A man of height 2 metre walks at uniform speed of 3 metre per second away from the lamp post of height 5 metre. Then the rate at which the length of his shadow increases is m/sec
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Solution
Let AB be the height lamp post and MN be the height of the man.
Its shadow will be y m.
We know that, AB=5 m,MN=2 m
Assuming, AM=x,MC=y
Then dxdt=3 m/s
Since, △ABC and △MNC are similar. ∴MNAB=MCAC ⇒25=yx+y ⇒y=23x
Differentiating w.r.t. to t, we get dydt=23×dxdt⇒dydt=23×3=2 m/sec