Question

A spotlight on the ground is shining on a wall 24m away. If a woman 2m tall walks from the spotlight toward the building at a speed of 0.6m/s, how fast is the length of her shadow on the building decreasing when she is 2m from the building?

Answer 1

Answer:

$\dfrac{dy}{dt}=-0.059\ m/s$

Explanation:

It is given that,

Distance between the spotlight and the wall, y = 24 m

Height of the woman, h = 2 m

The woman walks toward the building at the rate of 0.6 m/s, $\dfrac{dx}{dt}=0.6\ m/s$

In the attached figure, triangle ABC and MNC are similar. So,

$\dfrac{2}{y}=\dfrac{x}{24}$............(1)

$y=\dfrac{48}{x}$

When she is 2 meters from the building. So x = 24-2 = 22 m

$y=\dfrac{48}{22}=2.18\ m$

Differentiating equation (1) i.e.

$xy=48$

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

$22.\dfrac{dy}{dt}+2.18\times 0.6=0$

$\dfrac{dy}{dt}=-0.059\ m/s$

So, her shadow is decreasing at the rate of 0.059 m/s. Hence, this is the required solution.