Question

A man 1.7m tall walks at the rate of 2 meters per minute towards a street light that is 10m above the ground a.) At what rate is the tip of his shadow moving? b.) At what rate is the length of the shadow changing?

Answer 1

Let at an instant the position of man `NM` is `x` m from the light post when the tip of his shadow `T` is `y` m from the post and length of his shadow is `s` m.
Given that man is walking towards the light post at the rate `2m"/"s`, we have `(dx)/(dt)=2m"/min"`

Here height of the man `MN=1.7m` and

height of the light `LO=10`m.

By simple geometry `Delta TMNand Delta TLO` are similar

So by the property of similar triangles

`(TM)/(TS)=1.7/10`

`=>s/(s+x)=0.17`

`=>s=0.17(s+x)`

`=>83s=17x`

Differentiating w. r. to `t` we get

`83xx(ds)/(dt)=17xx(dx)/(dt)=17xx2m"/min"`

`(ds)/(dt)=34/83m"/min"`

(b) So the length of the shadow changing at the rate

`(ds)/(dt)=34/83m"/min"`

Now `y=s+x`

Differentiating this w. r. to `t` we get

`(dy)/(dt)=(ds)/(dt)+(dx)/(dt)`

`(dy)/(dt)=34/83m"/min"+2m"/min"=200/83m"/min"`

(a) Hence the length of the shadow changing at the rate

`(dy)/(dt)==200/83m"/min"`