## 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 $\kappa = |k|$.

Exercises.

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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### 1 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