ap calculus ab frq

Question

Consider the curve given by the equation (y^2)-(2(x^2)y)=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 d^2y/dx^2 at the point (1, –1).

Answer 1

a) Let's go ahead and do implicit differentiation and differentiate both sides in respect to x.

d/dx [ y² - 2x²y ] = d/dx [ 3 ]

2y*dy/dx - 4xy - 2x²*dy/dx = 0

dy/dx*(2y - 2x²) = 4xy

dy/dx = 4xy/(2y - 2x²)

b) Use the dy/dx equation to find the slope of this tangent line at (1, -1).

dy/dx = 4(1)(-1)/(2(-1) - 2(1)²) = 1

Now, use the point-slope formula in order to write out an equation for the line tangent at point (1, -1)

y - y₁ = m(x - x₁)

y + 1 = 1(x - 1)

y = x - 2

c) The points where the curve is horizontal is where the slope of the curve is zero. In other words, where dy/dx = 0. Set dy/dx = 0 and find the x-values in which dy/dx is zero.

0 = 4xy/(2y - 2x²)

0 = 4xy

x = 0

Now, plug in x = 0 into the original equation to find y.

y² - 2(0)²y = 3

y² = 3

y = ±√3

(0, √3) and (0, -√3)

d) Find d²y/dx² through implicit differentiation.

d/dx [dy/dx*(2y - 2x²)] = d/dx [4xy]

d²y/dx²*(2y - 2x²) + dy/dx*(2*dy/dx - 4x) = 4y + 4x*dy/dx

d²y/dx²*(2y - 2x² + 2) = dy/dx*(8x) + 4y

d²y/dx² = (dy/dx*(8x) + 4y)/(2y - 2x² + 2)

Now, let's evaluate it at point (-1, 1). Remember, dy/dx at point (-1, 1) is 1 according to problem b.

d²y/dx² = (1*(8*-1) + 4(1))/(2(1) - 2(-1)² + 2)

d²y/dx² = -2

Answer 2

For part a, differentiate both sides with respect to x, using the power and product rules. I'll use the notation y' to represent the derivative dy/dx, to make it easier to read.

2yy' - 2(yd(x²)/dx + x²y' = 0

divide through by 2

yy' - (yd(x²)/dx + x²y') = 0

Now solve for y'

yy' -(2xy + x²y') = 0

yy' - x²y' = 2xy

y'(y-x²) = 2xy

y' = 2xy/(y-x²)

I will post information for the remaining parts later when I have time!