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
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.
1 answer
1 answer