Question

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

$\text{39.}\mathbf{r}\left(t\right)=\cos 3t\mathbf{i}+\sin 4t\mathbf{j}+t\mathbf{k},t=5\tau -2$

Short answer

$\frac{dr}{d\tau }=⟨-15\sin (15\tau -6),20\cos (20\tau -8),5⟩$

Step-by-step solution

Step 1. Given data 

The given vector function is$\mathbf{r}\left(t\right)=\cos 3t\mathbf{i}+\sin 4t\mathbf{j}+t\mathbf{k},t=5\tau -2$

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,

$\begin{array}{r}r\left(t\right)=\cos 3ti+\sin 4tj+tk\\ r\left(t\right)=⟨\cos 3t,\sin 4t,t⟩\\ \frac{dr}{dt}=⟨-3\sin 3t,4\cos 3t,1⟩\end{array}$

It is given,$t=5\tau -2$

$\frac{dt}{d\tau }=5$

By the chain rule,

$$\begin{gathered} \frac{dr}{d\tau }=⟨-3\sin 3t,4\cos 4t,1⟩5 \\ =⟨-15\sin 3t,20\cos 4t,5⟩ \\ =⟨-15\sin 3(5\tau -2),20\cos 4(5\tau -2),5⟩ \\ =⟨-15\sin (15\tau -6),20\cos (20\tau -8),5⟩ \end{gathered}$$

Thus$\frac{dr}{d\tau }=⟨-15\sin (15\tau -6),20\cos (20\tau -8),5⟩$