What values determine the concavity of a parametric curve?

[sapiental]

Hello,

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 a lot in advance.

 

[courtrigrad]

If $$f''(x) > 0$$ then the curve is concave upward and vice versa.Looking at $$\frac{d^{2}y}{dx^{2}} = \frac{3(t^{2}+1)}{4t^{3}}$$

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: $$3t^{2} + 3 > 0$$ and $$4t^{3} > 0$$. Clearly, all values of t work for the first inequality, but only positive values of t work for the second inequality.