# Finding twice-differentiable function of $x$ of a parametric curve when $dx/dt = 0$

We're working on finding tangents of parametric curves and I feel like this problem isn't as hard as I'm making it out to be, but I am completely stumped. I am given this information:

Given $$x=3\cos(2t)$$ and $$y=3\sin(2t)$$ for $$0\leq t\leq \frac{\pi}{2}$$, find $$\frac{d^2y}{dx^2}$$ at $$t=\frac{\pi}{2}$$.

So what I did was rewrite it like this: $$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$$

If I solve for $$\frac{dx}{dt}$$, I get this: $$\frac{dx}{dt}=-6\sin(2t)=-6\sin(2(\frac{\pi}{2}))=-6\sin(\pi)=0$$

Which leads to dividing by zero. So what is the answer? What can I determine from this? I understand that the twice-differentiation determines the direction of concavity, and I understand that there would be a vertical tangent line, but would the answer just be "does not exist"? I've tried looking up how to do this, but every problem I've come across does not have $$\frac{dx}{dt} = 0$$, including all of our homework problems and practice problems I've found online. If anyone could help explain how to figure this out or at least point me in the right direction, it would be much appreciated. Thanks.

• Actually, you seem to be doing it correctly, perhaps you just need to solve everything for $t$ and plug in later, you might get a case of $0/0$ which you can solve with L'Hopital rule. Jul 9, 2015 at 6:33
• Well, I get $(-1/3)/sin^3(2t)$, which still goes to infinity. So perhaps that is indeed the answer. Jul 9, 2015 at 6:38
• If you like, you can get the same (infinity) answer by using the equation $x^2 + y^2 = 9$. Jul 9, 2015 at 6:47

As @Michael says in the comments - this system of parametric equations is equivalent to $x^2+y^2=9$ (restricted to $y\geq0$). This is a semicircle in the upper half plane and the reason for infinite $\frac{d^2y}{dx^2}$ (and actually infinite $\frac{dy}{dx}$) becomes apparent - the point $t=\frac{\pi}{2}$ corresponds to $(-3,0)$ at which the tangent line to the curve is vertical (hence $\frac{dy}{dx}=\infty$). But since $\frac{d^2y}{dx^2}$ tells us how $\frac{dy}{dx}$ would change if we were to increase $x$ slightly this too must be infinite since a slight increase in $x$ would give a finite value for $\frac{dy}{dx}$ and the difference between any finite value and infinity is always infinity.