0
$\begingroup$

Consider the path of a particle in a conservative force field represented by the vector-valued function $$r(t)= \left(4(\sin t−t \cos t), 4( \sin t+t \sin t), \frac{3}{2} t^2 \right).$$

A) Find the arc length function $s$. - Answered here.

B) Use your answer from part A to write $r(t)$ as $r(s)$.

Please explain how to find part B. Thank you!

$\endgroup$
  • $\begingroup$ Did you do part A)? $\endgroup$ – JimmyK4542 Jul 5 '14 at 4:11
  • $\begingroup$ Yeah, it's the integral of a complicated square root with sines, cosines, and t. Do you need it specifically or can you explain the principle? $\endgroup$ – Erin Peterson Jul 5 '14 at 4:17
  • $\begingroup$ Is the second component $4(\sin t + t\sin t)$ or $4(\cos t + t\sin t)$? If it was supposed to be the latter, then you should have gotten something nice for the arclength function. $\endgroup$ – JimmyK4542 Jul 5 '14 at 4:26
  • $\begingroup$ No, the second component is written properly as originally posted. $\endgroup$ – Erin Peterson Jul 5 '14 at 4:42
0
$\begingroup$

I'm going to assume that the second component was supposed to be $4(\cos t + t\sin t)$.

For part a), $r'(t) = (4t\sin t, 4t\cos t, 3t)$, so $|r'(t)| = \sqrt{(4t\sin t)^2+(4t\cos t)^2+(3t)^2} = 5t$.

Therefore, $s = \displaystyle\int_{0}^{t}|r'(u)|\,du = \int_{0}^{t}5u\,du = \dfrac{5}{2}t^2$. Solve for $t$ to get $t = \sqrt{\dfrac{2}{5}s}$.

So to get $r(s)$, take the expression for $r(t)$ and plug in $t = \sqrt{\dfrac{2}{5}s}$.

EDIT: Since you are sure that the second component is indeed $4(\sin t + t\sin t)$, then $|r'(t)|$ becomes messy and so does the resulting integral. However, the principle is the same. You use the arclength formula to write $s = \ell(t)$ (some increasing function of $t$). Then, to find $r(s)$, take the expression for $r(t)$ and substitute $t = \ell^{-1}(s)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.