2 sample AP calc AB questions (need help)

Test Preparation
1

<p>i need help on these two problems. It might be too much of a hassle to show your work and type it out so just give me the general idea of what to do in each part of the problem and mention the answer also. Thanks SO MUCH.</p>

<li>Consider a CURVE defined by the equation y + cosy = x + 1 for [0,360]</li>
</ol>

<p>(a) find dy/dx in terms of y
(b) write an equation for each vertical tangent to the CURVE
(c) find d2y/dx2 in terms of y</p>

<li>at time t, t >= 0, the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius. At t = o, the radius of the sphere is 1 and at t = 15, the radius is 2. (The volume V of a sphere with radius r is V = 4/3(pi)r^3</li>
</ol>

<p>(a) find the radius of the sphere as a function of t
(b) at what time t will the volume of the sphere be 27 times its voume at t = 0</p>

2

<p>1.
a) Simply differentiate both sides with respect to x. Nothing tricky.
b) Not sure what you mean by vertical tangent. If you mean "tangent", then use the point-slope form of a linear equation, x - x0 = (dy/dx)(y-y0).
c) When you differentiate your eqn from (a) again, you'll be left with some dy/dx's. Plug in your dy/dx from (a) to get it into terms of y.</p>

<p>2)
Edit: This question doesn't make a ton of sense.</p>

<p>Your teacher/book messed up or you mistyped something. It should read at time t>0, not t>=0, since it is impossible to be inversely proportional and have one of the terms ever equal 0. Just look at the curve y = 1/x; there's a reason x=0 and y=0 are asymptotes.</p>

<p>Also, it reads r and t are inversely proportional. That means as t gets large, r gets small. But you told me that sphere was expanding as time got large.</p>

3

<ol>
<li>Consider a CURVE defined by the equation y + cosy = x + 1 for [0,360]</li>
</ol>

<p>(a) find dy/dx in terms of y
(b) write an equation for each vertical tangent to the CURVE
(c) find d2y/dx2 in terms of y</p>

<p>SOLUTION:</p>

<p>a) Use implicit differentiation to find dy/dx in terms of y:</p>

<p>y(dy/dx) + cosy(dy/dx) = x (dy/dx) + 1(dy/dx) <-- take the derivative of everything</p>

<p>dy/dx + [-sin (y)*(dy/dx)] = 1 + 0</p>

<p>(dy/dx)[1 - sin(y)] = 1 <-- factor out dy/dx</p>

<p>dy/dx = 1/[1 - sin(y)]</p>

<p>b) There will be a vertical tangent when the slope is undefined. In order words, when dy/dx = DNE, which will occur when the denominator is equal to zero:</p>

<p>1 - sin(y) = 0
sin (y) = 1</p>

<p>And, on the unit circle in terms of degrees and with respect to your domain, that occurs at 90 and 270.</p>

<p>c) Now take the derivative of dy/dx, by using implicit differentiation again...</p>

<p>dy/dx = 1/[1 - sin(y)] --> d2y/dx2 = [(1-sin(y) )(0) - (1)(-cos(y)*(dy/dx))]/[(1 - sin(y) )^2]</p>

<p>d2y/dx2 = [cos (y) * (dy/dx)]/[(1 - sin (y))^2]</p>

<p>Now substitute in dy/dx from (a):</p>

<p>d2y/dx2 = [cos (y) * 1/(1 - sin(y) )]/[1 - sin(y) )^2]</p>

<p>d2y/dx2 = cos (y)/[(1 - sin(y) )^3]</p>

<ol>
<li>at time t, t >= 0, the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius. At t = o, the radius of the sphere is 1 and at t = 15, the radius is 2. (The volume V of a sphere with radius r is V = 4/3(pi)r^3</li>
</ol>

<p>(a) find the radius of the sphere as a function of t
(b) at what time t will the volume of the sphere be 27 times its voume at t = 0</p>

