How Do you do this Calculus AB Problem?

Test Preparation AP Test Preparation
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<p>A particle moves along the y-axis so that its velocity at any time t = 0 is given by v(t)=tcos(t). At time t = 0, the position of the particle is y =3.</p>

<p>a. for what values of t, 0<t<5, is the particle moving upward?</p>

<p>b. write an expression for the acceleration of the particle in terms of t.</p>

<p>c. write an expression for the position y(t) of the particle in terms of t.</p>

<p>d. for t>0, find the position of the particle the first time the velocity of the particle is zero.</p>

<p>i have answers, im just unsure how to get the solutions</p>

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<p>Something happened with the quick reply. Ignore that upper part. ^^^^</p>

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<p>This is an AB Problem? </p>

<p>a. Velocity is a vector. Therefore, when it is positive, it will move upward. tcos(t) = 0 at t = 0, pi/2, 3pi/2</p>

<p>Now, check the pockets. 0 < t < pi/2, pi/2 < t < 3pi/2, and 3pi/2 < t < 5</p>

<p>t will always be positive. Using the Unit Circle, the only non-negative pockets will be 0 < t < pi/2 and 3pi/2 < t < 5.</p>

<p>b. Derivative of velocity is accel.
Product rule
f ’ g + g ’ f
f = t, g = cos t
f ’ = 1 , g ’ = -sin t</p>

<p>cos t - t sin (t) </p>

<p>c. The integral of velocity is position but we’d have to use parts but… there is an intuitive way of doing this. We know that if we derived our answer for c, she would get tcos(t). </p>

<p>Now, observe what happens we we derive the answer in b).</p>

<p>We’d get -2sin(t) - tcos(t). That has the -tcos(t). So, if we can change our answer in a, we could get the derivative. First, we need to make it negative, so when we derive, we a positive tcos(t). </p>

<p>So we have -cos(t) + tsin(t). </p>

<p>Now, if we derived that, we’d get 2sin(t) +tcos(t).</p>

<p>So we need to add something that, when derived, cancels the 2sin(t). How about 2cos(t)? Thus. If we derive our new answer, -cos(t) + 2cos(t) + tsin(t) = cos(t) + tsin(t), we get -sin(t) + sin(t) + tcos(t) = tcos(t). Thus, our answer is **y = cos(t) + tsin(t) + C **. At t = 0, y = 3, thus, 3 = cos(0) + 0sin(0) + C –> 3 = 1 + C , 2 = C, thus the real answer is ** y= cos(t) + tsin(t) + 2 **</p>

<p>Parts = Easier xD. </p>

<p>d. Easy enough. Just plug in the first value we got from a into our new function, pi/2. </p>

<p>y = cos(pi/2) + (pi/2) sin (pi/2) + 3 = pi/2 + 3 or (pi +6)/2</p>

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<p>THANKS A ton. I greatly appreciate your help</p>

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<p>For part (c) on the AB test, I’d expect them to probably accept an answer similar to</p>

<p>y(t) = 3 + integral(xcosx evaluated from 0 to t) dx.</p>

<p>After all, that is an expression for y as a function of t. :)</p>

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<p>Yeah, part c has to be different on the AB exam, because integration by parts isn’t tested on the AB exam.</p>

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<p>MathProf and Anomaly:
If you two are right, d would be impossible to solve.
I think my method works out. No parts, just intuition.</p>

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<p>calc is gross</p>

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<p>^Calc is S e^xy, where S = Integral sign.</p>

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<p>Actually, it would still be solvable on a calculator-active section.</p>

<p>Since the first time that v(t) = 0 where t > 0 occurs at t = pi/2, so y(pi/2) = 3 + integral(xcosx from 0 to pi/2) dx, which can be done with the graphing calculator.</p>

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<p>If it’s on the calculator-active part, then you can just let your TI-89 integrate the tcos(t). </p>

<p>Though, I suppose if you had a TI-84 and the likes, maybe. The question is probably calculator-active section, since I doubt they would expect you to pull off what I did. xD</p>

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<p>It won’t let me edit…</p>

<p>Also, another problem with your idea…</p>

<p>integral(xcosx evaluated from 0 to t) dx.</p>

<p>Once evaluated</p>

<p>cos(t) + tsin(t) - ( cos(0) + (0)sin(0) )</p>

<p>cos(t) + tsin(t) - 1</p>

<p>Which would make your answer wrong.</p>

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<p>You forgot the “3+” at the beginning. :)</p>