Since some people seem to be having trouble with optimization, I'm going to help by presenting some steps and various example problems.
Steps:
1. Read each problem slowly and carefully. Read the problem at least three times before trying to solve it. Sometimes words are tricky and unimportant. Find out exactly what the problem is asking. If you misread the problem or hurry through it, you have NO chance of solving it correctly (coming from my experience...).
2. If appropriate, draw a sketch or diagram of the problem to be solved. Pictures are a great help in organizing and sorting out your thoughts.
3. Define variables to be used and carefully label your picture or diagram with these variables.
4. Write down all equations which are related to your problem or diagram. Clearly label the equation which you are asked to maximize or minimize. In most problems, you are given an equation that you're optimizing and an equation that you solve for one variable.
Before differentiating, make sure that the optimization equation is a function of only one variable.
Example 1: Build a rectangular pen with three parallel partitions using 500 feet of fencing. What dimensions will maximize the total area of the pen ?
Solution 1: Where x is the width and y the length
The total amount of fencing is given to be
500 = 5 (width) + 2 (length) = 5x + 2y ,
so that
2y = 500 - 5x
or
y = 250 - (5/2)x .
We wish to MAXIMIZE the total AREA of the pen
A = (width) (length) = x y .
However, before we differentiate the right-hand side, we will write it as a function of x only. Substitute for y getting
A = x y
= x ( 250 - (5/2)x)
= 250x - (5/2)x2 .
Now differentiate this equation, getting
A' = 250 - (5/2) 2x
= 250 - 5x
= 5 (50 - x )
= 0
for
x=50 .
Now since we have the x, we must plug 50 into the equation in which we solved for the y giving y = 125.
So, If
x=50 ft. and y=125 ft. ,
then
A = 6250 ft.2
is the largest possible area of the pen.
OK!!! That was a lot and I dont feel like doing another example...SO, that's my math!!
SEE YA!
Saturday, January 2, 2010
Second of the Holiday Posts
Hope everybody had a good New Year's...
I'll start with Integration for this post.
Sooo, Integration is the area under a curve, and riemann sums are approxamations of this area using rectangles or trapezoids
You can use LRAM, RRAM, MRAM, or Trapezoidal to do these approxamations.
LRAM is when you start estimating from the left side by drawing rectangles from the x-axis up to the graph and then go over to the right and go down.
Equation form: delta x [f(a) + f(a + delta x) + ... + f(b - delta x)]
RRAM is when you start estimating from the right side by drawing rectangles from the end of the interval of the graph being looked at to where you want to go down with the rectangle.
Equation form: deltax [f(a + delta x) + f(b)]
MRAM is the midpoints and basically a combination of RRAM and LRAM.
Equation form: delta x [f(mid point 1) + f(mid point 2) + f(mid point n + 1)]
Trapezoidal Riemann Sums are the most accurate of the 4 because with the trapezoid, you can get closer to the curve.
Equation form: delta x / 2 [f(a) + 2f(a + delta x) + 2f(a + 2delta x) + ... + f(b)]
Oh, I forgot. A and B are also [a,b], which is the interval the curve is on. Delta x is equal to (b - a) / n.
Well I guess I'll go into the exact integration on the third post. Again, I hope everybody had a nice two weeks off and see you all on Monday. :)
I'll start with Integration for this post.
Sooo, Integration is the area under a curve, and riemann sums are approxamations of this area using rectangles or trapezoids
You can use LRAM, RRAM, MRAM, or Trapezoidal to do these approxamations.
LRAM is when you start estimating from the left side by drawing rectangles from the x-axis up to the graph and then go over to the right and go down.
Equation form: delta x [f(a) + f(a + delta x) + ... + f(b - delta x)]
RRAM is when you start estimating from the right side by drawing rectangles from the end of the interval of the graph being looked at to where you want to go down with the rectangle.
Equation form: deltax [f(a + delta x) + f(b)]
MRAM is the midpoints and basically a combination of RRAM and LRAM.
Equation form: delta x [f(mid point 1) + f(mid point 2) + f(mid point n + 1)]
Trapezoidal Riemann Sums are the most accurate of the 4 because with the trapezoid, you can get closer to the curve.
Equation form: delta x / 2 [f(a) + 2f(a + delta x) + 2f(a + 2delta x) + ... + f(b)]
Oh, I forgot. A and B are also [a,b], which is the interval the curve is on. Delta x is equal to (b - a) / n.
Well I guess I'll go into the exact integration on the third post. Again, I hope everybody had a nice two weeks off and see you all on Monday. :)
Friday, January 1, 2010
A post..Jan. 1, 2010!!
Ok, so to review a couple of things (at 3 o'clock in the morning..insomnia).
this is the second derivative test.
x^3 - 6x^2 + 12x
take the first derivative:
3x^2 -12x + 12
take the second derivative:
6x - 12
Set equal to 0 and solve:
x=2
Set up intervals:
(-inf., 2) U (2, inf.)
Plug in values within each interval:
-ve and +ve
And if the question asked you about the concavity, it would be concave up because you are going from decreasing to increasing.
So the above example problem is just basic, and I'm pretty sure if you were to get this problem on a test, you'd kill it. Now, maybe something a little more tedius? Perhaps?
Related Rates maybe...possibly?? Okay, so to start off on the steps:
1. Identify all of the variables and equations.
2. Identify what you want to find.
3. Sketch and label.
4. Write an equations involving your variables.
5. Take the derivative with in terms of time.
6. Substitute the derivative in and solve.
So say a spherical ball has a radius that in increasing at a rate of 4 cm/s. What is the change in volume when the radius equals 8 cm?
I get dr/dt = 4
r = 8
Since you are looking for dv/dt, you take the derivative of the volume formula. We know by definition that the volume of a spherical object is 4/3 pi r^3.
So:
dv/dt = 4/3 pi 3r^2 dr/dt
now plug in your respective data.
dv/dt = (4/3)(pi)3(8 cm)^2(4 cm/s)
Solve:
dv/dt = 1024pi cm^2/s.
And thats your answer!!! think you've got it! It really helps if you have a picture, but unfortunately I am at a loss to do so. See ya later!
this is the second derivative test.
x^3 - 6x^2 + 12x
take the first derivative:
3x^2 -12x + 12
take the second derivative:
6x - 12
Set equal to 0 and solve:
x=2
Set up intervals:
(-inf., 2) U (2, inf.)
Plug in values within each interval:
-ve and +ve
And if the question asked you about the concavity, it would be concave up because you are going from decreasing to increasing.
So the above example problem is just basic, and I'm pretty sure if you were to get this problem on a test, you'd kill it. Now, maybe something a little more tedius? Perhaps?
Related Rates maybe...possibly?? Okay, so to start off on the steps:
1. Identify all of the variables and equations.
2. Identify what you want to find.
3. Sketch and label.
4. Write an equations involving your variables.
5. Take the derivative with in terms of time.
6. Substitute the derivative in and solve.
So say a spherical ball has a radius that in increasing at a rate of 4 cm/s. What is the change in volume when the radius equals 8 cm?
I get dr/dt = 4
r = 8
Since you are looking for dv/dt, you take the derivative of the volume formula. We know by definition that the volume of a spherical object is 4/3 pi r^3.
So:
dv/dt = 4/3 pi 3r^2 dr/dt
now plug in your respective data.
dv/dt = (4/3)(pi)3(8 cm)^2(4 cm/s)
Solve:
dv/dt = 1024pi cm^2/s.
And thats your answer!!! think you've got it! It really helps if you have a picture, but unfortunately I am at a loss to do so. See ya later!
Thursday, December 31, 2009
post 20
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).
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).
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!
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