<p>I'm not so sure about how to do this one, because it's asking to find the radius as a function of t, and I'm used to having a Related Rates question as for the "rate at with r is changing at t" or something like that. I'll think about it a little more and if I come up with a solution I'll post it -- I'm sure someone else here knows, however.</p>

4

<p>


</p>

<p>Problem no 2 is partially a related rate question (because, well, it deals with ... related rates), but to find the r(t) function, you need integration. Since the rate at which v changes with respect to time is proportional to to the reciprocal of the sphere's radius, we can write:</p>

<p>dV/dt = k / r</p>

<p>We also know that V = 4/3(pi)r^3, which means, dV / dt = 4(pi)r^2 dr / dt. Set them equal to each other, we get:</p>

<p>4(pi)r^2 dr / dt = k / r
4(pi)r^3 dr = k dt</p>

<p>Take the integral of both side.</p>

<p>(pi)r^4 = kt + C</p>

<p>r^4 = kt/(pi) + A (A is C / (pi). Since it's a constant as well, it's all right to write it that way). </p>

<p>Plug in t = 0, we get r = 1, then, A = 1. Plug in t = 15, we get k = pi. Then, the equation r(t) is (t+1)^(1/4).</p>

<p>To check if this is correct, just express V in the term of t, which is, <a href="t+1">4(pi)/3</a>^(3/4), then take the derivate of it. Its derivative is (pi) / [(t+1)^(1/4)], which is proportional to the reciprocal of the radius. SO we got that requirement covered. Plug in t = 0, and t = 15 to the r(t) equation, and you'd find that the sphere's radius is 1 & 2 respectively).</p>

<p>


</p>

<p>dV / dt is inversely proportional to the radius, NOT to the time.</p>

5

<p>Oh. Well, that makes a ton more sense :D.</p>

6

<p>LOL, yeah. You gotta read carefully when doing problems. Concept-wise, I had the capability to ace through ALL of the math tests, but always made stupid mistakes in reading the questions. To make matters worse, partial credits are not given to most of these problems :(</p>

7

<p>Ah, okay. Thanks, asbereth. I don't know integrals, yet, so I guess it makes sense that I wouldn't know how to do that problem. :p</p>

8

<p>Aaahhhh. Yeah. I overlooked the "AB" part of the thread's title. IIRC, when I was in AB Calculus, we started doing anti-derivative in around mid-December. Since you're math inclined, I bet you will love integrals. Especially when it comes to practical applications.</p>

9

<p>ah dang. don't know how to do integral just yet. thanks a bunch for your responses, you guys!</p>

10

<p><strong><em>I'VE GOT TWO MORE SAMLE AP PROBLEMS HERE</em></strong>******
i've included some work that i've done. please help me with finishing the rest. i'm just stuck. </p>

<ol>
<li>A particle starts at time t = 0 and moves on a number line so that its position at time t is given by x(t) = (t-2)^3 (t-6). (i assume x is a function of t)</li>
</ol>

<p>a. When is the particle moving to the right?
i got (-infinite,2) U (5,infinite) </p>

<p>b. when is the particle at rest?
i got t = 2,5</p>

<p>c. when does the particle change direction?
same... t= 2,5?</p>

<p>d. What is the farthest to the left of the origin that the particle moves?
no idea</p>

<ol>
<li>A particle moves along the x-axis so that its velocity at time t, [0,5] <-the interval,
is given by v(t) - 3(t-1)(t-3). At time t =2, the position of the particle is x(2) = 0.</li>
</ol>

<p>a. find the minimum acceleration of the particle.
(what i have) i took the equation's derivative (6t - 12) at v'(o) which equals = -12</p>

<p>b. find the total distance traveled by the particle
do we need to know integrals or riemann sums for this(we didn't go over that topic in class)</p>

<p>c. find average velocity of the particle over interval [0,5]
no idea how to do this.</p>

11

<p>bump........................</p>