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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