Question

A person 1.8m tall is walking away from a lamppost 6m high at the rate 1.3m/s. At what rate is the end of the person's shadow moving away from the lamppost?

Answer 1

`= 13/7 m/s approx 1.86 m/s`

Explanation:

the point of what follows and the choice of variables in the drawing is this. we know that `dot x = 1.3`. we have been asked to find the rate at which the end of the shadow is moving away from the lamppost - that's `dot y` !

so similar triangles tells us that

`6/y = 1.8/(y - x)`

`y = (6x)/4.2 = 10/7 x`

and now for the little bit of calculus, taking the deriv of each side with respect to time

`implies dot y = 10/7 dot x`
`= 10/7 * 1.3 = 13/7 m/s approx 1.86 m/s`