Suppose y is defined implicitly as a function of x by x^2+Axy^2+By^3=1 where A and B are constants to be determined. Given that this curve passes through the point (3,2) and that its tangent at this point has slope -1, find A and B.
calculus - Reiny, Wednesday, May 4, 2011 at 6:08pm
take the derivative:
2x + (Ax)(2y)dy/dx + Ay^2 + 3By^2 dy/dx = 0
dy/dx = (-2x - Ay^2)/(2Axy + 3By^2)
at (3,2) dy/dx = -1
-1 = (-6 - 4A)/(12A + 12B)
4A + 6B = 3
also in original,
9 + 12A + 8B = 1
3A + 2B = -2
triple the last one, and subtract ....
sub back in to get B = 17/10
(Was expecting "nicer" numbers. I might have made an arithmetic error, check my arithmetic)