Posted by **Amber** on Tuesday, April 24, 2012 at 12:05am.

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?

- Calculus -
**Steve**, Tuesday, April 24, 2012 at 12:24am
height y = 12 - 16t^2

shadow at x, similar triangles, so

x/12 = (x-4)/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

- Calculus - PS -
**Steve**, Tuesday, April 24, 2012 at 4:50am
Oops.

x=12 when y=8

make the change and percolate it through the rest of the solution.

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