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.
Saturday, December 26, 2009
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!
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