Aside from the links on Edline, are there any places online that we can find help specifically in AP testing?
Or rather, not exactly that. Are there online places where we can take practice AP tests? (as well as the ones we take in class)
My mom's been on me trying to figure out where I can get more tests from (I'm pretty sure it's not gonna help if they don't have the corrections there though..)
Oh BTW: 666th post! =]
Wednesday, February 24, 2010
Questions
Once again remember that you can ask questions on the blog if you need help with your corrections!
Sunday, February 21, 2010
post 27
Okay, after having a good mardi gras break i'm going to have to do this blog, so for this particular one i am going to go over related rates and linearization.
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unkown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Example: 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
alright, so you put down the equation, y=2x^3-x+4.
Then you take the derivative of that, so you get dy/dt=6x^2(dx/dt)-(dx/dt)
then you plug in to find that dy/dt=6(2)^2(-1)-(-1)
and that is further simplified to, dy/dt=-23.
Linearization:
f(x)=f(c)+f'(c)(x-c)
example: Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
example 2: use differentials to approximate: sq root(16.5)
steps:
1: identify an equation--- f(x)=sq root(x)
2:f(x)+f;(x)dx--- sqrt(x)+ (1/(2sqrt(x)))(dx)
3:determine dx-- .5
4:determine x--- 16
5:plug in--- sqrt(16)+(1/2sqrt(16))(.5)= 4.0625
error= .0005
since we have to post something that we also have trouble with in our blogs, i'm going to go ahead and say that i could use some refreshing on angle of elevation, whoever wants to help with that can do that..
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unkown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Example: 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
alright, so you put down the equation, y=2x^3-x+4.
Then you take the derivative of that, so you get dy/dt=6x^2(dx/dt)-(dx/dt)
then you plug in to find that dy/dt=6(2)^2(-1)-(-1)
and that is further simplified to, dy/dt=-23.
Linearization:
f(x)=f(c)+f'(c)(x-c)
example: Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
example 2: use differentials to approximate: sq root(16.5)
steps:
1: identify an equation--- f(x)=sq root(x)
2:f(x)+f;(x)dx--- sqrt(x)+ (1/(2sqrt(x)))(dx)
3:determine dx-- .5
4:determine x--- 16
5:plug in--- sqrt(16)+(1/2sqrt(16))(.5)= 4.0625
error= .0005
since we have to post something that we also have trouble with in our blogs, i'm going to go ahead and say that i could use some refreshing on angle of elevation, whoever wants to help with that can do that..
Post.
Limits: if you can’t just plug in the number it approaches (which is when you get a zero on the bottom), you have to manipulate algebra and cancel out anything that can be and then plug in the number. If that doesn’t work, you then can try to use your table function. Then if That doesn’t work, it is DNE.
Optimization: Identify primary and secondary. Solve secondary for 1 variable (if applicable) and plug into primary. Take derivative of primary to solve for other variable and plug into the secondary equation to find the other variable. (Confusing I know…but it’s not that difficult…patience)
Example:
The sum of two numbers is 120. The product of the two is 7200. Find the minimum values
Set up equations: x + y = 240
xy = 7200
Solve the secondary for, let’s pick, y. So, y = 240-x.
Plug into primary
x(240 – x) = 7200
Take Derivative
1(240 – x) + (-1)(x) = 0
Solve for x.
x=120
Plug into when we solved y.
y=120
Simple enough right? So apply that same principle to other problems like when finding dimensions and area.
Related Rates: Everything is in reference to time. You have to be able to identify your formulas and everything that’s given out of words. Get what I’m saying? And remember that speed can’t be negative (made that mistake a few times). Also, don’t forget your units!!!!
Idk if this will be on the AP, but can someone explain percent error again? I vaguely remember it..but you know how those things are...
Optimization: Identify primary and secondary. Solve secondary for 1 variable (if applicable) and plug into primary. Take derivative of primary to solve for other variable and plug into the secondary equation to find the other variable. (Confusing I know…but it’s not that difficult…patience)
Example:
The sum of two numbers is 120. The product of the two is 7200. Find the minimum values
Set up equations: x + y = 240
xy = 7200
Solve the secondary for, let’s pick, y. So, y = 240-x.
