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November 28, 2014

November 28, 2014

Posted by **Isiah** on Thursday, October 13, 2011 at 10:20pm.

The top of the ladder is approaching the ground at the rate of____________ feet/minute. [Enter a positive number only.]

- Calculus -
**Steve**, Friday, October 14, 2011 at 11:53amLet the ladder make an angle t with the ground.

Let the distance from the base of the fence = x

Let the end of the ladder be at height h.

tan(t) = 6/x

sin(t) = h/23

tan^2 = sin^2/cos^2

36/x^2 = (h/23)^2 / (1 - (h/23)^2)

flip it upside down:

x^2/36 = (1-h^2/529)/(h^2/529)

x^2/36 = 529/h^2 - 1

2x/36 dx = -2(529)/h^3 dh

when x = 6

tan(t) = 6/6 = 1

so, h = 23/√2

2(6)/36 * 7 = -2(529)/(23^3/2√2) dh

84/36 = -1058/4302 dh

dh = -9.49 ft/s

or,

9.49 ft/s downward

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