# Arc Length with Vector-Valued Functions, Part B

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!

• Did you do part A)? – JimmyK4542 Jul 5 '14 at 4:11
• 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? – Erin Peterson Jul 5 '14 at 4:17
• 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. – JimmyK4542 Jul 5 '14 at 4:26
• No, the second component is written properly as originally posted. – Erin Peterson Jul 5 '14 at 4:42

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 $t = \ell^{-1}(s)$.