Hello everyone... again this week we took two more AP tests but this time they were made by someone different... so lets go over some concepts shall we:
First derivative test:
The problem will give you a function. They will ask you to look for max, min, or if the function is increasing or decreasing. You take the derivative of the function and set it equal to zero. Then you solve for the x values which are your critical points. Then you set the critical points up into intervals between negative infinity and infinity. Then you plug in numbers between those intervals into the first derivative and then solve.
Second derivative test:
Again, the problem will give you a function but this time they may want to know where the function is concave up, concave down, or where there is a point of inflection. You take the derivative of the function twice and set it equal to zero. You then solve for the critical points once again and set them up into intervals between negative infinity and infinity. Plug in numbers between those intervals into the second derivative and solve.
Tangent line:
The problem will give you a function and most of the time an x value. If no y value is given, you plug the x value into the original equation and solve for it. Then you take the derivative of the function and plug in the x value to find the slope. Once that is done, you set it up into point- slope form, y-y1=slope(x-x1).
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 bottom, the limit is zero.
If the degree on the top is the same as the degree on the bottom, you divide the coefficients and get the limit.
Things i do not understand:
I am still having problems with questions that give me a graph and i have to break the graph up into individual triangles and rectangles and then find the area.
I also have problems integrating trig functions.
I also have trouble when a problem mentions something is decaying or stuff like that. I just need to know where to start.
Have a great weekend :)
Saturday, March 13, 2010
Thursday, March 11, 2010
Update
Class,
I will be back on Mon the 22. We will review for 3 days before I
have to leave for MAO state. Start compiling your list of questions
now. These can come from any test you have taken or concepts in
general. After that we will begin taking real AP Exams. Also, we will
prepare to take a full length practice exam at LSU. Here are some pictures
of the baby!
Tuesday, March 9, 2010
Post 29
Late again. The past few weeks, we’ve taken AP tests twice a week, the calculator and non calculator portion of the multiple choice portion. Lately, it just seems that I’m overwhelmed with all of the AP tests and getting the same stuff wrong over and over. It also seems like the tests keep getting harder and harder, but B-rob said the ones we are doing right now are harder than the actual AP (I sure hope so).
In stats since Friday, Alex has sat down with the calculus side of the class and tried to go over the previous AP test as best he could. It’s helping a good bit. I had forgotten a few things since B-rob left and since she sat down and went over our APs with us. One of the things I realized is that you’ll use substitution more times than not. Alex was going over a problem Friday in class that I didn’t even know substitution was supposed to be used for. I thought I could just take the integral normally, but I forgot to put a 1/8 outside of the problem. In many integration problems I’ve been seeing that I haven’t been getting the right answers, I guess this is why. I’m looking for it right now, but I can’t find the problem, but I think it went something like this:
the integral of cos(8x) from 0 to 3
For this problem, I just would’ve tried to find a way to do it without using substitution, but Alex did. He set u = 8x and du = 8, then since the 8 wasn’t origionally in the problem, he put it in, and put a 1/8 on the outside to balance it out. This is technically substitution, I was just forgetting this step in my head.
I’m doing better with particle problems, but things I’m confused with are problems like integrating (sin(x))^2
Sunday, March 7, 2010
Ash's 29th Post
So, I just want to say one thing before I start:
This last AP test seemed lake the HARDEST out of all of them so far.
That is all.
Also, this is going to be more of a clarification blog than an explaining blog..
On the non-calculator portion, I can get so far on some...and then just lose it, then on others I'm sooo lost
1. xSpi/4 cos(2t)dt
Step 1: Look at your answer choices
Well, A) and C) are out of the picture because, obviously, the integral of cos isn't cos...D) can also be taken out because the pi/4 wold not just disappear..
Step 2: Integrate
-sin(2pi/4)
-sin(2x)
Step 3: Solve
I have no idea how I did this (TRIG CHART), but the answer is sin(2x)-1/2
Can anyone explain how I got to the last part? Trig always scares me...
5. It's a graph with "the graph of the piecewise linear function, for 0 less than or equal to (LTET) x LTET 8 is shown above. what is the value of 8S0 f(x)dx" Can anyone explain it to me how to approach this and do it?
6. If f is continuous for a *LTET* x *LTET* b then at any point x=c, a LTET c LTET b, which of the following is true?
The answer is lim x->c f(x)=f(c)
Why? How can you figure that out?
7. if f(x) = x^2 sqrt(3x+1), then f'(x)=?
