What is the largest circle that fits in $\sin(x)?$

Imagine dropping a circle into the trough of $\sin(x)$. Would it reach the bottom or get wedged between two points on the curve? Depends on the size of the circle.

So, what is the radius of the largest circle that will reach the bottom of the curve $y=\sin(x)$?

This problem was inspired by this similar one I found in a calculus textbook: Find the radius of the largest circle that will reach the bottom of the curve $y=x^2$ without getting stuck. I was intrigued by the answer of $r=1/2$.

I have tried attacking this problem from several angles but all have failed. Perhaps an exact numerical solution is not obtainable, I don't know. All help will be appreciated. Thanks!

The largest circle that fits the (sufficiently smooth) curve at given point is called the osculating circle, its radius is the inverse of the curvature. In case of $\sin x$, the curvature is

$$\frac{|\sin x|}{(1+\cos^2 x)^{\frac{3}{2}}}$$

which happens to be $1$ exactly at points, where $|\sin x| = 1$. Hence, the answer is the inverse, namely $1$.

To argue it actually fits, consider the following curves: \begin{align} t &\mapsto \Bigg(t,\ \sin\Big(t+\frac{3}{2}\pi\Big)\Bigg) = (t,-\cos t), \tag{$\spadesuit$}\\ t &\mapsto \Bigg(\sin t,\ \sin\Big(t+\frac{3}{2}\pi\Big)\Bigg) = (\sin t,-\cos t). \tag{$\clubsuit$} \end{align}

The first is the sine, and the second is the osculating circle at $t = 0$. However, $|t| \geq |\sin t|$ so the $(\spadesuit)$ is stretched more than $(\clubsuit)$, hence the circle is always (weakly) above the sine.

I hope this helps $\ddot\smile$

The first derivative of $y = - \sqrt {r^2 - x^2}= - (r^2 - x^2)^{1/2}$ is $y' = dy/dx = x \,(r^2 - x^2)^{-1/2}.$ The second derivative is $$y'' = r^2 \, (r^2 - x^2)^{-3/2}= \frac{r^2}{(r^2 - x^2)^{3/2}}.$$ At $x=0$ this is $1/r.$

The second derivative of $y = \sin x$ at $x = -\pi/2$ is $1.$

Alright, so the answer for sine is $r=1.$ For $y=x^2$ it is $1/2$ because the second derivative of $x^2$ is 2.

You should read up on curvature of plane curves; http://en.wikipedia.org/wiki/Curvature#Curvature_of_plane_curves

Some aspects of this are not obvious...If I start with the pointy end of an ellipse, this all works correctly, matching curvature by matching second derivative of graph. If I try that at the shallow side part of an ellipse, the matching circle is actually outside the ellipse, or may intersect it at other points. You picked problems without extra complications of that sort. In particular, the absolute value of the "curvature" is maximal at the points in question, for both the sine wave and the parabola. That is, locally, enough information. Together with some careful pictures in the large, that is all that is needed.

Nice question.

If we are dropping the circle so that its center is right above $\frac{3\pi}{2}$ then the coordinates of the center of the dropped circle will be $(x_0,y_0) = (3\pi/2,r-1)$ where $r$ is the radius of the circle. Now, the distance from this point to any point $(x,y)$ on the curve of $\sin$ is $$ \sqrt{(x_0 - x)^2 + (y_0 - y)^2} = \sqrt{ (3\pi/2-x)^2 + (r-1)^2 - 2(r-1)\sin x + \sin^2 x}, $$ and this can be no less than $r$. The resulting inequality is $$ 2r(1 + \sin x) \leq (3\pi/2-x)^2 + 1 + 2\sin x + \sin^2 x $$ which leads to $ r \leq ((3\pi/2-x)^2 + (1+\sin x)^2)/2(1+\sin x)$. Then the answer should be \begin{align} r &= \inf \{ ((3\pi/2-x)^2 + (1+\sin x)^2)/2(1+\sin x) \mid x \in (0,\pi) \} \\ &= \frac{1}{2}\inf \{ (3\pi/2-x)^2/(1+\sin x) + (1+\sin x) \mid x \in (0,\pi) \}. \end{align} Let the function in the infimum be $f$. Differentiating yields $$ \frac{-2(3\pi/2 - x)(1+\sin x) - \cos x(3\pi/2-x)^2}{(1+\sin x)^2} + \cos x $$ and if you solve for where this is $0$ you get $$ \cos x((1 + \sin x)^2 - (3\pi/2-x)^2) = 2(3\pi/2-x)(1+\sin x). $$ One solution is at $x = 3\pi/2$, using L'hopital's rule, $$ \lim_{x \to 3\pi/2} f(x) = \frac{1}{2} \lim_{x \to 3\pi/2} \frac{-2(3\pi/2-x)}{\cos x} = \frac{1}{2} \lim_{x \to 3\pi/2} \frac{2}{-\sin x} = 1. $$ This is $r$ (it just remains to show that the critical point is the minimum for $f$, and the first derivative test verifies this).