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!

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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook