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Worked Examples · Example 14

Q.A boy of height 1 m is walking towards a lamp post of height 5 meters at the rate of 0.5 m/sec. Then find the rate at which the length of the shadow of the boy is decreasing.

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Set up similar triangles for boy and lamp post to relate shadow length ss to the boy's distance xx from the post, then differentiate w.r.t. time.

Similar triangles (post height H=5H=5, boy height h=1h=1): Hx+s=hs\dfrac{H}{x+s}=\dfrac{h}{s}, where xx = boy's distance from the post and ss = shadow length. Rate: dsdt\dfrac{ds}{dt}.

  1. Let xx be the distance of the boy from the lamp post and ss the length of his shadow. The tip of the shadow, the boy's head, and the lamp top give similar triangles:

5x+s=1s.\dfrac{5}{x+s}=\dfrac{1}{s}.

  1. Cross-multiply: 5s=x+s ⇒ 4s=x ⇒ s=x4.5s=x+s\ \Rightarrow\ 4s=x\ \Rightarrow\ s=\dfrac{x}{4}.
  2. Differentiate with respect to time tt: dsdt=14dxdt.\dfrac{ds}{dt}=\dfrac{1}{4}\dfrac{dx}{dt}. …

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