Calculus
posted by Amber .
A flower pot is dropped from a point 12 feet above the ground,so that it will land 4 feet from the base of a 12 foot lamppost. At what rate is the shodow of the flower pot moving along the ground when the pot is 8 feet above the ground?

height y = 12  16t^2
shadow at x, similar triangles, so
x/12 = (x4)/y
or,
xy = 12x  48
y dx/dt + x dy/dt = 12 dx/dt
y=8 when t=1/2
x=6 when y=8
(12  16t^2) dx/dt + x (32t) = 12 dx/dt
at t=1/2,
8 dx/dt  96 = 12 dx/dt
dx/dt = 96/4 = 24 ft/s 
Oops.
x=12 when y=8
make the change and percolate it through the rest of the solution.