I can get this far, but get completely stuck in everything =/
1. x^2(3x+1)^1/2
2. 2x(3x+1)^1/2 + 1/2(3x+1)(3)(x^2)
ahh...can someone tell me what to do next? =/
10. This is another graph, but with velocity and distance. Can anyone explain it to me? I've never gotten graphs and I don't expect to...maybe there's a trick?
Thanks guys :)
Hope everyone had a good weekend!
This last AP test seemed lake the HARDEST out of all of them so far.
That is all.
Also, this is going to be more of a clarification blog than an explaining blog..
On the non-calculator portion, I can get so far on some...and then just lose it, then on others I'm sooo lost
1. xSpi/4 cos(2t)dt
Step 1: Look at your answer choices
Well, A) and C) are out of the picture because, obviously, the integral of cos isn't cos...D) can also be taken out because the pi/4 wold not just disappear..
Step 2: Integrate
-sin(2pi/4)
-sin(2x)
Step 3: Solve
I have no idea how I did this (TRIG CHART), but the answer is sin(2x)-1/2
Can anyone explain how I got to the last part? Trig always scares me...
5. It's a graph with "the graph of the piecewise linear function, for 0 less than or equal to (LTET) x LTET 8 is shown above. what is the value of 8S0 f(x)dx" Can anyone explain it to me how to approach this and do it?
6. If f is continuous for a *LTET* x *LTET* b then at any point x=c, a LTET c LTET b, which of the following is true?
The answer is lim x->c f(x)=f(c)
Why? How can you figure that out?
7. if f(x) = x^2 sqrt(3x+1), then f'(x)=?
I can get this far, but get completely stuck in everything =/
1. x^2(3x+1)^1/2
2. 2x(3x+1)^1/2 + 1/2(3x+1)(3)(x^2)
ahh...can someone tell me what to do next? =/
10. This is another graph, but with velocity and distance. Can anyone explain it to me? I've never gotten graphs and I don't expect to...maybe there's a trick?
Thanks guys :)
Hope everyone had a good weekend!
post 29
This week we took ap tests yet again, and corrected them. so i am going to do my blog on the first derivative test. the first derivative test is used to find absolute maximums and minimums, to do these, first yhou take the derivative of the equation given to you, then you set it equal to zero and solve for x, then you set up intervals using the x values, then you plug in numbers within the intervals into the derivative to find where your relative maximums and minimums are, then you plug the x values into the original equation to find the absolute maximums and minimums.
So here is an example problam i had:
Find the absolute maximums and minimums of the function f(x)=x^2-8x+4
First, you take the derivative of it:
f'(x)= 2x-8
Then you set it equal to zero and solve for x.
2x-8=0 2x=8 x=4
Then you set up the intervals
(-infinity, 4) U (4, infinity)
then you plug in a number found within the interval into the first derivative to find relative max's and min's.
(-infinity, 4)= negative number
(4, infinity)= positive number
so at x=4 there is a minimum
after that, you plug in that x value into the original equation to find the absolute max's and min's. (in this case it is only a min)
x=4 is the absolute minimum (4,-12)
i could use some help on substitution..
So here is an example problam i had:
Find the absolute maximums and minimums of the function f(x)=x^2-8x+4
First, you take the derivative of it:
f'(x)= 2x-8
Then you set it equal to zero and solve for x.
2x-8=0 2x=8 x=4
Then you set up the intervals
(-infinity, 4) U (4, infinity)
then you plug in a number found within the interval into the first derivative to find relative max's and min's.
(-infinity, 4)= negative number
(4, infinity)= positive number
so at x=4 there is a minimum
after that, you plug in that x value into the original equation to find the absolute max's and min's. (in this case it is only a min)
x=4 is the absolute minimum (4,-12)
i could use some help on substitution..
