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.(adsbygoogle = window.adsbygoogle || []).push({});

(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.)

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# Related Rates Calculus

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