Question

It is night. Someone who is 4 feet tall is walking away from a street light at a rate of 8 feet per second. The street light is 12 feet tall. The person casts a shadow on the ground in front of them. How fast is the length of the shadow growing when the person is 3 feet from the street light?

Answer 1

Answer: 4 ft/s

Explanation:

Given

height of man$=4 ft$

speed of person $v=8 ft/s$

height if street light$=12 ft$

Let x be the distance between person and street light and y be the length of his shadow

From diagram

as the two triangle ADE and ABC are similar therefore we can say that

$\frac{4}{12}=\frac{y}{x+y}$

$\frac{1}{3}=\frac{y}{x+y}$

$x+y=3y$

$x=2y$

differentiate above Equation w.r.t time we get

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

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{8}{2}=4 ft/s$

Answer 2

The rate at which the length of the shadow growing when the person is 3 feet from the street light is 4ft/s

Find the diagrammatic representation of the statement in question attached

Using the similar triangle theorem;

$\frac{4}{y}=\frac{12}{x+y}$

Cross multiply

$4(x+y)=12y\\4x + 4y= 12y\\4x=12y-4y\\4x=8y$

Divide both sides by 4:

$\frac{4x}{4} =\frac{8y}{4}\\x=2y$

Differentiate both sides of the equation with respect to t

$\frac{dx}{dt} =2\frac{dy}{dt}$

$\frac{dx}{dt}$ is the rate at which the person is walking away from the street light

$\frac{dy}{dt}$ is the rate at which the length of the shadow is growing

Given that $\frac{dx}{dt}$ = 8ft/s

$\frac{dy}{dt}=\frac{\frac{dx}{dt} }{2} \\ \frac{dy}{dt} = \frac{8}{2}\\\frac{dy}{dt}=4ft/s$

Hence the rate at which the length of the shadow growing when the person is 3 feet from the street light is 4ft/s

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