This week in calculus we took some more AP exams. I did pretty well on the non-calculator portion but the calculator portion was a different story.
Things i know:
1. Limit rules:
if the degree on the top is bigger than the degree on the bottom, the limit is infinity
if the degree on the top is smaller than the degree on the botton, then limit is zero.
if the degree on the top is the same as the degree on the bottom, you divide the coefficients to get the limit.
2. First and Second derivative test:
Whenever a question is asking for relative max or min or whether the function is increasing or decreasing, you use the first derivative test. You take the derivative of the function, set the derivative equal to zero and solve for the x values (critical points). Then you set the critical points into intervals between negative infinity and infinity. Then you plug a number between those intervals to see whether the function is increasing or decreasing or if they are max or mins.
For absolute max or mins, you do the same steps, but if they give you a point, you have to plug those in as well. For absolute max and mins, you plug your values into the original function to find a y value. The point with the highest y values is an absolute max and the lowest y values is an absolute min.
For the second derivative test, you will find change in concavity and points of inflection. You take the derivative of the function twice and solve for critical values and set them up into intervals and plug in numbers between the intervals into the second derivative. This will tell you where there are changes in concavity and points of inflection.
Things i do not know:
1. Number 27 and 28 on the non-calculator portion
2.Number 35 on the calculator portion.
3. Number 39 and 40 on the calculator portion
4. Number 42 and 43 on the calculator portion
5.Number 44 and 45 on the calculator portion.
Have a great extended weekend and GO SAINTS :)
Saturday, February 6, 2010
Friday, February 5, 2010
Ash's 25th Post Part I
Okay, so, my computer is failing; I have viruses and whatnot coming out of the wazoo.
I've been trying to fix it, but it's not working and I'm about to throw it across the room.
That being said, I just wanted to say that I AM going to do my comments from this week and I AM going to do this blog, but I refuse to go on my email account while I have this virus.
I'm running a new program right now to try to get rid of it, so I'll update this later when/if my computer is working again.
I've been trying to fix it, but it's not working and I'm about to throw it across the room.
That being said, I just wanted to say that I AM going to do my comments from this week and I AM going to do this blog, but I refuse to go on my email account while I have this virus.
I'm running a new program right now to try to get rid of it, so I'll update this later when/if my computer is working again.
Post # 25
Alright, so this week started off okay...
We took the first test on the slope field on Monday and it was only out of six points [i think] and we got the corrections to that on Thursday!
Then Tuesday we took the non-calculator portion of the AP exam.
QUESTIONS ON THIS PACKET:
number 9: I dont even know where to being. please explain!
number 15: How is x=2 a local minimum?
number 24: I tried it, but I get lost in my work!...help!
number 25: I don't know why I can't do this one, but every time I do it I don't get any of the possible answers.
number 27: same as 25
number 28: I still don't understand, or grasp, how to take the derivitive or integral of LN
WHAT I CAN POSSIBLY HELP WITH:
number 3: super easy, just careless mistakes...factor out an X and then use the quotient rule.
number 4: same thing, factor and then cancel out the (x-4)'s, so you're left with x-3 so plug in 4 leaving you with 1 for your answer!
Wednesday we took the calculator portion of the AP exam.
QUESTIONS ON THIS PACKET:
number 35: HELP..that's all I can say!
the rest, I'm going to try to work on!
WHAT I CAN POSSIBLY HELP WITH:
number 31: you're finding the limit as x is approaching zero to tan pi over six plus h minus tan pie over six, all over h. H H H H H H H H H H H when you see H in a problem it's the definition of a deriviative, so you just take the part that has H plus something, in this case H plus (pie over six) and take the derivative of it giving you sec^2(x) plug in (pie over six) for x giving you 4/3!!
number 32: if you're given 30-100 for A and 50-100 for B, and asked to find what 30-50 is then you can just see what's the difference of both. for instance, 100-30=70 and 100-50=50..take the two things you just found and subtract them because that's what the problem does so it's 70-50=20 which gives you the same answer as 50-30=20! Therefore, A-B is your answer!
THANKS JOHN for helping me with all of the other problems!!!
Thursday, Mrs. Robinson had a scare when she was having too much pain, so we went in A3. Some of us worked on corrections. Thank God I had some help!
Friday, Mrs. Robinson is not there and some of us worked on corrections again.
SOOO...MONDAY NIGHT...THAT CORRECTIONS PARTY AT RICKY'S .. SOMEONE SHOULD CALL ME! =)
~ElliE~
We took the first test on the slope field on Monday and it was only out of six points [i think] and we got the corrections to that on Thursday!
Then Tuesday we took the non-calculator portion of the AP exam.
QUESTIONS ON THIS PACKET:
number 9: I dont even know where to being. please explain!
number 15: How is x=2 a local minimum?
number 24: I tried it, but I get lost in my work!...help!
number 25: I don't know why I can't do this one, but every time I do it I don't get any of the possible answers.
number 27: same as 25
number 28: I still don't understand, or grasp, how to take the derivitive or integral of LN
WHAT I CAN POSSIBLY HELP WITH:
number 3: super easy, just careless mistakes...factor out an X and then use the quotient rule.
number 4: same thing, factor and then cancel out the (x-4)'s, so you're left with x-3 so plug in 4 leaving you with 1 for your answer!
Wednesday we took the calculator portion of the AP exam.
QUESTIONS ON THIS PACKET:
number 35: HELP..that's all I can say!
the rest, I'm going to try to work on!
WHAT I CAN POSSIBLY HELP WITH:
number 31: you're finding the limit as x is approaching zero to tan pi over six plus h minus tan pie over six, all over h. H H H H H H H H H H H when you see H in a problem it's the definition of a deriviative, so you just take the part that has H plus something, in this case H plus (pie over six) and take the derivative of it giving you sec^2(x) plug in (pie over six) for x giving you 4/3!!
number 32: if you're given 30-100 for A and 50-100 for B, and asked to find what 30-50 is then you can just see what's the difference of both. for instance, 100-30=70 and 100-50=50..take the two things you just found and subtract them because that's what the problem does so it's 70-50=20 which gives you the same answer as 50-30=20! Therefore, A-B is your answer!
