How to Find the Area Under y = cos(x) Above y = k?

[carlodelmundo]

Homework Statement

Let A be the area of the region in the first quadrant under the graph of y = cos (x) and above the line y = k for 0 <= k <= 1.

a.) Determine A in terms of k.

b) Determine the value of A when k = 1/2.

c) If the line y = k is moving upward at the rate of ( 1 / pi ) units per minute, at what rate is the area, A, changing when k = 1/2 ?

Homework Equations

Fundamental Theorem of Calculus

The Attempt at a Solution

Here's my work insofar:

http://carlodm.com/calc/prob2.jpg

For c.) I have no idea on how to tackle this problem. Should I derive my area formula in terms of dt?

Thanks

 

[Dick]

It would really help if you could make a smaller scan. You are doing fine up to c). First you need to find dA/dk correctly. A(k)=sin(arccos(k))-k*arccos(k). It looks like its almost right, except why are you mixing x's and k's. Shouldn't they all be k's?

 

[carlodelmundo]

Thank you Dick for your help. Here's my new work:

http://carlodm.com/calc/prob4.JPG

I have a problem. I have the rate of change of area with respect to time in terms of x. I'm given a rate of change in terms of y. I thought to myself that maybe I can just "cheat" and plug in dK/dt for dx/dt, but isn't this wrong?

Basically, can you give me tips to solve for the rate of change of y? I'm stumped.

 

[Dick]

You are sort of confusing x and k. x is the variable you are integrating over. The upper limit is arccos(k). Area should just come out as a function of k. I get that dA/dk=-arccos(k). dA/dt=dA/dk*(dk/dt).

 

[carlodelmundo]

Thanks a lot, Dick!

You're right about me confusing x and k. I thought all related rates problems derived in terms of t, but now I can see that that's not always the case. That is a very elegant solution in my opinion.

Here is my revised work:

http://carlodm.com/calc/prob6.PNG

Is my answer -1/3 Units^2 / min correct?

 

[Dick]

Looks good to me!