<p>Ok so so to get all those people studying for the AB on cc together we can have a nice little study group.
We can review concepts that we are unsure on, discuss questions that we dont understand.
and other stuff = )</p>
<p>ohhh calculus!! Um.. should we explain concepts that are commonly forgotten like the mean value theorem and the intermediate value theorem??</p>
<p>yea lets do it.</p>
<p>I'll start I guess...
Mean Value Theorem: [f(b)-f(a)]/[b-a] = f`(c)
In other words the slope of the line should equal the derivative. Find the slope, then find the derivative of the equation. Then set the numericsl value (the slope) = to the derivative and solve for c. </p>
<p>I forgot intermediate value theorem lol..</p>
<p>
[QUOTE]
Mean Value Theorem: [f(b)-f(a)]/[b-a] = f`(c)
In other words the slope of the line should equal the derivative. Find the slope, then find the derivative of the equation. Then set the numericsl value (the slope) = to the derivative and solve for c.
[/QUOTE]
</p>
<p>Also, f must be continuous and differentiable from a to b. ETS loves to ask questions where a continuous but NOT differentiable function is given as it will end up trapping quite a few people.</p>
<p>The IVT says that given a continuous fuction over the interval (a,b) and a number N that falls between (f(a),f(b) there will be some f(c) that equals N. To put it plainly, it's the mean value theorem, but only for values of f(x), not f'(x)</p>
<p>The only theorem my teacher really taught us was the Fundamental Theorem.</p>
<p>To clarify on the MVT, it basically says that if there is secant line that passes through the endpoints of some interval, there must be some point on that interval such that the line tangent at that point has the same slope as the secant line.</p>
<p>I'll continue with the review
Acceleration = d/dx Velocity
Velocity = d/dx Position
Speed = |Velocity|
The position function for an object falling (neglecting air resistance) is 1/2 gt^2 + vt + h</p>
<p>g in the position function is acceleration (which is constant) right? On the AP test is it 9.8 or 10 m/s. Or do we use the other scale for gravity.</p>
<p>So anyways, I was doing this problem and am sort of stuck on it. The topic is finding area between curves. Here's the problem:
Find the area of the region between the two curves-
The curve y=x³ and the x-axis, from x= -1 to x=2.</p>
<p>So first I drew it. Then I integrated it which eventually gives me 1/4 + 4 = 17/4. Now 17/4 is the right answer, but if you look at the graph the integral on the left side should be negative shouldnt it? Because it's negative so it should be -1/4 + 4 = 15/4. </p>
<p>If someone could help that'll be awesome</p>
<p>O and btw how difficult are the PR problems compared to the actual thing?</p>
<p>g is -9.8 m/s or -32 ft/s (acceleration due to gravity)</p>
<p>Your problem isn't an area between two curves, it's just a typical integration problem.</p>
<p>integral from -1 to 2 of x^3 dx
= x^4/4 from -1 to 2
= 2^4/4 - (-1)^4/4
= 15/4</p>
<p>idk about PR's problems</p>
<p>I got the same answer as you ... but PR says its 17/4..</p>
<p>Ok so i'll assume PR made a mistake.
Help with this problem: <a href="http://i287.photobucket.com/albums/ll157/rb9109/Calc.jpg%5B/url%5D">http://i287.photobucket.com/albums/ll157/rb9109/Calc.jpg</a></p>
<p>The answer is (A). You have to recognize that the the limit is in the form of a definition of a derivative:
f'(x) = lim<a href="f(x+h)%20-%20f(x)">h>>0</a> / h</p>
<p>In this case you have </p>
<p>lim<a href="ln(e+h)%20-%201">h>>0</a> / h</p>
<p>you know that the Ln[e] = 1;</p>
<p>so now look at the first part, according to the definition, you need to have f(x+h), you have ln(e+h), so f(x) in this case would have to be ln(x),a nd since instead of x you have e, the x in this case is e; this works out for the secong part too, where in the definition you have f(x+h) - f(x); you proved the f(x+h), but now for the -f(x); as I said before you know that ln(e) is 1; and since you established that x = e in this case, and f(x) is ln(x); the entire thing wroks out :)</p>
<p>I hope this helps!!</p>
<p>o ok makes sense. Thanks</p>
<p>Can anyone recall the growth/decar formula?
I remember it's like y = Ce^kt i think?
And then there's a specific formula for k always (like 1/2ln something something)</p>
<p>O and how often do growth/decay problems show up? And i'm guessing when they do show up they're usually in the frq.</p>
<p>Growth decay is y = Ce^kt
C is the initial amount, t is time, and k is a constant. Its value changes based on the problem.</p>
<p>@jamesford and asc3nd</p>
<p>at the y=x^3 problem, and it asking area between the two curves, PR is correct. Draw out the graph if you need to. The area from [-1,0] would be, if you integrated from [-1, 0], negative (-1/4) because it is below the x-axis, and the area from [0,2] would be 2 and above the x-axis. Because it asks for the area BETWEEN curves you can't integrate the whole thing together.</p>
<p>Rather, you have to use formula of integral of upper curve minus lower curve in the respective regions.</p>
<p>So it's (I'll use S for the integral sign)</p>
<p>(S 0-x^3 dx from -1 to 0) + (S x^3-0 dx from 0 to 2) = 17/4</p>
<p>Taggert, I see what you did. That's a tricky one. Most people wouldn't think of the x axis as a "curve."</p>
<p>is anyone there can explain the volume revolve probs to me?
I cannot imagine the whole picture and always get confused about which one minus which one...thanks =)</p>
<p>Can somebody go over Newton's Law of Cooling? Do we even have to know that? We learned it when I wasn't there for a week but somebody said we don't need to know it.</p>
<p>@ LesleyG: type out an example problem and I'll solve it for you. I'm just not sure what you want.</p>
<p>If it's just revolving a curve, then think of it as a lot of cylinders stacked on top of one another. The distance between the axis of revolution and the function is the radius of each circle (and thus it changes depending on its position on the function). The volume of the "shell" can be calculated by subtracting the volume of the inner shape from the volume of the outer shape.</p>
<p>Newton's Law of Cooling on an AP Calculus test? Uh... if it's on the test, I'm sure it will be explained thoroughly.</p>
<p>Regarding the cylindrical shell method, google for some videos/animations describing it.</p>
<p>i.e. Antiderivatives</a> / Volumes - 6</p>
<p><a href="http://talk.collegeconfidential.com/ap-tests-preparation/503162-someone-please-double-check-my-logic-real-quick.html%5B/url%5D">http://talk.collegeconfidential.com/ap-tests-preparation/503162-someone-please-double-check-my-logic-real-quick.html</a>, is useful. Check the third post.</p>