The 2010 Calculus BC Study Thread

Test Preparation AP Test Preparation
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<p>a. dv/dt = 1/t
dv = dt/t
v = ln(t) + C
-2 = ln(1) + C
C = -2
v(t) = ln(t) - 2</p>

<p>b. dx/dt = ln(t) - 2
dx = [ln(t) - 2]dt</p>

<p>We’ll want to integrate ln(t) by parts. udv = uv - ∫vdu
We’ll say that ln(t) = u and 1 = dv.</p>

<p>x = ln(t) * t - ∫(t/t)dt - 2t + C
x = tln(t) - t - 2t + C
x = tln(t) - 3t + C</p>

<p>4 = 1*ln(1) - 3(1) + C
4 = -3 + C
C = 7
x(t) = tln(t) - 3t + 7</p>

<p>c. When the particle is farthest to the left, the velocity is at a minimum. There is no relative minimum of the graph, but on the interval [1, ∞], the minimum exists at v(1). To find the position, we just plug into x(t). However, we’re already told that x(1) = 4, so good for us. The position of the particle is 4 units.</p>

<p>I’m an AB student who’s been reading ahead just for fun, so I hope that I did the integration by parts right (and for that matter, I hope I solved the problem correctly, hahaha).</p>

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<p>parts (a) and (b) are correct. Check part (c), v(t) is the derivative of x(t) so therefore there must be max/min for x(t). Hint: When is ln(t) - 2=0? The particle is farthest left when the velocity is negative(position goes left) and then becomes 0.</p>

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<p>Oopsies, I did part c while on the phone so I may have been a bit distracted. Well, t = e^2, so the position should be 7 - e^2, assuming I’m doing the math right in my head.</p>

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<p>correct keasbey nights, you’re pretty good for an ab student, are you going to take the calc bc exam?</p>

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<p>Thank you! My school won’t allow it, as the corresopnding exam for every course must be taken in order to receive the weighted GPA boost. I would have taken BC, but it didn’t fit into my schedule. I’m a junior anyway, so taking the AB exam this year and the BC exam next year (I’ll probably dual enroll in calc 2 and take the exam anyway) will give me an extra exam toward state AP scholar and an extra lifeline for college credit in case I don’t get a 5 on the AB exam.</p>

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<p>Thats too bad, you’re school should let you take the bc exam. Anyways at least you can take ap physics c next year, taking calc ab this year will prepare you with the math you need. </p>

<p>Here’s another free response question: (Calculator allowed)
Consider the region, R, bounded by the x-axis, y-axis, y=1/(x^2+1), and y=ln(x).</p>

<p>a) Find the area of the region R.</p>

<p>b)Set up but do not evaluate an integral expression using discs to find the volume of the object formed by rotating the region R around the y – axis.</p>

<p>c)Set up but do not evaluate an integral expression to find the volume of the object with the region R as its base and all square cross sections perpendicular to the x – axis.</p>

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<p>a. Because it’s bounded by the y-axis, we know that one of the limits will be x = 0. Set 1/(x^2 + 1) = ln(x) to find where they intersect. Use the calculator to figure out that x = about 1.40132162, which we will call “s”. So our bounds of integration overall are [0, s].</p>

<p>HOWEVER, ln(x) intersects the x-axis at x = 1. The shape of these graphs indicates that we’ll have to use two integrals, one with the bounds [0, 1], and one with the bounds [1, s].</p>

<p>∫[1 / (x^2 + 1)]dx [0,1] + ∫[1 / (x^2 + 1)] - ln(x) dx [1, s]</p>

<p>Run that **** through the calculator to find out that your area is about 0.879.</p>

<p>b. We’re rotating around x = 0, which will require the equations to be written in f(y) style because we’re forced to use discs. SO:</p>

<p>x^2 + 1 = 1/y
x^2 = (1/y) - 1
x = sqrt((1/y) - 1)</p>

<p>x = e^y</p>

<p>Also, we’re bound by x = 0 (the y-axis). One of our new limits of integration is y = 0, the x-axis. Anyway, set them equal to each other to find where they intersect, which is y = 0.3374518, which we’ll call “r”. Also, if you set sqrt((1/y) - 1) = 0, we find out that y = 1 is another bound. So, here’s our crazy integral.</p>

