# definition – Discrepancies in defining the curvature of a Cartesian function

I was thinking about how one might be able to define how curved a function was. I thought of two ways:

Method 1

The angle of a tangent of a function is $$arctanleft(frac{dy}{dx}right)$$. The rate that the angle of the tangent changes is, thus, $$frac d{dx}left(arctanleft(frac{dy}{dx}right)right)=frac{d^2y}{dx^2}arctan’left(frac{dy}{dx}right)=frac{frac{d^2y}{dx^2}}{1+left(frac{dy}{dx}right)^2}=frac{f”(x)}{1+(f'(x))^2}$$
Hence the curvature of a function is $$frac{f”(x)}{1+(f'(x))^2}$$.

Method 2

We can define the curvature of a circle $$(x-a)^2+(y-b)^2=r^2$$ as having curvature $$frac1r$$. For each point on a function, we find $$a,b,r$$ that approximates the function well around that point, i.e. such that $$(x-a)^2+(y-b)^2=r^2$$ intersects at the point, and has the same first and second derivative as the function at that point, and calculate $$frac1r$$ from there.

The first derivative of a circle is $$frac d{dx}left((x-a)^2+(y-b)^2right)=frac d{dx}r^2Rightarrow frac d{dx}left((x-a)^2right)+frac d{dx}left((y-b)^2right)=0Rightarrow 2frac d{dx}left(x-aright)(x-a)+2frac d{dx}left((y-b)right)(y-b)=0Rightarrow x-a+frac {dy}{dx}(y-b)=frac {dy}{dx}(y-b)=-(x-a)Rightarrowfrac {dy}{dx}=-frac{x-a}{y-b}$$

The second derivative of a circle is $$frac d{dx}left(frac {dy}{dx}right)=frac d{dx}left(frac{a-x}{y-b}right)Rightarrow frac{d^2y}{dx^2}=frac{frac d{dx}(a-x)(y-b)-frac d{dx}(y-b)(a-x)}{(y-b)^2}=frac{-(y-b)-frac d{dx}(y-b)(a-x)}{(y-b)^2}=-frac{y-b+frac{dy}{dx}(a-x)}{(y-b)^2}=-frac{y-b+frac{a-x}{y-b}(a-x)}{(y-b)^2}=-frac{(y-b)^2+(a-x)^2}{(y-b)^3}=-frac{r^2}{(y-b)^3}$$

Thus, we have the following system of equations:

$$begin{cases} (x-a)^2+(y-b)^2=r^2 \ -frac{x-a}{y-b}=y’ \ -frac{r^2}{(y-b)^3}=y” \ end{cases}$$

To solve this equation, we first write $$(y-b)^2=r^2-(x-a)^2$$ and square both sides of the bottom two equations to get:

$$begin{cases} (y-b)^2=r^2-(x-a)^2 \ frac{(x-a)^2}{(y-b)^2}=left(y’right)^2 \ frac{r^4}{(y-b)^6}=left(y”right)^2 \ end{cases}$$

Now replace $$(y-b)^2=r^2-(x-a)^2$$ and multiply the bottom two equations by $$(y-b)^2$$ and $$(y-b)^6$$ respectively:

$$begin{cases} (x-a)^2=left(y’right)^2left(r^2-(x-a)^2right) \ r^4=left(y”right)^2left(r^2-(x-a)^2right)^3 \ end{cases}$$

We can unpack the second equation:

$$(x-a)^2=left(y’right)^2left(r^2-(x-a)^2right)=left(y’right)^2r^2-left(y’right)^2(x-a)^2$$
$$left(1+left(y’right)^2right)(x-a)^2=left(y’right)^2r^2$$
$$(x-a)^2=frac{left(y’right)^2}{1+left(y’right)^2}r^2$$

This leaves us with the equation $$r^4=left(y”right)^2left(r^2-(x-a)^2right)^3$$. Substituting $$(x-a)^2=frac{left(y’right)^2}{1+left(y’right)^2}r^2$$ gives us

$$r^4=left(y”right)^2left(r^2-frac{left(y’right)^2}{1+left(y’right)^2}r^2right)^3$$
$$r^4=left(y”right)^2left(frac1{1+left(y’right)^2}r^2right)^3$$
$$r^4=frac{left(y”right)^2}{left(1+left(y’right)^2right)^3}r^6$$
$$frac{left(y”right)^2}{left(1+left(y’right)^2right)^3}r^6-r^4=0$$
$$r^4left(frac{left(y”right)^2}{left(1+left(y’right)^2right)^3}r^2-1right)=0$$
(It seems that the quadruple root at $$r=0$$ is extraneous, so we can divide by $$r^4$$)
$$frac{left(y”right)^2}{left(1+left(y’right)^2right)^3}r^2-1=0$$
$$frac{left(y”right)^2}{left(1+left(y’right)^2right)^3}r^2=1$$
$$r^2=left(frac{left(y”right)^2}{left(1+left(y’right)^2right)^3}right)^{-1}$$
$$frac1r=sqrt{frac{left(y”right)^2}{left(1+left(y’right)^2right)^3}}=frac{y”}{left(1+left(y’right)^2right)^frac32}$$
Hence the curvature of a function is $$frac{f”(x)}{left(1+left(f'(x)right)^2right)^frac32}$$.

The two methods that I used to define curvature ended up giving different formulae for curvature, a result I was not expecting. According to the Wikipedia article for curvature the formula from Method 2 is the generally accepted formula for finding the curvature of . The definition seems to originally come from so-called “osculating circles”, which seems remarkably similar to what I did in method 2. What I have yet to find, however, is why my first method gave a different result from my second result?

(Something I found note-worthy is that the integral of $$frac{f”(x)}{left(1+left(f'(x)right)^2right)^frac32}$$ is $$frac{f'(x)}{sqrt{1+left(f'(x)right)^2}}$$. Note that the denominator, $$sqrt{1+left(f'(x)right)^2}$$, is the derivative of the arc length of $$f(x)$$. The second formula seems somehow related to arc length and the first one isn’t. Not quite sure what to make of it.)