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Second derivative with parametric equations

  1. Sep 8, 2015 #1
    http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx

    On this page the author makes it very clear that:

    $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

    provided ##\frac{dx}{dt} \neq 0##.

    In example 4, ##\frac{dx}{dt} = -2t##, which is zero when ##t## is zero. In simplifying ##\frac{dy}{dx}## the author even divides the numerator and denominator by ##t## which is only possible if ##t \neq 0##.
    This is all consistent with the requirement ##\frac{dx}{dt} \neq 0##.
    The author then obtains an expression for the second derivative in terms of ##t##, plugs in zero, and finds out that the second derivative is zero at ##t = 0##.
    How is this consistent with the assumption that ##\frac{dx}{dt} \neq 0##? What's going on here?
     
  2. jcsd
  3. Sep 8, 2015 #2

    DEvens

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    Education Advisor
    Gold Member

  4. Sep 8, 2015 #3

    Mark44

    Staff: Mentor

    He doesn't actually "plug in" zero. Toward the bottom of the example he shows this work:
    $$\frac{d^2y}{dx^2} = \frac{-\frac 1 2 (35t^4 + 15t^2) }{-2t} = \frac 1 4 (3t^3 + 15t)$$
    The expression in the middle is undefined if t = 0, but as t approaches zero, the value of this expression approaches zero as well. The 2nd and 3rd expressions are exactly the same except at a single point -- the origin. He could have elaborated on this fact a bit more.

    The graph he shows doesn't make it clear that there is a "hole" at (0, 0). At this point both dy/dx and d2y/dx2 are undefined.
     
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