Arc Length with Vector Functions — Question 4

Question 4 – Solution

Strategy. Integrate speed from the specified base parameter, invert the resulting monotone function, and substitute.

Step 1: Arc-length function $$\mathbf r'(t)=\left\langle 3,4,0\right\rangle,\qquad \|\mathbf r'(t)\|=5.$$ Therefore $$s(t)=\int_1^t5\,du=5(t-1).$$ Since $t\ge 1$, we have $s\ge 0$, and solving for $t$ gives $$\boxed{t(s)=1+\frac{s}{5}}.$$

Step 2: Substitute $$\begin{align*} \mathbf R(s)&=\mathbf r\left(1+\frac{s}{5}\right)\\ &=\boxed{\left\langle 3+\tfrac 35s,\ 4+\tfrac 45s,\ 0\right\rangle},\qquad s\ge 0. \end{align*}$$

Step 3: Unit-speed verification $$\mathbf R'(s)=\left\langle \tfrac 35,\tfrac 45,0\right\rangle,$$ so $$\|\mathbf R'(s)\|=\sqrt{\frac 9{25}+\frac{16}{25}}=1.$$ Thus an increase of one unit in $s$ corresponds to exactly one unit of distance along the curve.