Curves and Vector-Valued Functions

So far, the only functions we have dealt with are \(f:\mathbb{R}\rightarrow\mathbb{R}\), sending real numbers to real numbers; to be blunt, apart from the fact that the concept of a limit was rigorously defined using \(\epsilon\)-\(\delta\), there was little difference from what we learned in high school. We now generalize this and examine genuinely new content. The direction of generalization is to increase dimension, where there are two directions: raising the dimension of the domain and raising the dimension of the codomain. In this post we first treat the latter, namely functions \(\mathbf{r}:\mathbb{R}\rightarrow\mathbb{R}^n\) that send one real number to several real numbers. These are vector-valued functions that trace a curve in space with a single parameter, and can be viewed as the locus of a point moving in time; consequently, differentiation acquires the physical meaning of velocity and acceleration, and integration measures the length of a curve.

Vector Spaces

Since a curve is a function that takes one real number and returns a point consisting of several real numbers (that is, a function into \(\mathbb{R}^n\)), we first briefly review the structure of this space. \(\mathbb{R}^n\) is a vector space, and the elements of this space are called vectors. A vector is written as an ordered tuple

\[\mathbf{a}=(a_1, \ldots, a_n)\qquad a_i\in\mathbb{R}\]

and there are two basic operations between vectors. One is addition, adding two vectors \(\mathbf{v}=(v_1,\ldots,v_n)\) and \(\mathbf{w}=(w_1,\ldots,w_n)\),

\[\mathbf{v}+\mathbf{w}=(v_1+w_1,\ldots,v_n+w_n)\]

and the other is scalar multiplication, multiplying by a real number \(c\),

\[c\mathbf{v}=(cv_1,\ldots,cv_n).\]

These operations directly carry over the vector operations seen in the coordinate plane to \(n\) dimensions, and the commutative, associative, and distributive laws naturally hold. In particular, the following vectors

\[\mathbf{e}_1=(1,0,\ldots, 0,0),\ldots, \mathbf{e}_n=(0,0,\ldots, 0,1)\]

are called the standard basis vectors, and using them, any vector can be written as

\[\mathbf{v}=\sum_{i=1}^n v_i \mathbf{e}_i.\]

Since the space we deal with is Euclidean space, we also use the inner product and norm here. The inner product of two vectors \(\mathbf{v},\mathbf{w}\in\mathbb{R}^n\) is defined as the sum of coordinate-wise products,

\[\mathbf{v}\cdot \mathbf{w}=v_1w_1+\cdots+v_nw_n\]

and from this, the norm, that is, the magnitude of a vector, is given by \(\lVert \mathbf{v}\rVert=\sqrt{\mathbf{v}\cdot \mathbf{v}}\). The inner product is also used to measure the angle between two vectors; in particular, when \(\mathbf{v}\cdot \mathbf{w}=0\), the two vectors are said to be orthogonal.

Meanwhile, in three dimensions \(\mathbb{R}^3\), the cross product is additionally defined. For two vectors \(\mathbf{v},\mathbf{w}\in\mathbb{R}^3\), the cross product \(\mathbf{v}\times \mathbf{w}\) is the vector that is orthogonal to both \(\mathbf{v}\) and \(\mathbf{w}\), whose direction follows the right-hand rule, and whose magnitude equals the area of the parallelogram formed by \(\mathbf{v}\) and \(\mathbf{w}\). In coordinates, it is computed as

\[\mathbf{v}\times \mathbf{w}=(v_2w_3-v_3w_2,\ v_3w_1-v_1w_3,\ v_1w_2-v_2w_1).\]

In this post, we mainly use the cross product as a tool to produce orthogonal vectors.

Vector-Valued Functions

Then, first, we can define the following.

Definition 1 A function on an interval \(I \subseteq \mathbb{R}\) assigning to each \(t\) the point \(\mathbf{r}(t) = (x_1(t), \ldots, x_n(t)) \in \mathbb{R}^n\), \(\mathbf{r}\colon I \rightarrow \mathbb{R}^n\), is called a vector-valued function or parametrized curve, and each \(x_i\) is called its component function.

Intuitively, this is a rule by which a different point in the vector space corresponds as \(t\) varies, and can be thought of as expressing the locus of a point moving in time.

