Question

A light on the ground shines on a wall 20 meters away. If a man 2 meters tall walks from the light toward the wall at the rate 2.5 m/s, how fast is the length of his shadow decreasing when he is 10 meters away from the wall?

Answer 1

Answer:

The height of the shadow is decreasing at the rate of $\frac{dy}{dt}=-1\:{\frac{m}{s} }$.

Step-by-step explanation:

According with the graph (below) the small triangle (with the man as height), and the big triangle (with height along the wall) are similar, and hence the ratio of adjacent side length to opposite side length (with respect to the light) will be the same for both. This translates to

$\frac{x}{2}=\frac{20}{y}  \\\\xy=40$

Differentiating this equation with respect to t, use the product rule, we get

$xy=40\\\\\frac{d}{dt}(xy)=\frac{d}{dt}(40)\\\\\frac{dx}{dt}y+x\frac{dy}{dt}=0$

We want to find how fast is the length of his shadow decreasing when the man is 10 meters away from the wall, so we solve for $\frac{dy}{dt}$

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

At the moment when the man is 10 meters from the wall, we have

$x=20-10=10$,

$\frac{dx}{dt} = 2.5 \:{\frac{m}{s} }$,

using the relation of similar triangles we can find y, $10y=40\\y=4$

$\frac{dy}{dt}=-\frac{4}{10} (2.5)=-1\:{\frac{m}{s} }$

The height of the shadow is decreasing at the rate of $\frac{dy}{dt}=-1\:{\frac{m}{s} }$.