What is the related rates problem for a ladder leaning against a wall?

[crybllrd]

Homework Statement

A ladder of 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away at 2 ft per second.

(a) How fast is the top of the ladder moving down the wall when the base is seven feet from the wall?

(b) Consider the triangle formed by the side of the house, ladder and ground. Find the rate at which the area of the triangle is changing when the ladder is 7 feet from the wall.

(c) Find the rate at which the angle between the ladder and wall is changing when the base of the ladder is 7 ft from the wall.

The Attempt at a Solution

(a) x=7ft y= 24 (pythag thm) dx/dt=2ft/s dy/dt=?

x^{2}+y^{2}=25^{2}

2x\frac{dx}{dt}+2y\frac{dy}{dt}=0


\frac{dy}{dt}=\frac{2x\frac{dx}{dt}}{-2y}


\frac{dy}{dt}=\frac{-7}{12}ft/s

I feel like part A is correct. Part B is where I'm stuck:
(b)

A=\frac{1}{2}xy

\frac{dA}{dt}=\frac{1}{2}\frac{dx}{dt}\bullet\frac{dy}{dt}

Here I was going to plug in dx/dt and dy/dt from part (a) but it gives me the same solution as part (a). What should I do?

(c) Not sure about (c) either, assuming I need to start with sine theta, but I want to get on top of part B first.

 

[Quinzio]

Part b: you need to apply derivation rules correctly, then reuse what you found in part a.

 

[crybllrd]

Ah jeez, I always make a mistake like that. OK, now I have dA/dt= 527/24 ft^2/sec.

Alright, now on to part (c):

sin\theta=\frac{O}{H}=\frac{x}{y}


cos\theta\frac{d\theta}{dt}=\frac{y\frac{dx}{dt}-x\frac{dy}{dt}}{y^{2}}

Can I just calculate Theta from the triangle, plug in everything then solve for dT/dt?