Calculus
posted by Isiah on .
A 23 foot ladder is leaning on a 6 foot fence. The base of the ladder is being pulled away from the fence at the rate of 7 feet/minute. How fast is the top of the ladder approaching the ground when the base is 6 from the fence? [Note: The ladder is protruding over the top of the fence.]
The top of the ladder is approaching the ground at the rate of____________ feet/minute. [Enter a positive number only.]

Let 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 = (1h^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