Find points where tangent line is horizontal

[Dustobusto]

Homework Statement

Find the points on the graph of x³-y³=3xy-3 where the tangent line is horizontal

Homework Equations

y = f(x) so implicit differentiation must be used when taking the derivative of y

(xy)' = xy' + y

The Attempt at a Solution

So if the tangent line is horizontal the slope = 0 and the derivative of y' must be equal to zero.

So first taking the derivative I end up with

3x²-3y²y' = 3xy' + 3y

Factor out the three and divide by three on both sides to cancel them out. That leaves me with

x²-y²y' = xy' + y

Subtract y on both sides, add y²y' on both sides and get

x²-y = xy' + y²y'

Factor out the y' to get

x²-y = y'(x+y²)

divide by x+y on both sides and

y' = (x²-y)/(x+y²)

So at this point, I set y' equal to zero and solve for x? Don't I remove the y variables or something like that?

 

[milesyoung]

So at this point, I set y' equal to zero and solve for x? Don't I remove the y variables or something like that?

Remove it how?

You're looking for points (x,y) where dy/dx = 0. You have dy/dx = some fraction. When is a fraction zero?

 

[Dustobusto]

Remove it how?

You're looking for points (x,y) where dy/dx = 0. You have dy/dx = some fraction. When is a fraction zero?

Well I understand part of what you're saying.

I can solve the y' so that x² = y and then go back into the equation of y' and replace y with x² and demonstrate that it equals zero. Do I plug x² into the original y and solve for x? Or take the derivative and solve for x?

 

[Dick]

Well I understand part of what you're saying.

I can solve the y' so that x² = y and then go back into the equation of y' and replace y with x² and demonstrate that it equals zero. Do I plug x² into the original y and solve for x? Or take the derivative and solve for x?

Second choice. Substitute y=x^2 into the original equation and find x. Having a horizontal tangent at (x,y) means it has to satisfy the original equation AND y'=0. Two equations.

 

[Dustobusto]

OK, so x³-y³ = 3xy -3

Substitute,

x³ -(x²)³ = 3(x)(x²) - 3

x³ - x⁶ = 3x³-3

Subtract 3x³ both sides and get

-2x³-x⁶ = -3

So at this point, I'm not sure what's factorable and what's not. I could factor out an x3 and get

x³(-2-x³) = -3 but I'm not sure how much that helps me. Do I need to use the quadratic formula at some point?

NOPENOPENOPE I think I got it. I believe x = √3 cubed root rather

 

[ehild]

I think I got it. I believe x = √3 cubed root rather

Not quite. And what about the other root?

ehild

 

[Dustobusto]

Not quite. And what about the other root?

ehild

Yeah I'm starting to realize that wasn't right

x³ - x⁶ = 3x³-3

is that at least on the right track?

 

[Dick]

Yeah I'm starting to realize that wasn't right

x³ - x⁶ = 3x³-3

is that at least on the right track?

Fine so far. Move everything to one side and try to factor if. It's a quadratic in u=x^3.