1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find points where tangent line is horizontal

  1. Jun 9, 2013 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

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

    (xy)' = xy' + y

    3. 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?
    Last edited: Jun 9, 2013
  2. jcsd
  3. Jun 9, 2013 #2
    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?
  4. Jun 9, 2013 #3
    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?
  5. Jun 9, 2013 #4


    User Avatar
    Science Advisor
    Homework Helper

    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.
    Last edited: Jun 9, 2013
  6. Jun 10, 2013 #5
    OK, so x3-y3 = 3xy -3


    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 whats factorable and whats 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
  7. Jun 10, 2013 #6


    User Avatar
    Homework Helper

    Not quite. And what about the other root?

  8. Jun 10, 2013 #7
    Yeah I'm starting to realize that wasn't right

    x3 - x6 = 3x3-3

    is that at least on the right track?
  9. Jun 10, 2013 #8


    User Avatar
    Science Advisor
    Homework Helper

    Fine so far. Move everything to one side and try to factor if. It's a quadratic in u=x^3.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted