Question

A 15-meter ladder is leaning against a wall. The distance y(t) between the top of the ladder and the ground is decreasing at a rate of 6 meters per minute. At a certain instant t_0 start subscript, 0, end subscript, the bottom of the ladder is a distance x(t_0) start subscript, 0, end subscript, of 12 meters from the wall. What is the rate of change of the angle \theta(t) between the ground and the ladder at that instant?

Answer 1

Answer:

Explanation:

Given  

length of ladder L=15 feet

velocity with which top is falling is 6 m/min

when bottom of ladder is at a distance of 12 ft away from wall then top of ladder from bottom is given by  

$y^2=15^2-12^2$

$y=9 m$

From diagram

$\sin \theta =\frac{y}{15}$

differentiating

$\cos \theat \times \frac{d\theta }{dt}=\frac{dy}{dt}\times \frac{1}{15}$

$\frac{d\theta }{dt}=\frac{6}{15}\times \frac{15}{12}$

$\frac{d\theta }{dt}=0.5 rad/min$