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For this question I already determined the 1st and 2nd derivatives using: $\dfrac{dy}{dx}$ and $\dfrac{d^2y}{dx^2}$.

The last part of the question asks me to determine for which values of $t$ is the curve concave upward. The answer in the book simply says: "The curve is CU when $[\sec(t)]^3<0 \Rightarrow \sec(t)<0\Rightarrow \cos(t)<0 \Rightarrow \frac{\pi}2 < t < \frac{3\pi}2$".

I tried doing the second derivative test but I got lost on that and didn't come up with anything. Then I graphed $-\dfrac 34\sec(t)^3$ which is the second derivative I calculated. However I wasn't able to see how they came up with that solution still.

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2 Answers 2

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Just from what you've said, you have the second derivative and you know the second derivative test, so with your second derivative, the curve is concave up when $$-\frac{3}{4}\sec^3t>0.$$ Dividing both sides by $-\frac{3}{4}$ reverses the inequality, so, dividing by 4, $$\sec^3t<0,$$ which is what the book said. Since $$\sec t=\frac{1}{\cos t},$$ and since $x$ and $x^3$ have the same sign, $$\sec^3t<0\implies\cos t<0\implies \frac{\pi}{2}<t<\frac{3\pi}{2},$$ at least within the given interval $0< t<2\pi$.

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Since the first part of the problem asked for the second derivative as a function of $t$, the solution by Isaac is the one that requires the least additional effort. However, note that $$\left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2=\sin^2 t+\cos^2 t =1.$$

We recognize this curve as an ellipse. The full ellipse is traced out, starting at $(0,3)$, and moving clockwise, as $t$ ranges over the interval $[0,2\pi)$. Since $0$ is not in the domain of $t$, there is a hole in our ellipse, albeit a tiny one.

The bottom half of the ellipse is the part that is concave up. This is traced out as $t$ goes from $\pi/2$ to $3\pi/2$. The interval where we have concavity upwards is partly a matter of definition, either $\pi/2 \lt t \lt 3\pi/2$ or $\pi/2 \le t \le 3\pi/2$. So the familiar geometry of the ellipse provides a check on the parametric calculation.

Comment: As was pointed out, you had to calculate $\dfrac{d^2y}{dx^2}$ anyway, probably by computing $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ first, then $\dfrac{dy}{dx}$. Then you needed to do some further differentiation for the second derivative. That's good, one needs to have full control over these things.

However, from the ordinary equation of the ellipse, we can in one line note that $$\frac{2x}{2^2} +\frac{2y}{3^2}\frac{dy}{dx}=0.$$ It is now easy to find $\dfrac{dy}{dx}$ in terms of $x$ and $y$, and hence, by easy substitution, in terms of $t$.

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+1. I'd been debating adding a note to my answer about it clearly being an ellipse that traced from the top, clockwise, so that it was concave up for $\frac{\pi}{2}<t<\frac{3\pi}{2}$ almost by inspection, but you've covered it in much better depth than I would have. –  Isaac Jul 17 '11 at 6:05

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