calculussssssssss
Optimization:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
Finding absolute max/min:
1. First derivative test
2. Plug critical values into the original function to get y-values
3. Plug endpoints into the original function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
Volume by disks:
Pi bSa [R(x)]^2 dx
Volume by washers:
pi bSa (top equation)^2 – (bottom equation)^2 dx
Limit Rules:
1. if the degree of the top is bigger than the degree of the bottom, the limit is infinity.
2. if the degree of the top is smaller than the degree of the bottom, the limit is 0.
3. if the degree of the top is equal to the degree of the bottom, the limit is the coefficient of the leading term of the top divided by the coefficient of the leading term of the bottom equation.
First derivative test:
-Take the derivative of the original function
-Solve for x (the values will be your critical values)
-Set those values up into intervals between negative infinity and infinity
- Plug in numbers between the intervals into the function
-This will show you when the function is increasing, decreasing, and you will find max's and mins
Second derivative test:
-Take the derivative of the original function twice
-Solve for x values(critical values)
-Set up into intervals between infinity and negative infinity
-Plug in values between the intervals into the function
-This will show you where the graph is concave up and down, and where there is a point of inflection.
i am not good at some integrals and ooptimization.
Sunday, April 25, 2010
Post #36
Tangent Lines:
To find a tangent line, you need a slope and a point.
The problem usually gives an equation and a x-value.
To find the slope, take the derivative of the equation and plug in the x-value given.
To find the point, use the x-value and plug it into the original equation to find the y-value.
(x,y) will be your point.
From there, put the point and slope in point-slope form, which is y-y1 = m (slope) (x-x1)
Example: Find the line tangent to x^3 - 2x^2 + 4x at x=1
Take derivative and plug in x: 3x^2 - 4x +4
3(1)^2 - 4(1) +4 = 3
The slope is m =3
Plug in x to find y: (1)^3 - 2(1)^2 +4(1) = 3
Point slope form: y-3 = 3 (x-1)
The answer choices may have the equation written in a different form.
Another option is solved for y: y-3 = 3x -3 = y = 3x
You can also be asked to just find the slope of the line tangent to a center graph. In this case the answer would be 3.
Normal lines:
The equation of a normal line is solve by using the same steps as the equation of the tangent line except instead of using the slope, you have to use the perpendicular slope, which is the negative reciprocal of the slope.
Using the same example:
Take derivative: 3x^2 - 4x +4
Plug in: 3(1)^2 -4(1) +4 = 3
Now take the negative reciprocal of the slope: -1/3
Your point will be the same as it was for the tangent line (1,3)
Now you plug into point slope using the negative reciprocal slope
y-3 = -1/3 (x-1)
Again, this problem may be written differently such as 3y-9 = -1 (x-1)
3y - 9 = -x +1
x+3y = 10
L'Hopital's Rule:
This is used to find the limit when it originally equals 0/0. The mistake is often made by saying the limit is nonexistent, but L'hopital's rule states the take the derivative of the top and the derivative of the bottom then plug in the number given to find the limit. If it still does not work, you have to continue to take the derivative of each separately until it does.
Example: Lim as y->0 sin y / y
Sin (0) / 0 = 0/0 so L'hopital's rule can be applied
the derivative of sin y is cos y
the derivative of y is 1
now it is the lim y ->0 cos y/ 1
cos 1 = 1
1/1 = 1
The limit is 1.
I can use help on finding limits as they approach the left or the right only because I seem to have forgotten how to do those.
To find a tangent line, you need a slope and a point.
The problem usually gives an equation and a x-value.
To find the slope, take the derivative of the equation and plug in the x-value given.
To find the point, use the x-value and plug it into the original equation to find the y-value.
(x,y) will be your point.
From there, put the point and slope in point-slope form, which is y-y1 = m (slope) (x-x1)
Example: Find the line tangent to x^3 - 2x^2 + 4x at x=1
Take derivative and plug in x: 3x^2 - 4x +4
3(1)^2 - 4(1) +4 = 3
The slope is m =3
Plug in x to find y: (1)^3 - 2(1)^2 +4(1) = 3
Point slope form: y-3 = 3 (x-1)
The answer choices may have the equation written in a different form.
