Alright everyone, let's practice for the AP Calc BC exam

Test Preparation AP Test Preparation
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<p>I have searched through all the topics and was unable to find one of these, so I guess I will start one. I will post a question, someone answers it and post another question for the next poster and so on. This will be a nice informal last min prep for the upcoming AP exam. I want to see everyone get 5s, so lets go.</p>

<p>I'll start:</p>

<p>Find the speed of a particle defined by x(t) = t^2 + 1 and y(t) = 2t at point (10,6) (3 dec places)</p>

<p>Edit: ChaosTheory has a point, posting explanations will be helpful</p>

2

<p>dy/dt = 2
dx/dt = 2t</p>

<p>10 = t^2 + t
t = 3</p>

<p>dy/dt (3) = 2
dx/dt (3) = 6</p>

<p>ds/dt = ((dx/dt(3))^2 + (dy/dt(3))^2)^0.5
= (6^2 + 2^2)^0.5
= 6.3245
= 6.325</p>

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<p>Please post explanations.</p>

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<p>Will the poster of the problem verify the solution?</p>

<hr>

<p>To what limit does the sequence S_n = (3+n)/(3^n) converge as n approaches infinity?</p>

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<p>Solution: sqrt((dy/dt)^2 + (dx/dt)^2) @t = 3,
ans = 6.325</p>

<p>(Sorry about my slowness, my internet acts weird at times)</p>

6

<p>New question:</p>

<p>Find int of convergence:</p>

<p><a href="(x%20+%203)%5En">sigma from 0 to inf</a> /(n^2 *5^n)</p>

<p>@goodusername: You will probably see at lease one FRQ involving parametrics.</p>

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<p>Answer to #7:</p>

<p>Use the ratio test [Excuse my excessive use of parentheses]:</p>

<p>lim (n -> ∞) abs(((x+3)^(n+1))/((n+1)^2<em>5^(n+1))/(((x+3)^n)/(n^2</em>5^n)))
= lim (n -> ∞) abs((x+3)^(n+1)<em>n^2</em>5^n)/((x+3)^n<em>(n+1)^2</em>5^(n+1)))</p>

<p>Here, a massacre occurs and many terms cancel to leave</p>

<p>= lim (n -> ∞) ((x+3)*(n^2)/(5(n+1)^2)) = lim (n -> ∞) [n^2/(5(n+1)^2)] * |x+3|</p>

<p>Note that lim (n -> ∞) [n^2/(5(n+1)^2)] = 1/5.</p>

<p>= |x+3|/5 < 1
= |x+3| < 5
x < 2 or x > -8. </p>

<p>The interval converges absolutely between -8 and 2. I don't want to do the endpoints. :D Am I slightly correct?</p>

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<p>ChaosTheory has the numbers correct.</p>

<p>Eval the endpoints, for x = 2 you get [sigma] 1/n^2 which converges, and for x = -8, you get [sigma] -1/n^2, which also converges.</p>

<p>[-8,2] is the final answer</p>

<p>Next question:</p>

<p>∫1/(x+1)(x+2)dx =</p>

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<p>W00T! Now I'm guaranteed to get a 0.5 on the exam!</p>

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<p>Answer to #8 - </p>

<p>Use partial fraction decomposition to separate the integral.
1/(x+1)(x+2) = A/(x+1) + B/(x+2)
1 = A(x+2) + B(x+1)
1 = Ax + 2A + Bx + B
1 = x(A + B) + 2A + B
A + B = 0 (There are 0 x's in 1.)
2A + B = 1</p>

<p>Subtracting 2A + B from A + B, we have (-A) = -1, therefore A = 1. Plugging back into the equation A + B = 0, we see that B = -1. </p>

<p>So now we separate the original integral:</p>

<p>∫ 1/(x+1)(x+2)dx = ∫ (1/(x+1))dx - ∫ (1/(x+2))dx </p>

<p>= ln|x+1| - ln|x+2| + C.
= ln|(x+1)/(x+2)| + C.</p>

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<p>For the answer to this question: "To what limit does the sequence S_n = (3+n)/(3^n) converge as n approaches infinity?"</p>

<p>You get 0 (3^n grows much more quickly than n, and we can verify this by L'H rule: [lim n>inf] 1 / (3^n *ln3) = 0)</p>

<p>Edit: Chaos is right on the dot for the part fractions question.</p>

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<p>That is correct. Verify my solution to #8. Who will post the next problem?</p>

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<p>ln[x+1]-ln[x+2]....................for the previous convergence question, what does it have to do so that a point in the interval is closed again? does it have to converge at that point or equal to 1, i'm confused........</p>

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<p>haha oops i was figuring it out while someone was doing it</p>

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<p>@mantiz24:</p>

<p>Well, when you get the 2 values for the end points for the interval of convergence, you need to test both values by plugging it back to the original series. For the example posted above: if you plug in -8 for x, you get ((x + 3)^n) /(n^2 *5^n) = ((-8 + 3)^n) /(n^2 *5^n) = ((-5)^n) /(n^2 *5^n) = -1/n^2 (which converges by the alternating series test). As for the other end point, you plug in again at x = 2, ((2 + 3)^n) /(n^2 *5^n) = ((5)^n) /(n^2 *5^n) = 1/n^2, which converges because it is a p series with P > 1. Hope this helps!</p>

<p>Next problem:</p>

<p>find infliction points of x^3 + x^2 + 1</p>

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<p>I can't even do arithmetic today. That's what happens when you don't get enough sleep. :(</p>

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<p>Yea same here, I keep screwing up on practice tests because I have been up late studying. We all ought to sleep at like 8:00PM at tuesday night</p>

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<p>I keep making the dumbest mistakes possible.
i.e. finding dy/dx instead of ds/dt on that one question, saying that x=3 if 6x+2=0, etc.
:(</p>

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<p>thread here:
<a href="http://talk.collegeconfidential.com/showthread.php?p=4091396#post4091396%5B/url%5D"&gt;http://talk.collegeconfidential.com/showthread.php?p=4091396#post4091396&lt;/a&gt;&lt;/p>

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<p>@PatrickK I checked that thread out. It has some pretty good resources, but not that many practice questions. Thanks though.</p>