# A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a spe?

A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along a straight path. How fast is the tip of his shadow moving when he is 30 ft from the pole?

### 2 Answers

- az_lenderLv 78 years agoFavorite Answer
Let S denote the length of the shadow.

When he's F feet from the pole, you have

6/15 = x/(F+x), where x = S - F

2(F+x) = 15x

2F = 13x, so

S = F + (2F/13) = 28F/13

dF/dt = 4 ft/s

dS/dF = 28/13

dS/dt = (dS/dF) (dF/dt) = (112/13) ft/s

= about 8.6 ft/s

- cidyahLv 78 years ago
See the figure:

http://s1169.photobucket.com/user/chibuckt/media/I...

X = Xp+Xs

dXp/dt = 4 ft/s (given)

dX/dt = ?

Using similar triangles, 6/Xs = 15/X

6/Xs = 15/(Xp+Xs)

6(Xp+Xs) = 15Xs

15Xs = 6Xp+6Xs

9Xs = 6Xp

Xs = (6/9) Xp

Xs + Xp = (6/9) Xp + Xp

X = (15/9) Xp

dX/dt = (15/9) dXp/dt

dX/dt = (15/9)(4) = 6.667 ft/s

The tip of his shadow is moving 6.667 ft/s regardless of how far away he is from the base of the pole.