THANKS JOHN for helping me with all of the other problems!!!
Thursday, Mrs. Robinson had a scare when she was having too much pain, so we went in A3. Some of us worked on corrections. Thank God I had some help!
Friday, Mrs. Robinson is not there and some of us worked on corrections again.
SOOO...MONDAY NIGHT...THAT CORRECTIONS PARTY AT RICKY'S .. SOMEONE SHOULD CALL ME! =)
~ElliE~
Monday, February 1, 2010
News 2/1/10
Financial due dates:
Feb 9th - 50 due for state
Feb 26 - $100 due for nationals
March 22 - $50 due for state
March 28 - $100 due for nationals
Other news:
We will resume Bake Sale in March and will have two to make up for the lack of bake sales in the past few months. Life has simply been too hectic for me to plan everything :)
Feb 20th Angie Roussel will be bringing the kids to Catholic High. Permission slips went home today.
Feb 27 - Carrie Turnbull will be taking the kids to LSU. A 10 fee is charged and the kids get a t-shirt.
GREAT NATIONALS NEWS!!!!!!
The price of nationals has been drastically reduced. I petitioned the state to charter the bus so we could save money on taxes and ensure that the bus was filled better. The state has agreed to pay 3,000 of the bus and several other schools have already signed on to ride the bus. This has resulted in the new price being 1050 for nationals.
Therefore, each person will have to raise $650. With the potential breakdown as follows
Pi week usually brings in $100 per person
Spring Fest brings in $100 per person
Our tournament brings in $100 per person
In addition, I have letters typed up for plants. They must be submitted by the families that are affiliated. Students can pick those up ASAP. In the past we have secured several thousand from the plants and are hoping to repeat this.
If we win Log 1 (we are 40 points from 1st) this will bring in an addition $150 per person attending.
We will have a doughnut sale if a parent is willing to pick up the doughnuts in Feb. We will also have a lunch over Easter. I will see where we stand when I get back and will plan more fundraisers as needed. We have until July to get the full amount in. However, if by May 15th the student has the full 1050 in their account then they will not have to make the final $100 dollar payment.
We have already booked the rooms and the bus and deposits have been paid. I am also going to be asking for ideas on what the kids want to do. We will spend three days in Washington DC doing activities. Most of the activities are free so we are expecting to keep the cost of excursions down.
OKAYYY, so my blog wouldn't post lastnight so i just saved it and i'm posting
it nowwww.
This past week in calculus has been prettty good for me actually, i think i've made more A's this week alone than i did with all of advanced math and calculus combined, but that's just because i'm good a guessing multiple choiceeeee.
i'm just going to go over some old stuff incase it comes up soon because things tend to do that in calc.
discontinuities
1. Jump
-The limit does not exist
-The function is continuous everywhere except at the jump. Therefore, if we are
talking about the function as a whole we say that it is not continuous.
2. Removable- when the graph is not defined at a point..(open circle)
-The limit exists
-The function is continuous everywhere except at that point. Therefore, if we are
talking about the function as a whole we say that it is not continuous.
3. Infinate – An asymptote
-The limit cab exist.
-The function is continuous everywhere except at the asymptote.
4. Oscillation – An extreme oscillating graph
-The limit doesn't exist
-The function is not continuous.
WHAT TO DO WHEN YOU SEE THESE KEY WORDS:
Maximum
Minimum
Critical values
Increasing
Decreasing
Take derivative
set equal to 0
solve for x.
Set up intervalsusing these numbers.
Plug in numbers on each interval.
If it changes from positive to negative it is a maximum. If it changes from negative to positive it is a minimum. All potential maximums and minimums are called critical values. If the interval is positive the
interval is increasing. If the interval is negative then the interval is decreasing.
help?? um can someone explain linearization, i kinda remember what it is and what to do but not really so and explanation would be awesomee.
it nowwww.
This past week in calculus has been prettty good for me actually, i think i've made more A's this week alone than i did with all of advanced math and calculus combined, but that's just because i'm good a guessing multiple choiceeeee.
i'm just going to go over some old stuff incase it comes up soon because things tend to do that in calc.
discontinuities
1. Jump
-The limit does not exist
-The function is continuous everywhere except at the jump. Therefore, if we are
talking about the function as a whole we say that it is not continuous.
2. Removable- when the graph is not defined at a point..(open circle)
-The limit exists
-The function is continuous everywhere except at that point. Therefore, if we are
talking about the function as a whole we say that it is not continuous.
3. Infinate – An asymptote
-The limit cab exist.
-The function is continuous everywhere except at the asymptote.
4. Oscillation – An extreme oscillating graph
-The limit doesn't exist
-The function is not continuous.
WHAT TO DO WHEN YOU SEE THESE KEY WORDS:
Maximum
Minimum
Critical values
Increasing
Decreasing
Take derivative
set equal to 0
solve for x.
Set up intervalsusing these numbers.
Plug in numbers on each interval.
If it changes from positive to negative it is a maximum. If it changes from negative to positive it is a minimum. All potential maximums and minimums are called critical values. If the interval is positive the
interval is increasing. If the interval is negative then the interval is decreasing.
help?? um can someone explain linearization, i kinda remember what it is and what to do but not really so and explanation would be awesomee.
Sunday, January 31, 2010
Post 24
So this week we did more ap tests. We did a non calculator portion and a calculator portion just like we did last week. Only this time we couldn't get points for making corrections. I was worried because last time on the calculator portion I made a zero. I ended up getting lucky and I got an 8 and a ten on the tests. Tomorrow we have the slope field short answer problem. These are supposed to be really easy and they're the kind of short answer we want to see on the real ap.