Post
blogggggggggggyyyyyyyyy
1. the variables x and y are functions of t and related by the equation y=2x^3-x+4 when x=2, dy/dt=-1 find dy/dt when x=2
y=2x^3-x+4
take deriv: dy/dt=6x^2(dx/dt)-(dx/dt)
plut in: dy/dt=6(2)^2(-1)-(-1)
dy/dt=-23
2. given f(x)=2x^2-7x-10, find the absolute maximum of f(x) on [-1,3]
f(x)=2x^2-7x-10
4x-7=0
=7/4
2(7/4)^2-7(7/4)-10= -129/8
2(-1)^2-7(-1)-10= -1
2(3)^2-7(3)-10= -13
answer is -1 for absolute max
3. chain rule-work from outside in
sin^2(x^2)
2(sin (x^2))(cos(x^2)(2x)
4xsin(x^2)cos(x^2)
4. average value of f(x)=1/x from x=1 to x=e is
1/e-1[ln (absoulte value of e)- ln (absoulte value of 1)]
1/e-1[1-0]
1/e-1
5. a particle's position is given by s=t^3-6t^2+9t what is the acceleration at time t=4
t^3-6t^2+9t
take deriv: 3t^2-12t+9
take second deriv: 6t-12
plug in 4: 6(4)-12= 12
i need help on something like
d/dx (integral from 2x to 5x)cost dt
and
like number 44 on the last calculator portion about a radioactive isotpe, y, decays..
1. the variables x and y are functions of t and related by the equation y=2x^3-x+4 when x=2, dy/dt=-1 find dy/dt when x=2
y=2x^3-x+4
take deriv: dy/dt=6x^2(dx/dt)-(dx/dt)
plut in: dy/dt=6(2)^2(-1)-(-1)
dy/dt=-23
2. given f(x)=2x^2-7x-10, find the absolute maximum of f(x) on [-1,3]
f(x)=2x^2-7x-10
4x-7=0
=7/4
2(7/4)^2-7(7/4)-10= -129/8
2(-1)^2-7(-1)-10= -1
2(3)^2-7(3)-10= -13
answer is -1 for absolute max
3. chain rule-work from outside in
sin^2(x^2)
2(sin (x^2))(cos(x^2)(2x)
4xsin(x^2)cos(x^2)
4. average value of f(x)=1/x from x=1 to x=e is
1/e-1[ln (absoulte value of e)- ln (absoulte value of 1)]
1/e-1[1-0]
1/e-1
5. a particle's position is given by s=t^3-6t^2+9t what is the acceleration at time t=4
t^3-6t^2+9t
take deriv: 3t^2-12t+9
take second deriv: 6t-12
plug in 4: 6(4)-12= 12
i need help on something like
d/dx (integral from 2x to 5x)cost dt
and
like number 44 on the last calculator portion about a radioactive isotpe, y, decays..
Post #too many
We meet again.
I'm just going to review the easy stuff that I keep forgetting to look over!
I. L'Hopital's Rule:
lim f(x)/f(x)=0
x-->0
*When this occurs it means to take the derivative of top and bottom then plug in for the limit. f^1(x)/g^(x)
EXAMPLE:
lim 3(x-4)/x^2-14 =0/0
x-->4
3/2x = 3/2(4)=3/8
II. Derivatives of Integrals:
EXAMPLE:
F(x)= integral from 0 to x^2 sin(t)
sin(x^2)(2x) *the 2x is the derivative of the bounds
III. Definition of Derivative:
*take derivative and plug in number
EXAMPLE:
lim sin(pi+g)-sin(pi)/g
g-->0
It is like sinx @ pi, so take deriv which is cos and plug in pi
cos(pi)=-1
IV. ln integration of a fraction:
EXAMPLES:
*What I put in parenthesis should be absolute value!
integral of 2x+6/x^2+6x
= ln (x^2+6x) +C
integral of cos(x)/sin(x)
= ln (sinx) +C
V. NEED HELP:
Although this only involves taking a derivative, can someone help me on
If f(x)=sin^2x, find f^111(x)
I'm just going to review the easy stuff that I keep forgetting to look over!
I. L'Hopital's Rule:
lim f(x)/f(x)=0
x-->0
*When this occurs it means to take the derivative of top and bottom then plug in for the limit. f^1(x)/g^(x)
EXAMPLE:
lim 3(x-4)/x^2-14 =0/0
x-->4
3/2x = 3/2(4)=3/8
II. Derivatives of Integrals:
EXAMPLE:
F(x)= integral from 0 to x^2 sin(t)
sin(x^2)(2x) *the 2x is the derivative of the bounds
III. Definition of Derivative:
*take derivative and plug in number
EXAMPLE:
lim sin(pi+g)-sin(pi)/g
g-->0
It is like sinx @ pi, so take deriv which is cos and plug in pi
cos(pi)=-1
IV. ln integration of a fraction:
EXAMPLES:
*What I put in parenthesis should be absolute value!
integral of 2x+6/x^2+6x
= ln (x^2+6x) +C
integral of cos(x)/sin(x)
= ln (sinx) +C
V. NEED HELP:
Although this only involves taking a derivative, can someone help me on
If f(x)=sin^2x, find f^111(x)
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
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
EVT: a continuous function on a closed interval [a,b] must have both a mnimmum and a maximum on the interval
Rolle's: Let F be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number "c" in (a,b) such that f(c)=0.
