pwrinkle

I'm having problems doing this word problem and one similar to it.

A street light is mounted at the top of a 15 foot tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

I don't think I am doing the correct relation between the rates. I'm forming a relation using similar triangles and then deriving it and solving for the rate of change of the shadow.

If m=distance between man and wall, and s = distance between man and tip of shadow i form this relation.

15/(m+s)=6/s
use cross multiplicatoin
15s=6m+6s
9s=6m
derive to get rates of change
9s'=6m'
already know m'=5ft/s
9s'=6(5)
s'=3.333
3.333ft/s is wrong.

Could anyone please help me?

thanks

molloby

Here's how I'd do it.

theta is the angle between the line from the top of the lamp post to the tip of the shadow.

x is the displacement of the man from the lamp post.

s is the displacement of the shadow from the lamp-post.

theta = tan^-1(9/x) (triangle (0,15),(x,6),(0,6))

tan(theta) = 15/s (triangle (0,15),(s,0),(0,0))

s = 15/tan(theta)
s(x) = 15x/9

dx/dt = 5

ds/dt = (15/9)(dx/dt)
=75/9
=8.33 ft/s

I think you got your similar triangles mixed up. But there you go.

a-j

Here it is without the trig:
Same variables:
x is the displacement of the man from the lamp post.
s is the displacement of the shadow from the lamp-post.

Using similar triangles:
6/(s-x) = 15/s
6s=15(s-x)
s=(15/9)x

ds/dx = 15/9

ds/dt = (ds/dx)(dx/dt)
ds/dt = (15/9)(5 ft/s)
ds/dt = 8.333 ft/s

molloby

Or you could do it that way...

Or use the relation 15/9=s/x from two triangles I used before.

I like the trig though, who needs patterns when you've got brute force? Also it means you don't have to do a similarity proof and I haven't done one of those in years.