Arc Length with Vector Functions — Question 4

Question 4 – Solution

Strategy Integrate speed from the stated base point, solve for $t$, and substitute.

Arc-length coordinate From Question 1, $\|\vec r'(t)\|=5$. Hence $s(t)=\int_0^t5\,du=5t$, so $t=s/5$.

Reparameterization $$\boxed{\vec R(s)=\left\langle 3\cos(s/5),3\sin(s/5),4s/5\right\rangle},\qquad s\ge 0.$$

Verification $$\vec R'(s)=\left\langle -\frac 35\sin(s/5),\frac 35\cos(s/5),\frac 45\right\rangle, \quad \|\vec R'(s)\|=\sqrt{\frac 9{25}+\frac{16}{25}}=1.$$ Thus an increase of one unit in $s$ represents exactly one unit of distance along the helix.