculate the right Riemann Sum for q(x)=-x^2-3x on the interval [-2,1] divided into 4 subintervals
so first, you have to find your delta x which you get from using the formula b-a/n and in this specific example, would be 1+2/4, which is 3/4.
Next you just plug into the formula
3/4[f(-1.25)+f(-.5)+f(.25)+f(1)
so then you plug in those values into the formula they gave you to get:
3/4[-(35/16)-1.25-(11/16)-2]
which is then simplified to -147/32
and that is it. for the other ones you just plug into the other formulas to get the answer, but it is worked basically the same way.
yay, i'm finally done all my reflections
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