Chapter 4, Problem 68RE

To determine

Find the value of $dw/dsats=0.$

Answer to Problem 68RE

  $\frac{dw}{ds}=\frac{\cos (\sqrt{r}-2)\sqrt{64-r^2}}{2\sqrt{r}}$ .

Explanation of Solution

Given information: $i\text{f}w=\sin (\sqrt{r}-2)\text{and}r=8\sin (s+\frac{\pi }{6})$

Calculation:

  $\begin{array}{l}w=\sin (\sqrt{r}-2)\\ \frac{dw}{dr}=\frac{d(\sin (\sqrt{r}-2)}{dr}=\frac{\cos (\sqrt{r}-2)}{2\sqrt{r}}\\ r=8\sin (s+\frac{\pi }{6})\\ \frac{dr}{ds}=\frac{d(8\sin (s+\frac{\pi }{6}))}{ds}=8\cos (s+\frac{\pi }{6})\\ \frac{dr}{ds}=\sqrt{8^2{\cos }^2(s+\frac{\pi }{6})}\\ \frac{dr}{ds}=\sqrt{8^2(1-{\sin }^2(s+\frac{\pi }{6})})\\ \frac{dr}{ds}=\sqrt{8^2-8^2{\sin }^2(s+\frac{\pi }{6})}\\ \frac{dr}{ds}=\sqrt{8^2-r^2}\\ \frac{dw}{ds}=\frac{dw}{dr}\times \frac{dr}{ds}=\frac{\cos (\sqrt{r}-2)\sqrt{64-r^2}}{2\sqrt{r}}\\ \frac{dw}{ds}=\frac{\cos (\sqrt{r}-2)\sqrt{64-r^2}}{2\sqrt{r}}\end{array}$