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

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
`(2)/(9) m/s` or 0.22 m/s.

Step-by-step 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

`(12)/(20) = (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

`(20-x)/(20) = (2)/(y)` which gives
`1 - (1)/(20)*x = 2* (1)/(y)` finding the derivative of both sides gives

`-(1)/(20)dx =-2*(1)/(y^2)dy` hence we have by dividing by dt gives
`-(1)/(20)(dx)/(dt) =-2*(1)/(y^2)(dy)/(dt)`

However we know that
`(dx)/(dt) = 0.8 m/s`

Therefore
`-(0.8)/(20) = -0.18(dy)/(dt)`

The rate of decrease of her shadow
`(dy)/(dt)` is given by

`(dy)/(dt) = (0.8)/(3.6) =(2)/(9) m/s` or 0.222 m/s.