Math 344: Calculus III

13.3 Arc Length and Curvature

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13.3.1 Length Along a Space Curve

Recall that the length of a plane curve with parametric equations

$$ \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \begin{align} x = f(t) \qquad\qquad\qquad y = g(t) \end{align} $$
on the interval $a\le t\le b$, where $f'(t)$ and $g'(t)$ are continuous is given by

$$ L = \displaystyle\int_a^b \sqrt{\left[f'(t)\right]^2 + \left[g'(t)\right]^2}\,dt = \displaystyle\int_a^b \sqrt{\left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt}\right)^2}\,dt $$

In three dimensions we compute the length of a curve or trajectory the same way. Given the parametric equations of a curve

$$ \begin{align*} x &= f(t) = 4\cos(t) \\ y &= g(t) = 4\sin(t) \\ z &= h(t) = t \end{align*} $$
we can write the vector function in terms of these parametric functions

$$ \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle = \begin{bmatrix} f(t) \\ g(t) \\ h(t) \end{bmatrix} = \begin{bmatrix} 4\cos(t) \\ 4\sin(t) \\ t \end{bmatrix} $$

and divide up the trajectory in a finite number of sub-trajectories by partitioning the domain interval $a=t_0 \lt t_1 \lt t_2 \dots \lt t_n=b$,

compute the tangent vector $\mathbf{r}'(t) = \begin{bmatrix} f'(t) \\ g'(t) \\ h'(t) \end{bmatrix} = \begin{bmatrix} -4\sin(t) \\ 4\cos(t) \\ 1 \end{bmatrix}$

and place the tail of each tangent vector at the corresponding point on the curve.

The length of the tangent vector may be too long or too short so we scale the tangent vector by $\Delta t$. Adding the lengths of these vectors gives a Riemann sum that approximates the length of the curve

$$ \begin{align*} L &\approx \sum_{j=1}^n \left|\mathbf{r}'(t_j)\right|\Delta t \\ \\ &\approx \sum_{j=1}^n \sqrt{\left[f'(t_j)\right]^2 + \left[g'(t_j)\right]^2 + \left[h'(t_j)\right]^2}\ \Delta t \\ \end{align*} $$
Computing the limit as $n\rightarrow\infty$; that is $\Delta t\rightarrow 0$, gives us the definite integral

$$ \begin{align*} L &= \displaystyle\lim_{\Delta t\rightarrow 0} \sum_{j=1}^n \sqrt{\left[f'(t_j)\right]^2 + \left[g'(t_j)\right]^2 + \left[h'(t_j)\right]^2}\ \Delta t \\ \\ &= \displaystyle\int_a^b \sqrt{\left[f'(t)\right]^2 + \left[g'(t)\right]^2 + \left[h'(t)\right]^2}\ dt \\ \\ &= \displaystyle\int_a^b \left|\mathbf{r}'(t)\right|\,dt \end{align*} $$

13.3.2 Smooth Curves

A parameterization of a curve $\mathbf{r}(t)$ is called smooth on an interval $I=[a,b]$ if

• $\mathbf{r}$ is differentiable on interval $I$
• $\mathbf{r}'(t)$ is continuous on interval $I$
• $\mathbf{r}'(t)\neq \mathbf{0}$ on interval $I$

If a function is differentiable and the derivative is continuous on an interval $I=[a,b]$ we say that the curve is continuously differentiable on interval $I$. We denote a function that is continuously differentiable on an interval $I=[a,b]$ by

$$ \mathbf{r}\in\mathbf{C}^1[a,b] $$
Here $\mathbf{C}^1[a,b]$ is the vector space of continuously differentiable curves parameterized over the interval $I=[a,b]$. Thus $\mathbf{r}'(t)$ is a continuous function on the interval $I$, denoted by

$$ \mathbf{r}'\in\mathbf{C}[a,b] $$
Under these conditions, we can discuss two very useful ideas, the first is the parametrization of a curve with respect to arc length and the second is curvature . We need to understand the first before tackling the second, so let us start there.

13.3.3 Arc Length Parametrization

We may define the arc length function as

$$ s(t) = \int_a^t \left|\mathbf{r}'(\tau)\right|\,d\tau $$

Here $\tau$ is just a dummy variable for the integral, we needed to pick something other than $t$ or $s$.

This function measures the length along the curve of $\mathbf{r}$ in the interval from $[a,t]$. If you imagine a car driving on a road from one point to another, $s(t)$ would tell you how many miles are added to the measurement on the vehicle's odometer as $t$ (the time traveled) increases. If we have a formula for $s(t)$, we can use this expression to write $\mathbf{r} = \mathbf{r}(t(s))$. This expression gives a one-to-one correspondence to the increase of the parameter (independent variable) of the vector function and the distance traveled along the space curve.

Example 13.3.1

Given the helix $\mathbf{r}(t) = t\,\ihat + 3\cos t\,\jhat + 3\sin t\,\khat$, reparametrize the curve with respect to arc length measured from $(0,3,0)$ in the direction of increasing $t$.