The limit of a vector-valued function is defined in the same way as the limit of a one-dimensional function. That is, whenever any \(\epsilon>0\) is given, if there always exists a suitable \(\delta>0\) such that

\[0<\lvert t-t_0\rvert<\delta\implies \lVert \mathbf{r}(t)-\mathbf{v}\rVert<\epsilon\]

holds, we say this vector function \(\mathbf{r}\) converges as \(t\rightarrow t_0\) to the vector \(\mathbf{v}\). Then by the Cauchy–Schwarz inequality

\[\lvert\mathbf{a}\cdot\mathbf{b}\rvert\leq\lVert\mathbf{a}\rVert\lVert\mathbf{b}\rVert\]

we have

\[\lvert x_i(t)-v_i\rvert=\lvert(\mathbf{r}(t)-\mathbf{v})\cdot\mathbf{e}_i\rvert\leq\lVert\mathbf{r}(t)-\mathbf{v}\rVert\]

so if \(\mathbf{r}(t)\rightarrow\mathbf{v}\) then each component also converges, \(x_i(t)\rightarrow v_i\). Conversely, if all components converge then

\[\lVert\mathbf{r}(t)-\mathbf{v}\rVert^2=\sum_i(x_i(t)-v_i)^2\rightarrow 0\]

so \(\mathbf{r}(t)\rightarrow\mathbf{v}\). That is, convergence of a vector function is equivalent to convergence of each component function, and in particular continuity and differentiability of a vector function are both well defined component-wise.

Proposition 2 If each component function of \(\mathbf{r}(t) = (x_1(t), \ldots, x_n(t))\) is differentiable at \(t\), then \(\mathbf{r}\) is also differentiable at \(t\) and \(\mathbf{r}'(t) = (x_1'(t), \ldots, x_n'(t))\).

Proof

For the difference quotient \((\mathbf{r}(t+h) - \mathbf{r}(t))/h\), the \(i\)-th component is exactly \((x_i(t+h) - x_i(t))/h\). Since the limit of each component exists and equals \(x_i'(t)\), we have \(\mathbf{r}'(t) = (x_1'(t), \ldots, x_n'(t))\).

Moreover, by a similar argument the product rule for scalar functions carries over to vector products.

Proposition 3 (Differentiation Rules) If \(\mathbf{u}, \mathbf{v}\colon I \rightarrow \mathbb{R}^n\) are differentiable, \(f\colon I \rightarrow \mathbb{R}\) is differentiable, and \(\varphi\colon J \rightarrow I\) is a differentiable real function, then

\[(f \mathbf{u})' = f' \mathbf{u} + f \mathbf{u}', \qquad (\mathbf{u} \cdot \mathbf{v})' = \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}', \qquad (\mathbf{u} \circ \varphi)'(t) = \varphi'(t) \mathbf{u}'(\varphi(t))\]

and when \(n = 3\), for the cross product, \((\mathbf{u} \times \mathbf{v})' = \mathbf{u}' \times \mathbf{v} + \mathbf{u} \times \mathbf{v}'\).

Proof

Writing everything out component-wise reduces each identity to the product rule and chain rule for scalar functions. For example, the inner product is \(\mathbf{u} \cdot \mathbf{v} = \sum_i u_i v_i\), so \((\mathbf{u} \cdot \mathbf{v})' = \sum_i (u_i' v_i + u_i v_i') = \mathbf{u}' \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{v}'\); the cross product and scalar multiplication can be shown in the same way since each component is a sum of products of the form \(u_i v_j\). The composition rule follows by applying §Differentiation, ⁋Theorem 4 (Chain rule) to each component.

One useful corollary of this proposition is that if \(\lVert \mathbf{u}(t)\rVert\) is constant, then \(\mathbf{u} \cdot \mathbf{u} = \lVert \mathbf{u}\rVert^2\) is also constant, so \((\mathbf{u} \cdot \mathbf{u})' = 2 \mathbf{u} \cdot \mathbf{u}' = 0\), namely \(\mathbf{u} \perp \mathbf{u}'\). That is, the rate of change of a vector of constant length is always perpendicular to that vector, and the fact that in circular motion the position vector and velocity are perpendicular is a special case of this.

Velocity and Acceleration

As we saw above, vector functions are a good tool for representing physical phenomena. We now explain this physical intuition.

