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Taken the equation of an ellipse

$$x^2/a^2 + y^2/b^2 = 1$$

The parametric equation of an ellipse is usually given as

$\begin{array}{c} x = a\cos(t)\\ y = b\sin(t) \end{array}$

Let's rewrite this as the general form (*assuming a "friendly" shape, i.e. only one point for each radial vector at angle T)

$\begin{array}{c} x = r(t)\cos(t)\\ y = r(t)\sin(t) \end{array}$

where $r(t)$ is the radius at angle $t$. This mimic the idea of a sin/cos pair with variable radius (which is, indeed, an ellipse)

So how can $r(t)=a$ and $r(t)=b$ at the same time?

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  • $\begingroup$ The only way it can be is if $a = b$. The "friendly" shape you wrote doesn't work for a general ellipse because you always scale $x$ and $y$ by the same factor. An ellipse very specifically requires you to scale the $x$ and $y$ coordinate of the unit circle by different amounts. $\endgroup$
    – Zeno
    Jun 25 at 10:03
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    $\begingroup$ You have used the same variable $t$ to refer to two different angles: eccentric angle and ordinary polar angle. If you overloaded same object with multiple meanings, it is very easy to arrive at paradoxical conclusions. $\endgroup$ Jun 25 at 10:25
  • $\begingroup$ @Zeno That's not true? Imagine you center yourself in the center of the ellipse and sweep its "circumference" with a vector. At any point in time, this vector will have a (variable) radius (magnitude) r(t) and if you take the cos and sen of t at that radius (i.e. multiplied by the magnitude of r(t)) you get X,Y of the point on the "circumference". $\endgroup$ Jun 25 at 10:49
  • $\begingroup$ @achillehui Ahhh, that's where the penny drops. Can you formalize the answer and maybe explain the relationship between the polar and eccentric angle so that I can accept it? $\endgroup$ Jun 25 at 10:54
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Let's look at the suggested form: $$\left\lbrace \begin{aligned} x &= r(\theta) \cos(\theta) \\ y &= r(\theta) \sin(\theta) \\ \end{aligned} \right.$$ As Achille Hui mentioned in a comment to the question, this can seem paradoxical, because $\theta$ is not the ordinary polar angle here. We all know that the expected form is $$\left\lbrace \begin{aligned} x &= a \cos(\theta) \\ y &= b \sin(\theta) \\ \end{aligned} \right.$$

Note, "seem paradoxical". Whether one finds it paradoxical or not only depends on what one thinks $\theta$ is. The suggested form is written with $\theta$ as the proper polar angle; it is not that in the expected form, because here, $\theta \ne \arctan(y/x)$ (assuming $a \ne b$, and excepting the four points where $x = 0$ or $y = 0$).

In the expected form, $\theta$ is just an angular parameter, eccentric angle, which is related to the actual polar angle $\varphi$ via, $$\theta = \arctan\left(\frac{a y}{b x}\right) = \arctan\left(\frac{a}{b}\tan\varphi\right) \iff \varphi = \arctan\left(\frac{b}{a}\tan\theta\right)$$

In a polar coordinate system, this ellipse is described by $$r(\varphi) = \frac{a b}{\sqrt{(b \cos\varphi)^2 + (a \sin \varphi)^2}} $$

You can always convert from polar to Cartesian coordinates, getting $$\left\lbrace\begin{aligned} x &= \frac{a b}{\sqrt{(b \cos \varphi)^2 + (a \sin \varphi)^2}} \cos\varphi \\ y &= \frac{a b}{\sqrt{(b \cos \varphi)^2 + (a \sin \varphi)^2}} \sin\varphi \\ \end{aligned}\right.$$ where $\varphi$ is the correct polar angle; $\varphi = \arctan(y/x)$.

The difference between polar angle $\varphi$ and eccentric angle $\theta$ is subtle, but important. It is also quite annoying for us non-mathematicians, because we do not always notice the difference, and get bitten by it.

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  • $\begingroup$ Perhaps it would be more correct to use $p = \max(a, b) / \sqrt{\max(a^2, b^2) - \min(a^2, b^2)}$, or add a comment saying "assuming $a \ge b$ when calculating $p$" or some such, but I think it is already so messy it is hard to understand. If you have cleanups to suggest or apply, please go ahead; this sure could use some! $\endgroup$ Jun 25 at 13:21
  • $\begingroup$ I disagree with your first polar equation, which relates to the pole at a focus, not at the center. $\endgroup$
    – user65203
    Jun 25 at 14:05
  • $\begingroup$ @YvesDaoust: Ouch! Thanks for pointing it out. I'll just remove it, as the second form is the useful one, and also removes the problem I was having with $p$. $\endgroup$ Jun 25 at 15:21
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A classical mistake: $t$ is not the polar angle !

From

$$x=a\cos t,\\y=b\sin t\ $$

you draw the parametric polar equation

$$r(t)=\sqrt{a^2\cos^2t+b^2\sin^2t},$$ $$\theta(t)=\arctan\left(\dfrac ba\tan t\right),$$

or after elimination of $t$,

$$r(\theta)=\sqrt{a^2\cos^2\left(\arctan\left(\frac ab\tan\theta\right)\right)+b^2\sin^2\left(\arctan\left(\frac ab\tan\theta\right)\right)}.$$

Ugly, isn't it ?

enter image description here

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Points set

$$\left\lbrace \begin{aligned} x &= r(\theta) \cos(\theta) \\ y &= r(\theta) \sin(\theta) \\ \end{aligned} \right.\tag 1$$

is an identity concerning any point for all curves.

$$\left\lbrace \begin{aligned} x &= a \cos(\theta) \\ y &= b \sin(\theta) \\ \end{aligned} \right.\tag 2$$ is parametrization of a particular curve, the ellipse where $ \theta $ is not its polar coordinate but of another point that connects projections of segments between circles of radius $(a,b).$

$$\left\lbrace \begin{aligned} x &= a \cos^2(\theta) \\ y &= b \sin^2(\theta) \\ \end{aligned} \right.\tag 3$$

is polar parametrization of a particular case of straight line.

Identities should not be pulled into particular situations, as the context in which the separate entities are first defined is lost or taken out.

In common parlance ( no insinuation implied ) one would say.. it is not known what is being talking about.

Reading (1) and (3) together throws it out of context:

$$ \frac{y}{x}=\frac{a}{b}\; ! $$

which is not the straight line we started with in (3).

Geometric derivation of a particular case of the straight line in polar parameterized form is mentioned even if elementary. For a line making intercepts $(a,b)$ passing through $P(x,y)$

enter image description here $$ \frac{x}{OP}=\frac{OP}{a},\quad OP= a \cos \theta $$

$$ \frac{y}{OP}=\frac{OP}{b},\quad OP= b \sin \theta $$

$$ (x,y)= ( a \cos ^2\theta, b \sin ^2\theta). $$

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