On the non-calculator portion, we had things like limits as x goes to infinity, definition of derivatives, equations of tangent lines, linearization, integrals and slope fields. On the calculator portion, I guessed mostly. I’m still not very good with my calculator yet and what it can do for me.
Infinity rules:
1. If the degree of the top is greater than the degree of the bottom, then the limit is going to infinity.
2. If the degree at the top is less than the degree at the bottom, then the limit is going to zero.
3. If the degree of the top is the same as the degree of the bottom, set the coefficients to a fraction.
For definitions of derivatives they may not always give you an exact definition. If they do not, you have to take an abstract derivative. An abstract derivative is a derivative taken about an equation made up.
For equations of tangent lines, you need two points (x,y) and a slope. If these aren’t given to you, you need to find them. To find them, you set your equation equal to zero and find x and y. To find slope, take the first derivative, plug in your x and/or y values and solve. After you have this done, plug into point-slope form, and you’ll have your equation of a tangent line.
Linerization is really easy, you do all the same steps at tangent lines, except you plug in the decimal they give you into x and solve giving you a number.
I always think slope fields are tricky, but they’ve shown to be easy. I don’t know w hat my problem is.
On the non-calculator portion, we had things like limits as x goes to infinity, definition of derivatives, equations of tangent lines, linearization, integrals and slope fields. On the calculator portion, I guessed mostly. I’m still not very good with my calculator yet and what it can do for me.
Infinity rules:
1. If the degree of the top is greater than the degree of the bottom, then the limit is going to infinity.
2. If the degree at the top is less than the degree at the bottom, then the limit is going to zero.
3. If the degree of the top is the same as the degree of the bottom, set the coefficients to a fraction.
For definitions of derivatives they may not always give you an exact definition. If they do not, you have to take an abstract derivative. An abstract derivative is a derivative taken about an equation made up.
For equations of tangent lines, you need two points (x,y) and a slope. If these aren’t given to you, you need to find them. To find them, you set your equation equal to zero and find x and y. To find slope, take the first derivative, plug in your x and/or y values and solve. After you have this done, plug into point-slope form, and you’ll have your equation of a tangent line.
Linerization is really easy, you do all the same steps at tangent lines, except you plug in the decimal they give you into x and solve giving you a number.
I always think slope fields are tricky, but they’ve shown to be easy. I don’t know w hat my problem is.
I've lost count.
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
So for implicit derivatives...its pretty easy. You are going to be using this when you have like y^2 + x^2 = 4. It will ask for dy/dx. So what you do is you take the derivative like normal...but whenever you take the derivative of y you write dy/dx. So the above would be 2y (dy/dx) + 2x = 0. Now you solve for dy/dx. To do this, minus over 2x and then divide by 2y. So the answer would be
dy/dx = -x/y.
Some people have been asking about related rates!!!!!!!!!!!!! SO HERE IS an EXAMPLE!!!
Um let's see...
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
Okay so first derivative refers to slope of position..which is velocity.
Second derivative is slope of velocity...or acceleration.
Third derivative would be the slope of acceleration which I believe is jerk.
Past that, its just the rate of change of whatever was before...so...yeah.
Anyway, think this is around 285 words so...see you guys tomorrow.
Posted by XxDohxX at 8:30 PM 0 comments
Reactions:
post of holidays..3
IMPLICIT DERIVATIVES:
1. Take the derivative like normal.
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
steps to SECOND IMPLICIT DERIVATIVES
1. Take the derivative of the first derivative
2. Put d^2y/dx^2 for dy/dx
3. Simplify
4. Plug in values
so an example using the same equation is:
we take the first implicit derivative first
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
You have to identify what you are looking for and what you are given. Not only does this make it easier on you, it's kind of necessary, especially when you want those points on free response questions (or so I'm told). You also have to realize that when you are doing related rates, you have to put dy/dt or dx/dt or whatever whenever you are taking the derivative of some variable in relation to time (hence the t). So given that:
Given xy = 4
you want to know what dy/dt equals given x = 8 and dx/dt=10.
Take the derivative (product rule):
dx/dt y + dy/dt x = 0
Plug in everything:
dy/dt = -10y/8
= -5y/4
Obviously, I'm not doing perfect, and I don't know everythin, so here's my trouble areas:
1. I can't remember when to use washers and disks-and i switch up the equations.
2. Slope Fields!!!! I get the one we did on the free response,,,just none like on the practice tests..if that makes any sense.
3. my e integrating skills have something to be desired...
any suggestions?
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
So for implicit derivatives...its pretty easy. You are going to be using this when you have like y^2 + x^2 = 4. It will ask for dy/dx. So what you do is you take the derivative like normal...but whenever you take the derivative of y you write dy/dx. So the above would be 2y (dy/dx) + 2x = 0. Now you solve for dy/dx. To do this, minus over 2x and then divide by 2y. So the answer would be
dy/dx = -x/y.
Some people have been asking about related rates!!!!!!!!!!!!! SO HERE IS an EXAMPLE!!!
Um let's see...
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
Okay so first derivative refers to slope of position..which is velocity.
Second derivative is slope of velocity...or acceleration.
Third derivative would be the slope of acceleration which I believe is jerk.
Past that, its just the rate of change of whatever was before...so...yeah.
Anyway, think this is around 285 words so...see you guys tomorrow.
Posted by XxDohxX at 8:30 PM 0 comments
Reactions:
post of holidays..3
IMPLICIT DERIVATIVES:
1. Take the derivative like normal.
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
steps to SECOND IMPLICIT DERIVATIVES
1. Take the derivative of the first derivative
2. Put d^2y/dx^2 for dy/dx
3. Simplify
4. Plug in values
so an example using the same equation is:
we take the first implicit derivative first
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
You have to identify what you are looking for and what you are given. Not only does this make it easier on you, it's kind of necessary, especially when you want those points on free response questions (or so I'm told). You also have to realize that when you are doing related rates, you have to put dy/dt or dx/dt or whatever whenever you are taking the derivative of some variable in relation to time (hence the t). So given that:
Given xy = 4
you want to know what dy/dt equals given x = 8 and dx/dt=10.
