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October 30, 2014

October 30, 2014

Posted by **Jeff** on Monday, October 26, 2009 at 3:22am.

- calculus -
**Reiny**, Monday, October 26, 2009 at 8:15amAt a time of t seconds,

let the height of the flag be h feet, then the distance to the top of the pole is 40-h feet and length of rope from the top of the pole to the car is 80+h feet.

let the car be x feet from the base of the pole.

given: dx/dt = 3 ft/s

fing: dh/dt , when h = 20

I see Pythagoras here,

x^2 + 40^2 = (80+h)^2

2xdx/dt = 2(80+h)dh/dt

dh/dt = xdx/dt/(80+h)

when h = 20

x^2 + 1600 = 10000

x = 8400

x = √8400

finally

dh/dt = √8400(3)/100

= 2.47 feet/second

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