Q.32. 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 ______ rad/s when its lower end is 4 m away from the wall.
A. -0.1
B. -0.025
C. 0.1
D. 0.025
Answer: D. 0.025
Explanation :-
According to the figure, x2+y2 = 25 —-(i)
Note that According to the figure, x^2 + y^2 = 25 …(i)
Note that \cos\theta = \frac{\textcircled{O}}{\textcircled{B}} = \frac{x}{5}
\therefore x=5\cos\theta \therefore (i) \Rightarrow 25\cos^2\theta + y^2 = 25Differentiating w.r.t. ‘t’, we get
-50\cos\theta\sin\theta\frac{d\theta}{dt} + 2y\frac{dy}{dt} = 0 25\sin\theta\cos\theta\frac{d\theta}{dt} = y\frac{dy}{dt} 25\sin\theta\cos\theta\frac{d\theta}{dt} = y(-0.1) \frac{dy}{dx} = -10 \text{ cm/s} = -0.1 \text{ m/s}25\sin\theta\cos\theta\frac{d\theta}{dt} = -(0.1)y …(ii)
At x=4, \cos\theta=\frac{4}{5}, \sin\theta=\frac{3}{5}, and y=3
(ii) \Rightarrow 25 \times \frac{3}{5} \times \frac{4}{5} \times \frac{d\theta}{dt} = -0.3\Rightarrow \frac{d\theta}{dt} = -0.025, i.e., the angle is decreasing at the rate of 0.025 rad/s.