MVT: If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f '(c)=f(b)-f(a)/b-a
Rolle's: Let F be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b), then there is at least one number "c" in (a,b) such that f(c)=0.
MVT: If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f '(c)=f(b)-f(a)/b-a
And i still have trouble with integrals
post 29
this week we did more AP tests and graded them. and corrected them and what not. here's some stuff to go over for you:
Substitution:
1. State your u and du
2. Integrate
3. Plug u back in
also a lot of the time, you need to substitue in a number. when doing this, simply put it in front of the problem when you are done.
LRAM-delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-delta x[f(a+delta x)+...f(b)]
MRAM-delta x[f(mid)+f(mid)+...]
TRAM-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*for TRAM, delta x is [b-a/# of subintervals]
volume by disks:
pi S [function]^2 dx.
solve it by taking the integral of it and then pluging in the numbers they give you.
REMEMBER TO GRAPH
volume by washers:
pi S [top function]^2 - [bottom function]^2 dx.
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
REMEMBER TO GRAPH
what i need help with is ANGLE OF ELEVATION.
Substitution:
1. State your u and du
2. Integrate
3. Plug u back in
also a lot of the time, you need to substitue in a number. when doing this, simply put it in front of the problem when you are done.
LRAM-delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-delta x[f(a+delta x)+...f(b)]
MRAM-delta x[f(mid)+f(mid)+...]
TRAM-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*for TRAM, delta x is [b-a/# of subintervals]
volume by disks:
pi S [function]^2 dx.
solve it by taking the integral of it and then pluging in the numbers they give you.
REMEMBER TO GRAPH
volume by washers:
pi S [top function]^2 - [bottom function]^2 dx.
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
REMEMBER TO GRAPH
what i need help with is ANGLE OF ELEVATION.
post 29
here go teh blogs. again.
The limit rules are:
1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2) If the highest exponent is on the top then the limit is infinity.
3) But if the highest exponent is on the bottom then the limit is 0.
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
Tangent lines:
The problem will give you a function and an x value. Sometimes they may give you a y value; if not then you plug the x value into the original function and solve for y to get the y value. Next, you take the derivative of the function and plug in x to get the slope. After that, you plug everything into point-slope form.
First derivative test:
For the first derivative test, you are solving for max and mins and may be trying to see where the graph is increasing and decreasing. You take the derivative of the function and and set it equal to zero and solve for the x values (critical points). Then you set those points up into intervals between negative infinity and infinity. Then, you plug in numbers between those intervals to see if it is positive or negative.
Second derivative test:
For the second derivative test, you are solving to see whether the graph is concave up, concave down, or where there is a point of inflection in the graph. You take the derivative of the function twice and set it equal to zero and solve for the x values. You set those values up into intervals between negative infinity and infinity. You then plug in numbers between those intervals to see if it is positive or negative. If it is positive, it is concave up. If it is negative it is concave down. Where there is a change in concavity, there is a point of inflection.
im not good at e integration
The limit rules are:
1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2) If the highest exponent is on the top then the limit is infinity.
3) But if the highest exponent is on the bottom then the limit is 0.
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
Tangent lines:
The problem will give you a function and an x value. Sometimes they may give you a y value; if not then you plug the x value into the original function and solve for y to get the y value. Next, you take the derivative of the function and plug in x to get the slope. After that, you plug everything into point-slope form.
First derivative test:
For the first derivative test, you are solving for max and mins and may be trying to see where the graph is increasing and decreasing. You take the derivative of the function and and set it equal to zero and solve for the x values (critical points). Then you set those points up into intervals between negative infinity and infinity. Then, you plug in numbers between those intervals to see if it is positive or negative.
Second derivative test:
For the second derivative test, you are solving to see whether the graph is concave up, concave down, or where there is a point of inflection in the graph. You take the derivative of the function twice and set it equal to zero and solve for the x values. You set those values up into intervals between negative infinity and infinity. You then plug in numbers between those intervals to see if it is positive or negative. If it is positive, it is concave up. If it is negative it is concave down. Where there is a change in concavity, there is a point of inflection.
im not good at e integration
Post 29
For this weeks blog I am going to review some older things.
