A man of height 6 ft walks at a uniform speed of 9 ft/sec from a lamp fixed at 15 ft height. The length of his shadow is increasing at the rate of

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KollegeApply · Accepted Answer

6 ft/sec 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 ft be the length of his shadow CE. \text { Since the triangles ABE and CDE are similar }, AB/CD = AE/CE \Rightarrow $\frac{15}{6}$ = x + y/y \Rightarrow x/y = $\frac{15}{6}$ - 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 = $\frac{2}{3}$ $\times$ 9 \Rightarrow dy/dt = 6 \text { ft }/\sec [Solution diagram: 44323_solution_1.png]

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