Applications Of Derivatives Question 76
Question: 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
Options:
A) $ -\frac{4}{125}rad/s $
B) $ -\frac{2}{25}rad/s $
C) $ -\frac{1}{625}rad/s $
D) None of these
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Answer:
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 $ .
BETA
