Question

In Exercises 35-39 a vector function $\mathbf{r}\left(t\right)$and scalar function localid="1649670265501" $t=f\left(\tau \right)$are given. Find localid="1649670261624" $\frac{d\mathbf{r}}{d\tau }$.

localid="1649670268998" $35.r\left(t\right)=⟨t,t^2,t^3⟩,t=t^3+1$

Short answer

$\frac{dr}{d\tau }=⟨3{\tau }^2,6{\tau }^2({\tau }^3+1),9{\tau }^2{({\tau }^3+1)}^2⟩$

Step-by-step solution

Step 1. Given data 

The given vector function is$r\left(t\right)=⟨t,t^2,t^3⟩,t=t^3+1$

We have to find$\frac{d\mathbf{r}}{d\tau }$

Step 2. Use chain rule

If $r\left(t\right)$is a vector function and $t=f\left(\tau \right)$, a scalar function. Then the chain rule states that $\frac{dr}{d\tau }=\frac{dr}{dt}\cdot \frac{dt}{d\tau }$

$\begin{array}{rl}r\left(t\right)& =⟨t,t^2,t^3⟩\\ \frac{dr}{dt}& =\frac{d}{dt}⟨t,t^2,t^3⟩\\ & =⟨\frac{d}{dt}\left(t\right),\frac{d}{dt}\left(t^2\right),\frac{d}{dt}\left(t^3\right)⟩\\ & =⟨1,2t,3t^2⟩\end{array}$

We have,$t=t^3+1$

$\begin{array}{r}\frac{dt}{d\tau }=\frac{d}{d\tau }\left({\tau }^3+1\right)\\ =3{\tau }^2\end{array}$

By the chain rule,

$\begin{array}{rl}\frac{dr}{d\tau }& =\frac{dr}{dt}\cdot \frac{dt}{d\tau }\\ & =⟨1,2t,3t^2⟩\cdot 3{\tau }^2\\ & =⟨1,2({\tau }^3+1),3{({\tau }^3+1)}^2⟩\cdot 3{\tau }^2\\ & =⟨3{\tau }^2,6{\tau }^2({\tau }^3+1),9{\tau }^2{({\tau }^3+1)}^2⟩\end{array}$