Posted by kait on .
a woman 6ft tall walks away from a light 10ft above the ground. if her shadow lengthens at the rate of 2ft/sec. how fast is she walking?

diffirential calculus 
Reiny,
did you make a diagram?
let her distance from the lightpole be x ft
let the length of her shadow be y ft
by similar triangles ...
6/y = 10/(x+y)
6x + 6y = 10y
6x = 4y
6 dx/dt = 4 dy/dt
we are given: dx/dt + dy/dt = 2 ft/sec
dx/dt = 2  dy/dt
so
6(2  dy/dt) = 4 dy/dt
12  6dy/dt = 4dy/dt
12 = 10dy/dt
dy/dt = 12/10 = 1.2 ft/sec