Question

(1 point) A spotlight on the ground is shining on a wall 20m20m away. If a woman 2m2m tall walks from the spotlight toward the building at a speed of 0.8m/s,0.8m/s, how fast is the length of her shadow on the building decreasing when she is 8m8m from the building

Answer 1

Answer:

The answer to the question is;

When she is 8 m from the building fast the length of her shadow on the building is decreasing at $\frac{2}{9} m/s$ or 0.22 m/s.

Explanation:

We have

Distance of the spotlight from the building = 20 m

Distance  of woman from the building when her speed is measured = 8 m

Height of the woman = 2 m

Actual speed of the woman = 0.8 m/s

Comparing the distance of the woman from the spotlight and the wall from the spotlight, we have when the woman is 8 m from the building she is 12 m from the spotlight    

Therefore we have

$\frac{12}{20} = \frac{2}{y}$ where y is the shadow cast by the woman on the building = 10/3

When the woman is x distance from the building, she is 20 - x meters from the spotlight

Therefore the above equation can be written  as

$\frac{20-x}{20} = \frac{2}{y}$  which gives $1 - \frac{1}{20}*x = 2* \frac{1}{y}$ finding the derivative of both sides gives

$-\frac{1}{20}dx =-2*\frac{1}{y^2}dy$ hence we have by dividing by dt gives $-\frac{1}{20}\frac{dx}{dt} =-2*\frac{1}{y^2}\frac{dy}{dt}$

However we know that $\frac{dx}{dt} = 0.8 m/s$

Therefore $-\frac{0.8}{20} = -0.18\frac{dy}{dt}$

The rate of decrease of her shadow $\frac{dy}{dt}$ is given by

$\frac{dy}{dt} = \frac{0.8}{3.6} =\frac{2}{9} m/s$ or 0.222 m/s.