Area Between a Line and a Curve
Date: Sat, 5 Nov 1994 17:13:46 -0500 From: Adam Portley Subject: math problem Here is a problem that has been bugging me: "The horizontal line y=c intersects the curve y=2x-3x^3 in the first quadrant. Area A is bounded by y=c, x=0, and the curve. Area B is bounded above by the curve and below by y=c. Find c such that (area of A)=(area of B)." I can do this using integrals, but I don't want to solve any @!%$! cubics. There must be an easier way to find c. AP ps I am a senior at Athens High School.
From: Dr. Ken Date: Mon, 7 Nov 1994 17:33:47 -0500 (EST) AP! Nice problem! At first glance, it seems so easy, and then you think it's virtually unsolvable. Try this, though: You need to get a handle on the limits of integration, i.e. what values of x you can plug in to f(x) to get c. I'll call these c1 and c2. Then they satisfy the equation 2c1 - 3c1^3 = 2c2 - 3c2^3. So 2(c1 - c2) - 3 (c1^3 - c2^3) = 0, and since x^3 - y^3 = (x-y)(x^2+xy+y^2), we can factor out a (c1 - c2) from this equation (it yields only the trivial root). Then you get a quadratic equation relating c1 and c2, and you can solve for c1 in terms of c2. It's nasty, but it works. I haven't taken this any farther, but I think this is going to help. If it doesn't, write back and we'll see what we can do. -Ken "Dr." Math
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