A man 2 metres tall walks away from a lamp post 5 metres height at the rate of 4.8 km/hr. The rate of increase of the length of his shadow is

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KollegeApply · Accepted Answer

Let AB be the lamp post. Suppose at any time t, the man CD be at a distance of x km from the lamp post and y m be the length of his shadow CE. \text { Since triangles ABE and CDE are similar }, AB/CD = AE/CE \Rightarrow $\frac{5}{2}$ = x + y/y \Rightarrow x/y = $\frac{5}{2}$ - 1 \Rightarrow x/y = $\frac{3}{2}$ \Rightarrow y = $\frac{2}{3}$x \Rightarrow dy/dt = $\frac{2}{3}$\left( dx/dt \right) \Rightarrow dy/dt = $\frac{2}{3}$ $\times$ 4 . 8 \Rightarrow dy/dt = 3 . 2 \ km/hr [Solution diagram: 44319_solution_1.png]

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