Take the derivative (product rule):
dx/dt y + dy/dt x = 0
Plug in everything:
dy/dt = -10y/8
= -5y/4
Obviously, I'm not doing perfect, and I don't know everythin, so here's my trouble areas:
1. I can't remember when to use washers and disks-and i switch up the equations.
2. Slope Fields!!!! I get the one we did on the free response,,,just none like on the practice tests..if that makes any sense.
3. my e integrating skills have something to be desired...
any suggestions?
Posting...#24
Area between curves: formula: bSa top equation-bottom equation
example: Find the area of the region enclosed by y=-4x^2+41x+94 and y=x-2 between x=1 and x=7.
Steps:
1. draw a picture
2. subtract the two equations: (-4x^2+41x+94-(x-2)-don't square, it's not volumee! (-4x^2+41x+94-(-x+2)
3.combine like terms and integrate: S(-4x^2+41x+94-(-x+2))(-4/3)x^3+20x^2+96x
4. plug in 1 and 6 and subtract: (3584/3)-(344/3)= 1080
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
STEPS:1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
I don't know why but lately i been having trouble with your integration so if yall got tips are a better way of explaining it thanks.
example: Find the area of the region enclosed by y=-4x^2+41x+94 and y=x-2 between x=1 and x=7.
Steps:
1. draw a picture
2. subtract the two equations: (-4x^2+41x+94-(x-2)-don't square, it's not volumee! (-4x^2+41x+94-(-x+2)
3.combine like terms and integrate: S(-4x^2+41x+94-(-x+2))(-4/3)x^3+20x^2+96x
4. plug in 1 and 6 and subtract: (3584/3)-(344/3)= 1080
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
STEPS:1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
I don't know why but lately i been having trouble with your integration so if yall got tips are a better way of explaining it thanks.
Ash's 24th Post
Uhhmm...fun fact for this week...I don't think I remember one..sorry!
I'm gonna explain a few of the problems on the AP that we had to correct ourselves :)
2. lim sin(pi/2+h)-1
h->0 h
Mrs. Robinson explained this to me, so I'll try to explain it back to y'all.
What's the first thing you see? Definition of a Derivative, right? Right!
You're on the right track!
So what do you do? Take the derivative of sin!
sin(pi/2)
=
cos(pi/2)
=
0!!
That's your answer :)
Number 7!!
A relative maximum value of the function (lnx)/(x) is....
Okay, I think I got Trina's help on this one, but it looked fairly simple when I just looked at it :D
You take the derivative of lnx/x
=
(1)-(lnx)/(x^2) = 0
Set your top equal to zero since that's all that you're worrying about anyway
1-lnx=0
Solve for x
lnx=1
x=e
y=lne/e = 1/e
TaDa!!
I need extreme help on problems like number 19!
Suppose the graph of f is both increasing and concave up on A less than/equal to X less than/equal to B. Then, using the same number of subdivisions, and with L, R, M, and T denoting, respectively, left, right, midpoint, and trapezoid sums, it follows that
A) R <= T <= M <= L
B) L <= T <= M <= R
C) R <= M <= T <= L
D) L <= M <= T <= R
E) NOTA
Also, I think I'm afraid of fractions. I know how to work some of them, but...I dunno =/ Can someone clarify these please?
For fractions:
Derivative: quotient rule, got it :)
Integration: ????
Solving for x when it's in the fraction: ????
that's all I can think of at the moment :)
Oh, btw...I'm SUPER terrified of that Free Response tomorrow v.v
I'm gonna explain a few of the problems on the AP that we had to correct ourselves :)
2. lim sin(pi/2+h)-1
h->0 h
Mrs. Robinson explained this to me, so I'll try to explain it back to y'all.
What's the first thing you see? Definition of a Derivative, right? Right!
You're on the right track!
So what do you do? Take the derivative of sin!
sin(pi/2)
=
cos(pi/2)
=
0!!
That's your answer :)
Number 7!!
A relative maximum value of the function (lnx)/(x) is....
Okay, I think I got Trina's help on this one, but it looked fairly simple when I just looked at it :D
You take the derivative of lnx/x
=
(1)-(lnx)/(x^2) = 0
Set your top equal to zero since that's all that you're worrying about anyway
1-lnx=0
Solve for x
lnx=1
x=e
y=lne/e = 1/e
TaDa!!
I need extreme help on problems like number 19!
Suppose the graph of f is both increasing and concave up on A less than/equal to X less than/equal to B. Then, using the same number of subdivisions, and with L, R, M, and T denoting, respectively, left, right, midpoint, and trapezoid sums, it follows that
A) R <= T <= M <= L
B) L <= T <= M <= R
C) R <= M <= T <= L
D) L <= M <= T <= R
E) NOTA
Also, I think I'm afraid of fractions. I know how to work some of them, but...I dunno =/ Can someone clarify these please?
For fractions:
Derivative: quotient rule, got it :)
Integration: ????
Solving for x when it's in the fraction: ????
that's all I can think of at the moment :)
Oh, btw...I'm SUPER terrified of that Free Response tomorrow v.v
Post Number Twenty Four
Limits:
If the bottom is 0 don’t assume the limit does not exist right away, first factor and cancel or use your calculator.
Related Rates:
1. Identify all variables and equations.
2. Identify what you are looking for.
3. Make a sketch and label.
4. Write an equations involving your variables.
*You can only have one unknown so a secondary equation may be given
5. Take the derivative with respect to time.
6. Substitute in derivative and solve.
For example, 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.Since everything is given you can skip straight to the derivative.dy/dt=6x^2dx/dt - dx/dtNow plug in all your givens in order to find dy/dt.dy/dt=6(2)^2(-1) - (-1)dy/dt= -23
Volume by disks:
The formula is pi S[r(x)]^2dx
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.