Definition 4 For a curve \(\mathbf{r}(t)\), we call \(\mathbf{r}'(t)\) the velocity, its magnitude \(\lVert \mathbf{r}'(t)\rVert\) the speed, and \(\mathbf{r}''(t)\) the acceleration. At points where \(\mathbf{r}'(t) \neq 0\), we call

\[\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\lVert \mathbf{r}'(t)\rVert}\]

the unit tangent vector.

Velocity points in the direction of the curve’s progression and its magnitude is how fast the point moves, while the unit tangent vector retains only the direction of velocity. The condition for this to be well defined is that the parametrized curve \(\mathbf{r}\) is a \(C^1\)-regular curve, which is a curve satisfying both of the following conditions.

1. \(\mathbf{r}\) is \(C^1\). That is, the velocity vector \(\mathbf{r}'(t)\) exists continuously.
2. \(\mathbf{r}\) is regular. That is, there is no point where \(\mathbf{r}'(t)\) is \(0\).

Then, in particular, by the second condition, the unit tangent vector \(\mathbf{T}\) is well defined everywhere, and by the first condition, \(\mathbf{T}(t)\) is continuous. In this post, unless stated otherwise, we take continuously differentiable regular curves as the default.

Given a moving object, one of the things we wish to know is the distance traveled over time. Just as in one dimension, this can be obtained by integrating the speed; rigorously, one approximates the curve by a polygon connecting points on the curve, forms for a partition

\[a = t_0 < \cdots < t_m = b\]

the sum of line-segment lengths

\[\sum_k \lVert \mathbf{r}(t_k) - \mathbf{r}(t_{k-1})\rVert\]

and refines the partition to show that this converges to a Riemann sum of \(\lVert \mathbf{r}'(t)\rVert\). Compressing this process yields the following definition.

Definition 5 The arc length of a continuously differentiable curve \(\mathbf{r}\colon [a, b] \rightarrow \mathbb{R}^n\) is

\[L = \int_a^b \lVert \mathbf{r}'(t)\rVert \dd{t}\]

The integrand \(\lVert \mathbf{r}'(t)\rVert\) is continuous, hence integrable (§Integration, ⁋Theorem 10). Now, we call

\[s(t) = \int_a^t \lVert \mathbf{r}'(\tau)\rVert \dd{\tau}\]

the arc length measured from the starting point; by the fundamental theorem of calculus, \(s'(t) = \lVert \mathbf{r}'(t)\rVert > 0\), so \(s\) is an increasing function, and we can solve for \(t\) in terms of \(s\) to reparametrize the curve by arc length.

Proposition 6 If a curve is parametrized by arc length \(s\), then it has unit speed. That is, \(\lVert d\mathbf{r}/\dd{s}\rVert = 1\).

Proof

By the chain rule, \(d\mathbf{r}/\dd{t} = (d\mathbf{r}/\dd{s})(\dd{s}/\dd{t})\) and \(\dd{s}/\dd{t} = \lVert \mathbf{r}'(t)\rVert = \lVert d\mathbf{r}/\dd{t}\rVert\), so \(\lVert d\mathbf{r}/\dd{s}\rVert = \lVert d\mathbf{r}/\dd{t}\rVert / (\dd{s}/\dd{t}) = 1\).

The parametrization obtained in this way is called arc length parametrization.

Curvature

How much a curve bends is measured by how quickly the direction of the unit tangent vector changes as we proceed along the curve. However, if the curve moves quickly, the rate of change will also increase, so to examine how much the shape of the curve is bent we must make this speed the same. Moreover, intuitively, this is obtained by differentiating the unit tangent vector (already the derivative of the original curve) once more, so we must now consider twice continuously differentiable curves. That is, derivatives up to the acceleration vector \(\mathbf{r}''(t)\) exist continuously.

Definition 7 The curvature of a twice continuously differentiable regular curve is the magnitude of the rate at which the unit tangent vector changes with respect to arc length:

\[\kappa = \left\lVert \frac{d\mathbf{T}}{\dd{s}}\right\rVert\]

When \(d\mathbf{T}/\dd{s} \neq 0\), the unit vector in that direction

\[\mathbf{N} = \frac{d\mathbf{T}/\dd{s}}{\lVert d\mathbf{T}/\dd{s}\rVert}\]

is called the unit normal vector.