Another option is solved for y: y-3 = 3x -3 = y = 3x
You can also be asked to just find the slope of the line tangent to a center graph. In this case the answer would be 3.
Normal lines:
The equation of a normal line is solve by using the same steps as the equation of the tangent line except instead of using the slope, you have to use the perpendicular slope, which is the negative reciprocal of the slope.
Using the same example:
Take derivative: 3x^2 - 4x +4
Plug in: 3(1)^2 -4(1) +4 = 3
Now take the negative reciprocal of the slope: -1/3
Your point will be the same as it was for the tangent line (1,3)
Now you plug into point slope using the negative reciprocal slope
y-3 = -1/3 (x-1)
Again, this problem may be written differently such as 3y-9 = -1 (x-1)
3y - 9 = -x +1
x+3y = 10
L'Hopital's Rule:
This is used to find the limit when it originally equals 0/0. The mistake is often made by saying the limit is nonexistent, but L'hopital's rule states the take the derivative of the top and the derivative of the bottom then plug in the number given to find the limit. If it still does not work, you have to continue to take the derivative of each separately until it does.
Example: Lim as y->0 sin y / y
Sin (0) / 0 = 0/0 so L'hopital's rule can be applied
the derivative of sin y is cos y
the derivative of y is 1
now it is the lim y ->0 cos y/ 1
cos 1 = 1
1/1 = 1
The limit is 1.
I can use help on finding limits as they approach the left or the right only because I seem to have forgotten how to do those.
POST 36
Q:An equation of the line tangent to the graph of y=cos(2x) at x=pi/4 is
tangent line steps: take derivative, plug in x to get slope, use point slope formula, and if not given y plug x value into original
A: cos(2x)
-2sin(2x)
-2sin2(pi/4)
-2sin(pi/2)
-2=slope
cos(2(pi/4))
cos(pi/2)=0
y=-2(x-pi/4)
3. Let f be a function defined for all real numbers x. If f^1(x)=(4-x^2)/x2, then f is decreasing on the interval *(4-x^2 is really the absolute value of 4-x^2)
first derivative test!
f^1(x)=(4-x^2)/x2 =0
4-x^2=0
-x$2=4
x=+ or - 2
(-infinity,-2) (-2,2) (2,infinity)
plug in -3=-ve/decreasing
plug in 0=-ve/decreasing
plug in 3=+ve/increasing
So decreasing on the interval (-infinity,2)
The limit rules are:
1. if the degree of the top is bigger than the degree of the bottom, the limit is infinity.
2. if the degree of the top is smaller than the degree of the bottom, the limit is 0.
3. if the degree of the top is equal to the degree of the bottom, the limit is the coefficient of the leading term of the top divided by the coefficient of the leading term of the bottom equation.
The steps for optimization are:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
The steps to find maximums and minimums are:
1. First derivative test
2. Plug critical values into the original function to get y-values
3. Plug endpoints into the original function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
The steps for related rates are:
1.Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
DONT KNOW LIMITS!!!!!!!!!!!!!!!!
tangent line steps: take derivative, plug in x to get slope, use point slope formula, and if not given y plug x value into original
A: cos(2x)
-2sin(2x)
-2sin2(pi/4)
-2sin(pi/2)
-2=slope
cos(2(pi/4))
cos(pi/2)=0
y=-2(x-pi/4)
3. Let f be a function defined for all real numbers x. If f^1(x)=(4-x^2)/x2, then f is decreasing on the interval *(4-x^2 is really the absolute value of 4-x^2)
first derivative test!
f^1(x)=(4-x^2)/x2 =0
4-x^2=0
-x$2=4
x=+ or - 2
(-infinity,-2) (-2,2) (2,infinity)
plug in -3=-ve/decreasing
plug in 0=-ve/decreasing
plug in 3=+ve/increasing
So decreasing on the interval (-infinity,2)
The limit rules are:
1. if the degree of the top is bigger than the degree of the bottom, the limit is infinity.
2. if the degree of the top is smaller than the degree of the bottom, the limit is 0.
3. if the degree of the top is equal to the degree of the bottom, the limit is the coefficient of the leading term of the top divided by the coefficient of the leading term of the bottom equation.
