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Strange tangent for parametric

  1. Jan 21, 2015 #1
    For the parametric equations x = t^3 - 3t and y = t^3 - 3t^2 I got that the graph has a vertical tangent when t is = to postive or negative one. And it is horizontal at t = 2. However, this implies that at the point (x,y) = (2, -4) the graph has both a vertical and horizontal tangent. How is this possible? Thanks
     
  2. jcsd
  3. Jan 21, 2015 #2

    jedishrfu

    Staff: Mentor

    How are you computing the tangents?
     
  4. Jan 21, 2015 #3
    https://mail.google.com/mail/u/0/?ui=2&ik=263fc3780d&view=fimg&th=14b0f97b7ad0fc5a&attid=0.1&disp=inline&safe=1&attbid=ANGjdJ9rFhryHaxukeJKxVFkvC2oxwR748Jb1IOpuC3WKs59gfmKcuR1B_n8TTjb5NhFNTaqw8vUgJYVuaXRo63FYUiD7TCzRXcrnGNGm3MDzoHF8zC1VlC2vVLwC3o&ats=1421902241584&rm=14b0f97b7ad0fc5a&zw&sz=w1273-h532
     
  5. Jan 21, 2015 #4

    Mark44

    Staff: Mentor

    Just show us your work, not an image of it, especially one that doesn't render.
     
  6. Jan 21, 2015 #5

    Mark44

    Staff: Mentor

    When t = 2, x = 2 and y = -4, just as you say. And dy/dx = 0 when t = 2, so the tangent is horizontal at (2, -4). Why do you think that the tangent is vertical when t = 2?
     
  7. Jan 22, 2015 #6
    My friend from math class actually explained it to me, but thank you for your help. At the point (2, -4), the graph crosses itself, so as a result there are two tangents at that spot.
     
  8. Jan 22, 2015 #7

    Mark44

    Staff: Mentor

    Right, but they occur for two different values of t.

    When t = -1, (x, y) = (2, -4) and the tangent is vertical.
    When t = 2 you get (2, -4) again, but this time with a horizontal tangent.

    To find the values of t, I solved the equation t3 - 3t = 2, or t3 - 3t - 2 = 0, which in factored form is (t + 1)2(t - 2) = 0.
     
  9. Jan 22, 2015 #8
    Thank you so much, I understand now
     
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