<p>TheMathProf:</p>
<p>Thanks!! That makes me feel much better. =)</p>
<p>Ok pearl let me get this straight.</p>
<p>You had E^-x sin x
F`(X) = E^-x cos x - E^-x sinx = 0
-E^-x (-cosx + sinx) = 0
So you said since E^x never equals 0, Cosx = sinx.
Let me guess you chose pi / 4 ? (this falls from 0, pi and thats why I am asking). pi / 4 in one period of the function is the farthest right though. 5pi/4 is what you were looking for. BUT you will get points for it for sure.</p>
<p>And oh yeah. My method works to for the cross section one. I just didn't divide D / 2. I left it in terms of D so I got 57Pi or w/e the answer was. But I like your method better. Mine was messy...</p>
<p>I can't believe I missed 2C. When they said where was it greatest I put JUST 150 or w/e (i never read the initial 5000.... so....).</p>
<p>Oh I have a question. What did yall put for 3c. and 3d. I got -1 for 3c. iirc. I dont remember 3d. to well.. but it was easy but I am not sure.</p>
<p>I think I checked pi/4 and 3pi/4 but I knew that neither was a minimum, I checked the endpoints and they werent mins either, so I moved on to the next question...i never really put an answer because i never considered 5pi/4</p>
<p>I think 3c is -2...w'(3)=f(g(3)) g'(3)</p>
<p>Wait for 3c, isn't w'(3) just f(3)? because they said w(x) = integrand f(x). so if you find w'(x) isn't it just f(x) without the integrand sign?</p>
<p>How did you guys do D. I said that the slope of g^-1
was (1/g^-1) so (y-2) = (1/5)(x-1)</p>
<p>Also, for the tan approx. one is it ok that I did </p>
<p>f(x) + f'(x)(delta x) </p>
<p>30 + 2(.04) = 30.8?</p>
<p>damn i totally got 3d wrong cause i did (1/g^-1) first but then i changed it</p>
<p>i got -2 for w'(3)</p>
<p>w'(3) = -2.</p>
<p>In the fundamental theorem of calculus, whenever your limit of integration isn't a nice clean value like x, you need to take the derivative of that limit of integration to multiply (because of the Chain Rule).</p>
<p>If F(x) is an antiderivative of f(x), then if j(x) = Integral from g(x) to h(x) of f(x) dx = F(g(x)) - F(h(x)) from the Second Fundamental Theorem.</p>
<p>Then j'(x) = F'(g(x))<em>g'(x) - F'(h(x))</em>h'(x) = f(g(x))<em>g'(x) - f(h(x))</em>h'(x).</p>
<p>where in this particular problem g(x) is literally g(x) and h(x) = 1, so h'(x) = 0.</p>
<p>Then evaluate at x = 3, and voila...</p>
<p>for 3d i first found that g^-1(2)=1 ( g(1)=2, so g^-1(2)=1) </p>
<p>we get the coordinates (2,1)</p>
<p>then we needed to find g^-1'(2), so i said g^-1'(2)=1/g'(g^-1(2))=1/g'(1)=1/5</p>
<p>so y-1=(1/5)(x-2)</p>
<p>I screwed up on:</p>
<p>4a) Used incorrect values for sin and cos of 5pi/4 so it ended up looking identical to pi/4 to me.</p>
<p>4b) I think I got most of the answer, but I panicked and didn't actually solve for the final number.</p>
<p>6c) I may have misinterpreted the second derivative (inverted the exponent) or not have thought of using e^foo to get rid of the natural log.</p>
<p>Okay I still don't understand 3c it's a stupid problem. I don't understand why you can't just take away the integrand sign, and I swear I have never seen a problem like that before.</p>
<p>^^ Sounds like someone's teacher shorted them on the Fundamental Theorem of Calculus. It's come up a number of times before on the AP Exam, and will continue to come up so long as people keep missing it. :)</p>
<p>The idea that the antiderivative of the derivative ALWAYS leads back to where you started is just a little too simplistic.</p>
<p>TheMathProf -- you sound so like my math teacher....................!!!! Xo</p>