The steps for optimization are:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
The steps to find maximums and minimums are:
1. First derivative test
2. Plug critical values into the original function to get y-values
3. Plug endpoints into the original function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
The steps for related rates are:
1.Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
DONT KNOW LIMITS!!!!!!!!!!!!!!!!
Post # 36
Okay, its getting down to the wire here and we have the AP test in like a week and a few days...So i'm thinking I should just say the things that are on the top of my head and hopefully it helps them stick in my head!
So, first, something I FINALLY understand: TANGENT LINES.
1. Look at the information given, its usually an x-value, and an equation.
2. Solve for the y-value, which means plug in the x-value into ORIGINAL equaiton.
3. Solve for slope, plug in the x-values to the derivative of the equaiton.
Then put it in point slope form.
Now, lets go over simple things like implicit derivatives.
-So, first you need to recognize that its an implicit derivative, you know it is bc it is going to have an x and y variable and its going to be equal to a number.
-You do the same thing as you would for a regular derivative except everytime you derive y, you put dy/dx behind it, noting you took the derivative of y.
Now, its getting really close to the ap test, and thanks to so much practicing, i'm finally beginning to get confident. If only i can keep my smarts on the multiple choice parts, i can possibly make a passing score!
Questions: umm, when you have an integral and you neeed to figure out what to substitute, and there is no trig function, what do you put as your u?
So, first, something I FINALLY understand: TANGENT LINES.
1. Look at the information given, its usually an x-value, and an equation.
2. Solve for the y-value, which means plug in the x-value into ORIGINAL equaiton.
3. Solve for slope, plug in the x-values to the derivative of the equaiton.
Then put it in point slope form.
Now, lets go over simple things like implicit derivatives.
-So, first you need to recognize that its an implicit derivative, you know it is bc it is going to have an x and y variable and its going to be equal to a number.
-You do the same thing as you would for a regular derivative except everytime you derive y, you put dy/dx behind it, noting you took the derivative of y.
Now, its getting really close to the ap test, and thanks to so much practicing, i'm finally beginning to get confident. If only i can keep my smarts on the multiple choice parts, i can possibly make a passing score!
Questions: umm, when you have an integral and you neeed to figure out what to substitute, and there is no trig function, what do you put as your u?
36
Alright here are some things that keep popping up, but we just forget.
LIMIT RULES:
1. top degree is larger than the bottom degree-->the limit approaches infinity
2. bottom degree is larger than the top degree-->the limit approaches zero
3. and if the top and bottom degree are equal-->then you make a fraction out of the coefficients
FIRST DERIVATIVE TEST:
deals with increasing and decreasing/max and mins
1. derivative of the function
2. solve for x values which are critical values
3. make those numbers into intervals between -ve infinity and infinity
4. when seeing if +ve or -ve plug those numbers into the derivative
5. if -ve it is decreasing and +ve it is increasing, changing from -ve to to positive you have a min and going from +ve to -ve you have a max, absolute maxs and minx deal with the highest and lowest value
SECOND DERIVATIVE TEST:
deals with concave up and down/points of inflection
1. take the derivative of the function twice
2. set equal to 0 and solve for x (these are not really critical values, more like points of interest)
3. set up intervals just like the first derivative test
4. plug in also like the first derivative test but plug into the second derivative
5. -ve is concave down and +ve is concave up, points of inflection is where the concavity changes like going from -ve to +ve
EQUATION OF A TANGENT LINE:
1. take the derivative
2. plug in x value
3. if not given a y value, plug into the original equation to get the y value
4. then plug those numbers into point slope form: y − y1 = m(x − x1)
5. *if wanting the normal line, the slope is just the reciprocal
AND slope field stuff:
I'm just going to review how to draw it.
