Posted by **Katie** on Sunday, February 6, 2011 at 6:12pm.

A street light is hung 18 ft. above street level. A 6-foot 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

- Calculus -
**Damon**, Sunday, February 6, 2011 at 6:24pm
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

- Calculus -
**Katie**, Sunday, February 6, 2011 at 6:40pm
Okay, I see what I did wrong. I did dy/dt = 6 instead of dy/dt = .5. Thanks a lot

- Calculus -
**helper**, Sunday, February 6, 2011 at 6:41pm
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

- Calculus -
**crystal S**, Saturday, November 26, 2011 at 3:03am
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?

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