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Question
A man, 2m tall, walks at the rate of `1 2/3` m/s towards a street light which is `5 1/3`m above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is `3 1/3`m from the base of the light?
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Solution
Let AB is the height of street light post and CD is the height of the man such that
AB = `5 1/3 = 16/3 "m"` and CD = 2 m
Let BC = x length (the distance of the man from the lamp post) and CE = y is the length of the shadow of the man at any instant.
From the figure, we see that
ΔABE ~ Δ DCE ......[By AAA Similarity]
∴ Taking ratio of their corresponding sides, we get
`"AB"/"CD" = "BE"/"CE"`
⇒ `"AB"/"CD" = ("BC" + "CE")/"CE"`
⇒ `(16/3)/2 = (x + y)/y`
⇒ `8/3 = (x + y)/y`
⇒ 8y = 3x + 3y
⇒ 8y – 3y = 3x
⇒ 5y = 3x
Differentiating both sides w.r.t, t, we get
`"dy"/"dt" = 3 * "dx"/dt"`
⇒ `"dy"/"dt" = 3/5 * "dx"/"dt"`
⇒ `"dy"/"dt" = 3/5 * ((-5)/3)` ......[∵ man is moving in opposite direction]
= – 1 m/s
Hence, the length of shadow is decreasing at the rate of 1 m/s.
Now let u = x + y .....(u = Distance of the tip of shadow from the light post)
Differentiating both sides w.r.t. t, we get
`"du"/"dt" = "dx"/"dt" + "dy"/dt"`
= `(- 1 2/3 - 1)`
= `-(5/3 + 1)`
= `- 8/3`
= `-2 2/3` m/s
Hence, the tip of the shadow is moving at the rate of `2 2/3` m/s towards the light post and the length of shadow decreasing at the rate of 1 m/s.
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