calculus
posted by laquisha on .
A street light is at the top of a 17 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 6 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 40 ft from the base of the pole?

Draw a rightangled triangle ABC, B the right angle, and the vertical AB = 17 feet
Somewhere between B and C draw a vertical DE = 6 feet, the height of the woman
let BD = x and DC = y, the length of her shadow
given: dx/dt = 6 ft/s
find: dy/dt
(actually d(x+y)/dt, I'll come back to that later)
by similar triangles 6/y = 17/(x+y)
.
.
.
11y = 6x
then 11dy/dt = 6dx/dt
dy/dt = 6(6)/11 = 36/11 ft/s
so no matter where she is, the length of her shadow is increasing at 36/11 ft/s
but the woman herself is moving at 6 ft/sec
so her shadow is moving at 6+36/11 or
102/11 ft/s