A man of height 2 m walks at a uniform speed of 6 km/hr away from a lamp post of 6 m high. find the rate at which the length of the shadow is increasing.
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Let the length of shadow be DE = ym
and let the distance between the man and the lamp post be x meter.
Let AB be the height of the lamp post and CD be the height of man.
∴ AB = 6m and CD = 2m
Therefore, by similar triangles theorem,
AB = CD
BE DE
6 = 2
x + y y
6y = 2x + 2y
⇒x=2y⟶(1)
Differentiating equation (1) w.r.t. time, t, we have
dx = 2dy
dt dt
∵ dx = 6km/hr
dt
∴ 2dy = 6
dt
dy = 3km/hr
dt
The rate at which the length of his shadow increases is 3 km/hr
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