Umm my ap score seems to be dropping every single test, so I need help.
Mainly with ln and e integration and derivatives, related rates, substitution integration, remembering all the area/volume formulas, and stuff I need to figure out on my own. Even when I know what to do and know the way I have to do it I have a lot of trouble applying it and never get the answer. SO HELP.
If the bottom is 0 don’t assume the limit does not exist right away, first factor and cancel or use your calculator.
Related Rates:
1. Identify all variables and equations.
2. Identify what you are looking for.
3. Make a sketch and label.
4. Write an equations involving your variables.
*You can only have one unknown so a secondary equation may be given
5. Take the derivative with respect to time.
6. Substitute in derivative and solve.
For example, 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.Since everything is given you can skip straight to the derivative.dy/dt=6x^2dx/dt - dx/dtNow plug in all your givens in order to find dy/dt.dy/dt=6(2)^2(-1) - (-1)dy/dt= -23
Volume by disks:
The formula is pi S[r(x)]^2dx
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.
Umm my ap score seems to be dropping every single test, so I need help.
Mainly with ln and e integration and derivatives, related rates, substitution integration, remembering all the area/volume formulas, and stuff I need to figure out on my own. Even when I know what to do and know the way I have to do it I have a lot of trouble applying it and never get the answer. SO HELP.
post 24
Well another week down and its getting closer and closer to graduation. This week we did more with AP tests.
The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u4. substitute back in
Rieman's sums
LRAM- left hand approximation delta x[f(a)+f(a+delta x)+...f(b)]
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+2delta x)+...f(b)]
The steps for taking and implicit derivative are:
1. take the derivative of both sides
2. everytime the derivative of y is taken note it with dy/dx or y'
3.solve for dy/dx
One of my biggest problems is doing some of the math on the non-calculator portion that causes simple errors in my math.
The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u4. substitute back in
Rieman's sums
LRAM- left hand approximation delta x[f(a)+f(a+delta x)+...f(b)]
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+2delta x)+...f(b)]
The steps for taking and implicit derivative are:
1. take the derivative of both sides
2. everytime the derivative of y is taken note it with dy/dx or y'
3.solve for dy/dx
One of my biggest problems is doing some of the math on the non-calculator portion that causes simple errors in my math.
Post #24
So, this week in calculus we went over moreeeeee ap tests. I did improve! I know amazing.
So a problem:
On which interval(s) does the function f(x)=x^4-4x^3+4x^2+6 increase?
Key word: increase, which means to take first derivative test
x^4-4x^3+4x^2+6
4x^3-12x^2+8x
4x(x^2-3x+2)
(x-2)(x-1)
x=0, 1, 2
(-infinity, 0) (0,1) (1,2) (2, infinity)
-1: 4(-1)^3-12(-1)^2+8(-1) = - ve
.5: 4(.5)^3-12(.5)^2+8(.5) = + ve
1.5: 4(1.5)^3-12(1.5)^2+8(1.5) = - ve
3: 4(3)^3-12(3)^2+8(3) = + ve
So 02.
touble with:
I'm having trouble with problems from the non calculator section. I just don't know where to begin sometimes. But I'm mainly still having trouble with washers and the stuff like that.
So a problem:
On which interval(s) does the function f(x)=x^4-4x^3+4x^2+6 increase?
Key word: increase, which means to take first derivative test
x^4-4x^3+4x^2+6
4x^3-12x^2+8x
4x(x^2-3x+2)
(x-2)(x-1)
x=0, 1, 2
(-infinity, 0) (0,1) (1,2) (2, infinity)
-1: 4(-1)^3-12(-1)^2+8(-1) = - ve
.5: 4(.5)^3-12(.5)^2+8(.5) = + ve
1.5: 4(1.5)^3-12(1.5)^2+8(1.5) = - ve
3: 4(3)^3-12(3)^2+8(3) = + ve
So 0
touble with:
I'm having trouble with problems from the non calculator section. I just don't know where to begin sometimes. But I'm mainly still having trouble with washers and the stuff like that.
post 24
This week in calculus we took some more ap test. one of the things i am comfortable doin on these test is the growth problems with these problems you use the formula A(not) = e^kt. You know you are using this formula when you see words such as growth over x amount of years, bacteria, decay, birth rate, etc.
33) An investment of 4000 dollars grows at the rate of 320e^0.08t dollars per year after t years. Its value after 10 years is approximately ?
You then plug into your a (not) formula and get your answer. YOu are basically given everything so all you have to plug in is your (t). you then plug in 10 for t and get that it equals $8902. Giving your answer to be B.
ONe thing that i am still having trouble with is taking intergrals of like absolute value and intergrals with substitution.
33) An investment of 4000 dollars grows at the rate of 320e^0.08t dollars per year after t years. Its value after 10 years is approximately ?
You then plug into your a (not) formula and get your answer. YOu are basically given everything so all you have to plug in is your (t). you then plug in 10 for t and get that it equals $8902. Giving your answer to be B.
ONe thing that i am still having trouble with is taking intergrals of like absolute value and intergrals with substitution.
week 24
this week in calculus we did the same thing as the past couple weeks, been working on practice ap tests, and going over the corrections, binder grades and such. i'm just gonna review today.
Optimization is used for finding maximum/minimum area.
STEPS FOR OPTIMIZATION:
1. Identify primary and secondary equations. **
2. Solve the secondary equation for any variable, then plug that back into primary equation.
3. Take the derivative of the primary equation after plugging in
4. set ^that equal to zero
5. solve for the OTHER variable.
6. Plug ^that back into the secondary equation.
7. solve for the last missing variable.
**Primary equation has variable that is being maximizied/minimized. Secondary equation is the other one, haha.
How to find area under a curve.
S (Top Equation)-(Bottom Equation) on interval [a,b].
To find a and b set 2 equations equal to each other and solve.
