Question

A street light is mounted at the top of a 15-ft-tall pole. A man 6 feet tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast (in ft/s) is the tip of his shadow moving when he is 45 feet from the pole

Answer 1

The tip pf the shadow is moving with speed 25/3 ft/s.

Explanation:

height of pole = 15 ft

height of man = 6 ft

x = 45 ft

According to the diagram, dx/dt = 5 ft/s.

Now

`(y-x)/(y)=(6)/(15)\\\\15 y - 15 x = 6 y \\\\y = (5)/(3) x\\\\(dy)/(dt) = (5)/(3)(dx)/(dt)\\\\(dy)/(dt)=(5)/(3)* 5 =(25)/(3) ft/s`

Answer 2

25/3 ft/s

Explanation:

Height of pole , h=15 ft

Height of man, h'=6 ft

Let BD=x

BE=y

DE=BE-BD=y-x

All right triangles are similar

When two triangles are similar then the ratio of their corresponding sides are equal.

Therefore,

`(AB)/(CD)=(BE)/(DE)`

`(15)/(6)=(y)/(y-x)`

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

`5y-5x=2y`

`5y-2y=5x`

`3y=5x`

`y=(5)/(3)x`

Differentiate w.r.t t

`(dy)/(dt)=(5)/(3)(dx)/(dt)`

We have dx/dt=5ft/s

Using the value

`(dy)/(dt)=(5)/(3)(5)=(25)/(3)ft/s`

Hence, the tip of his shadow moving with a speed 25/3 ft/s when he is 45 feet from the pole.