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April 2, 2015

April 2, 2015

Posted by **Stacy** on Wednesday, February 22, 2012 at 4:02pm.

- Calculus -
**Reiny**, Wednesday, February 22, 2012 at 5:10pmLet the ladder be x ft from the wall, and y ft up along the wall

x^2 + y^2 = 20^2

2x dx/dt + 2y dy/dt = 0

x dx/dt + y dy/dt = 0

when x=16 , y^2 = 400-256 ---> y = 12

and dy/dt = -2 ft/sec

16(dx/dt) + 12(-2) = 0

dx/dt = -24/16 = -3/2

the ladder is moving away from the wall at 1.5 ft/s

- Calculus -
**Reiny**, Wednesday, February 22, 2012 at 5:11pmlast two lines should have been

16(dx/dt) + 12(-2) = 0

dx/dt = 24/16 = 3/2

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