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Consider the following planar curve, given in terms of the coordinates $(X,Z)$. (This is the curve formed by the intersection of the surface $z=\frac{x^2}{a^2}+\frac{y^2}{b^2}$ with the vertical plane $-x\sin\theta+y\cos\theta=0$, and then setting $Z=z$ and $x=X\cos\theta$).

$$Z=X^2\left[\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right]$$

I want to find the curvature of this planar curve at the point $Z=X=0$, and then to find it's maximum and minimum values as $\theta$ varies, where $a\gt b\gt0$.

Here is my attempt so far: we can find a parametrisation for the above curve since it is of the form $Z=Z(X)$, which yields the parametrisation $X\to\begin{pmatrix}X\\Z(X)\\ \end{pmatrix}=:\mathbf c$. Then we can find $\dot{\mathbf c}=\begin{pmatrix}1\\Z'(X)\\ \end{pmatrix}$ and $\ddot{\mathbf c}=\begin{pmatrix}0\\Z''(X)\\ \end{pmatrix}$.

So then, recalling that for such a parametrisation we have $\kappa(X)=\frac{Z''(X)}{[1+(Z'(X))^2]^{3/2}}$, we get that:

$$\kappa(X)=\frac{2\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)}{\left[1+4X^2\left(\frac{\cos^2\theta}{a^2}\frac{\sin^2\theta}{b^2}\right)\right]^{3/2}}$$

So that at $X=Z=0$,

$$\kappa(X)=2\left(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}\right)$$

From which it is fairly simple to find the maximum and minimum.

My query: looking over the solutions it seems that I have the right idea, except that there, curvature is given with respect to the parameter $\theta$ as opposed to $X$. What is the significance of choosing $\theta$ over $X$ as the parameter with which to parametrise the given curve? Is there a right or wrong way to choose a parameter when several are involved? Should I have written $\kappa(X,\theta)$ instead of choosing either $X$ or $\theta$? As far as I can see, if I chose to parametrise in terms of $\theta$ over $X$ I wouldn't get the same $c$ it's subsequent derivatives.

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As you have stated your problem, $\theta$ is not a valid choice of parameter for the curve obtained by intersecting the surface with the vertical plane $-x \sin(\theta) + y \cos(\theta)$: the curve lies in that plane, $\theta$ is constant on that plane, so $\theta$ is constant on that curve, and a quantity which is constant on a curve cannot be a parameter for that curve. If you change the value of $\theta$, then the vertical plane changes, and its intersection with the surface is an entirely different curve.

But to answer your question, while there are correct parameters for a curve (such as $X$) and incorrect parameters for that curve (such as $\theta)$, there is no single correct choice of parameterization of a curve. For example, if the curve is parameterized as $t \mapsto (x(t),y(y))$ for $a<t<b$, and if $f : (c,b) \to (a,b)$ is any smooth bijection with smooth inverse, then the curve can also be parameterized as $s \mapsto (x(f(s)),y(f(s)))$.

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  • $\begingroup$ That makes perfect sense, and I failed to see it like that, thanks. With regards to the expression I have as $\kappa(X)$ am I right in saying that I have calculated this correctly? Or should this be $\kappa(\theta)$? This is what I have in my solutions ($\kappa(\theta))$, but from what you have said, and having moved to find $\kappa(X)$ it doesn't seem it would ever have made sense to find $\kappa(\theta)$. $\endgroup$ Apr 22 '18 at 20:21
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    $\begingroup$ Rather than go through your calculation, I'll just point out that the expression $\kappa(X)$ does not make sense. Think about it like this. Since for each $\theta$ you have a different curve, and since you are evaluating the curvature of that curve at the point $X=0$, $Z=0$, it follows that $\kappa$ depends on $\theta$, it has no dependence on $X$. So, since $\kappa$ is a function of $\theta$, it should be written $\kappa(\theta)$. $\endgroup$
    – Lee Mosher
    Apr 22 '18 at 21:10

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