Saturday, March 13, 2010
Post #30
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 :)
Thursday, March 11, 2010
Update
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
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
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
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
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)
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
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
post 29
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
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