Also, if the area is on y, solve for X.
if it's on x, solve for Y.
ok so i have a couple questions.
-can someone go over the rules for a limit approaching zero?
-rules for integral of functions like sin cos tan cot sec csc?
-someone can you explain how to find points of inflection, max & min, concavity, all of the ways to find things for graphs like that.
-ok this might be a stupid question, but whenever it says find the slope of the equation of the tangent line to [a,b], idk what to do!? help!
Optimization is used for finding maximum/minimum area.
STEPS FOR OPTIMIZATION:
1. Identify primary and secondary equations. **
2. Solve the secondary equation for any variable, then plug that back into primary equation.
3. Take the derivative of the primary equation after plugging in
4. set ^that equal to zero
5. solve for the OTHER variable.
6. Plug ^that back into the secondary equation.
7. solve for the last missing variable.
**Primary equation has variable that is being maximizied/minimized. Secondary equation is the other one, haha.
How to find area under a curve.
S (Top Equation)-(Bottom Equation) on interval [a,b].
To find a and b set 2 equations equal to each other and solve.
Also, if the area is on y, solve for X.
if it's on x, solve for Y.
ok so i have a couple questions.
-can someone go over the rules for a limit approaching zero?
-rules for integral of functions like sin cos tan cot sec csc?
-someone can you explain how to find points of inflection, max & min, concavity, all of the ways to find things for graphs like that.
-ok this might be a stupid question, but whenever it says find the slope of the equation of the tangent line to [a,b], idk what to do!? help!
Post #24
This week in calculus, we took the AP test Monday and Tuesday and reviewed them the rest of the week.
Some problems that I learned how to do this week were
If a particle moves on a line according to the law s=t^5 + 2t^3, then the number of times it reverses direction is
A. 4 B. 3 C. 2 D. 1 E. 0
The key words in this problem is reverses direction. Reverse direction only means how many maxs or mins there are the the problem.
To find the max and min, take the derivative and set it equal to zero to find the zeros. Then set up intervals and test the intervals with the zeros found. In order to have a max or min or a change in direction in this case, the interval must go from negative to positive or positive to negative.
Derivative: 5t^4 + 6t^2=0
t^2(5t^2+6) = 0
t=0, +- the square root of -6/5
Since you can't have an imaginary number, the second two zeros do not count so the only one you have to test is
(-infinity, 0) and (0, infinity)
To test the intervals, plug in a number between them.
-1: 5(-1)^4 + 6(-1)^2 = positive
1: 5(1)^4 + 6(1)^2 = positive
Since both intervals are positive, there are no maxs or mins and therefore no reverse direction so the answer is E. 0.
Suppose f(x) - x^2 +x / x if x is not equal to 0 and f(0) = 1. Which of the following statements is (are) true of f?
I. f is defined at x=0
II. lim as x->0 f(x) exists.
III. f is continuous at x=0
A. I only B. II only C. I and II only D. None of the statements are true. E. All are true.
For this problem, the key words are the roman numerals. You would have to check each to see if it is true.
I. Defined means that f is there at x=0. To test this, take the find the limit at 0. First you will have to factor the top equation.
x(x+1) / x The x's cancel and you are left with x+1. If you plug in 0, you get 1. Therefore, f is defined at x=0 so I is TRUE.
II. This is basically stating the same thing as I. Factor, and plug in 0. You get 1, so the limit of f(x) exists at 0. TRUE.
III. Since you canceled the x's and found a removable at x=0, this function would not be continuous BUT since they state in the question, f(0)=1, it closes in the removable, making this a TRUE statement.
The answer is E. All are true.
f(x) = the integral of [0,x^2+2] the square root of 1+cos t dt. Then f'(x) =
Since the derivative and the integral cancel each other out, for this problem all you have to do is plug in b for and multiply that by the derivative of b. So the square root of 1+cos (x^2+2) X 2x or 2x the square root of 1+cos(x^2+2).
I am having a harder time with the calculator portion of the AP than I am the non-calculator portion. I also need help with related rates and area under a curve.
Don't forget to look over the short answer for tomorrow.
Some problems that I learned how to do this week were
If a particle moves on a line according to the law s=t^5 + 2t^3, then the number of times it reverses direction is
A. 4 B. 3 C. 2 D. 1 E. 0
The key words in this problem is reverses direction. Reverse direction only means how many maxs or mins there are the the problem.
To find the max and min, take the derivative and set it equal to zero to find the zeros. Then set up intervals and test the intervals with the zeros found. In order to have a max or min or a change in direction in this case, the interval must go from negative to positive or positive to negative.
Derivative: 5t^4 + 6t^2=0
t^2(5t^2+6) = 0
t=0, +- the square root of -6/5
Since you can't have an imaginary number, the second two zeros do not count so the only one you have to test is
(-infinity, 0) and (0, infinity)
To test the intervals, plug in a number between them.
-1: 5(-1)^4 + 6(-1)^2 = positive
1: 5(1)^4 + 6(1)^2 = positive
Since both intervals are positive, there are no maxs or mins and therefore no reverse direction so the answer is E. 0.
Suppose f(x) - x^2 +x / x if x is not equal to 0 and f(0) = 1. Which of the following statements is (are) true of f?
I. f is defined at x=0
II. lim as x->0 f(x) exists.
III. f is continuous at x=0
A. I only B. II only C. I and II only D. None of the statements are true. E. All are true.
For this problem, the key words are the roman numerals. You would have to check each to see if it is true.
I. Defined means that f is there at x=0. To test this, take the find the limit at 0. First you will have to factor the top equation.
x(x+1) / x The x's cancel and you are left with x+1. If you plug in 0, you get 1. Therefore, f is defined at x=0 so I is TRUE.
II. This is basically stating the same thing as I. Factor, and plug in 0. You get 1, so the limit of f(x) exists at 0. TRUE.
III. Since you canceled the x's and found a removable at x=0, this function would not be continuous BUT since they state in the question, f(0)=1, it closes in the removable, making this a TRUE statement.
