Calculus
posted by Katie .
A street light is hung 18 ft. above street level. A 6foot tall man standing directly under the light walks away at a rate of 3 ft/sec. How fast is the tip of the man's shadow moving?
I know I would've to set up a proportion.
18 / 6 = x + y / y
x = distance of man from light
y = length of shadow
x + y = tip of shadow

You mean
18 / 6 = (x + y) / y
we know dx/dt, we need dy/dt
then the tip moves at dx/dt + dy/dt
18 y = 6x + 6 y
12 y = 6 x
12 dy/dt = 6 dx/dt
dy/dt = .5 dx/dt
so dy/dy = 3/2 = 1.5
and the sum
3+1.5 = 4.5 ft/sec 
Okay, I see what I did wrong. I did dy/dt = 6 instead of dy/dt = .5. Thanks a lot

x = distance of man from base of light
y = length of shadow
y/(y+x) = 6/18
Solve for y
18y = 6(y + x)
18y = 6y + 6x
12y = 6x
y = 6/12 x = 1/2 x
Find dy/dx of 1/2 x
dy/dx = 1/2
Find derivative with respect to t
dy/dt = 1/2 dx/dt
x is increasing 3 ft/sec
dx/dt = 3 ft/sec
dy/dt = 1/2 dx/dt
dy/dt = 1/2 (3)
dy/dt = 3/2 = 1.5 ft/sec
Shadow moving at the rate of 1.5 ft/sec 
A street light is at the top of a 13 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 7 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?