Calculus with Parametric Curves: Concavity?

Question

For the following problem, we are asked to find dy/dx, [d^2 * y] / [d * x^2], and to state for which values of t is the curve concave upward.


1. x = t^3 -12t, y = t^2 – 1

(the first equation reads as x equals t cubed minus 12t; the second equation reads as y equals t squared minus 1).

For this problem I got:

2t / [3t^2 – 12] for dy/dx;

[-6t^2 – 24] / [(3t^2 – 12)^3] for [d^2 * y] / [d * x^2];

after setting [d^2 * y] / [d * x^2] equal to 0 I had:

6t^2 + 24 = 0, and thus t^2 = -4;

Assuming I did all of this right so far, does this mean that the function does not concave upward anywhere because the square root of negative 4 is a non-real answer, or does is still concave up somewhere? Please explain which is the case and why, and if I did something wrong earlier in the problem (i.e. an algebraic error somewhere), please explain how to do the problem step-by-step.

Answer 1

Congratulations. You got it mostly right and you didn't make many stupid algebra mistakes. (The only one I found was the denominator for d2y/dx2 should have an exponent of 2, not 3.) You just overlooked a couple of points.

I suggest you check out:
http://www.math.hmc.edu/calculus/tutorials/secondderiv/

and pay careful attention to all the provisos, particularly the ones about "If f'(c) exists" and "if the function f is twice differentiable"

If that didn't help enough, here is a little more help.

If dy/dx = 2t / (3t^2 – 12), then dy/dx is not defined when x^2 = 4. That is, the denominator is 0 and so you can't divide by it.

Similarly, if d2y/dx2 = (-6t^2 – 24) / ((3t^2 – 12)^2), the denominator is 0 at t^2 = 4 so, again, this derivative is not defined at these two points.

Therefore, these two points could be inflection points without violating the theorems.

As for how to do this specific problem, the first thing to notice is that x is an odd function of t and y is an even function of t. That means that a plot of will be symmetric about the Y axis.

The second thing to notice is that when t gets large, x(t) grows as t^3 while y(t) will grow as t^2. That means that y(x) will look like x^(2/3) which is concave down.

By symmetry, the same is true when t gets very small (i.e. goes very negative.

So the area of interest is in the middle, where the 12t term has an influence.

I suggest you evaluate x and y for t = 0, +/-1, +/-2, +/-3, and +/-4. (By symmetry, you only have to calculate for t >= 0 and then adjust the signs for the t < 0 values.) Then plot and look. I think you will see the answer immediately.

HTH.