positive slopes is /
negative slope is \
for a zero slope is a horizontal line
for an undefined slope is a vertical line
LIMIT RULES:
1. top degree is larger than the bottom degree-->the limit approaches infinity
2. bottom degree is larger than the top degree-->the limit approaches zero
3. and if the top and bottom degree are equal-->then you make a fraction out of the coefficients
FIRST DERIVATIVE TEST:
deals with increasing and decreasing/max and mins
1. derivative of the function
2. solve for x values which are critical values
3. make those numbers into intervals between -ve infinity and infinity
4. when seeing if +ve or -ve plug those numbers into the derivative
5. if -ve it is decreasing and +ve it is increasing, changing from -ve to to positive you have a min and going from +ve to -ve you have a max, absolute maxs and minx deal with the highest and lowest value
SECOND DERIVATIVE TEST:
deals with concave up and down/points of inflection
1. take the derivative of the function twice
2. set equal to 0 and solve for x (these are not really critical values, more like points of interest)
3. set up intervals just like the first derivative test
4. plug in also like the first derivative test but plug into the second derivative
5. -ve is concave down and +ve is concave up, points of inflection is where the concavity changes like going from -ve to +ve
EQUATION OF A TANGENT LINE:
1. take the derivative
2. plug in x value
3. if not given a y value, plug into the original equation to get the y value
4. then plug those numbers into point slope form: y − y1 = m(x − x1)
5. *if wanting the normal line, the slope is just the reciprocal
AND slope field stuff:
I'm just going to review how to draw it.
positive slopes is /
negative slope is \
for a zero slope is a horizontal line
for an undefined slope is a vertical line
36
The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals*
stuff i dont understand. intergrals of trig functions. need help with these
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals*
stuff i dont understand. intergrals of trig functions. need help with these
post 36
This past week we took another AP test to get ready for the upcoming real AP test. Some old stuff to refresh some memories are as follows.
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
e integration:very easy whatever e is raised to is your u, the derivative of u is du
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
delta x=b-a/number of subintervals
The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
e integration:very easy whatever e is raised to is your u, the derivative of u is du
The steps for substitution are:
1. Find u and du
2. set u equal to whatever isn't the derivative
3. take the derivative of u
4. substitute back in
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
delta x=b-a/number of subintervals
Post #36
Complex Derivatives y=ln(e^x) (Chain Rule)
First off one should should identify the steps of your problem. In this case they would be:
1. Natural Log
2. e^x
you problem should be (1/(e^x)).(e^x)'
then you find the derivative of e^x which is e^x . x' (x'=1)
so your final problem should be (1/(e^x)).(e^x)
After this you have to simplify algebraically, giving you (e^x)/(e^x) ,which equals 1.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
First off one should should identify the steps of your problem. In this case they would be:
1. Natural Log
2. e^x
you problem should be (1/(e^x)).(e^x)'
then you find the derivative of e^x which is e^x . x' (x'=1)
so your final problem should be (1/(e^x)).(e^x)
After this you have to simplify algebraically, giving you (e^x)/(e^x) ,which equals 1.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Post # 36
Hellooo everyone! I hope everybody had a great time at prom and prom mania... i know i did :)... anyways let's go over some simple stuff first:
First derivative test:
You will be given a function, so you have to take the derivative of the function once and set it equal to zero. Then solve for the x values (critical points) and set them up into intervals between negative infinity and infinity. Then plug in values between those intervals into the first derivative to see where there are max and mins or if the function is increasing or decreasing.
Second derivative test:
You take the derivative of the function twice and you set it equal to zero. Solve for your critical points once more and set them up into intervals between negative infinity and infinity. Plug in numbers between those intervals into the second derivative to see where the function is concave up, concave down, or where there is a point of inflection.
Tangent lines:
You will be given a function and an x value. To find the y value you have to plug the x value into the original function and solve. Then take the derivative of the function and plug in the x value to find the slope. Then plug everything into point-slope form (y-y1=slope(x-x1)).
Limit Rules:
If the degree on top is bigger than the degree on the bottom, the limit will be infinity
If the degree on top is smaller than the degree on the bottom, the limit will be zero
If the degree on the top is the same as the degree on the bottom, divide the coefficients to get the limit
Things i still struggle with:
# 2 on the calculator portion of the free response
# 7 on the calculator multiple choice
#11 on the calculator MC
First derivative test:
You will be given a function, so you have to take the derivative of the function once and set it equal to zero. Then solve for the x values (critical points) and set them up into intervals between negative infinity and infinity. Then plug in values between those intervals into the first derivative to see where there are max and mins or if the function is increasing or decreasing.
