Question

3. Consider a particle moving along the path r(t) = (t + cost, t - sint). (a) (6 pts) Show that ||a(t)|| is constant, but a(t) is not constant. MATH 010A: a(t) = r''(t). (b) (4 pts) MATH 010A: skip Let C be the curve parameterized by r(t) for 0 ? t ? 2?. Set up an integral that measures the length of C. Your answer should be in the form $\int_0^{2\pi} f(t) dt$. You do not have to evaluate this integral.

          3. Consider a particle moving along the path r(t) = (t + cost, t - sint).
(a) (6 pts) Show that ||a(t)|| is constant, but a(t) is not constant. MATH 010A: a(t) = r''(t).
(b) (4 pts) MATH 010A: skip Let C be the curve parameterized by r(t) for 0 ? t ? 2?.
Set up an integral that measures the length of C. Your answer should be in the form
$\int_0^{2\pi} f(t) dt$. You do not have to evaluate this integral.

3. Consider a particle moving along the path r(t) = (t + cost, t - sint).
(a) (6 pts) Show that ||a(t)|| is constant, but a(t) is not constant. MATH 010A: a(t) = r”(t).
(b) (4 pts) MATH 010A: skip Let C be the curve parameterized by r(t) for 0 ? t ? 2?.
Set up an integral that measures the length of C. Your answer should be in the form
∫0^2π f(t) dt. You do not have to evaluate this integral.

Added by Melissa W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Joseph Liao

09/03/2020

Step 1

First, we find the velocity vector $v(t) = r'(t)$: $v(t) = (\frac{d}{dt}(t + \cos t), \frac{d}{dt}(t - \sin t)) = (1 - \sin t, 1 - \cos t)$. Next, we find the acceleration vector $a(t) = v'(t) = r''(t)$: $a(t) = (\frac{d}{dt}(1 - \sin t), \frac{d}{dt}(1 - \cos t)) Show more…

Show all steps

Thanks for your feedback!

3. Consider a particle moving along the path $r(t) = (t + \cos t, t - \sin t)$. (a) (6 pts) Show that $||a(t)||$ is constant, but $a(t)$ is not constant. MATH 010A: $a(t) = r''(t)$. (b) (4 pts) MATH 010A: skip Let C be the curve parameterized by r(t) for $0 \le t \le 2\pi$. Set up an integral that measures the length of C. Your answer should be in the form $\int_0^{2\pi} f(t) dt$. You do not have to evaluate this integral.