## What is Curvature? (II)

[ Background required: rudimentary vector calculus. ]

The aforementioned definition of curvature is practical but a little aesthetically displeasing. Specifically, one seeks a definition which is independent of the parametrization of the curve. The advantage of such a definition is not merely conceptual clarity, but also the ease at which we can generalise to higher dimensions.

Suppose the curve is given by $\mathbf x(t) = (f(t), g(t))$, where all vectorial values are given in boldface. One imagines this as a particle travelling along the curve such that at time t, its position is at x(t). Recall that the velocity of the particle is then given by $\mathbf{x}'(t) = (f'(t), g'(t))$ which is parallel to the gradient of the curve at x(t). Thus the speed of the particle is the magnitude of its velocity, $||\mathbf{x}'(t)|| = \sqrt{f'(t)^2 + g'(t)^2}$, from which it’s clear that the distance travelled by the particle during a time interval [a,b] is given by: $s = \int_a^b || \mathbf{x}'(t) || dt = \int_a^b \sqrt{f'(t)^2 + g'(t)^2} dt.$

Example: consider the parabola $y = x^2$ parametrized via $(x, x^2)$. The arc length of the curve from (0,0) to (1,1) is given by $s = \int_0^1 \sqrt{1 + 4x^2} dx$ which is $\frac 1 4 (2\sqrt 5 + \sinh^{-1}(2))$, i.e. around 1.479, which is only slightly longer than the straight line joining the two points.

Exercise: find the length of the curve $y = x^{3/2}$ from (0,0) to (1,1).

Exercise: for the curve in polar coordinates $r = f(\theta)$, find the formula of the arc-length.

Now we’re ready to state our definition of curvature “intrinsically” (i.e. without reference to external coordinates or parametrization). At each point of time t, take the unit tangent vector at the curve, i.e. let T(t) be the unit vector along $\mathbf{x}'(t)$. Since regardless of parametrization, $\mathbf{x}'(t)$ is always parallel to the tangent at the curve at x(t), we see that T(t) depends only on the point x(t) and not on the parametrization t. [ E.g. if we let t = 2u, then the new parametrization x(u) would be “twice as fast”: from u=0 to u=1, the particle would have traversed the path from t=0 to t=2. ]

Next, we fix a standard parametrization of the curve: namely via the arc length s, i.e. we assume that the particle is travelling along the curve at unit speed.

Example. Consider the semi-circle in the upper-half plane $y = \sqrt{25 - x^2}$. Let’s calculate its parametrization by arc-length s, via the hard way. [ The easy way is via polar coordinates. ] Starting from the point (-5, 0), the arc-length to the point $(a, f(a))$ is given by: $\int_{-5}^a \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx = \int_{-5}^a \frac {5} {\sqrt{25-x^2}} dx = 5(\frac\pi 2 + \sin^{-1} (\frac a 5)).$

This gives the parametrization $x(s) = 5\sin(\frac s 5 - \frac \pi 2) = -5 \cos \frac s 5$, $y(s) = 5\sin \frac s 5$, where $0 \le s \le 5\pi$. This is consistent with our intuition that the arc-length is a linear function of the angle θ.

Exercise. Take the curve $y= \frac 2 3 x^{3/2}$. Parametrize the curve by arc-length, starting from the origin (0, 0).

Exercise. Take the curve defined by the polar equation $r = e^\theta$ and parametrize it by arc-length, starting from $\theta = 0$.

Exercise. (*) Prove that if $\mathbf{x}(s) = (f(s), g(s))$ is parametrized by arc-length, then $||\frac{d\mathbf{x}}{ds}|| = 1$. [ Note: if your concept is right, this should be a 2-line proof. We will need this result later. ]

Now we’re ready to unveil the new definition of curvature.

Intrinsic Definition of Curvature. Let $\mathbf{T}(s)$ be the unit tangent vector of the curve $\mathbf{x}(s)$, parametrized by arc-length. The curvature $\kappa$ of a curve is defined by $||\frac{d\mathbf{T}}{ds}||$.

While we’re at it, we also define the curvature vector to be $\frac{d\mathbf{T}}{ds}$. Our most pressing need is to check that this definition of curvature is consistent with our last. To be specific, for a function $y = f(x)$ we need to show that after parametrizing by arc-length that: $||\frac{d\mathbf{T}}{ds}|| = \frac{f''(x)}{(1+f'(x)^2)^{3/2}}.$

To do that, first rotate and translate the figure so that the graph is tangent to the x-axis at the origin (i.e. $f(0) = f'(0) = 0$). The RHS remains unchanged thanks to the way we derived it. The LHS is intrinsically defined, so it has nothing to do with orientation. Now we have: $\frac{d\mathbf{T}}{dx} = \frac{d\mathbf{T}}{ds} \cdot \frac{ds}{dx}$ and $\frac{ds}{dx} = \sqrt{1 + f'(x)^2}$.

