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Question
A man 1.6 m tall walks at the rate of 0.3 m/sec away from a street light that is 4 m above the ground. At what rate is the tip of his shadow moving? At what rate is his shadow lengthening?
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Solution
Let AB represent the height of the street light from the ground. At any time t seconds, let the man represented as ED of height 1.6 m be at a distance of x m from AB and the length of his shadow EC be y m.
Using the similarity of triangles, we have `4/1.6`
= `(x + y)/y`
⇒ 3y = 2x
Differentiating both sides w.r. to t, we get `3 (dy)/(dt) = 2 (dx)/(dt)`
`(dy)/(dt) = 2/3 xx 0.3`
⇒ `(dy)/(dt)` = 0.2
At any time t seconds, the tip of his shadow is at a distance of (x + y) m from AB.
The rate at which the tip of his shadow moving = `((dx)/(dt) + (dy)/(dt))` m/s = 0.5 m/s
The rate at which his shadow is lengthening = `(dy)/(dt)` m/s = 0.2 m/s
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