Question

A floodlight at ground level is illuminating a building that is 100m away. A man 1.8m tall is walking from the floodlight to the building along a path that is perpendicular to the building. If he is walking at the rate of 1m/s, at what rate is his shadow on the building decreasing when he is 60 m from the building Expert Answer

Answer 1

Answer:0.1125 m/s

Explanation:

Given

Height of Person is 1.8 m

walking rate towards building is 1 m/s

From Similar triangle concept

$\frac{40}{100}=\frac{1.8}{y}$

y=4.5 m

Now suppose at any distance x between man and building

$\frac{100-x}{100}=\frac{1.8}{y}$

$1-\frac{x}{100}=\frac{1.8}{y}$

Differentiate w.r.t to time

$-\frac{\mathrm{d} x}{\mathrm{d} t}=-\frac{1.8}{y^2}\frac{\mathrm{d} y}{\mathrm{d} t}$

and $\frac{\mathrm{d} x}{\mathrm{d} t}=1 m/s$

and at x=60 y=4.5 m

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{y^2\times 1}{1.8\times 100}$

$\frac{\mathrm{d} y}{\mathrm{d} t}=0.1125 m/s$