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^2-y
dy/dx = (-2xy^2-y)/(x+2yx^2)
how do I finish from here? I need to know where the slop of the tangent line equals -1.
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.
So I put -1 in for dy/dx and got this equation: -2x^2y+2xy-x+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?
Yes.
I got this for y.
2x^2y-2xy+x-y=0
2x^2y-2xy+x=y
2x^2-2x+x=1 (divide through by y)
2x^2-x-1=0
what do I from here? It doesn't factor evenly. I'm stuck!
To find the points on the curve where the slope of the tangent line is -1, you need to set the derivative you found equal to -1 and solve for the corresponding values of x and y.
So you have:
(-2xy^2 - y)/(x + 2yx^2) = -1
Multiply both sides by x + 2yx^2 to get rid of the denominator:
-2xy^2 - y = -x - 2yx^2
Rearrange the terms:
2yx^2 - y + 2xy^2 + x = 0
Combine like terms:
2yx^2 + 2xy^2 + x - y = 0
Factor out y:
y(2x^2 + 2xy - 1) + x = 0
Now, you have two cases to consider:
Case 1: y = 0
If y = 0, then the equation becomes x = 0.
So one point on the curve where the slope of the tangent line is -1 is (0, 0).
Case 2: 2x^2 + 2xy - 1 + x = 0
To solve this equation for x and y, you may need to use numerical methods or approximation techniques. One common approach is to use a graphing calculator or software to graph the equation and find its intersection points with the curve x^2y^2 + xy = 2.
Once you have the approximate values of x and y, you can verify if the slope of the tangent line at those points matches -1.