# Orientation of a parametrically-defined circle

I have answered all questions on the sample exam for my Calculus, but for this one I'm not sure.

Determine the curve described by the parametric equations $x = 1 + 2\sin 3t$ and $y = 2\cos 3t - 4$ and its orientation.

Here is my solution:

Solving for $\sin 3t$ and $\cos 3t$, we have \begin{align} x &= 1 + 2\sin 3t & y &= 2\cos 3t - 4 \\ 2\sin 3t &= x - 1 & 2\cos 3t &= y + 4 \\ \sin 3t &= \frac{x-1}{2} & \cos 3t &= \frac{y+4}{2} \end{align} Using the identity $\cos^2 \theta + \sin^2 \theta = 1$, we have \begin{align} \cos^2 3t + \sin^2 3t &= 1 \\ \left(\frac{y+4}{2}\right)^2 + \left(\frac{x-1}{2}\right)^2 &= 1 \\ \frac{(x-1)^2}{4} + \frac{(y+4)^2}{4} &= 1 \\ (x-1)^2 + (y+4)^2 &= 4 \end{align} which is an equation of a circle of radius 2 that has its center at (in rectangular coordinates) $(1,-4)$.

For the orientation, I asserted that it is oriented clockwise because $x$ [$y$] increases [decreases] from $0$ to $\frac{\pi}{6}$, decreases [decreases] from $\frac{\pi}{6}$ to $\frac{\pi}{3}$, decreases [increases] from $\frac{\pi}{3}$ to $\frac{\pi}{2}$, and increases [increases] from $\frac{\pi}{2}$ to $\frac{2\pi}{3}$.

Questions:

1. Do you know the orientation from the changes in the values of $x$ and $y$?
2. What effect does interchanging $x$ and $y$ (assigning the cosine and the sine function to $x$ and $y$, respectively) to the orientation of the circle? I conjectured that it reverses the orientation of a curve. Is this true?
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You know that the usual parametrization with $(\cos\,t,\sin\,t)$ has the circle going anticlockwise, yes? You should be able to see what happens if you swap out the sine and cosine. –  Ｊ. Ｍ. Aug 12 '12 at 11:55
You can see the direction of the tangent vector at, e.g., $A=(x_0+r,y_0)=(1+2,-4)$. –  enzotib Aug 12 '12 at 12:33