<p>pi<em>∫(e^y)^2 dy [0, r] + pi</em>∫(sqrt((1/y) - 1))^2 dy [r, 1]</p>

<p>I’m getting 2.845 as the integral.</p>

<p>c. ∫[1 / (x^2 + 1)]^2 dx [0,1] + ∫[[1 / (x^2 + 1)] - ln(x)]^2 dx [1, s]</p>

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<p>Very good, I think you’ll get a five on the calc ab test.</p>

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<p>Let A(x) be the area of the rectangle inscribed under the curve y=e^[-2(x^2)] with vertices at (-x,0) and (x,0), x >or= 0.
a)Find A(1).
b)What is the greatest value of A(x)? Justify your answer.
c) What is the aveage value of A(x) on the interval 0<or= x<or= 2?</p>

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<p>a. Alright, so we know that the bounds are going to be x = -1, x = 1, y = 0, and y = e^[-2(1^2)] = e^(-2). Our integral is going to be:</p>

<p>∫[e^(-2)] on the interval [-1, 1].</p>

<p>This turns out to be 2e^(-2), which is about 0.271.</p>

<p>b. We’re going to want to maximize the area function - i.e., finding where its rate of change equals zero. A rectangle is height times width.</p>

<p>Height: e^[-2(x^2)]
Width: 2x</p>

<p>A(x) = 2x * e^[-2(x^2)]
A’(x) = (2 - 8x^2) * e^(-2x^2) = 0</p>

<p>We find solutions - critical points - at -1/2 and 1/2. We can never make e^anything = 0, and solving the first term for x yields -1/2 and 1/2. Plugging in -1/2 back into A(x) gives us a negative answer; therefore, x = 1/2 gives us the greatest rectangle value, and that value is 0.607.</p>

<p>c. Average value integral: 1/(b-a) * ∫f(x)dx on the interval [a, b]
1/2 * ∫2x * e^[-2(x^2)] dx on the interval [0, 2]
Average value = 0.250.</p>

<p>Again, I hope I’m doing these BC problems right. Excellent question, good sir.</p>

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<p>I’m trying to self-study BC while in AB, but I’m not sure how the elegance and depth of my textbook compare to the AP test material. Princeton Review simplifies everything down to practical tips and methods. Is PR going to be close to the real thing?</p>

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<p>I recommend the College Board Website: [AP</a> Central - Exam Questions](<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board)
Go to Calculus BC and do the problems there, the PR book is good(I’m using it now), but I feel they differ in the Free Response section. The real FRQ’s are a bit harder (imho).</p>

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<p>A) A spherical snowball melts so that its surface area shrinks at the constant rate of 10 square cm per minute. What is the rate of change of volume when the snowball is 12 centimeters in diameter?
B) The snowball is packed most densely nearest the center. Suppose that, when it is 12 centimeters in diameter, its density x centimeters from the center is given by d(x) = 1/(1+x^.5) grams per cubic centimeter. What is the total number of grams (mass) of the snowball then?</p>

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<p>thanks steel</p>

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<p>@henryclay:
A) dS/dt=-10=8pi<em>r</em>dr/dt, when r=6cm(diameter=12), then dr/dt=-5/24pi.
dV/dt=4pi<em>r^2</em>dr/dt, substituting proper variables and we get dV/dt=-30 cm^3/min.</p>

<p>B)first integrate d(x) from 0 to 6 and we get D=2.4225. And we know that Density=mass/volume, so solving for mass, m=DV. V=4pi/3*r^3, let r=6 and V=288pi.
Finally subsitute and mass= 2193.5 grams.</p>

<p>Please correct me if I’m wrong, I did this in a hurry.</p>

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<p>Today my class took a real released AP Calc exam and I got a four, with absolutley no studying… woohoo! Hopefully that means with more preparation/review for the real thing, I should have a five.</p>

<p>My class is actually DONE with the course, and now all we’re doing is reviewing. It’s such a relief, due tothe fact that the rest of my classes are so far behind… :/</p>

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<p>Given g(3)=-, g’(3)=1, g’‘(3)=2, and given that f(x) = g[g(x)+(x)(g’(x)]m determine f’(3)</p>

<p>I hate these questions.</p>

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<p>Just out of curiosity, how far are you guys in the course? Right now, we’re doing series at my school</p>

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<p>@jgraider - if you’re doing series, you should be on the home stretch… that’s the last thing we did before we finished the course.</p>

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<p>We finished everything last Wednesday.</p>