# Find the second derivative of the parametric curve $x = 2\sin t$, $y = 3\cos t$

I need help trying to find when the parametric curve is concave upward.

$x = 2\sin(t)$

$y = 3\cos(t)$

I successfully got $-\frac {3}{2}\tan(t)$ as the first derivative. However, I thought the second derivative was supposed to be $-\frac{3}{2}\sec^2(t)$.

What am I doing wrong? I know it should be $-\frac {3}{4}\sec^3(t)$ and is concave upward whenever that expression is positive, but I don't know how to get there.

Thank you

• The second derivative for parametrics is not taken by taking the derivative of the first derivative assuming the latter just to be a "function. Here is a link with an example that may be helpful: qc.edu.hk/math/Certificate%20Level/… – imranfat Mar 1 '16 at 0:07
• @imranfat Thank you for clarifying that – chris Mar 1 '16 at 0:12
• We want $\frac{d}{dx}(\frac{dy}{dx})$. By the Chain Rule this is $\frac{d}{dt}(\frac{dy}{dx})$ divided by $\frac{dx}{dt}$. – André Nicolas Mar 1 '16 at 0:38

We are given the parametric equations \begin{align*} x(t) & = 2\sin t\\ y(t) & = 3\cos t \end{align*} By the Chain Rule, $$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$ At points where $dx/dt \neq 0$, we obtain $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ Since \begin{align*} x'(t) & = 2\cos t\\ y'(t) & = -3\sin t \end{align*} we obtain $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-3\sin t}{2\cos t} = -\frac{3}{2}\tan t$$ By the Chain Rule, $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{dy}{dx}}{dt} \cdot \frac{dt}{dx}$$ provided that $dt/dx$ exists.
Observe that since $x = 2\sin t$, $$\frac{x}{2} = \sin t$$ Differentiating each side of equation with respect to $x$ yields $$\frac{1}{2} = \cos t \frac{dt}{dx}$$ Solving for $dt/dx$ yields $$\frac{1}{2\cos t} = \frac{dt}{dx}$$ Since $$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(-\frac{3}{2}\tan t\right) = -\frac{3}{2}\sec^2t$$ we conclude that $$\frac{d^2y}{dx^2} = \frac{\frac{dy}{dx}}{dt} \cdot \frac{dt}{dx} = \left(-\frac{3}{2}\sec^2t\right)\left(\frac{1}{2\cos t}\right) = -\frac{3}{4}\sec^3t$$ Your error was not applying the Chain Rule when you took the second derivative.