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Horizontal/Vertical Tangents

  1. Sep 15, 2008 #1
    Give the (x,y) coordinate of each point where a horizontal/vertical tangent for the curve: x=(3t)/(1+t^3), y=(3t^2)/(1+t^3). from -inf to inf.

    Can someone please guide me through this?! I'm not sure where to begin.
  2. jcsd
  3. Sep 15, 2008 #2
    Chain rule?

    [tex]\frac{dy}{dx}[/tex] = [tex]\frac{dy}{dt}[/tex] [tex]\frac{dt}{dx}[/tex]
  4. Sep 16, 2008 #3
    Horizontal tangents => dy/dx = 0

    Vertical tangents => dy/dx = undef.
  5. Sep 16, 2008 #4
    Horizontal Tangent: dy/dt= (3t^2)/(1+t^3)= -3t(t^3-2)/(t^3+1)^2=0. How do I solve?
  6. Sep 16, 2008 #5
    Assuming you did the derivative correctly, setting the numerator equal to zero will give you the horizontal tangents as long as the denominator is a non-zero.

    Setting the denominator of dy/dx equal to zero will give you an undefined slope, meaning the tangent line is vertical.
    Last edited: Sep 16, 2008
  7. Sep 16, 2008 #6
    horizontal tan dy/dt=o=3t(t^3-2)/(1+t^3)^2, t=0, -2^1/3, 2^1/3.?
  8. Sep 16, 2008 #7
    Looks right.
  9. Sep 16, 2008 #8
    and so my coordinated would be (x,y)=(0,0), (-2^1/3,0), and (2^1/3,0)??
  10. Sep 16, 2008 #9
    You solved for the parameter value t for when the derivative is 0. However x and y are both functions of t, so you still need to find x and y
  11. Sep 16, 2008 #10
    ahhh, I see. (0,0)(3*2^1/3)/(1+(2^1/3)^3),0) and -3*2^1/3/(1-(2^1/3)^3),0) ???
  12. Sep 16, 2008 #11
    "horizontal tan dy/dt..."

    The coordinates of this curve are defined in terms of a parameter t.

    The tangent to the curve is not dy/dt, it is dy/dx. To find dy/dx, there are two possibilities.

    You solve one of the given equations for either x or y (depending on which equation you choose), subtitute into the other equation and get y in terms of x, then differentiate to get dy/dx. It would be messy to do so in this case.

    Or, use the chain rule.

    [tex]\frac{dy}{dx}[/tex] = [tex]\frac{dy}{dt}[/tex] [tex]\frac{dx}{dt}^{-1}[/tex]

    = [tex]\frac{-3t^{4} + 6t}{(1 + t^{3})^{2}}[/tex] [tex]\frac{(1 + t^{3})^{2}}{-6t^{3} + 3t}[/tex] =...

    Simplify and set = 0; solve for t; sub into the original equations to get (x,y).
  13. Sep 16, 2008 #12
    Yes, but edziura, as you said:

    [tex]\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

    So you can individually solve dy/dt and dx/dt for the points of vertical and horizontal tangency.
  14. Sep 16, 2008 #13
    I agree.
  15. Sep 16, 2008 #14
    Now, I'm a bit confussed. What did I do wrong? I did dy/dt and dx/dt separately.
  16. Sep 17, 2008 #15
    "horizontal tan dy/dt=o=3t(t^3-2)/(1+t^3)^2, t=0, -2^1/3, 2^1/3."

    t = 0 and t = 2^1/3 are correct, but not t = -2^1/3; you are finding an odd root, not an even one.
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