What is Curvature? (I)

[ 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 y = f(x) 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 \mathbb R^2. Recall that the first derivative f'(x) = \frac {dy}{dx} tells us the gradient of the curve at a point. To be precise, at the point x = x_0, the value f'(x_0) is exactly the gradient of the curve at (x_0, f(x_0)).

Now we can differentiate the function again: f''(x) = \frac {d^2 y}{dx^2}. 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 f''(x_0) tells us how much the curve turns at (x_0, f(x_0)), as can be seen from the graph of y = ax^2 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 y = f(x) at the point (x_0, f(x_0)) is given by the value of f''(x_0).

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 y = f(x) passes through the origin and is tangent to the x-axis (i.e. f(0) = f'(0) = 0). Then the curvature is defined by f''(x). 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 x = 0 makes an angle of θ with the x-axis, then \tan\theta = f'(0). To rotate this an angle of -\theta, we use the transformation

\begin{pmatrix} u \\ v\end{pmatrix} = \begin{pmatrix} \cos \theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \implies \begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix} \begin{pmatrix} u \\ v \end{pmatrix}.

Substituting into y = f(x) and partial differentiating with respect to u (noting that θ is constant) gives:

\sin\theta + \cos\theta \frac{dv}{du} = f'(\cos\theta \cdot u - \sin\theta\cdot v) (\cos\theta - \sin\theta \frac{dv}{du}).

[ Side remark: note that \frac{dv}{du} = 0 at (u,v) = (0,0) since \tan\theta = f'(0), which is what we want. ]  Now, differentiating again and putting u=0, v=0, we get:

\cos\theta \frac{d^2 v}{du^2} = f'(0)(-\sin\theta \frac{d^2 v}{du^2}) + f''(0)(\cos\theta - \sin\theta \frac{dv}{du})^2.

Simplifying with \frac{dv}{du}=0 and \tan\theta = f'(0) gives the curvature:

\frac{d^2 v}{du^2} = f''(0) \cos^3\theta = \frac{f''(0)}{(1 + f'(0)^2)^{3/2}},

which is typically what you see in the definition of curvature. So in short, to define the curvature of the graph of y = f(x) at the point (x_0, f(x_0)), we use the formula

k = \frac{f''(x_0)}{(1 + f'(x_0)^2)^{3/2}}.

This is also known as the signed curvature. The unsigned curvature is usually denoted by \kappa = |k|.


  1. Compute the curvature of the graph of y = x^2 at a general point.
  2. Prove that the curvature of the circle x^2 + y^2 = a^2 at any point is equal to \frac 1 a.
  3. Obtain the formula for the curvature when the curve is expressed parametrically as x = f(t), y = g(t). Use this to obtain a simpler proof of 2.
  4. Find the curvature of the logarithmic spiral whose polar equation is r = e^\theta, at various points.
  5. Use problem 3 to find the maximum and minimum curvatures of the ellipse \frac{x^2} 4 + y^2 = 1. 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.

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One Response to What is Curvature? (I)

  1. Rayna says:

    Reblogged this on Rayna's Blog and commented:
    A short article about how to derive the concept of “curvature” from second-order derivatives

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