# Why is this curve convex?

I am considering the curve traced by the equation $r=a\sin 3\theta$. Specifically as $\theta$ varies from $0$ to $\frac{\pi}{6}$, $r$ varies from $0$ to $a$. How do I conclude that the curve is convex in this domain? If we are dealing with cartesian coordinates for a function $y=f(x)$, the book says that the second derivative being non-negative tells us whether the function is convex or not. But is there such a test in polar coordinates for arbitrary curves? Secondly, if we allow $\theta$ to take all real values, $r$ becomes negative occasionally. Does this not contradict the meaning of r?

Thanks.

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Are you sure they want you to use polar coordinates? You could put $\theta$ on the x axis and r on the y axis. – Codie CodeMonkey Sep 26 '11 at 7:49
$y=3\sin\,x$ and $r=3\sin\,\theta$ are very different curves. One's transcendental and unbounded, while the other's algebraic and bounded. – J. M. Sep 26 '11 at 7:57
@DY: Yes, the equation of the curve is in polar coordinates. – Shahab Sep 26 '11 at 9:39
Did you guys by any chance cover how to test parametrically-defined curves for convexity? – J. M. Sep 26 '11 at 10:19

For simplicity assume $a=2$. In $(x,y)$-coordinates your curve then has the parametric representation $$\gamma:\quad \theta\ \mapsto \cases{x(\theta):=2\sin(3\theta)\cos\theta=\sin(4\theta)+\sin(2\theta) \cr y(\theta):=2\sin(3\theta)\sin\theta=\cos(2\theta)-\cos(4\theta)\cr}\qquad(0\leq\theta\leq 2\pi).$$ To study the convexity of $\gamma$ we have to look at the sign of its curvature $\kappa$ as a function of $\theta$. Now ${\rm sgn}(\kappa)={\rm sgn}(\dot x\ddot y-\ddot x\dot y)$, where the dot denotes differentiation with respect to the parameter $\theta$. The computation gives $$\dot x\ddot y-\ddot x\dot y=8\bigl(7+2\cos(6\theta)\bigr)>0\qquad\forall\theta\ .$$ It follows that the tangent vector $(\dot x,\dot y)$ always turns counterclockwise as $\theta$ increases. Since $(\dot x(0),\dot y(0))=(6,0)$ this implies that in the neighborhood of $\theta=0$ the curve describes a convex arc. Globally $\gamma$ is a threefoil.
$\frac{\dot{x}\ddot{y}-\ddot{x}\dot{y}}{(\dot{x}^2+\dot{y}^2)^{3/2}}$ is the counter-clockwise curvature. – robjohn Sep 26 '11 at 13:08
@Arjang: I was simply verifying Christian's statement that $\dot{x}\ddot{y}-\ddot{x}\dot{y}$ has the same sign as the counter-clockwise curvature (positive when the curve is veering left, i.e. moving counter-clockwise). Are you asking if more is needed? – robjohn Sep 26 '11 at 15:40