Question

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

$\text{38.}\mathbf{r}\left(t\right)=⟨\sin t,\cos t,2\sin 2t⟩,t=\sqrt{{\tau }^2+1}$

Short answer

$\frac{dr}{d\tau }=⟨\frac{\tau \cos \sqrt{{\tau }^2+1}}{\sqrt{{\tau }^2+1}},\frac{-\tau \sin \sqrt{{\tau }^2+1}}{\sqrt{{\tau }^2+1}},\frac{4\cos 2\sqrt{{\tau }^2+1}}{\sqrt{{\tau }^2+1}}⟩\mid$

Step-by-step solution

Step 1. Given data 

The given vector function is$\mathbf{r}\left(t\right)=⟨\sin t,\cos t,2\sin 2t⟩,t=\sqrt{{\tau }^2+1}$

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 $\frac{dr}{d\tau }=\frac{dr}{dt}\cdot \frac{dt}{d\tau }$
We have,$\begin{array}{rl}& r\left(t\right)=⟨\sin t,\cos t,2\sin 2t⟩\\ & \frac{dr}{dt}=r^{\mathrm{\prime }}\left(t\right)=⟨\cos t,-\sin t,4\cos 2t⟩\end{array}$

It is given$t=\sqrt{{\tau }^2+1}$

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

By the chain rule,

$\begin{array}{rl}& \frac{dr}{d\tau }=\frac{dr}{dt},\frac{dt}{d\tau }\\ & =⟨\cos t,-\sin t,4\cos 2t⟩\frac{\tau }{\sqrt{{\tau }^2+1}}\\ & =⟨\cos \sqrt{{\tau }^2+1},-\sin \sqrt{{\tau }^2+1},4\cos 2\sqrt{{\tau }^2+1}⟩\frac{\tau }{\sqrt{{\tau }^2+1}}\\ & =⟨\frac{\tau \cos \sqrt{{\tau }^2+1}}{\sqrt{{\tau }^2+1}},\frac{-\sin \sqrt{{\tau }^2+1}}{\sqrt{{\tau }^2+1}},\frac{4\cos 2\sqrt{{\tau }^2+1}}{\sqrt{{\tau }^2+1}}⟩\end{array}$

Thus$\frac{dr}{d\tau }=⟨\frac{\tau cos\sqrt{{\tau }^2+1}}{\sqrt{{\tau }^2+1}},\frac{-\tau sin\sqrt{{\tau }^2+1}}{\sqrt{{\tau }^2+1}},\frac{4cos2\sqrt{{\tau }^2+1}}{\sqrt{{\tau }^2+1}}⟩$