<p>Haha. Yeah. that's what my calc book basically says. I thiiiink. I have issues with reading things not written in TI-89 input notation. =P</p>
<p>bah. now to review the bc material. >.<</p>
<p>Solids of revolution just requires you to remember 2 (or 3 with the shells) methods and knowing when to choose which one.</p>
<p>Disk: choose when you can tell there will be no hole in the solid. Check to see if the base of the region to be revolved is on the axis of revolution, for example.</p>
<p>Washer and shell are both used when there will be a hole in the solid. The representative rectangle in the region of revolution for the disk and washer methods is PERPENDICULAR to the axis of revolution. The representative rectangle in the region of revolution for the shell methods is PARALLEL to the axis of revolution. It really depends on what the equations are that bound the region of revolution. If one of the regions was y = x^4+3x^2-2x+1, you would be best off not trying to solve for x and using that.</p>
<p>The method found in that other post (#3 on "check my logic") may work, but you're better off reasoning through the problem rather than memorizing another formula.</p>
<p>Ok so i finally went through a practice test in Princeton Review and got a raw score of 69 (1 freeking point away from a 5).
So anyway's I'm gonna type the problems up which I still can't understand:</p>
<p>Non-calculator section
lim(x->0) [Tan^3(2x)/x^3]
The answer is 8</p>
<p>S[x(5-x)^1/2]dx (Im using S as integral sign)
Answer is (-10/3)[(5-x)^(3/2)]+(2/5)[(5-x)^(5/2)]+C</p>
<p>Calculator section
d/dx<a href="S%20=%20integral%20sign">S(from 0 to x^2)(sin^2t)dt</a>
Answer is 2xsin^2(x^2)</p>
<p>f(x) is continuous and differentiable and f(x) = {ax^4 +5x; x<2 and bx^2 - 3x; x>2}. Find b. The problem is set up basically like a piecewise function.
Answer is a = 1/2 and b = 6</p>
<p>as far as I know, you only have to know the disk method for the AP Calculus AB test (pi * Integral R^2 -r^2 dx). And Newton's Cooling might appear on the AP test.</p>
<p>@asc3nd:</p>
<p>Am I correct in reading that your first problem is the limit as x approaches 0 of: the tangent of 2x, quantity cubed, over x cubed? I ran it through Maple (a mathematics program) and it ~27 steps to solve... <em>and</em> it used l'Hopital's rule. A problem like that definitely won't be on the AB test, unless there's a much simpler way of solving it (which I don't see right now). :</p>
<p>EDIT: AFAIK you'll need to know integration by parts, which isn't on the AB test, to solve the second problem.</p>
<p>for the AP Calc. AB test do we have to memorize the complicated Arc formulas?</p>
<p>@jasonlee576:</p>
<p>On the AB test the most you'll have to do is differentiate an inverse trig function. And it's just 3 formulas (since the cofunctions are just opposites of the other 3)!</p>
<p>Plus you can differentiate to find the formulas on the fly (I think TheMathProf has done this at least once on the forum), so.. don't fret :)</p>
<p>@ asc3nd:</p>
<p>To solve the third, use the chain rule, i.e.</p>
<p>u = x^2
du/dx = 2x</p>
<p>then:</p>
<h1>d/du[ Integral from 0 to u (sin^2[t])dt ] * du/dx</h1>
<h1>d/du[ Integral from 0 to u (sin^2[t])dt ] * 2x</h1>
<h1>sin^2 * 2x <- by the second FTC</h1>
<p>2x * sin^2[x^2]</p>
<p>EDIT: And if you didn't understand that, take a look at The</a> Second Fundamental Theorem of Calculus - the author does a problem like this towards the bottom of the page.</p>
<p>@asc3nd:</p>
<p>For your fourth problem, I believe that you take the derivatives of the two equations that define f(x), and then using those derivatives and the original two equations, you solve a linear system of equations blah blah blah. If that made sense :)</p>
<p>Hey I'm having some trouble with Question 3 Free Response (Form A) from the 2007 test. This test just seems alot harder then the other FR up on the collegeboard.</p>
