Question

A man is moving away from a tower $41.6m$ high at a rate of $2m/s$. If the eye level of the man is $1.6m$ 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 $30m$ 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

Answer 1

Answer: (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 $\frac{dx}{dt}=2m/s$
$AE=AB-EB=AB-CD$
$=41.6-1.6=40m$
We have to find $\frac{d\theta }{dt}$ when $x=30m$.
From $△AEC,\tan \theta =\frac{AE}{EC}=\frac{40}{x}$
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 $x=30m$,
$\frac{d\theta }{dt}=-\frac{80}{30^2+40^2}$
$=-\frac{4}{125}rad/s$