How do you find the coordinates of the stationary points to this implicit differentiation equation?
x^2y = y^2 + 2x
2xy + x^2yy' = 2yy' + 2
y'(x^2y-2y) = 2-2xy
y' =
2(1-xy)
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y(x^2-2)
if z = x^2y-y^2-2x
∂z/∂x = 2xy-2
∂z/∂y = x^2-2y
If they are both zero (the stationary point), then
xy=1, so y=1/x
x^2-2/x = 0
x^3-2=0
x = ∛2
so, y=1/∛2
To find the coordinates of the stationary points, we need to find the values of x and y that make the derivative y' equal to zero.
First, let's simplify the equation:
y'(x^2y - 2y) = 2 - 2xy
Divide both sides by (x^2 - 2) to isolate y':
y' = (2(1 - xy))/(y(x^2 - 2))
Now, set y' equal to zero and solve for x and y:
0 = (2(1 - xy))/(y(x^2 - 2))
To make it easier to solve, let's multiply both sides by y(x^2 - 2):
0 = 2(1 - xy)
Now we have:
1 - xy = 0
Solve for y:
y = 1/x
Substitute y = 1/x back into the original equation:
x^2(1/x) = (1/x)^2 + 2x
x = (1/x)^2 + 2x
Multiply both sides by x^2:
x^3 = 1 + 2x^3
Rearrange the equation to isolate x:
x^3 - 2x^3 - 1 = 0
-x^3 - 1 = 0
Now, you can solve this cubic equation to find the values of x. To do so, you can use numerical methods, such as Newton's method or the bisection method, or use a graphing calculator or software to find the x-values.
Once you have the x-values, plug them back into y = 1/x to find the corresponding y-values. These (x, y) coordinates will be the stationary points of the implicit differentiation equation.