Consider the curve given by the equation y^2-2x^2y=3

Question

Consider the curve given by the equation y²-2x²y=3.

a) Find dy/dx .

b) Write an equation for the line tangent to the curve at the point (1, –1).

c) Find the coordinates of all points on the curve at which the line tangent to the curve at that point is horizontal. d) Evaluate d²y/dx² at the point (1, –1)

Answer 1

Use implicit differentiation where the derivative of y² is 2y•dy/dx (chain rule) and the derivative of 2x²y is 2xy + 2x²dy/dx (product rule and chain rule):

To find the equation at (1, -1), plug in x = 1, y = -1 into the derivative to get the slope: dy/dx = 2(1)(-1)/(-1 - 1²) = -2/-2 = 1 and use the point-slope form of a line to write the equation:

y + 1 = 1(x - 1)

y = x - 2

To find the points where the tangent line is horizontal, make the derivative equal zero:

2xy/(y - x²) = 0 which is only true when 2xy = 0 so x = 0 or y = 0

Using the original function and plugging in x = 0:

y² - 2x²y = 3

y² - 2(0)²y = 3

y² = 3

y = ± √3

so (0, √3) and (0, -√3)

But using y = 0, there is no solution.

The second derivative is done the same way but requires implicit differentiation using the quotient rule:

dy/dx = 2xy/(y - x²)

d²/dx² = (after too much algebra to write out) = 2y(y² - 3x⁴)/(y - x²)³ then, of course, just plug in (1, -1) to get the value