Finding a local parameterization of a plane curve

Question

I'm attempting to find a parameterization of $\frac{x_1^2}{a^2} + \frac{x_2^2}{b^2} = 1$.

I find a tangent vector field: $X = \left( \frac{2x_2}{b^2}, -\frac{2x_1}{a^2} \right)$ (by taking the gradient, which is orthogonal to the surface, and rotating it by $-\frac{\pi}{2}$).

I'd like an $\alpha(t)$ such that $X(\alpha(t)) = \dot{\alpha}(t)$. Let $\alpha(t) = (\alpha_1(t), \alpha_2(t))$. Then, $X(\alpha(t)) = \left( \frac{2\alpha_2(t)}{b^2}, -\frac{2\alpha_1(t)}{a^2} \right)$, and we have the equations

\begin{align} \dot{\alpha_1}(t) = \frac{2\alpha_2(t)}{b^2} && (1)\\ \dot{\alpha_2}(t) = -\frac{2\alpha_1(t)}{a^2} && (2) \end{align}

Thus,

$$(1) \implies \alpha_2(t) = \frac{b^2 \dot{\alpha_1}(t)}{2} \implies \dot{\alpha_2}(t) = \frac{b^2 \ddot{\alpha_1}(t)}{2}\\ (2) \implies -\frac{2 \alpha_1(t)}{a^2} = \frac{b^2 \ddot{\alpha_1}(t)}{2} \implies \alpha_1(t) = \frac{a^2b^2 \ddot{\alpha_1}(t)}{4} \implies \alpha_1 = \cos{\left( \frac{ab}{2}t \right)}$$

Now, I go the other way, but instead just plug $\dot{\alpha_1}$ into equation $(1)$ to get $\alpha_2(t) = -\frac{ab^3}{4}\sin{\left( \frac{ab}{2}t \right)}$. There's an obvious lack of symmetry here—and when you plug these equations into $X$, the first coordinate works out (that is, $X_1 = \dot{\alpha_1}$), but the second coordinate does not.

I suspect something simple is going on here—it seems like this is a problem related to the "One equation per unknown" thing... but I have two equations, and two unknowns.

Can someone point me in the right direction?

Answer 1

Hint

The solution of the first ODE should read

$$\alpha_1(t)=c_1\cos \left(\frac{2t}{ab}\right)+c_2\sin \left(\frac{2t}{ab}\right).$$ (Thus you don't have broken symmetry.) Note also that you can't expect that a particular solution of such ODE solves a coupled system. That is, you have to consider the general solution.

In a similar way you should obtain

$$\alpha_2(t)=d_1\cos \left(\frac{2t}{ab}\right)+d_2\sin \left(\frac{2t}{ab}\right).$$

Now, it should be easy to get values for the constants.

Answer 2

You have: $$\frac{x_1^2}{a^2} + \frac{x^2_2}{b^2} = 1 \iff \left(\frac{x_1}{a}\right)^2+\left(\frac{x_2}{b}\right)^2 = 1.$$

You also know that: $$\cos^2t + \sin^2t = 1 \iff (\cos t)^2+(\sin t)^2 = 1.$$

Don't you want to try $$\frac{x_1}{a} = \cos t \quad \text{and}\quad \frac{x_2}{b} = \sin t ?$$ That is, ${\bf r}(t) = (a \cos t, b \sin t), 0 \leq t < 2\pi$.