Second derivative test:
You take the derivative of the function twice and you set it equal to zero. Solve for your critical points once more and set them up into intervals between negative infinity and infinity. Plug in numbers between those intervals into the second derivative to see where the function is concave up, concave down, or where there is a point of inflection.
Tangent lines:
You will be given a function and an x value. To find the y value you have to plug the x value into the original function and solve. Then take the derivative of the function and plug in the x value to find the slope. Then plug everything into point-slope form (y-y1=slope(x-x1)).
Limit Rules:
If the degree on top is bigger than the degree on the bottom, the limit will be infinity
If the degree on top is smaller than the degree on the bottom, the limit will be zero
If the degree on the top is the same as the degree on the bottom, divide the coefficients to get the limit
Things i still struggle with:
# 2 on the calculator portion of the free response
# 7 on the calculator multiple choice
#11 on the calculator MC
Prom Weekend Post
So Prom was all right... Prom Mania was better. Either way, had a long night so kind of tired even though I still slept 9 hours or so.
I guess I will just do what I did last week...and that was explain a few of the problems that I got wrong and my ways of fixing those issues...
Number 2, Non Calculator Portion
The integral from 0 to 1 of e to the -4x.
So this problem is fairly easy, you just integrate it like normal, and then plug in from 0 to 1. I integrated it and got e^-4 - 1 but I forgot to multiply in my (-1/4) that I had on the outside of the integral...silly mistakes are always the easiest to correct!
Number 9, Non Calculator Portion
If f(x) = ln(x + 4 + e^(-3x)), then f'(0) is ?
So for this problem it's obvious what needs to be done. We need to take the derivative, and then we need to plug in 0 and simplify. Well, when taking the derivative of this, I did the ln x rule, which is 1 / x. But I forgot that you have to multiply by x' or the derivative of x, so that changed my answer completely. After re doing it, I found my answer to be -2/5.
Anyway, other than that, I didn't have too many problems on this portion. The AP this week I found was relatively ... the same in terms of difficulty as the last one. As far as I know, most of you improved your score a bit so that's good. I figure if we keep improving our score by a little bit every week, everyone will be able to pass.
Also, I guess since I'm pretty well endowed with knowledge of Calculus thus far, if you guys need any help I will gladly help you with it. Just try to find some time to see me...whether it be recess, lunch, or even 7th our if Mrs. Robinson allows me to take some time to help you (which I'm sure she won't mind). Trust me, I won't be aggravated by you asking for help, because it's kind of my way of learning... I just happen to learn better by teaching others...I don't know, I've always learned that way.
So yeah, anyway...I'm off to work on my bazillion projects that are due this week...
I guess I will just do what I did last week...and that was explain a few of the problems that I got wrong and my ways of fixing those issues...
Number 2, Non Calculator Portion
The integral from 0 to 1 of e to the -4x.
So this problem is fairly easy, you just integrate it like normal, and then plug in from 0 to 1. I integrated it and got e^-4 - 1 but I forgot to multiply in my (-1/4) that I had on the outside of the integral...silly mistakes are always the easiest to correct!
Number 9, Non Calculator Portion
If f(x) = ln(x + 4 + e^(-3x)), then f'(0) is ?
So for this problem it's obvious what needs to be done. We need to take the derivative, and then we need to plug in 0 and simplify. Well, when taking the derivative of this, I did the ln x rule, which is 1 / x. But I forgot that you have to multiply by x' or the derivative of x, so that changed my answer completely. After re doing it, I found my answer to be -2/5.
Anyway, other than that, I didn't have too many problems on this portion. The AP this week I found was relatively ... the same in terms of difficulty as the last one. As far as I know, most of you improved your score a bit so that's good. I figure if we keep improving our score by a little bit every week, everyone will be able to pass.
Also, I guess since I'm pretty well endowed with knowledge of Calculus thus far, if you guys need any help I will gladly help you with it. Just try to find some time to see me...whether it be recess, lunch, or even 7th our if Mrs. Robinson allows me to take some time to help you (which I'm sure she won't mind). Trust me, I won't be aggravated by you asking for help, because it's kind of my way of learning... I just happen to learn better by teaching others...I don't know, I've always learned that way.
So yeah, anyway...I'm off to work on my bazillion projects that are due this week...
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