Posted by Amy on Sunday, February 5, 2012 at 4:20pm.
A ladder 15 feet long is leaning against a building so that end X is on level ground and end Y is on the wall. Point 0 is where the wall meets the ground. X is moving away from the building at a constant rate of 1/2 foot per second.
(a.) Find the rate in feet per second at which the length of 0Y is changing when X is 9 feet from the building.
(b.) Find the rate of change in square feet per second of the area of triangle X0Y when X is 9 feet from the building
If you can help answer even part of it, that would be super helpful! Thank you so much :x :x

Calculus  Reiny, Sunday, February 5, 2012 at 4:42pm
This is the classic question used by most textbooks to introduce rate of change.
using your definitions
x^2 + y^2 = 15^2
2x dx/dt = 2y dy/dt = 0
x dx/dt + y dy/dt = 0
a) when x = 9
81+y^2 = 225
y =√144 = 12
also given: dx/dt = 1/2
a) 9(1/2) + 12dy/dt = 0
dy/dt = 4.5/12 = .375
the top of the ladder is sliding down (the negative) at .375 ft/s
b) area = (1/2)xy
d(area)/dt = (1/2)(x dy/dt + y dx/t)
= (1/2)(9(.375) + 12(.5))
= 4.6875
At that moment the area is increasing at aprr. 4.69 ft^2/s
check my arithmetic.

Calculus  Anonymous, Sunday, November 25, 2012 at 9:17pm
This answer is wrong b/c dy/dt is negative
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