There are two types of integration are indefinite and definite. The answer for indefinite integration is an equation. But on the other hand the answer for a definite intergration is a number.
Indefinite Integration-Sx^n(dx)={(x^n+1)/(n+1)}+C
Definine Integration-bSa f(x)dx=F(b)-F(a)=number
The limit rules are:
1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2) If the highest exponent is on the top then the limit is infinity.
3) But if the highest exponent is on the bottom then the limit is 0.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
For something I am having trouble with is tangent inverse integration. If someone can help that would be great.
There are two types of integration are indefinite and definite. The answer for indefinite integration is an equation. But on the other hand the answer for a definite intergration is a number.
Indefinite Integration-Sx^n(dx)={(x^n+1)/(n+1)}+C
Definine Integration-bSa f(x)dx=F(b)-F(a)=number
The limit rules are:
1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2) If the highest exponent is on the top then the limit is infinity.
3) But if the highest exponent is on the bottom then the limit is 0.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
For something I am having trouble with is tangent inverse integration. If someone can help that would be great.
Post Number Twenty Nine
This week when we actually had a full calculus class we took ap tests. I’m assuming we did corrections in class Friday but I was not there so I only got back the non-calculator portion. I’ll talk about some of those problems later..
First, I’m going to type the Riemann sums formulas because I have yet to remember them!
The Riemann sums approximate 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
I understand exactly how to take implicit derivatives but for some reason I never get to right answer..
Implicit Derivatives:
take derivative like normal, of both sides of the equation.
any derivative taken for y, mark with dy/dx behind it.
solve for dy/dx
PVA
Position
Velocity
Acceleration
When moving down the list, take a derivative (down-derivative), while when moving up the list, integrate!
I don’t know how to do problems with the constant K or whatever..all help with that will be appreciated.
I also need some refreshment on average value, I thought I knew how to do those but apparently not.
I still need help with integration, especially with trig functions. And I don’t know how to do anything with trig functions being squared..
Some specific problems I’m having are:
Problems when you have to like draw triangles and such to find values.
Problems that ask if f is continuous.
Problems that you need to take a derivative of something like x^2 (square root of 3x+1)
Instantaneous rate of change..
e + 1 S 2 (4/x – 1) dx
And of course plenty more.
I need a tutor… good night everyone!
First, I’m going to type the Riemann sums formulas because I have yet to remember them!
The Riemann sums approximate 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
I understand exactly how to take implicit derivatives but for some reason I never get to right answer..
Implicit Derivatives:
take derivative like normal, of both sides of the equation.
any derivative taken for y, mark with dy/dx behind it.
solve for dy/dx
PVA
Position
Velocity
Acceleration
When moving down the list, take a derivative (down-derivative), while when moving up the list, integrate!
I don’t know how to do problems with the constant K or whatever..all help with that will be appreciated.
I also need some refreshment on average value, I thought I knew how to do those but apparently not.
I still need help with integration, especially with trig functions. And I don’t know how to do anything with trig functions being squared..
Some specific problems I’m having are:
Problems when you have to like draw triangles and such to find values.
Problems that ask if f is continuous.
Problems that you need to take a derivative of something like x^2 (square root of 3x+1)
Instantaneous rate of change..
e + 1 S 2 (4/x – 1) dx
And of course plenty more.
I need a tutor… good night everyone!
Post #29
Substitution:
1. Label the parts of your integral. (x^2)(2x) u= (x^2) du=(2x)
2. Integrate (u du). (1/2)(u^2) +c
3. Plug u back in. (1/2)((x^2)^2) +c
Your final answer should be (1/2)(x^4) +c or (x^4)/2 +c
For substitution to work, one must recognize derivative properties that applies the product rule, quotient rule, or chain rule. Say if a problem asks you to find the integral of (x^3)(x). You would have to bring out a 1/3 then multiply your x by 3, making it look like (1/3) integrate (x^3)(3x). Then you just follow yours steps of substitution, but make sure to leave the 1/3 out until the last step then you distribute it in.