<p>I dont really get the reasoning for Part b,c or d...I havent seen the MVT applied to B with that reasoning before. hellp thanks!</p>
<p>@knickknack:</p>
<p>For part b, it makes sense if you think about it. The question asks you to find h'(c) over an interval. That just sounds like MVT right? [(f(b)-f(a))/(b-a)]. They give you the interval (1,3), so you just apply the theorem. Even if you have no idea how to do the problem, given the things they give you and what you already know, you can probably find the answer to a question, seeing as the calc exam repeats the same ideas every year.</p>
<p>For part c, it is the second fundamental theorem of calculus. I'm sure you can figure it out from here on on your own. </p>
<p>For part d, first you have to find what g^-1(x) is. To do that, you have to know a little about inverses. If (x,y) is a point on a graph, then for the inverse, you switch x and y, making it (y,x). Such is the case for this problem. g(1)=2, so in the inverse of g, g^-1(2)=1. Then, you have to use the equation for the derivative of an inverse (it's something you just memorize). Then, just find the equation of the tangent line at x=2. </p>
<p>I hope these helped. I just got back from seeing Iron Man, and it's pretty late (1:11AM eSt). If you need further clarification, I'll be happy to help after some rest ^^</p>
<p>I took a Princeton Review yesterday and got a high 4. But today, I took a practice test from Sparknotes
and am in the range of a high 3, low 4.</p>
<p>Is Sparknotes suppose to be very hard?</p>
<p>is there integration by parts on the AP test?</p>
<p>@jasonlee576:</p>
<p>BC yes, AB no.</p>
<p>Integration by parts is BC only</p>
<p>i need help puttin the units in the free response questions
for the units, is there like steps to know what unit u have to use.
like the derivative is always squared? im not really sure so can someone help thanks</p>
<p>well it depends on what it's in respect of. say you're given x(t)= 5t^2 meters and your taking the derivative with respect of t, where t = t seconds.
v(t) = 10t meters/seconds. If you were finding acceleration, it would be a(t) = 10 meters/(seconds^2).</p>
<p>To add onto what swim said, make sure you don't write cm^2 or cm when you should've written cm^3. This is not hard, but it throws people off. Some words that suggest</p>
<p>Normal units (e.g. cm): length
Squared units: area, surface area
Cubed units: volume</p>
<p>I'm stuck on this problem:
lim (n->infinite) E(where i =1 to n) (1+(2i/n))^3(2/n)</p>
<p>The E is the thing used in Riemens sum. The thing on top of the E is n, and at the bottom is i=1.
I asked so many people and no one could get it. Any help would be appreciated.</p>
<p>You don't need to know the gravitational constant.</p>
<hr>
<p>When it comes to units, it helps to think about what you're looking at. If s(t) is given in meters, and the time variable is given in seconds, then v = ds/dt or the rate of distance / the rate of time, so your units are meters/second.</p>
<p>Similarly, a would be dv/dt, or the rate of velocity / the rate of time, so your units are (meters/second)/second or meters/second^2.</p>
<p>Sometimes, the units aren't quite as straightforward. My favorite example is from the 2005 AP Calculus AB Free Response, question 3 (Page</a> Not Found%5DPage">http://www.collegeboard.com/prod_downloads/ap/students/calculus/ap05_frq_calculus_ab.pdf)). Here T(x) represents the temperature (in degrees Celsius), and x represents the distance traveled in centimeters.</p>
<p>T'(x) or dT/dx is the rate of temperature / rate of centimeters, so the units would be degrees Celsius / centimeter.</p>
<hr>
<p>You can always re-derive the formula for the derivative of the inverse of a function. Let's say that you know both f(x) and f '(x). Let y = f(x). Then the inverse is x = f(y).</p>
<p>Then the derivative of the x = equation is 1 = f '(y)*y' (using the Chain Rule). So y' = 1/(f '(y)) = 1/(f '(f(x))</p>