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Homework Help: Parametric Curve Concavity

  1. Nov 28, 2006 #1

    My textbook says that to determine concavity we calculate the second derivative of the curve. This is a problem from my book,

    x = t^2 and y = t^3 - 3t

    the second derivative of this is (3(t^2+1))/(4t^3)

    I know all the steps to get to this point.. However, the book says that the curve is concave upward when t > 0 and concave downward when t < 0.

    Can somebody please explain to me what values this last statement refers to. Is there a general theorem/procedure that I can apply to any second derivative of a parametric curve to determine the concavity?

    Thanks alot in advance.
  2. jcsd
  3. Nov 28, 2006 #2
    If [tex] f''(x) > 0 [/tex] then the curve is concave upward and vice versa.

    Looking at [tex] \frac{d^{2}y}{dx^{2}} = \frac{3(t^{2}+1)}{4t^{3}} [/tex]

    it is greater than 0 when both numerator and denominator are positive (cant do negative over negative because the numerator will always be positive).

    So solve the following inequalities: [tex] 3t^{2} + 3 > 0 [/tex] and [tex] 4t^{3} > 0 [/tex]. Clearly, all values of t work for the first inequality, but only positive values of t work for the second inequality.
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