Posted by **chelsea** on Wednesday, May 4, 2011 at 4:34pm.

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 ....

5A=-9

A=-9/5

sub back in to get B = 17/10

(Was expecting "nicer" numbers. I might have made an arithmetic error, check my arithmetic)

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