Answer
a. $-2.5ft/sec$
b. $-\frac{3}{20} rad/sec$
Work Step by Step
Using the figure given in the book, label the height of the dock as $y$, the horizontal length from the dock to the boat as $x$ and the diagonal length from the boat to the top of the dock as $s$.
a. Using the figure, we are given that $s'=\frac{ds}{dt}=-2ft/sec, y=6ft, s=10ft$.
Using the relation $s^2=x^2+y^2$, we have $2ss'=2xx'+2yy'$ thus $2(10)(-2)=2xx'+2(6)(0)=2xx'$ and $x'=-\frac{20}{x}$. Since $x=\sqrt {10^2-6^2}=8$, we have $\frac{dx}{dt}=x'=-\frac{5}{2}=-2.5ft/sec$
b. At what rate is the angle $u$ changing at this instant (see the figure)?
Use $cos\theta=\frac{6}{s}$, we have $-sin\theta\frac{d\theta}{dt}=-\frac{6}{s^2}\frac{ds}{dt}$, which gives
$(\frac{8}{10})\frac{d\theta}{dt}=\frac{6}{10^2}(-2)$ and $\frac{d\theta}{dt}=-\frac{3}{20} rad/sec$
