Question

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

$\text{36.}\mathbf{r}\left(t\right)=t{e}^t\mathbf{i}+t\ln t\mathbf{j},t=e^{\tau }$

Short answer

$\frac{dt}{d\tau }=⟨e^{\tau }{e}^{\epsilon }(e^t+1),e^r(1+\tau )⟩$

Step-by-step solution

Step 1. Given data 

The given vector function is$\mathbf{r}\left(t\right)=t{e}^t\mathbf{i}+t\ln t\mathbf{j},t=e^{\tau }$

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 $\frac{dr}{d\tau }=\frac{dr}{dt}\cdot \frac{dt}{d\tau }$

$\begin{array}{rl}r\left(t\right)& =t{e}^{t}i+t\ln tj\\ r\left(t\right)& =⟨t{e}^t,t\ln t⟩\\ \frac{dr}{dt}& =\frac{d}{dt}⟨t{e}^t,t\ln t⟩\\ & =⟨t{e}^t+e^t,\frac{t}{t}+\ln t⟩\\ & =⟨t{e}^t+e^t,1+\ln t⟩\end{array}$

We have,$t=e^{\tau }$

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

Thus$\frac{dt}{d\tau }=⟨e^{\tau }{e}^{e^{\tau }}(e^{\tau }+1),e^{\tau }(1+\tau )⟩$