Question

A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground. When he is 10 feet from the base of the light, a) at what rate is the tip of his shadow moving?

Answer 1

`(25)/(3)ft/s`

Step-by-step explanation:

Height of man= 6ft

Height of light=1 5ft

Let BC=x and CD=y

BD=x+y

Triangle ABD and ECD are similar

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

`(AB)/(EC)=(BD)/(CD)`

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

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

`5y=2x+2y`

`5y-2y=2x`

`3y=2x`

Differentiate w.r.t t

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

We have
`(dx)/(dt)=5ft/s`

Substitute the value then we get

`2* 5=3(dy)/(dt)`

`(dy)/(dt)=(2* 5)/(3)=10/3ft/s`

Rate at which the tip of shadow is moving=
`(dx)/(dt)+(dy)/(dt)`

Rate at which the tip of shadow is moving=
`5+(10)/(3)=(15+10)/(3)ft/s`

Rate at which the tip of shadow is moving=
`(25)/(3)ft/s`