Answer
(a)$1 ft/s$
(b) $ 2$ ft/s
Work Step by Step
(a) Ref to fig Ex 32
Let $x_1$ be the length of the shadow.
Let $x_2$ be the distance of the man from the foot street light pole Let
$x=x_1+x_2$ .................. eq(1)
Reference to the right angled triangle in the figure
$\frac{6}{x_1}=\frac{18}{x}$ $\Longrightarrow$
$6x=18x_1$
$x=3x_1$ ........................... eq(2)
Differentiating with respect to t
$\frac{dx}{dt}=\frac{d (3x_1)}{dt} = 3\frac{d (x_1)}{dt} $
Putting $\frac{dx}{dt}=3 $ft/s in the above equation
$3 = 3\frac{d (x_1)}{dt} $
$\frac{d (x_1)}{dt} =1$ ft/s
(b) From equation (1) and equation (2)
$x_1+x_2=3x_1$Or $x_2=2x_1$ Or
$x_1=\frac{1}{2}x_2$
Putting in equation (2)
$x=3 (\frac{1}{2})x_2= \frac{3}{2}x_2$
Differentiating with respect to t
$\frac{dx}{dt}=\frac{d( \frac{3}{2}x_2)} {dt}= \frac{3}{2} \frac{d( x_2)} {dt}$
Putting $\frac{dx}{dt}=3$ ft/s in the above equation
$3= \frac{3}{2} \frac{d( x_2)} {dt}$
$ \frac{d( x_2)} {dt}=\frac{2}{3}\times 3=2$ ft/s