Table Issues
Using the Table, estimate f'(2.1).
f(2.0)=1.39
f(2.2)=1.73
f(2.4)=2.10
f(2.6)=2.48
f(2.8)=2.88
f(3.0)=3.30
This problem is really easy if you remember what a derivative is, which is a slope. All you need to do for this problem is find the slope of two points nearest to 2.1. After picking the two points, 2.0 and 2.2, you must use the slope formula, which is f(b)-f(a) over b-a or (f(b)-f(a))/(b-a). Using the chart, you plug in f(b)=1.73 and f(a)=1.39, while b=2.2 and a=2.0.
(f(b)-f(a))/(b-a)
(f(2.2)-f(2.0)/(2.2-2.0)
(1.73 - 1.39)/(.2)
(.34)/(.2)
slope= 1.70
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.
1. Label the parts of your integral. (x^2)(2x) u= (x^2) du=(2x)
2. Integrate (u du). (1/2)(u^2) +c
3. Plug u back in. (1/2)((x^2)^2) +c
Your final answer should be (1/2)(x^4) +c or (x^4)/2 +c
For substitution to work, one must recognize derivative properties that applies the product rule, quotient rule, or chain rule. Say if a problem asks you to find the integral of (x^3)(x). You would have to bring out a 1/3 then multiply your x by 3, making it look like (1/3) integrate (x^3)(3x). Then you just follow yours steps of substitution, but make sure to leave the 1/3 out until the last step then you distribute it in.
Table Issues
Using the Table, estimate f'(2.1).
f(2.0)=1.39
f(2.2)=1.73
f(2.4)=2.10
f(2.6)=2.48
f(2.8)=2.88
f(3.0)=3.30
This problem is really easy if you remember what a derivative is, which is a slope. All you need to do for this problem is find the slope of two points nearest to 2.1. After picking the two points, 2.0 and 2.2, you must use the slope formula, which is f(b)-f(a) over b-a or (f(b)-f(a))/(b-a). Using the chart, you plug in f(b)=1.73 and f(a)=1.39, while b=2.2 and a=2.0.
(f(b)-f(a))/(b-a)
(f(2.2)-f(2.0)/(2.2-2.0)
(1.73 - 1.39)/(.2)
(.34)/(.2)
slope= 1.70
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.
Post #29
We had another AP tests this week, so I am going to review some of those questions.
The problems I always forget how to do are the ones like 5 and 10.
5 says the graph of a piecewise linear function f, for 0
A. 1 B. 4 C. 8 D. 10 E. 13
I can't show the graph, but all you have to do is divide the graph into rectangles and triangles then find the area of each shape. After that, you add the areas together. Finally you subtract the area of the top of the graph and the area of the bottom of the graph.
The first shape is a triangle so (1/2) (2) (2) = 2
Next is a rectangle (2)(2) = 4
Then a triangle (1/2) (1)(2) = 1
4+2+1 = 7
Bottom of the graph is two triangles.
(1/2)(1)(2) = 1
(1/2) (2)(2) = 2
1+2 = 3
Now subtract: 7-3 =4
The answer is B. 4
And number 10 says A car's velocity is shown on the graph above. Which of the following gives the total distance traveled from t=0 to t=16 (in kilometers)?
A. 360 B. 390 C. 780 D. 1000 E. 1360
Its the same steps as number 5, except there is no graph below the axis so you do not have to subtract.
The first shape is a triangle (1/2) (4) (60) = 120
Rectangle (4)(30) = 120
Rectangle (4)(90) = 360
Triangle (1/2)(4)(90)= 180
120+120+360+ 180 = 780
The answer is C. 780
Another one I got wrong was number 24. the lim as x -> 0 tan(3x) + 3x / sin (5x) =
A. 0 B. 3/5 C. 1 D. 6/5 E. Nonexistent
At first glance, I thought this problem was the shortcut so the answer would be 3/5, but that does not work for this problem.
To find the limit as x-> 0, you usually plug in 0, but in this case, you get 0/0, you have to us L'hopitals rule. L'hopitals rule states take the derivative of the top and the derivative of the bottom until you no longer have 0/0.
tan (3x) + 3x = sec^2 (3x) (3) + 3 = 3 sec^2 (3x) + 3
sin (5x) = cos (5x) (5) = 5 cos (5x)
Now plug in 0: 3 sec^2 (0) +3 / 5 cos (0) = 6/5
The answer is D. 6/5.
I need help on the volume and area problems. I never know if I need to solve the equation for x or y or when I have to subtract the equations by a number so pretty much everything to do with those problems.
The problems I always forget how to do are the ones like 5 and 10.
