question_answer

A man is moving away from a tower 41.6m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is

A) \[-\frac{4}{125}rad/s\]

B) \[-\frac{2}{25}rad/s\]

C) \[-\frac{1}{625}rad/s\]

D) None of these

Correct Answer: A

Solution :

[a] Let CD be the position of man at any time t.

Let BD be x. Then\[EC=x\]. Let \[\angle ACE\] be\[\theta \].

Given \[AB=41.6m,CD=1.6m,\]

and \[\frac{dx}{dt}=2\,m/s\].

\[AE=AB-EB=AB-CD=41.6-1.6=40\,m\]

We have to find \[\frac{d\theta }{dt}\] where\[x=30\,m\].

From \[\Delta AEC,\tan \theta =\frac{AE}{EC}=\frac{40}{x}\]

Differentiating w.r.t. to t, \[{{\sec }^{2}}\theta \frac{d\theta }{dt}=\frac{-40}{{{x}^{2}}}\frac{dx}{dt}\]

or \[{{\sec }^{2}}\theta \frac{d\theta }{dt}=\frac{-40}{{{x}^{2}}}\times 2\]

or \[\frac{d\theta }{dt}=\frac{-80}{{{x}^{2}}}{{\cos }^{2}}\theta =-\frac{80}{{{x}^{2}}}\frac{{{x}^{2}}}{{{x}^{2}}+{{40}^{2}}}\]

\[=-\frac{80}{{{x}^{2}}+{{40}^{2}}}\].

When\[x=30m,\frac{d\theta }{dt}=-\frac{80}{{{30}^{2}}+{{40}^{2}}}=-\frac{4}{125}rad/s\].