The answer is E. All are true.
f(x) = the integral of [0,x^2+2] the square root of 1+cos t dt. Then f'(x) =
Since the derivative and the integral cancel each other out, for this problem all you have to do is plug in b for and multiply that by the derivative of b. So the square root of 1+cos (x^2+2) X 2x or 2x the square root of 1+cos(x^2+2).
I am having a harder time with the calculator portion of the AP than I am the non-calculator portion. I also need help with related rates and area under a curve.
Don't forget to look over the short answer for tomorrow.
post 24
some stuff i know....
LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
im not good at related rates and integrating a fraction.
LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
im not good at related rates and integrating a fraction.
Post #24
hello, good afternoon, and welcome. Everyone seems like they are having trouble with ln integration and e integration..so i'm here to help with what i possibly can.
First things first, ln integration is the opposite of a derivative..so a derivative of ln x would be 1/x. Now, think backwards, the integral of ln x would be 2lnx, or ln x^2, in simplest terms.
e Integration is something i can attempt to help someone with..but i'm not quite sure i do it the right way.
So, here it goes i think you substitute the exponent for u and then du is the derivative of the exponent. Then, you plug in and solve the integral as it comes.
S e^2x would be
d= 2x
du = 2
1/2 S e^u du
..then i get confused.
whatever, i can't explain this.
So, things i don't understand are really small things while i'm solving..so i'm not sure anyone can help me in advance.
First things first, ln integration is the opposite of a derivative..so a derivative of ln x would be 1/x. Now, think backwards, the integral of ln x would be 2lnx, or ln x^2, in simplest terms.
e Integration is something i can attempt to help someone with..but i'm not quite sure i do it the right way.
So, here it goes i think you substitute the exponent for u and then du is the derivative of the exponent. Then, you plug in and solve the integral as it comes.
S e^2x would be
d= 2x
du = 2
1/2 S e^u du
..then i get confused.
whatever, i can't explain this.
So, things i don't understand are really small things while i'm solving..so i'm not sure anyone can help me in advance.
Calculus Blog
Well, thinking that we have a test on slope fields as well as a few things with differential equations...I figured I should go through the steps of doing that.
Basically, if they ask you to consider a differential equation (also known as the equation that will give you a slope at a certain x and y), then they would ask to sketch the slope field on axes at given points. All you have to do is take each individual point and plug in the x and the y into the differential equation to get the slope at each point. Then you need to draw little lines...basically a /, |,_, or \ at each given point to indicate either positive slope, undefined slope, a slope of 0, or a negative slope respectively. Also, if you have a slope of 2 and a slope of 1, the slope of 2 should show up a bit more steep than the slope of 1.
Another type of question given with these is to find the particular solution to the differential equation given a specific condition, such that f(x) = c. Basically all you need to do is to work the equation around so that all the x's and dx are on one side and all of the y's and dy's are on the other side. Once you have that, integrate both sides of the equation. When it asks you about a particular solution, it's asking basically about the value of "C" whenever you integrate it so that it fits the conditions. So, we integrate both sides accordingly and put the +c on the side with the x. Now we need to solve the equation for y. After solving the equation for y, you must now solve for c after you plug in for the conditions it gave you (the x and the y value). After solving for c, you need to plug this back into our equation that was solved for y. Now simplify the equation and that is your answer :-).
Now that you have completed the particular solution...you may be asked to do something such as the the limit of it or asked maybe even a concavity or increasing/decreasing question (which you can do, now that you have the equation).
Hope this helps someone study if you look at it in time :o.
-John
Basically, if they ask you to consider a differential equation (also known as the equation that will give you a slope at a certain x and y), then they would ask to sketch the slope field on axes at given points. All you have to do is take each individual point and plug in the x and the y into the differential equation to get the slope at each point. Then you need to draw little lines...basically a /, |,_, or \ at each given point to indicate either positive slope, undefined slope, a slope of 0, or a negative slope respectively. Also, if you have a slope of 2 and a slope of 1, the slope of 2 should show up a bit more steep than the slope of 1.
Another type of question given with these is to find the particular solution to the differential equation given a specific condition, such that f(x) = c. Basically all you need to do is to work the equation around so that all the x's and dx are on one side and all of the y's and dy's are on the other side. Once you have that, integrate both sides of the equation. When it asks you about a particular solution, it's asking basically about the value of "C" whenever you integrate it so that it fits the conditions. So, we integrate both sides accordingly and put the +c on the side with the x. Now we need to solve the equation for y. After solving the equation for y, you must now solve for c after you plug in for the conditions it gave you (the x and the y value). After solving for c, you need to plug this back into our equation that was solved for y. Now simplify the equation and that is your answer :-).
Now that you have completed the particular solution...you may be asked to do something such as the the limit of it or asked maybe even a concavity or increasing/decreasing question (which you can do, now that you have the equation).
Hope this helps someone study if you look at it in time :o.
-John
Post #24
Finding the Area and volume under a curve
Int (Top Equation)-(Bottom Equation) on the interval [a,b]
You can find a and b by setting the equations equal to each other and solving because a and b are their intersections.
Two rules needed to know: If the area is on the y then a and b need to be y values and solved for x. If the area is on the x then a and b need to be x values and solved for y.
Example:
Find the area of the rigion bounded by f(x)=2-x^2 and g(x)=x.
2-x^2=x 2-x^2-x=0 (-x-1)(x+2) x=-1 x=2
int (Top)-(Bottom) dx [-1,2]
To find the top and bottom equation just graph them on your graphing calculator. You'll see that 2-x^2 is on top with x on the bottom.
int (2-x^2)-(x) dx [-1,2]
int (2-x^2-x) = 2x-(1/3)x^3-(1/2)x^2
Solve like an ordinary definite integral.
2(2)-(1/3)((2)^3)-(1/2)((2)^2)-[2(-1)-(1/3)((-1)^3)-(1/2)((-1)^2)]=(3/2)
Volume is little different because there are two ways to find the volume of a region, depending on the region itself. The two methods are discs and washers.
