AP Calculus AB & BC 2013-2014

Test Preparation AP Mathematics & Computer Science
261

<p>5) a) f’(x) = 0 for max/min. x = +/- 1. </p>

<pre><code> At x = -1, the sign of the derivative does not change so there is no max/min.

At x = 1, the sign of the derivative changes from negative to positive so there is a relative min there.

b) f’(1) - f’(-1) / 2 = 0/2 = 0

So the mean value theorem applies since it is continuous and differentiable and the secant line between the interval is 0. So there must be a point c, where the tangent line also equals 0 in the interval [-1.1].

c) h’(x) = f’(x) / f(x)
h’(3) = f ’ (3) / f(3) = 1/2 / 7 = 1/14

d) This will be a pain to type…

 Let u = g(x). du = g'(x)dx
 When x = -2, u = -1. When x = 3, u = 1.

 Integral from -2 to 3 of f'(g(x))g'(x)dx = integral from -1 to 1 of f'(u)du

 Integral from -1 to 1 of f'(u)du = f(u) from -1 to 1 = 2 - 8 = -6

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262

<p>Yep, bombed 1 and 6, but got at least 7 points on 2-5.</p>

<p>Hopefully that’s enough for a 5.</p>

263

<p>6) a) Looks kinda like a sin graph… :P</p>

<pre><code>b) m = (3-1)cos(0) = 2

y = 2x + b
1 = 2(0) + b    b = 1

y = 2x + 1
y = 2(0.2) + 1 = 1.4

c) dy/dx = (3-y)cosx
dy/(3-y) = cosxdx
-ln(3-y) = sinx + C
-ln(3-1) = sin(0) + C
-ln(2) = C
-ln(3-y) = sinx - ln(2)
ln(3-y) = ln(2) - sinx
e ^ (ln(2) - sinx) = 3 - y
y = 3 - e^(ln(2) - sinx)

I love these questions... :D

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264

<p>The % to get a 5 will most likely be similar to that of previous years (approximately 60-65% for a 5).</p>

265

<p>@SuperN0va‌ A much easier solution to 2© is differentiating y=2sin(2x) and plugging in x=pi/3, which gives the same answer, -2. This works because the distance between the graphs is 3-(3-2sin(2x)), which is 2sin(2x).</p>

266

<p>@mingeun1998‌ Hmm…Much more elegant. I didn’t even think of that. But why is the distance between the two graphs r<em>1 - r</em>2? That is based on the polar coordinate system – we are looking for rectangular coordinates which I thought requires translation into x and y. </p>

267

<p>I have a question: for 4(d), I put 25sqrt(41), but many people approximated and put the answer as 160. Will I get points deducted for putting 25sqrt(41)?</p>

268

<p>@minegeun1998 Where did you get the sqrt(41) since the numbers did work out… :P</p>

269

<p>Yeah – how could 50sqrt(41) come from sqrt(25625)?</p>

270

<p>@SuperN0va‌ Where did you get the idea that the distance can’t be based on a polar coordinate system (not trying to be offensive… but it sounds offensive)? The distance of two points are the same in either coordinate system, so it doesn’t matter.</p>

271

<p>Oh, did you do d = sqrt(A^2 + B^2)?</p>

<p>If you did that, you did WAY more work than you should’ve and most likely made some sort of mathematical error.</p>

272

<p>25sqrt(41)=sqrt(26525)…</p>

273

<p>@minegeun1998 - How did you get to sqrt(26525) since the numbers do work out nicely.</p>

274

<p>@minegeun1998 Yeah I guess that makes sense. I don’t know – I really don’t know polar stuff that well. Your method is far superior – I should have known mine was wrong by the amount of effort it required. </p>

275

<p>What is the better solution to the problem that I did by sqrt(A^2 + B^2)? I didn’t find it complicated at all though.</p>

276

<p>Oh, look at the top of page 18. That’s my solution and you get exactly 160.</p>

277

<p>Yeah it’s sqrt(100^2 + 125^2) = sqrt(25625), which is approximately the square root of 25600 = 160. </p>

278

<p>@kingofxbox99‌ According to SuperN0va’s solution… I had the same answer too.</p>

279

<p>If you do the d = sqrt(A^2 + B^2) method, you must use the chain rule to differentiate d I believe, not sure if you did that.</p>

280

<p>Btw, I’m not completely sure you can say that V = sqrt(va^2 + vb^2) I think you have to use distances than differentiate with respect to time. I’ve never seen it done anywhere using velocities in that formula, it’s always use distance then differentiate with respect to time to find velocity.</p>