At x=0, we thus have $\frac{ds}{dx} = 1$ and it suffices to compute $\frac{d\mathbf{T}}{dx}$. To do that, note that the vector $(1, f(x))$ is parallel to the tangent of the curve at the point $(x, f(x))$, so we can express the unit tangent vector in terms of x : $\mathbf{T}(x) = \frac 1 {\sqrt{f'(x)^2+1}} (1, f'(x))$.

Now just differentiate both sides wrt x and put x=0: we get $\left.\frac{d\mathbf{T}}{dx}\right|_{x=0} = (0, f''(0))$. So our definition is consistent with the earlier one. Notice that the curvature vector $(0, f''(0))$ is perpendicular to tangent vector (x-axis). This always holds. Generalisation

Armed with this, we can generalise the definition of curvature to a curve in arbitrary dimensions. Let $\mathbf{x} : \mathbb R \to \mathbb R^n$ be a curve in n-space, with components $\mathbf{x}(t) = (x_1(t), x_2(t), \dots, x_n(t))$. Everything follows as before:

• The velocity vector is $\mathbf{v}(t) = \mathbf{x}'(t) = (x_1'(t), x_2'(t), \dots, x_n'(t))$.
• The unit tangent vector of the curve at a point is $\mathbf{T} = \frac{\mathbf{x}'}{||\mathbf{x}'||}$.
• The arc-length along $a \le t \le b$ is $s = \int_a^b ||\mathbf{v}(t)|| dt$, or $s = \int_a^b \sqrt{\sum_{i=1}^n x_i'(t)^2} dt$.

So naturally, we also define the curvature vector to be $\frac{d\mathbf{T}}{ds}$ and the curvature to be its length. This definition is consistent with all our earlier definitions of curvature (for n=2 case).

Example. Consider the helix given by $\mathbf{x}(t) = (\cos t, \sin t, 2t)$. Then the velocity vector is $\mathbf{v}(t) = \mathbf{x}'(t) = (-\sin t, \cos t, 2)$, so the tangent vector is $\mathbf{T}(t) = \frac 1 {\sqrt 5} (-\sin t, \cos t, 2)$. To compute the curvature vector, we apply: $\frac{d\mathbb{T}}{ds} = \frac{d\mathbb{T}}{dt} \left(\frac {ds}{dt}\right)^{-1}$.

And $\frac {ds}{dt} = ||\mathbf{v}(t)|| = \sqrt 5$. Hence, the curvature vector is given by $\frac 1 5 (-\cos t, -\sin t, 0)$. and the curvature is 1/5. In other words, the helix has constant curvature, which is consistent with our geometric figure.

Exercise. Take the curve $\mathbf x(t) = (2e^t \cos t, 2e^t \sin t, e^t)$. Compute the curvature and curvature vector at a general point on the curve.

Application. Take a cylinder with base radius R and height H. A strake is a curved strip which bends around the cylinder, following the shape of its contour: (Image from britannica.com)

What should the radius r be for the strip to wrap around the cylinder effectively? The correct answer is that the curvature must match that of the strake. The graph of the helix is given by: $\mathbf{x}(\theta) = (R \cos\theta, R\sin\theta, h\theta)$, where $0 \le \theta \le 2\pi$ and $h = \frac{H}{2\pi}$. Hence:

• $\mathbf{x}'(\theta) = (-R\sin\theta, R\cos\theta, h)$;
• so $\frac{ds}{d\theta} = ||\mathbf{x}'(\theta)|| = \sqrt{R^2 + h^2}$;
• $\mathbf{T}(\theta) = \frac{\mathbf{x}'}{||\mathbf{x}'||} = \frac 1 {\sqrt{R^2 + h^2}} (-R\sin\theta, R\cos\theta, h)$;
• $\frac{d\mathbf{T}}{d\theta} = \frac 1 {\sqrt{R^2+h^2}} (-R\cos\theta, -R\sin\theta, 0)$;
• so curvature vector is $\frac{d\mathbf{T}}{ds} = \frac{d\mathbf{T}}{d\theta}\left(\frac{ds}{d\theta}\right)^{-1} = \frac 1 {R^2+h^2} (-R\cos\theta, -R\sin\theta, 0)$ and the scalar curvature is $\frac R {R^2 + h^2}$.

So the correct radius for r is $\frac {R^2 + h^2} R$ where $h = \frac{H}{2\pi}$.

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### 1 Response to What is Curvature? (II)

1. Rayna says:

Reblogged this on Rayna's Blog and commented:
A follow-up of an earlier article, this post looks at “curvature” using concepts of path and velocity. The last question on helix and strakes is very interesting.