AP Calculus Integration Homework Help?

Question

Consider the curve given by xy² - x³y = 6

a) show that
dy 3x²y - y²
-- = ---------
dx 2xy - x³

b) find all points on the curve whose x-coordinate is 1, and write an equation for the tangent line at each of these points.

c) find the x-coordinate of each point on the curve where the tangent line is vertical.

Answer 1

a) differentiate both sides with respect to x using the product rule and implicit differentiation:

x(2y)(dy/dx) + y^2 -x^3(1)(dy/dx) - 3x^2(y) = 0

Solve algebraically for (dy/dx) to finish part a)

b) xy^2 - x^3y = 6 plug in x =1, solve for y:
y^2 - y = 6 (y-3)(y+2) =0 y = -2, 3

equation for a tangent line:
y - y0 = m(x-x0)

m = dy/dx (from part a)
for (1, -2) m = [3*(1^2)(-2) - (-2)^2]/[2(1)(-2) - 1^3] = 10/3
y +2 = (10/3)(x-1)

for (1, 3) m = 0
y + 2 = 0(x-1) y = -2

c) tangent line is vertical when (dy/dx) is infinite...or the denominator of (dy/dx) = 0

2xy - x^3 = 0

But the point must also be on the curve itself so this equation must all be satisfied.
xy^2 - x^3y = 6

Use first equation to find y = x^2/2

Subsitute y = x^2/2 into 2nd eqn.
x^5/4 - x^5/2 = 6
x^5 - 2x^5 = 24
x^5 = -24
Use calculator if you want a better numerical value.