5 says the graph of a piecewise linear function f, for 0
A. 1 B. 4 C. 8 D. 10 E. 13
I can't show the graph, but all you have to do is divide the graph into rectangles and triangles then find the area of each shape. After that, you add the areas together. Finally you subtract the area of the top of the graph and the area of the bottom of the graph.
The first shape is a triangle so (1/2) (2) (2) = 2
Next is a rectangle (2)(2) = 4
Then a triangle (1/2) (1)(2) = 1
4+2+1 = 7
Bottom of the graph is two triangles.
(1/2)(1)(2) = 1
(1/2) (2)(2) = 2
1+2 = 3
Now subtract: 7-3 =4
The answer is B. 4
And number 10 says A car's velocity is shown on the graph above. Which of the following gives the total distance traveled from t=0 to t=16 (in kilometers)?
A. 360 B. 390 C. 780 D. 1000 E. 1360
Its the same steps as number 5, except there is no graph below the axis so you do not have to subtract.
The first shape is a triangle (1/2) (4) (60) = 120
Rectangle (4)(30) = 120
Rectangle (4)(90) = 360
Triangle (1/2)(4)(90)= 180
120+120+360+ 180 = 780
The answer is C. 780
Another one I got wrong was number 24. the lim as x -> 0 tan(3x) + 3x / sin (5x) =
A. 0 B. 3/5 C. 1 D. 6/5 E. Nonexistent
At first glance, I thought this problem was the shortcut so the answer would be 3/5, but that does not work for this problem.
To find the limit as x-> 0, you usually plug in 0, but in this case, you get 0/0, you have to us L'hopitals rule. L'hopitals rule states take the derivative of the top and the derivative of the bottom until you no longer have 0/0.
tan (3x) + 3x = sec^2 (3x) (3) + 3 = 3 sec^2 (3x) + 3
sin (5x) = cos (5x) (5) = 5 cos (5x)
Now plug in 0: 3 sec^2 (0) +3 / 5 cos (0) = 6/5
The answer is D. 6/5.
I need help on the volume and area problems. I never know if I need to solve the equation for x or y or when I have to subtract the equations by a number so pretty much everything to do with those problems.
Post # 29
This week in Calculus we took AP tests and that's pretty much it. So, i'm going to go over a few problems that we answered on the tests..
LIMITS:
lim x -> infinity (5x^2 - 3x+1) / (4x^2 +2x +5)
Because they have the same highest exponent, you divide the coefficiants.
Which means your answer is, 5/4.
If f(x) = 5/ x^2 +1 and g(x) = 3x, then g(f(2)) =
To solve this problem, you need to plug in the numbers according to the directions.
Which means, you take the function, g, and plug in the function, f, where all the x's are.
Then, you plug in a two where all the x's are and solve using algebra.
3(5/(x^2 +1))
3(5/2^2 + 1))
= 3(5/5) = 3(1) = 3
And those are basically the only ones i got right..soo, yeah.
I currently forgot how to do average value if they don't give you endpoints..and only one equation. I don't remember how to find them, and i don't remember what test it came off of.
But if anyone can help, it'd be GREATTTTT :)
LIMITS:
lim x -> infinity (5x^2 - 3x+1) / (4x^2 +2x +5)
Because they have the same highest exponent, you divide the coefficiants.
Which means your answer is, 5/4.
If f(x) = 5/ x^2 +1 and g(x) = 3x, then g(f(2)) =
To solve this problem, you need to plug in the numbers according to the directions.
Which means, you take the function, g, and plug in the function, f, where all the x's are.
Then, you plug in a two where all the x's are and solve using algebra.
3(5/(x^2 +1))
3(5/2^2 + 1))
= 3(5/5) = 3(1) = 3
And those are basically the only ones i got right..soo, yeah.
I currently forgot how to do average value if they don't give you endpoints..and only one equation. I don't remember how to find them, and i don't remember what test it came off of.
But if anyone can help, it'd be GREATTTTT :)
a post...
Ok. So I figured I might, just maybe explain yet another topic. Work for you? Good.
Okay,so if you don't know what average value is and/or how to do it, my question for you is, "Where have you been???" The formula for average value is as follows:
1/b-a S f(x) (The S being the integral sign once again.)
So say you're given a function and an interval f(x) = 3x^2 and [1,3].
Plug in...
1/(3-1) 3S1 3x^2
1/2 (27-1)
26/2
= 13===> look at that!!!
Reba's AmAzInG!!
