Question

A ladder 15ft long rest against a vertical wall. Its top slides down the wall while its bottom moves away along with the level ground at a speed of 2 ft/s . How fast is the angle between the top of the ladder and the wall changing when the angle is π/3 radians.

Answer 1

Ladder and wall can be regarded as sides of a right triangle

$At t = 0, OB=0 and OC =BC =15 ft. OB =2t ft$

$sin \theta=\frac{OB}{BC} \implies \theta = sin^{-1}\frac{OB}{BC}$

$\theta = sin^{-1}\frac{2t}{15} \implies d\theta=\frac{2}{\sqrt{225-4t^2}}dt$

$\frac{d\theta}{dt}=\frac{2}{\sqrt{225-4t^2}}$

Determining t, when $\theta=\frac{\pi}{3}$

$sin\frac{\pi}{3}=\frac{2t_1}{15} \implies t_1= \frac{15\sqrt{3}}{4}$

$\frac{d\theta}{dt}_{t=t_1}=\frac{2}{\sqrt{225-4t^2}}=\frac{2}{\sqrt{225-4(\frac{15 \sqrt{3}}{4})^2}}$

$=\frac{2}{\sqrt{225-225(\frac{{3}}{4})}}=\frac{4}{15}$ rad/s