Plug into primary
x(240 – x) = 7200
Take Derivative
1(240 – x) + (-1)(x) = 0
Solve for x.
x=120
Plug into when we solved y.
y=120
Simple enough right? So apply that same principle to other problems like when finding dimensions and area.
Related Rates: Everything is in reference to time. You have to be able to identify your formulas and everything that’s given out of words. Get what I’m saying? And remember that speed can’t be negative (made that mistake a few times). Also, don’t forget your units!!!!
Idk if this will be on the AP, but can someone explain percent error again? I vaguely remember it..but you know how those things are...
Ash's 27th Post
Apparently, my blogs are talkaboutable.
For all of you who oh so completely enjoy these and love talking about them, here's a cookie.
Anyway, on subject:
Some people are saying they don't get how to start Substitution, right? Well, let's try to at least get to where you can get SOME points for these.
Let's say you have a simple...ish equation:
1S0 2sinx*times*cosx-x^2dx
So, your u would be sinx
And your du would be the derivative of that: cosx!
So you plug those back into your equation to get:
1S0 2udu-x^2dx
Now, this is where I lose how to explain this. I can do it, I just can't explain it too well on the computer.
After that, you just integrate sing the u/du's, then plug back in the sin and cos, and then solve like normal integration....right?
I have a serious problem when it comes to testing.
I *most of the time* know what to do, but I just can't do it.
I get intimidated and totally freaked out.
For example, the huge derivatives or huge integrating. Does anyone have any tricks on how to work these?
Also, when going over it with friends I understand what to do, I just cannot do it on my own. Does anyone else have this problem? Or if you did and you overcame it, how??
I'm just terrified I'm going to bomb all of my AP practices tests and STILL go on and take the real one and completely FAIL at it =/
For all of you who oh so completely enjoy these and love talking about them, here's a cookie.
Anyway, on subject:
Some people are saying they don't get how to start Substitution, right? Well, let's try to at least get to where you can get SOME points for these.
Let's say you have a simple...ish equation:
1S0 2sinx*times*cosx-x^2dx
So, your u would be sinx
And your du would be the derivative of that: cosx!
So you plug those back into your equation to get:
1S0 2udu-x^2dx
Now, this is where I lose how to explain this. I can do it, I just can't explain it too well on the computer.
After that, you just integrate sing the u/du's, then plug back in the sin and cos, and then solve like normal integration....right?
I have a serious problem when it comes to testing.
I *most of the time* know what to do, but I just can't do it.
I get intimidated and totally freaked out.
For example, the huge derivatives or huge integrating. Does anyone have any tricks on how to work these?
Also, when going over it with friends I understand what to do, I just cannot do it on my own. Does anyone else have this problem? Or if you did and you overcame it, how??
I'm just terrified I'm going to bomb all of my AP practices tests and STILL go on and take the real one and completely FAIL at it =/
Posting...#27
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
e integration:whatever is raised to the e power will be your u and du will be the derivative of u.
For example:e^2x-1dxu=2x-1 du=2
rewrite the function as:1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
limits:Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
3. if top degree is less than bottom degree, the answer is 0
Im still having problems with integration but i do understand the e one since we went over that that one class
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
e integration:whatever is raised to the e power will be your u and du will be the derivative of u.
For example:e^2x-1dxu=2x-1 du=2
rewrite the function as:1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
limits:Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
3. if top degree is less than bottom degree, the answer is 0
Im still having problems with integration but i do understand the e one since we went over that that one class
post 27
Related Rates:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation involving your variables
5. Take the derivative with respect to time
6. Substitute in derivative and solve.
Example:
Air is being pumped into a spherical balloon at a rate of 4.5 cubic ft/min. Find the rate of change of the radius when the radius is 2 ft.