So, you may have noticed in some of my pervious blogs, every always asks me about the Riemann Sums....which are quite possibly the easiest!!! things we've learned..it's all about the formulas!!! go look!! :))))))
What should we do next? Well, there's implicit derivatives, which I'm pretty sure most of you are fully versed in. And then there's e integration, which I hope most of you know how to do.
Ms. Mustian would kill me if she saw the grammar in this blog by the way.....
oh...I have no clue what to say/do...so How ABOUT we go over the steps in finding the tangent line to an equation.
1. Identify what you're given (pt, x value, equation, etc)
2. To find a slope, take the derivative and plug in the x value.
3. If you need a y value, plug into the x value into the original to find said value.
4. You now have a point and a slope======so I'm thinking maybe, just maybe, you can put that in *gasp* POINT SLOPE FORM!!!
AMAZING I'M TELLING YOU!!
So, now it's time for me to jet..( i just love that word sometimes). So have a FANTABULOUS week...I hope...and if you need any help, I believe that I could offer some assistance, although I'm not John and/or Tir, so, you know how that goes....
ADIOS!!!
P.S. I have no questions at the moment pertaining to calculus. Now you may want to ponder the cure for cancer or some other worldly problem....GO FOR IT!!!!
See ya!
I believe this is the end of my blog!!
Or not...
Maybe I'm just psycho....Hey!!! THAT'S A SONG!! and now it's stuck in my head
Adios....wait, I already said that....oh well!!!
Bye! and have fun!!!
Okay,so if you don't know what average value is and/or how to do it, my question for you is, "Where have you been???" The formula for average value is as follows:
1/b-a S f(x) (The S being the integral sign once again.)
So say you're given a function and an interval f(x) = 3x^2 and [1,3].
Plug in...
1/(3-1) 3S1 3x^2
1/2 (27-1)
26/2
= 13===> look at that!!!
Reba's AmAzInG!!
So, you may have noticed in some of my pervious blogs, every always asks me about the Riemann Sums....which are quite possibly the easiest!!! things we've learned..it's all about the formulas!!! go look!! :))))))
What should we do next? Well, there's implicit derivatives, which I'm pretty sure most of you are fully versed in. And then there's e integration, which I hope most of you know how to do.
Ms. Mustian would kill me if she saw the grammar in this blog by the way.....
oh...I have no clue what to say/do...so How ABOUT we go over the steps in finding the tangent line to an equation.
1. Identify what you're given (pt, x value, equation, etc)
2. To find a slope, take the derivative and plug in the x value.
3. If you need a y value, plug into the x value into the original to find said value.
4. You now have a point and a slope======so I'm thinking maybe, just maybe, you can put that in *gasp* POINT SLOPE FORM!!!
AMAZING I'M TELLING YOU!!
So, now it's time for me to jet..( i just love that word sometimes). So have a FANTABULOUS week...I hope...and if you need any help, I believe that I could offer some assistance, although I'm not John and/or Tir, so, you know how that goes....
ADIOS!!!
P.S. I have no questions at the moment pertaining to calculus. Now you may want to ponder the cure for cancer or some other worldly problem....GO FOR IT!!!!
See ya!
I believe this is the end of my blog!!
Or not...
Maybe I'm just psycho....Hey!!! THAT'S A SONG!! and now it's stuck in my head
Adios....wait, I already said that....oh well!!!
Bye! and have fun!!!
Okayy so this week we basically did what we always do..
took AP's and corrected them.
i'm just going to go over some stuff we see on every test.
A particle's position is given by s=t^3-6t^2+9t what is the acceleration at time t=4
t^3-6t^2+9t
take deriv: 3t^2-12t+9
take second deriv: 6t-12
plug in 4: 6(4)-12= 12
Original - Position
1st Derivative - Velocity
2nd Derivative - Acceleration
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
graph, then put the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: 153PI
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
questions: the first 3 problems on the calculator portion=helppp.
umm a review on substitution would also be nicee.
took AP's and corrected them.
i'm just going to go over some stuff we see on every test.
A particle's position is given by s=t^3-6t^2+9t what is the acceleration at time t=4
t^3-6t^2+9t
take deriv: 3t^2-12t+9
take second deriv: 6t-12
plug in 4: 6(4)-12= 12
Original - Position
1st Derivative - Velocity
2nd Derivative - Acceleration
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
graph, then put the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: 153PI
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
questions: the first 3 problems on the calculator portion=helppp.
umm a review on substitution would also be nicee.
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