Related rates problem (calculus)

10. A street light is mounted at the top of a 5-meter-tall pole. A man 2 m talk walks away from the pole with the speed of 1.5 m/s along a straight path. How fast is the tip of his shadow moving when he is 10 m from the pole?

The answer is 3/(square root 5)

For diagram , click here:

http://www.flickr.com/photos/61664301@N0…

Let Xp be the distance from the person to the base of the street light. Let Xs be the length of his shadow. Let X be the distance between the base of the street light and the tip of his shadow.
X = Xp+Xs

dXp/dt = 1.5 m/s (given)

Using similar triangles,
Xs/ 2 = X / 5
Xs = (2/5) X
Xs = (2/5) ( Xp+Xs)
Xs (1- 2/5) = (2/5) Xp
(3/5) Xs = (2/5) Xp
Xs = (2/3) Xp

Xs+Xp = (2/3) Xp + Xp
X = (5/3) Xp
dX/dt = (5/3) dXp/dt
dX/dt = (5/3) (1.5) = (5/3)(3/2) = 5/2 = 2.5 m /s

let line BC is the pole & line PQ be the height of the man(Q&B are points on the ground)
If A is the tip of the shadow of man consider the similar triangles ABC & APQ

so u'll see PQ/BC =AQ/AB
PQ=2m
BC=5m
AQ=length of shadow=y
let BQ=x=distance moved by the man
AB=AQ+BQ=y+x

=>2/5=y/(x+y)
=>y=2x/3

rate of movement of tip A=rate of change of shadow length=dy/dt
=d(2x/3)/dt
=2/3(dx/dt) =2/3 (rate of movement of the man) :since x=distance walked by the man

=2/3 (1.5m/s)=1m/s