Question

In Exercises 21–23 you are given a vector function $r$and a scalar function $t=f(\tau )$. Compute $\frac{d\mathbf{r}}{d\tau }$in the following two ways:

(a) By using the chain rule $\frac{d\mathbf{r}}{d\tau }=\frac{d\mathbf{r}}{dt}\frac{dt}{d\tau }$.

(b) By substituting $t=f(\tau )$into the formula for$r$. Ensure that your two answers are consistent.

Short answer

The two answers are consistent, that is.

Step-by-step solution

Part (a) Step 1. Given data

The given vector function is

We have to find $\frac{dr}{d\tau }$in two ways,

Part (a) Step 2. Find drdτusing chain rule

By using the chain rule,$\frac{dr}{d\tau }=\frac{dr}{dt}\cdot \frac{dt}{d\tau }$

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

Here, results obtained in part (a) and part (b) are equal.

Therefore, the two answers are consistent.

Part (b) Step 1. Find drdτ by substituting t=f(τ)

By substituting $t=\sin \tau$in $r\left(t\right)$

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