Question

A ladder20 ft long has one end on the ground and the other end in contact with a vertical wall. The lower end slips along the ground. If the lower end of the ladder is 16 ft away from the wall, upper end is moving $\lambda$ times as fast as the lower end, then $\lambda$ is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{4}{3}$
• D. $\frac{5}{3}$

Answer 1

The correct option is C

$\frac{4}{3}$

Let OC be the wall. Let AB be the position of the ladder at any time t such that OA =x and OB=y. Length of the ladder AB =20 ft.
In $\Delta AOB$,

$x^2+y^2=(20{)}^2$
$\Rightarrow 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\therefore \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=\frac{-x}{\sqrt{400}-x^2}.\frac{dx}{dt}=-\frac{16}{\sqrt{400}-(16{)}^2}.\frac{dx}{dt}=-\frac{4}{3}\frac{dx}{dt}$
-ve sign indicates, that when X increases with time, y decreases. Hence, the upper end is moving $\frac{4}{3}$ times as fast as the lower end.