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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?
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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