Question

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.4 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Answer 1

the length of his shadow on the building is decreasing at the rate of 0.525 m/s

Explanation:

From the diagram attached below;

the man is standing at point D with his head at point E

During that time, his shadow on the wall is y = BC

ΔABC and Δ ADE are similar in nature; thus their corresponding sides have equal ratios; i.e

`(AD)/(AB) = (DE)/(BC)`

`(8)/(12) = (2)/(y)`

8y = 24

y = 24/8

y = 3 meters

Let take an integral look at the distance of the man from the building as x, therefore the distance from the spotlight to the man is 12 - x

∴

`(12-x)/(12)=(2)/(y)`

`1- (1)/(12)x = 2* (1)/(y)`

To find the derivatives of both sides ;we have:

`- (1)/(12)dx = 2* (1)/(y^2)dy`

`- (1)/(12) (dx)/(dt) = 2* (1)/(y^2) (dy)/(dt)`

During that time ;

`(dx)/(dt )= 1.4 \ m/s` and y = 3

So; replacing the value into above ; we have:

`-(1)/(12)(1.4) = - (2)/(9) (dy)/(dt)`

`(dy)/(dt) = \frac{\frac{ 1.4} {12 } }{ (2)/(9)}`

`(dy)/(dt) = {\frac{ 1.4} {12 } }*{ (9)/(2)}`

`(dy)/(dt) =0.525 \ m/s`

Thus; the length of his shadow on the building is decreasing at the rate of 0.525 m/s