Discs: (π)int [R(x)]² dx [a,b]
Example:
(π)int √(sinx)² dx [0,π]
(π)int sinx dx [0,π]
(π)(-cosx) [0,π] -cos(π)-(-cos(0))
π(1+1)=2π
Washers: (π)int (Top equation)²-(Bottom equation)² dx [a,b]
Example:
√(x) and (x²)
(π)int (√(x))² - ((x²))² [0,1]
((1/2)x^2) - ((1/5)x^5) 1/2(1)-(1/5)(1) - [1/2(0) -(1/5)(0)]= 3/10
(π)(3/10)= (3π)/10
Int (Top Equation)-(Bottom Equation) on the interval [a,b]
You can find a and b by setting the equations equal to each other and solving because a and b are their intersections.
Two rules needed to know: If the area is on the y then a and b need to be y values and solved for x. If the area is on the x then a and b need to be x values and solved for y.
Example:
Find the area of the rigion bounded by f(x)=2-x^2 and g(x)=x.
2-x^2=x 2-x^2-x=0 (-x-1)(x+2) x=-1 x=2
int (Top)-(Bottom) dx [-1,2]
To find the top and bottom equation just graph them on your graphing calculator. You'll see that 2-x^2 is on top with x on the bottom.
int (2-x^2)-(x) dx [-1,2]
int (2-x^2-x) = 2x-(1/3)x^3-(1/2)x^2
Solve like an ordinary definite integral.
2(2)-(1/3)((2)^3)-(1/2)((2)^2)-[2(-1)-(1/3)((-1)^3)-(1/2)((-1)^2)]=(3/2)
Volume is little different because there are two ways to find the volume of a region, depending on the region itself. The two methods are discs and washers.
Discs: (π)int [R(x)]² dx [a,b]
Example:
(π)int √(sinx)² dx [0,π]
(π)int sinx dx [0,π]
(π)(-cosx) [0,π] -cos(π)-(-cos(0))
π(1+1)=2π
Washers: (π)int (Top equation)²-(Bottom equation)² dx [a,b]
Example:
√(x) and (x²)
(π)int (√(x))² - ((x²))² [0,1]
((1/2)x^2) - ((1/5)x^5) 1/2(1)-(1/5)(1) - [1/2(0) -(1/5)(0)]= 3/10
(π)(3/10)= (3π)/10
24th post
This week we went over another AP test and i actually did a lot better on these compared to my other ones lol... so let's start out with the things i know:
Limit Rules:
When dealing with a limit and it asks what is the limit as x goes to infinity, you are going to use your limit rules:
1. If the degree on top is bigger than the degree on the bottom, the limit is infinity.
2. If the degree on top is smaller than the degree on the botton, the limit is 0.
3. If the degree on top is the same as the degree on the bottom, then you divide their coefficients to get the limit.
If they give you a limit problem and it asks you what the limit is as h goes to 0, then you have to take the derivative of whatever is behind the parenthesis on top of the fraction. This only happens when h is in the problem.
If they asks for the limit as x goes to some number, then you plug that number into the function and see what the answer is.
Another thing that shows up a lot is first and second derivative.
First derivative:
you find if a function is increasing or decreasing and you find max and mins
You take the derivative of the function and solve for x values which are critical values. Then you set those numbers into intervals between negative infinity and infinity. Then you plug those numbers into the derivative to find whatever you need to find.
Second derivative:
you find where the function is concave up or concave down and where there are points of inflection.
You take the derivative of the function twice and find the critical values. Set them up into intervals and plug numbers between those intervals into the second derivative to find what you are looking for.
Things i am still having trouble with(on the AP test we just finished):
1. Number 11 on the non-calculator portion.
2.Number 12 on the non-calculator portion.
3.number 16 on the non-calculator portion.
4. Number 19 on the non-calculator portion.
5. Number 24 and 25 on the non-calculator portion.
6. Number 28 on the non-calculator portion.
7. Number30 and 32 on the calculator portion
8.Number 35,36, and 37 on the calculator portion.
9. Number 41 and 42 on the calculator portion.
10. Number 45 on the calculator portion.
See you all tomorrow :)
Limit Rules:
When dealing with a limit and it asks what is the limit as x goes to infinity, you are going to use your limit rules:
1. If the degree on top is bigger than the degree on the bottom, the limit is infinity.
2. If the degree on top is smaller than the degree on the botton, the limit is 0.
3. If the degree on top is the same as the degree on the bottom, then you divide their coefficients to get the limit.
If they give you a limit problem and it asks you what the limit is as h goes to 0, then you have to take the derivative of whatever is behind the parenthesis on top of the fraction. This only happens when h is in the problem.
If they asks for the limit as x goes to some number, then you plug that number into the function and see what the answer is.
Another thing that shows up a lot is first and second derivative.
First derivative:
you find if a function is increasing or decreasing and you find max and mins
You take the derivative of the function and solve for x values which are critical values. Then you set those numbers into intervals between negative infinity and infinity. Then you plug those numbers into the derivative to find whatever you need to find.
Second derivative:
you find where the function is concave up or concave down and where there are points of inflection.
You take the derivative of the function twice and find the critical values. Set them up into intervals and plug numbers between those intervals into the second derivative to find what you are looking for.
Things i am still having trouble with(on the AP test we just finished):
1. Number 11 on the non-calculator portion.
2.Number 12 on the non-calculator portion.
3.number 16 on the non-calculator portion.
4. Number 19 on the non-calculator portion.
5. Number 24 and 25 on the non-calculator portion.
6. Number 28 on the non-calculator portion.
7. Number30 and 32 on the calculator portion
8.Number 35,36, and 37 on the calculator portion.
9. Number 41 and 42 on the calculator portion.
10. Number 45 on the calculator portion.
See you all tomorrow :)
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