given the equation x^3-y^3=12xy:

Question

a) find dy/dx

b) Find the points (x and y coordinates) where the graph of the equation has a horizontal tangent line.

(c) Find the points (x and y coordinates) where the graph of the equation has a vertical tangent line.

(d) Plot the graph of the equation (using plot implicit) and the tangent lines from parts (b) and (c). Use the x-domain [-7.5,7.5].

Answer 1

Use implicit differentiation:

(y' is dy/dx)

To find a horizontal tangent, you want the tangent line to have a slope of zero. So set y' equal to zero and solve.

Setting y' equal to, the solution is found by setting the numerator equal to zero:

3x² - 12y = 0 is a family of solutions, but you want only those that are on the graph of x³ - y³ = 12xy

To do that, manipulate 3x² - 12y = 0 into y = (1/4)x² and plug "(1/4)x²" into x³ - y³ = 12xy in place of "y":

x³ - [(1/4)x²]³ = 12x[(1/4)x²]

x³ - (1/64)x⁶ = 3x³

0 = (1/64)x⁶ + 2x³

0 = x³[(1/64)x³ + 2]

Set each piece equal to zero so x³ = 0 meaning x = 0 and (1/64)x³ + 2 = 0 which we solve as follows:

(1/64)x³ + 2 = 0

(1/64)x³ = - 2

x³ = - 128

x = cuberoot(-128) = -4cuberoot(2)

We find the y-values of these points by plugging the x-values into y = (1/4)x²

So the points of intersection are (0, 0) and (-4cuberoot(2), 4cuberoot(4)) which are the points where the tangent line is horizontal.

To find the places where the tangent line is vertical, set the denominator of the derivative equal to zero and solve in the same way