1. r=2ft; dv/dt = 4.5 ft^3/min
2. NEED TO FIND dr/dt.
3. Volume of sphere: v=4/3 pi r^3
4. derivative: dv/dt = 4 pi r^2 dr/dt
5. Plug in: 4.5 = 4pi(2)^2 dr/dt
6. 4.5=16 pi dr/dt
dr/dt = 9/32 pi ft/min
things i need help with is bacteria problems. I also need help with intergration of trig functions still. I need to start getting these problems right on my ap test
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation involving your variables
5. Take the derivative with respect to time
6. Substitute in derivative and solve.
Example:
Air is being pumped into a spherical balloon at a rate of 4.5 cubic ft/min. Find the rate of change of the radius when the radius is 2 ft.
1. r=2ft; dv/dt = 4.5 ft^3/min
2. NEED TO FIND dr/dt.
3. Volume of sphere: v=4/3 pi r^3
4. derivative: dv/dt = 4 pi r^2 dr/dt
5. Plug in: 4.5 = 4pi(2)^2 dr/dt
6. 4.5=16 pi dr/dt
dr/dt = 9/32 pi ft/min
things i need help with is bacteria problems. I also need help with intergration of trig functions still. I need to start getting these problems right on my ap test
Post #the holidays
Well sadly, the holidays are over. Anyways, lets get straight to the blog.
I'm going to review some of the basics.
FORMULAS FOR DERIVATIVES:
We can all take derivatives. They are really easy, but some are easily forgotten
a^x = lna(a^x)
e^x = e^x
lnx = 1/x
tanx = sec^2x
cscx = -cscxcotx
secx= secxtanx
cotx = -csc^2x
DEALING WITH GRAPHS:
1st derivative: Increasing, Decreasing, max and mins
2nd derivative: Concave up, Concave down, point of inflection.
RELATED RATES:
First, some of the formulas dealing with related rates.
CONE- V=3/4(pi)(h)(r^2)
CUBE- V=E^3
RECTANGLE- A=1/2(B)(H)
SQUARE- A=L x W P=2(L+W)
SPHER- V=1/3(pi)(r^3)
CIRCLE- A=2(pi)(r)
Now for an example:
The radius, r, of a circle is increasing at a rate of 3 centimeters per minute. Find the rate of change of area, A, when the radius is 5.
First I write down my given:
A =pir^2
dr/dt = 3
r = 5
dA/dt = ?
So take the derivative of the formula.
dA/dt=pi2r(dr/dt)
now plug in:
dA/dt = pi(2)(5)(3)
dA/dt = 30pi
I still mess up with integration!
I'm going to review some of the basics.
FORMULAS FOR DERIVATIVES:
We can all take derivatives. They are really easy, but some are easily forgotten
a^x = lna(a^x)
e^x = e^x
lnx = 1/x
tanx = sec^2x
cscx = -cscxcotx
secx= secxtanx
cotx = -csc^2x
DEALING WITH GRAPHS:
1st derivative: Increasing, Decreasing, max and mins
2nd derivative: Concave up, Concave down, point of inflection.
RELATED RATES:
First, some of the formulas dealing with related rates.
CONE- V=3/4(pi)(h)(r^2)
CUBE- V=E^3
RECTANGLE- A=1/2(B)(H)
SQUARE- A=L x W P=2(L+W)
SPHER- V=1/3(pi)(r^3)
CIRCLE- A=2(pi)(r)
Now for an example:
The radius, r, of a circle is increasing at a rate of 3 centimeters per minute. Find the rate of change of area, A, when the radius is 5.
First I write down my given:
A =pir^2
dr/dt = 3
r = 5
dA/dt = ?
So take the derivative of the formula.
dA/dt=pi2r(dr/dt)
now plug in:
dA/dt = pi(2)(5)(3)
dA/dt = 30pi
I still mess up with integration!
Post 27
So it’s 8:56 on Sunday night and I’m actually doing my blog on time. Look at that! ha. This week off was a nice break. I had a lot of time to do things, although I procrastinated until the very last minute. Good thing I was off today.
Anyway, two weeks ago right before B-rob left we had two days of review of the things that were bothering us the most. I blogged about this last week, but I saved some things to blog about this week as well. A few more things that refreshed my memory are how to do a derivative and integral in my calculator, definitions of derivatives, how to recognize product rules and chain rules, composite functions and composite functions with graphs, e integration, and ln integration.
