Question

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

$\text{37.}\mathbf{r}\left(t\right)=⟨\sec t,1/t,e^t\ln t⟩,t={\tau }^{-1}$

Short answer

$\frac{dr}{d\tau }=⟨\frac{-\sec \frac{1}{\tau }\tan \frac{1}{\tau }}{{\tau }^2},1,\frac{e^{\frac{1}{\tau }}}{{\tau }^2}(\ln \tau -\tau )⟩$

Step-by-step solution

Step 1. Given data 

The given vector function is$\mathbf{r}\left(t\right)=⟨\sec t,1/t,e^t\ln t⟩,t={\tau }^{-1}$

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,

$\begin{array}{rl}& r\left(t\right)=⟨\sec t,\frac{1}{t},e^t\ln t⟩\\ & \frac{dr}{dt}=⟨\sec t\tan t,\frac{-1}{t^2},\frac{e^t}{t}+e^t\ln t⟩\end{array}$

By the chain rule,

$\frac{dr}{d\tau }\begin{array}{rl}& =⟨\sec t\tan t,\frac{-1}{t^2},\frac{e^t}{t}+e^t\ln t⟩⟨\frac{-1}{{\tau }^2}⟩\\ & =⟨\sec \frac{1}{\tau }\tan \frac{1}{\tau },\frac{-1}{{\left(\frac{1}{\tau }\right)}^2},\frac{e^{\frac{1}{\tau }}}{\frac{1}{\tau }}+e^{\frac{1}{\tau }}\ln \left(\frac{1}{\tau }\right)⟩\left(\frac{-1}{{\tau }^2}\right)\\ & =⟨\frac{-\sec \frac{1}{\tau }\tan \frac{1}{\tau }}{{\tau }^2},{\tau }^2\cdot \frac{1}{{\tau }^2},\frac{-e^{\frac{1}{\tau }}}{{\tau }^2}(\tau +\ln e^{-1})⟩\\ & =⟨\frac{-\sec \frac{1}{\tau }\tan \frac{1}{\tau }}{{\tau }^2},1,\frac{-e^{\frac{1}{\tau }}}{{\tau }^2}(\tau -\ln \tau )⟩\\ & =⟨\frac{-\sec \frac{1}{\tau }\tan \frac{1}{\tau }}{{\tau }^2},1,\frac{e^{\frac{1}{\tau }}}{{\tau }^2}(\ln \tau -\tau )⟩\end{array}$

Thus $\frac{dr}{d\tau }=⟨\frac{-\sec \frac{1}{\tau }\tan \frac{1}{\tau }}{{\tau }^2},1,\frac{e^{\frac{1}{\tau }}}{{\tau }^2}(\ln \tau -\tau )⟩$