Question

A 5-ft-tall person walks away from the wall at a rate of 2 ft/sec. A spotlight is located on the ground 40 ft from the wall. How fast does the height of the person’s shadow on the wall change when the person is 10 ft from the wall?

Answer 1

Answer:

The rate of change of the height is - 4 ft/s

Solution:

As per the question:

Height of the person, y = 5 ft

The rate at which the person walks away, $\frac{dx}{dt} = 4\ ft/s$

Distance of the spotlight from the wall, x = 40 ft

Now,

To calculate the rate of change in the height, $\frac{dy}{dt}$ of the person when, x = 10 m:

From fig 1.

$\Delta ABC[\tex] ≈ [tex]\Delta PQC[\tex]Thus[tex]\frac{BC}{AB} = \frac{PQ}{QC}$

$\frac{y}{40} = \frac{5}{x}$

xy = 200                                                                       (1)

Differentiating the above eqn w.r.t time t:

$x\frac{dy}{dt} + y\frac{dx}{dt} = 0$

Thus

$\frac{dy}{dt} = - \frac{y}{x}\frac{dx}{dt}$              (2)

From eqn (1):

When x = 10 ft

10y = 200

y = 20 ft

Using eqn (2):

$\frac{dy}{dt} = - \frac{20}{10}\times 2 = - 4\ ft$