A ladder, 5 meters long; rests against a vertical wall. If its top slides downwards at the rate of 10 ~cm / s , then the angle between the ladder and the floor is decreasing at the rate of radians/second when it's lower end is 4 ~m away from the wall.

Step-by-step solution

According to the figure, x^2+y^2=25 Note that = OB AB = x 5 x=5 (i) 25 ^2 +y^2=25 Differentiating w.r.t. ' t ', we get aligned & -50 d dt +2 y d y dt =0 & 25 d dt =y d y dt & 25 d dt =y(-0.1) & [ d y ~d x =-10 ~cm / s =-0.1 ~m / s ] & 25 d dt =-(0.1) y & at x=4, = 4 5 , = 3 5 and y=3 & (ii) 25 3 5 4 5 d dt =-0.3 & d dt =-0.025 aligned i.e., the angle is decreasing at the rate of 0.025 rad / s