If x^{2} + y^3 - x y^2 = 5, find dy/dx in terms of x and y.
Using your answer for dy/dx, fill in the following table of approximate y-values of points on the curve near x=1, y=2.
Finally, find the y-value for x=0.96 by substituting x=0.96 in the original equation and solving for y using a computer or calculator.
2x + 3y^2 dy/dx - x(2y)dy/dx - y^2 = 0
dy/dx(3y^2 - 2xy) = y^2 - 2x
dy/dx = (y^2 - 2x)/(3y^2 - 2xy)
at (1,2), dy/dx = (4-2)/(12-4) = 1/4
For your last part, sub in the x value of .96, rearrange the equation so it looks like a standard cubic and then use this webpage to solve
http://www.1728.com/cubic.htm
To find dy/dx, we will use implicit differentiation. We differentiate both sides of the equation with respect to x and treat y as a function of x.
Differentiating the equation x^2 + y^3 - xy^2 = 5, we get:
2x + 3y^2 * (dy/dx) - y^2 - 2xy * (dy/dx) = 0.
Rearranging the equation and isolating dy/dx, we have:
(2x - y^2) / (2xy - 3y^2) = dy/dx.
The formula for dy/dx in terms of x and y is:
dy/dx = (2x - y^2) / (3y^2 - 2xy).
To fill in the table of approximate y-values of points on the curve near x=1, y=2, we substitute x=1 and y=2 into the equation:
dy/dx = (2(1) - (2)^2) / (3(2)^2 - 2(1)(2))
= (2 - 4) / (12 - 4)
= -2 / 8
= -1/4.
So, the approximate value of dy/dx at the point (1, 2) is -1/4.
Now, we need to find the y-value for x=0.96 by substituting x=0.96 in the original equation and solving for y:
(0.96)^2 + y^3 - (0.96)y^2 = 5.
Substituting these values into a computer or calculator, we can solve for y.