A man of height 2m walks directly away from a lamp of height 5m, on a level road at 3 m/s. The rate at which the length of his shadow is increasing is
2m/s
Let be the lamp and PQ be the man and OQ=x metre be his shadow and let MQ =y metre.
∴dydt=speed of the man
=3 m/s (given)
∵ΔOPQ and ΔOLM are similar
∴OMOQ=LMPQ⇒x+yx=52⇒y=32x∴dydt=32dxdt⇒3=32dxdt⇒dxdt=2m/s