Solution

First, we notice that the point $(0,3,0)$ is when $t=0$. Next, we need to find the arc length function

$$ s = s(t) = \int_0^t \left|\mathbf{r}'(\tau)\right|\,d\tau $$
We compute

$$ \begin{align*} \left|\mathbf{r}'(\tau)\right| &= \sqrt{\left[(t)'\right]^2 + \left[(3\cos t)'\right]^2 + \left[(3\sin t)'\right]^2} \\ \\ &= \sqrt{\left[(1)'\right]^2 + \left[-3\sin t\right]^2 + \left[3\cos t\right]^2} \\ \\ &= \sqrt{1^2 + 3^2\left(\sin^2 t + \cos^2 t\right)} \\ \\ &= \sqrt{10} \end{align*} $$
Hence,

$$ s = \int_0^t \sqrt{10}\,d\tau = \sqrt{10}t $$
which implies that $t = \frac{s}{\sqrt{10}}$, and we may reparametrize $\mathbf{r}$ as

$$ \mathbf{r}(t(s)) = \left(\frac{s}{\sqrt{10}}\right)\,\ihat + 3\cos \left(\frac{s}{\sqrt{10}}\right)\,\jhat + 3\sin \left(\frac{s}{\sqrt{10}}\right)\,\khat $$

13.3.4 Curvature

Definition

The curvature of a curve is defined in terms of its arc length and the unit tangent vector:

$$ \kappa = \left|\dfrac{d\mathbf{T}}{ds}\right| $$

While this is the definition, it is in general a difficult expression to work with, so we will develop a couple of alternatives parametrized with respect to $t$. First, we think about

$$ \dfrac{d\mathbf{T}}{dt} = \dfrac{d\mathbf{T}}{ds}\dfrac{ds}{dt} $$
and use this to rewrite $\kappa$.

$$ \kappa = \left|\dfrac{d\mathbf{T}}{ds}\right| = \left|\dfrac{\dfrac{d\mathbf{T}}{dt}}{\dfrac{ds}{dt}}\right| = \dfrac{\left|\mathbf{T}'(t)\right|}{\left|\mathbf{r}'(t)\right|}$$
The expression in the denominator is found by applying the Fundamental Theorem of Calculus to the arc length function

$$ \dfrac{d}{dt} \left[s(t)\right] = \left(\int_a^t \left|\mathbf{r}'(\tau)\right|\,d\tau\right)' = \left|\mathbf{r}'(t)\right| $$

Example 13.3.2

Find the curvature of a circle of radius $a$.

Solution

The parametrization of a circle of radius $a$ is

$$ \mathbf{r}(t) = a\cos t\,\ihat + a\sin t\,\jhat $$
We can easily find that $\mathbf{r}'(t) = -a\sin t\,\ihat + a\cos t\,\jhat$ and hence $\left|\mathbf{r}'(t)\right| = a$. This means that the unit tangent vector is

$$ \mathbf{T}(t) = \dfrac{\mathbf{r}'(t)}{\left|\mathbf{r}'(t)\right|} = -\sin t\,\ihat + \cos t\,\jhat $$
and

$$ \mathbf{T}'(t) = -\cos t\,\ihat - \sin t\,\jhat $$
The magnitude of this vector $\left|\mathbf{T}'(t)\right| = 1$ and so the curvature is

$$ \kappa(t) = \left|\dfrac{d\mathbf{T}}{ds}\right| = \dfrac{1}{a} $$
The meaning of this is very important. First, circles have constant curvature. This matches our intuition. Furthermore, for large circles (with large $a$) curvature is small . The larger the circle, the less it curves at a particular point. This is why the Earth seems locally flat to us. The radius of the Earth is approximately $6,300 \text{ km}$. The corresponding curvature is $1.58\times 10^{-4}$. Compared to something like a basketball whose radius is $12 \text{ cm}$ and curvature is $8.\overline{3}$. That is a difference of 5 orders of magnitude.

Another arc length formula, which does not require the direct computation of $\mathbf{T}$ is

$$ \kappa(t) = \dfrac{\left|\mathbf{r}'(t)\times\mathbf{r}''(t)\right|}{\left|\mathbf{r}'(t)\right|^3} $$
This formula utilizes the first and second derivatives of $\mathbf{r}$, and may be easier to compute for some problems. Use the curvature formula that you believe is the most convenient.

Example 13.3.3

Find the curvature function for

$$ \mathbf{r}(t) = t\,\ihat + t^2\,\jhat + e^t\,\khat $$

Solution

Since we want a general curvature formula for any $t$, we will compute

$$ \kappa(t) = \dfrac{\left|\mathbf{r}'(t)\times\mathbf{r}''(t)\right|}{\left|\mathbf{r}'(t)\right|^3} $$
Let us start by finding each quantity, computing the cross product, then finding the necessary magnitudes.

