[ Background required: calculus, specifically differentiation. ]

In this post, we will give a little background intuition on the definition of curvature. One possible approach is given in wikipedia, ours is another. Note that this is not IMO-related (my apologies for that), but an interesting supplement to your usual calculus.

Let be a smooth (meaning: infinitely differentiable, which we will always assume implicitly from now onwards) function, whose graph then forms a one-dimensional curve in . Recall that the first derivative tells us the gradient of the curve at a point. To be precise, at the point , the value is exactly the gradient of the curve at .

Now we can differentiate the function again: . In secondary school calculus, we have little use for this value, except to determine whether a curve is a local maximum, minimum or stationary point. However, it’s quite clear on an intuitive level that tells us how much the curve turns at , as can be seen from the graph of for different values of *a*: as *a* > 0 increases, the curve bends more and more at the origin (0,0).

Curvature (First Attempt at Definition). The curvature of the graph for at the point is given by the value of .

**Exercise**. Let *C* be the circle with centre (0, *r*) and radius *r*, where *r*>0. Prove that the curvature of the circle at the point (0, 0) is 1/*r*.

This looks appealing since a circle ought to “curve less” as its radius increases. But now we encounter a problem: under our above definition, the second derivative changes as we vary the point on the circle (try it!), whereas our intuition tells us that the curvature of a circle ought to be constant since it seems to be “curving uniformly” throughout the circumference.

Curvature (Second Attempt at Definition). Suppose passes through the origin and is tangent to the x-axis (i.e. ). Then the curvature is defined by . In the general case, if we wish to define the curvature at a point, rotate and translate the curve so that the point maps to the origin and the curve is tangent to the x-axis. Now apply the previous definition.

Let’s explicitly calculate the curvature for a general point. Since translation doesn’t affect the 2nd derivative, we might as well assume the point is at origin. If the tangent at makes an angle of *θ* with the *x*-axis, then . To rotate this an angle of , we use the transformation

Substituting into and partial differentiating with respect to *u* (noting that *θ* is constant) gives:

[ Side remark: note that at since , which is what we want. ] Now, differentiating again and putting *u*=0, *v*=0, we get:

Simplifying with and gives the curvature:

which is typically what you see in the definition of curvature. So in short, to define the **curvature** of the graph of at the point , we use the formula

This is also known as the signed curvature. The **unsigned curvature** is usually denoted by .

**Exercises**.

- Compute the curvature of the graph of at a general point.
- Prove that the curvature of the circle at any point is equal to .
- Obtain the formula for the curvature when the curve is expressed parametrically as . Use this to obtain a simpler proof of 2.
- Find the curvature of the logarithmic spiral whose polar equation is , at various points.
- Use problem 3 to find the maximum and minimum curvatures of the ellipse . Locate where they occur.

In the next installation, we shall examine a conceptually clearer definition of curvature. But we will need the reader to know a bit of vector calculus.

Reblogged this on Rayna's Blog and commented:

A short article about how to derive the concept of “curvature” from second-order derivatives