Ok, so for using my calculator. For the calculator portion the calculator can be used for every problem to help, so I need to remember to graph in it as much as possible. For an integral in the calculator, I plug it into y=, although you can do it with the math function. The way I do it is I plug the integral into my y=, I graph it, hit second calc, and go all the way down to integral. Once I press that button, I just plug in my bounds and it gives me my integral. The way to do it with the math function is to hit math, fnint (equation, x, bound 1, bound 2). For this one, I always forget that x, or I don’t use enough parentheses.
Ok, so I’m still a little shakey when it comes to tables and composite functions. When there is a composite function, treat it as a chain rule and go from there. That was my biggest problem, I just did the composite without taking the derivative of it.
For integration with fractions, before substitution is considered, two things should be looked for: natural logs and tangent inverses. A natural log is when the top is the derivative of the bottom. A tangent inverse is when it is a number over x^(something) + 1.
I’m still not that great when it comes to particle problems.
Anyway, two weeks ago right before B-rob left we had two days of review of the things that were bothering us the most. I blogged about this last week, but I saved some things to blog about this week as well. A few more things that refreshed my memory are how to do a derivative and integral in my calculator, definitions of derivatives, how to recognize product rules and chain rules, composite functions and composite functions with graphs, e integration, and ln integration.
Ok, so for using my calculator. For the calculator portion the calculator can be used for every problem to help, so I need to remember to graph in it as much as possible. For an integral in the calculator, I plug it into y=, although you can do it with the math function. The way I do it is I plug the integral into my y=, I graph it, hit second calc, and go all the way down to integral. Once I press that button, I just plug in my bounds and it gives me my integral. The way to do it with the math function is to hit math, fnint (equation, x, bound 1, bound 2). For this one, I always forget that x, or I don’t use enough parentheses.
Ok, so I’m still a little shakey when it comes to tables and composite functions. When there is a composite function, treat it as a chain rule and go from there. That was my biggest problem, I just did the composite without taking the derivative of it.
For integration with fractions, before substitution is considered, two things should be looked for: natural logs and tangent inverses. A natural log is when the top is the derivative of the bottom. A tangent inverse is when it is a number over x^(something) + 1.
I’m still not that great when it comes to particle problems.
Post 27
I will start by explaining the area between curves. The formula is bSa top equation-bottom equation
1. draw the picture of the graphs
2. subtract the two equations from each other
3. put the like terms together and integrate the result
Next I will talk about trig inverse integration. The trig inverse integration formulas are: (sr=square root)
1. S du/sr(a^2-u^2)=-1/sr(u)arcsin u/a +C
2. S du/a^2+u^2=1/du(a)arctan u/a +C
3. S du/u sr(u^2-a^2)=1/du(a)arcsec lul/a +C
Another thing I will talk about is substitution. Substitution takes the position of derivative rules when integrating. The steps for substitution are:
1. Find a derivative of something else inside of the integral.
2. Set u equal to the non derivative found in the integral
3. Then take the derivative of u
4. Substitute back in
5. Solve
Example problem:S (2x^2+5) (4x) dx u=2x^2+5 du=4x dx
S u du
1/4 u^2+C
1/4 (2x^2+5)^2+C
Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
1. draw the picture of the graphs
2. subtract the two equations from each other
3. put the like terms together and integrate the result
Next I will talk about trig inverse integration. The trig inverse integration formulas are: (sr=square root)
1. S du/sr(a^2-u^2)=-1/sr(u)arcsin u/a +C
2. S du/a^2+u^2=1/du(a)arctan u/a +C
3. S du/u sr(u^2-a^2)=1/du(a)arcsec lul/a +C
Another thing I will talk about is substitution. Substitution takes the position of derivative rules when integrating. The steps for substitution are:
1. Find a derivative of something else inside of the integral.
2. Set u equal to the non derivative found in the integral
3. Then take the derivative of u
4. Substitute back in
5. Solve
Example problem:S (2x^2+5) (4x) dx u=2x^2+5 du=4x dx
S u du
1/4 u^2+C
1/4 (2x^2+5)^2+C
Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
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