Question

The harmonic motion of a particle is given by f(t) = 2 cos(3t) + 3 sin(2t), 0 ≤ t ≤ 8. (a) When is the position function decreasing? (Round your answers to one decimal place. Enter your answer using interval notation.) Correct: Your answer is correct. (b) During how many time intervals is the particle's acceleration positive? 4 Correct: Your answer is correct. time intervals (c) At what time is the particle at the farthest distance away from its starting position in the negative direction? (Round your answer to one decimal place.) t = 5.34 Correct: Your answer is correct. How far away is it from its original position? (Round your answer to the nearest integer.) 7 Correct: Your answer is correct. (d) At what time is the particle moving the fastest? (Round your answer to one decimal place.) t = 4.7 Correct: Your answer is correct. At what speed is the particle moving the fastest? (Round your answer to the nearest integer.) -5 Incorrect: Your answer is incorrect.

Answer 1

For the last part, you have to find where $f'(t)$ attains its maximum over $0\le t\le8$. We have

$f'(t)=-6\sin3t+6\cos2t$

so that

$f''(t)=-18\cos3t-12\sin2t$

with critical points at $t$ such that

$-18\cos3t-12\sin2t=0$

$3\cos3t+2\sin2t=0$

$3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0$

$\cos t(3\cos^2t-9\sin^2t+4\sin t)=0$

$\cos t(12\sin^2t-4\sin t-3)=0$

So either

$\cos t=0\implies t=\dfrac{(2n+1)\pi}2$

or

$12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi$

where $n$ is any integer. We get 8 solutions over the given interval with $n=0,1,2$ from the first set of solutions, $n=0,1$ from the set of solutions where $\sin t=\dfrac{1+\sqrt{10}}6$, and $n=1$ from the set of solutions where $\sin t=\dfrac{1-\sqrt{10}}6$. They are approximately

$\dfrac\pi2\approx2$

$\dfrac{3\pi}2\approx5$

$\dfrac{5\pi}2\approx8$

$\sin^{-1}\dfrac{1+\sqrt{10}}6\approx1$

$2\pi+\sin^{-1}\dfrac{1+\sqrt{10}}6\approx7$

$2\pi+\sin^{-1}\dfrac{1-\sqrt{10}}6\approx6$