$$ \begin{align*} \mathbf{r}'(t) &= 1\,\ihat + 2t\,\jhat + e^t\,\khat \\ \\ \mathbf{r}''(t) &= 0\,\ihat + 2\,\jhat + e^t\,\khat \end{align*} $$
Now we want the cross product

$$ \begin{align*} \mathbf{r}'(t)\times\mathbf{r}''(t) &= \left(1\,\ihat + 2t\,\jhat + e^t\,\khat\right)\times\left(0\,\ihat + 2\,\jhat + e^t\,\khat\right)\\ \\ &= \begin{vmatrix} \ihat & \jhat & \khat \\ 1 & 2t & e^t \\ 0 & 2 & e^t \end{vmatrix} \\ \\ &= \begin{vmatrix} 2t & e^t \\ 2 & e^t \end{vmatrix}\,\ihat - \begin{vmatrix} 1 & e^t \\ 0 & e^t \end{vmatrix}\,\jhat + \begin{vmatrix} 1 & 2t \\ 0 & 2 \end{vmatrix}\,\khat \\ \\ &= \left(2te^t - 2e^t\right)\,\ihat + \left(e^t - 1\right)\,\jhat + 2\,\khat \end{align*}$$
We now need the magnitude expressions for the numerator and denominator.

$$ \begin{align*} \left|\mathbf{r}'(t)\times\mathbf{r}''(t)\right| &= \sqrt{ \left(2te^t - 2e^t\right)^2 + \left(e^t - 1\right)^2 + 2^2} \\ \\ &= \sqrt{\left(4t^2e^{2t} - 8te^{2t} + 4e^{2t}\right) + \left(e^{2t} - 2e^t + 1\right) + 4} \\ \\ &= \sqrt{4t^2e^{2t} - 8te^{2t} + 5e^{2t} - 2e^t + 5} \\ \\ \left|\mathbf{r}'(t)\right| &= \sqrt{(1)^2 + (2t)^2 + (e^t)^2} \\ \\ &= \sqrt{e^{2t} + 4t^2 + 1} \end{align*} $$
Hence,

$$ \kappa(t) = \dfrac{\sqrt{4t^2e^{2t} - 8te^{2t} + 5e^{2t} - 2e^t + 5}}{\left(e^{2t} + 4t^2 + 1\right)^{\frac{3}{2}}} $$

13.3.5 Normal and Binormal Vectors

Along a space curve, we consider three main vectors of interest. We have discussed the tangent vector $\mathbf{T}(t)$ extensively already, but there are two others of note: the normal vector $\mathbf{N}(t)$ and the binormal vector $\mathbf{B}(t)$. These vectors can be thought of in a few ways, but one of particular usefulness is thinking of them as a "mobile" coordinate system traveling along the curve. Consider the figure Here you can clearly see the coordinate axes with respect to the curve. There are applications, such a traveling airplane, where this notion of the coordinate system may prove useful. For now, let us just consider the geometry of the vectors. First, we start with the tangent vector. It gives us the direction of the instantaneous velocity along the curve. The normal vector represents (part of) the acceleration direction of the curve. If you place your hand in the position to use the right hand rule, with your index finger extended forward, your thumb outward and you middle finger normal to your palm. You can see how each of the digits on your hand correspond to the tangent (index), normal (middle), and binormal (thumb) vectors.

The normal vector is defined as

$$ \mathbf{N}(t) = \dfrac{\mathbf{T}'(t)}{\left|\mathbf{T}'(t)\right|} $$
since we want a vector that is always orthogonal to $\mathbf{T}$. Previously, for a parametrization of a smooth curve $\mathbf{r}$ we specified that $\mathbf{r}'(t)\neq \mathbf{0}$. This condition contributes to the definition of $\mathbf{N}$ in the following way:

Example 13.3.4

Suppose that $\mathbf{u}(t)$ is a vector function such that $\left|\mathbf{u}(t)\right| = c \in\mathbb{R}$ for all $t$, then $\mathbf{u}'(t)$ is orthogonal to $\mathbf{u}(t)$.

Solution

We start with

$$ \mathbf{u}(t)\cdot\mathbf{u}(t) = \left|\mathbf{u}(t)\right|^2 = c^2 $$
If we differentiate this expression with respect to $t$ and apply the product rule, we have

$$ 0 = \dfrac{d}{dt}\left[c^2\right] = \dfrac{d}{dt}\left[\mathbf{u}'(t)\cdot\mathbf{u}'(t)\right] = \mathbf{u}'(t) \cdot \mathbf{u}(t) + \mathbf{u}(t) \cdot \mathbf{u}'(t) = 2\mathbf{u}'(t)\cdot\mathbf{u}(t) $$
which implies that $\mathbf{u}(t)$ and $\mathbf{u}'(t)$ are orthogonal vectors (since $\mathbf{u}'(t)\neq\mathbf{0}$).

This is why we define $\,\mathbf{N}(t)$ like we do, as the unit vector in the direction of $\,\mathbf{T}'(t)$. The above fact guarantees that for $\mathbf{r}'(t)\neq 0$ we have an orthogonal vector.

The binormal vector is required to complete the collection of three vectors in 3D space. It is defined as

$$ \mathbf{B}(t) = \mathbf{T}(t)\times\mathbf{N}(t) $$
It has the properties that it is orthogonal to both the unit tangent and unit normal vectors. Additionally, it is itself a unit vector since the cross product of two orthogonal unit vectors is a unit vector.

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