Find the point on the curve x^3 + y^3 = 6xy where the tangent line is parallel to the line x + y = 10.

Question

I found sloe and implicit differentiation whats once we set derivative equal to slope

Answer 1

Doing the implicit differentiation of x³ + y³ = 6xy we get:

3x² +3y²•dy/dx = 6y + 6x•dy/dx

dy/dx(3y² - 6x) = 6y - 3x²

dy/dx = (6y - 3x²)/(3y² - 6x)

dy/dx = (2y - x²)/(y² - 2x)

The slope of x + y = 10 is -1 so we can equate the derivative to -1

(2y - x²)/(y² - 2x) = -1 or

2y - x² = 2x - y²

To solve:

y² - x² = 2x - 2y

(y - x)(y+x) = -2(y - x)

Divide through by (y - x) but in so doing we need to realize that we are potentially losing a solution if y = x, which indeed, in checking the original equation we are. But let's proceed.

y + x = -2

y = -x - 2

So everywhere on that line, the slope is -1, we just need to find the places on that line that are also on x³ + y³ = 6xy, So we can substitute in for y, "-x - 2" to get:

x³ + (-x - 2)³ = 6x(-x - 2)

x³ + (- x³ - 6x² -12x - 8) = -6x² -12x

-8 = 0 therefore there is no solution for this.

Now, let's go back to the other solution we lost when we divided by zero, y = x. It is clear to see that when y = x, our derivative equation set equal to -1 is true:

2y - x² = 2x - y²

So let's use that y = x substitution in our original equation x³ + y³ = 6xy

x³ + (x)³ = 6x(x)

2x³ = 6x²

dividing through by 2x² (again realizing that we could be loosing the solution x = 0) we get

x = 3

In this case the solution x = 0 results in the point (0, 0) and the derivative is not defined there so we can ignore this.

So the solution is x = 3. To find y, we just substitute in x = 3 into x³ + y³ = 6xy to get 27 + y³ = 18y. Solving, we get 3 solutions, y = -4.854, y = 3, and y = 1.854. The solution we are looking for is where y = x so (3, 3)

Verifying, we can plug (3, 3) into our derivative to verify the result is -1

dy/dx = (2y - x²)/(y² - 2x)

dy/dx = (2(3) - (3)²)/((3)² - 2(3)) = (6 - 9)/(9 - 6) = -3/3 = -1