Arc Length with Vector Functions — Question 3

Question 3 – Solution

Strategy. Differentiate first and group symmetric components; the radicals collapse to a polynomial.

Step 1: Differentiate $$\mathbf r'(t)=\left\langle 1-t^2,2t,1+t^2\right\rangle.$$ Thus $$\begin{align*} \|\mathbf r'(t)\|^2 &=(1-t^2)^2+(2t)^2+(1+t^2)^2\\ &=(1-2t^2+t^4)+4t^2+(1+2t^2+t^4)\\ &=2+4t^2+2t^4\\ &=2(1+t^2)^2. \end{align*}$$

Step 2: Take the square root $$\|\mathbf r'(t)\|=\sqrt 2\,|1+t^2|.$$ Since $1+t^2>0$ for every real $t$, this becomes $$\|\mathbf r'(t)\|=\sqrt 2(1+t^2).$$

Step 3: Integrate $$\begin{align*} L&=\sqrt 2\int_0^2(1+t^2)\,dt\\ &=\sqrt 2\left[t+\frac{t^3}{3}\right]_0^2\\ &=\boxed{\frac{14\sqrt 2}{3}}. \end{align*}$$ The speed is strictly positive because $1+t^2>0$, so the curve is regular throughout the interval.