What is one way to prove that there exists no ellipse that matches the exact curvature of the sin wave?

Question

Preferably by not graphing both and showing they don't match visually. By the sin wave, I mean just plain old y=sinx.

Answer 1

Hint: An ellipse is formed by collecting all of the points where the distances from two foci add to a constant value.

While it is possible to approximate the sine curve using an ellise, the sine curve will soon spread apart from the ellipse.

Answer 2

An ellipse has all possible slopes, from $-\infty$ to $\infty$. The slope of a sine wave is limited from $-1$ to $1$.

Answer 3

Both the sine wave and (a part of) an ellipse are given by analytic functions. Hence they are either identical or do not coincide over any open interval.

Answer 4

For an ellipse passing through a particular point, the first 4 (I think that's right) derivatives at that point will uniquely determine the ellipse. By comparing the the first 5 derivatives of sin at a particular point you could show sin diverges from the 'local ellipse' in the 5th derivative.

Answer 5

By reduction to the absurd: Assume that the opposite were true. But first, let us shift the graphic of the sine function by $\dfrac\pi2$ to the right, so that its shape around $x=0$ might resemble that of a half ellipse centered in the origin. Now, on one hand, the shifted sine function is nothing else than $\cos x$ since $\sin\bigg(x+\dfrac\pi2\bigg)=\cos x$. On the other hand, an ellipse centered in the origin has the property that $\bigg(\dfrac xa\bigg)^2+\bigg(\dfrac yb\bigg)^2=1$. Here, $y=\cos x$. Making $x=0$, we deduce that $b=1$, since $\cos0$ is $1$. Then, by subtracting $\cos^2x$ from both sides, and extracting the radical, we have $\sin x=\dfrac xa$ for x around $0$, implying that the sine function is linear around the origin$(!)$. But $\sin x=2\sin\dfrac x2\cos\dfrac x2$ for all x, implying $\dfrac xa=2\dfrac x{2a}\sqrt{1-\bigg(\dfrac x{2a}\bigg)^2}$ for all x in the vicinity of $0$, which is absurd. QED.

Answer 6

As I said in a comment, if you're looking for an open interval where the curves are identical, you can just express both as functions $f(x),g(x)$, examine $f(x)-g(x)$, and see if there exists a choice of $a,b>0$ that produces $f-g=0$. But that's not how we define curvature, atleast not in the class I took.

Curvature is a well defined property of a curve passing through a point: $\kappa = 1/R$ where R is the radius of the circle that best approximates the curve at a point. What your question boils down to is whether or not there is a point where this best approximating ("osculating") circle is the same for an ellipse and a sine.

Just thought I'd point that out, since most answers thus far has implied that curvature is a property of an INTERVAL and not a point, which is a hugely important distinction. Read more about osculating circles at http://en.wikipedia.org/wiki/Curvature, they're very interesting objects.