Posted by Riley on Thursday, April 2, 2009 at 1:14pm.
Find all points on the curve x^2y^2+xy=2 where the slope of tangent line is 1.
This is what I got.
x^2y^2+xy=2
x^2*2ydy/dx + 2x*y^2 + x*dy/dx + y =0
dy/dx(2yx^2+x)= 2xy^2y
dy/dx = (2xy^2y)/(x+2yx^2)
how do I finish from here? I need to know where the slop of the tangent line equals 1.

calculus  Count Iblis, Thursday, April 2, 2009 at 2:15pm
You can insert dy/dx = 1 at this point:
x^2*2ydy/dx + 2x*y^2 + x*dy/dx + y =0
You then get an equation relating x and y. You also know that x and y are on te curve, so they satisfy the equation of the curve:
x^2y^2+xy = 2
So, you have two equations for the two unknowns x and y.

calculus  Riley, Thursday, April 2, 2009 at 2:26pm
So I put 1 in for dy/dx and got this equation: 2x^2y+2xyx+y=0. Then would I solve for either x or y and substitute back in to the original equation to get a value for x and y?

calculus  Count Iblis, Thursday, April 2, 2009 at 5:14pm
Yes.

calculus  Anonymous, Thursday, April 2, 2009 at 8:26pm
I got this for y.
2x^2y2xy+xy=0
2x^2y2xy+x=y
2x^22x+x=1 (divide through by y)
2x^2x1=0
what do I from here? It doesn't factor evenly. I'm stuck!
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