Calculus with Vector Functions — Question 9

Question 9

Let $$\mathbf r(t)=\left\langle t,t^2,t^3\right\rangle,\qquad t\ge 0,$$ and reparametrize it by $t=s^2+1$.

Tasks

1. Write the composite vector function $\mathbf R(s)=\mathbf r(s^2+1)$.

2. Compute $\mathbf R'(s)$ directly.

3. Compute it again using the vector chain rule and reconcile the fact that $\mathbf R'(0)=\mathbf 0$ while $\mathbf r'(1)\ne\mathbf 0$.

Question 9 – Solution

Strategy. Compare direct componentwise differentiation with $(\mathbf r\circ g)'=\mathbf r'(g)g'$.

Step 1: Composition $$\boxed{\mathbf R(s)=\left\langle s^2+1,(s^2+1)^2,(s^2+1)^3\right\rangle}.$$

Step 2: Direct differentiation $$\begin{align*} \mathbf R'(s) &=\left\langle 2s,\ 2(s^2+1)(2s),\ 3(s^2+1)^2(2s)\right\rangle\\ &=\boxed{\left\langle 2s,4s(s^2+1),6s(s^2+1)^2\right\rangle}. \end{align*}$$

Step 3: Chain-rule check Since $$\mathbf r'(t)=\left\langle 1,2t,3t^2\right\rangle,\qquad g(s)=s^2+1,$$ we obtain $$\begin{align*} \mathbf R'(s)&=\mathbf r'(g(s))g'(s)\\ &=\left\langle 1,2(s^2+1),3(s^2+1)^2\right\rangle(2s), \end{align*}$$ which is identical to the direct result.

At $s=0$, $g'(0)=0$, so $\mathbf R'(0)=\mathbf 0$ even though $$\mathbf r'(1)=\left\langle 1,2,3\right\rangle\ne\mathbf 0.$$ There is no contradiction: the new parameter momentarily stops changing the old parameter. The geometric curve is regular at $t=1$, but this particular parametrization is not regular at $s=0$.