Answer
$\frac {5}{6} $ft/s
Work Step by Step
Let $x$ denote the distance of the top of the ladder from the ground.
$x= 5$ ft
$\frac{dx}{dt}=-2$ft/s
Let y denote the distance of the foot of the ladder from the wall.
$\frac{dy}{dt}=? $
Length of the ladder $=13 $ft
By Pythagoras theorem
$x^2+y^2=13^2=169 $ .............. eq (3)
Taking derivative of the above equation with respect to t
$\frac{d(x^2+y^2)} {dt}=\frac {d (169)} {dt}$
$\frac{d(x^2+y^2)} {dx} \frac{dx} {dt} + \frac{d(x^2+y^2)} {dy} \frac{dy} {dt} =0$
$2x \frac{dx} {dt} +2y \frac{dy} {dt} =0$ Or
$x \frac{dx} {dt} +y \frac{dy} {dt} =0$ .......... eq (4)
Putting $x=5, y=12, \frac{dx}{dt} =-2$ in equation (4)
$5 \times (-2) +12 \frac{dy} {dt} =0$
$\frac{dy}{dt}=\frac {5}{6} ft/s$
