Chapter 4.6, Problem 21E

(a)

To determine

To find : How fast the boat approaching the dock when $10ft$ rope are out.

Answer to Problem 21E

The answer: boat approaching at $2.5\text{ft}/\text{sec}$

Explanation of Solution

Given information:

The rope is hauled at the rate of $2ft/\sec$

And the graph is:

  

Calculation:

  $r=$ amount of rope out (hypotenuse of the triangle).

  $x=$ distance of dinghy from dock

Put the given values,

  $r=10$ at the moment.

  $\begin{array}{c}\frac{dr}{dt}=-2\\ x^2+36=r^2\\ \Rightarrow x=\sqrt{r^2-36}\\ \Rightarrow \frac{dx}{dt}=(1/2){\left(r^2-36\right)}^{-1/2}(2r)\frac{dr}{dt}\\ =\frac{r}{\sqrt{r^2-36}}\frac{dr}{dt}\\ \Rightarrow \frac{dx}{dt}=\frac{r}{\sqrt{r^2-36}}\frac{dr}{dt}\\ =\frac{10}{\sqrt{{(10)}^2-36}}(-2)\\ =-2.5\end{array}$

Therefore the required boat approaching at $2.5\text{ft}/\text{sec}$ the dock when $10ft$ rope are out.

(b)

To determine

To find: The rate of angle $\theta$ changing at the moment.

Answer to Problem 21E

The answer: decreasing by $0.15\text{ radians/sec}$

Explanation of Solution

Given information:

The rope is hauled at the rate of $2ft/\sec$

And the graph is:

  

Calculation:

  $r=$ amount of rope out (hypotenuse of the triangle).

  $x=$ distance of dinghy from dock

Put the given values,

  $r=10$ at the moment.

  $\begin{array}{c}\theta =\arccos \frac{6}{10}\\ \approx {53.13}^{\circ }\end{array}$

  $\begin{array}{c}\frac{dr}{dt}=-2\\ x^2+36=r^2\\ \Rightarrow x=\sqrt{r^2-36}\\ \Rightarrow \frac{6}{r}=\cos \theta \\ \Rightarrow -6r^{-2}\frac{dr}{dt}=-\sin \theta \frac{d\theta }{dt}\\ \Rightarrow \frac{d\theta }{dt}=\frac{6\csc \theta }{r^2}\frac{dr}{dt}\\ \Rightarrow \frac{d\theta }{dt}=\frac{6\csc \theta }{r^2}\frac{dr}{dt}\\ =\frac{6\csc {53.13}^{\circ }}{{(10)}^2}(-2)\\ =-0.15\end{array}$

Therefore the required rate of angle $\theta$ changing at the moment is decreasing by $0.15\text{ radians/sec}$