Related Rates Calculus: Solving a Challenging Problem with Ladders and Slopes

[EbolaPox]

I'm in AP Calculus BC and I've had no trouble with any of Calc I or II (I self studied it all previously), but this one related rates problem that seemed relatively simple got me.
(This is also in the Schaum's Outline Calculus book if anyone has that. My teacher gave it to us on a worksheet of practice problems)

Problem:
A ladder 20 ft long leans against a house. Find the rate at which A: the top of the ladder is moving downward if its foot is 12 ft from the house and moving away at a rate of 2 ft/sec. B: The slope of the ladder decreases.

Part A is incredibly easy. Part B is where I'm having trouble. I drew my triangle with a base of x and a height of y and a hypotenuse of 20. I know dx/dt = 2 ft/sec. I solved that dy/dt = 3/2 ft/sec, which is correct (I have the answers also.)
Here is my work so far.

x^2 + y^2 = 20^2
Derive for rates of change
2x(dx/dt) + 2y(dy/dt) = 0
x(dx/dt) = -y(dy/dt)

I then thought "Dy/dx would be the slope of the ladder. Part B wants the rate of change of the slope, so it wants the second derivative."
So, I said -x/y = dy/dx.

I took the derivative of that and got
-(y(dx/dt) - x(dy/dt))/y^2 = Second derivative.
I put in values I had (x = 12 and y =16).

-(16*2) - (12(3/2))/(16^2) . This gave me -14/256. The correct answer is -25/72.

Anyone have any suggestions? (Note: i got the answer if I said dy/dx = -y/x, but that's not true. I was just playing around with it and found that to be correct.)

 

[amcavoy]

I'm in AP Calculus BC and I've had no trouble with any of Calc I or II (I self studied it all previously), but this one related rates problem that seemed relatively simple got me.
(This is also in the Schaum's Outline Calculus book if anyone has that. My teacher gave it to us on a worksheet of practice problems)

Problem:
A ladder 20 ft long leans against a house. Find the rate at which A: the top of the ladder is moving downward if its foot is 12 ft from the house and moving away at a rate of 2 ft/sec. B: The slope of the ladder decreases.

Part A is incredibly easy. Part B is where I'm having trouble. I drew my triangle with a base of x and a height of y and a hypotenuse of 20. I know dx/dt = 2 ft/sec. I solved that dy/dt = 3/2 ft/sec, which is correct (I have the answers also.)
Here is my work so far.

x^2 + y^2 = 20^2
Derive for rates of change
2x(dx/dt) + 2y(dy/dt) = 0
x(dx/dt) = -y(dy/dt)

I then thought "Dy/dx would be the slope of the ladder. Part B wants the rate of change of the slope, so it wants the second derivative."
So, I said -x/y = dy/dx.

I took the derivative of that and got
-(y(dx/dt) - x(dy/dt))/y^2 = Second derivative.
I put in values I had (x = 12 and y =16).

-(16*2) - (12(3/2))/(16^2) . This gave me -14/256. The correct answer is -25/72.

Anyone have any suggestions? (Note: i got the answer if I said dy/dx = -y/x, but that's not true. I was just playing around with it and found that to be correct.)

You know that the slope is:

m=\frac{y}{x}=\frac{\sqrt{400-x^2}}{x}

so differentiate with respect to x...

\frac{dm}{dx}=\frac{\frac{dm}{dt}}{\frac{dx}{dt}}

you know that dx/dt=2. Do you see why this gives you the answer?

Alex