Note that by Proposition 6, \(\mathbf{T} = d\mathbf{r}/\dd{s}\) is a unit vector. Then by the observation examined right after Proposition 3 (Differentiation Rules), that the rate of change of a vector of constant length is perpendicular to that vector, we have \(d\mathbf{T}/\dd{s} \perp \mathbf{T}\). That is, the unit normal vector \(\mathbf{N}\) is always perpendicular to the tangent, and points to the inside into which the curve is bending.

The above definition applies to curves in general \(n\)-dimensional space, but in \(3\)-dimensional space the cross product allows us to compute this more conveniently. In particular, the following formula can be applied directly without arc length parametrization, which is far more convenient.

Proposition 8 (Curvature Formula) In \(3\)-dimensional space, the curvature of a second-order regular curve \(\mathbf{r}(t)\) is

\[\kappa = \frac{\lVert \mathbf{r}' \times \mathbf{r}''\rVert}{\lVert \mathbf{r}'\rVert^3}\]

and in particular, the curvature of a plane curve \(y = f(x)\) is

\[\kappa = \frac{\lvert f''\rvert}{(1 + f'^2)^{3/2}}\]

Proof

Let \(v = \lVert \mathbf{r}'\rVert = \dd{s}/\dd{t}\); then \(\mathbf{r}' = v\mathbf{T}\). By the product rule,

\[\mathbf{r}'' = v'\mathbf{T} + v\mathbf{T}'(t)\]

and from the chain rule \(\mathbf{T}'(t) = (d\mathbf{T}/\dd{s})v\) and \(\lVert d\mathbf{T}/\dd{s}\rVert = \kappa\), we have \(\lVert \mathbf{T}'(t)\rVert = \kappa v\). Therefore

\[\mathbf{r}' \times \mathbf{r}'' = (v\mathbf{T}) \times (v'\mathbf{T} + v\mathbf{T}'(t)) = v^2(\mathbf{T} \times \mathbf{T}'(t))\]

Here the second equality follows from \(\mathbf{T}\times \mathbf{T}=0\). Now \(\mathbf{T}\) is a unit vector, and by the observation right after Proposition 3 (Differentiation Rules) that the derivative of a vector of constant length is perpendicular to that vector, we have \(\mathbf{T} \perp \mathbf{T}'(t)\), so \(\lVert \mathbf{T} \times \mathbf{T}'(t)\rVert = \lVert \mathbf{T}'(t)\rVert = \kappa v\), and

\[\lVert \mathbf{r}' \times \mathbf{r}''\rVert = v^3 \kappa = \lVert \mathbf{r}'\rVert^3 \kappa\]

yielding the desired equality. The formula for a plane curve \(y = f(x)\) follows by setting \(\mathbf{r}(x) = (x, f(x),0)\) and applying the \(3\)-dimensional formula directly.

Differentiating \(\mathbf{r}' = v\mathbf{T}\), which appeared in the proof, once more gives \(\mathbf{r}'' = v'\mathbf{T} + v\mathbf{T}'(t)\). At points where \(\kappa \neq 0\), since \(d\mathbf{T}/\dd{s} = \kappa\mathbf{N}\), we have \(\mathbf{T}'(t) = \kappa v\mathbf{N}\), so there it is written as \(\mathbf{r}'' = v'\mathbf{T} + \kappa v^2\mathbf{N}\). This equation is deeply meaningful in itself: the acceleration splits into a tangential component and a normal component. Let us record this as a separate proposition.

Proposition 9 (Decomposition of Acceleration) At points of a second-order regular curve where \(\kappa \neq 0\), the acceleration decomposes as

\[\mathbf{r}'' = \frac{\dd{v}}{\dd{t}}\mathbf{T} + \kappa v^2\mathbf{N}\]

Here \(v = \lVert \mathbf{r}'\rVert\) is the speed.

The tangential component \(\dd{v}/\dd{t}\) indicates how much the speed changes, and the normal component \(\kappa v^2\) indicates how much the direction bends. In uniform motion, \(\dd{v}/\dd{t} = 0\), so the acceleration is purely in the normal direction (that is, centripetal acceleration alone), and its magnitude is \(\kappa v^2\).

Example 10 (Helix) For the helix \(\mathbf{r}(t) = (\cos t, \sin t, t)\), we have \(\mathbf{r}'(t) = (-\sin t, \cos t, 1)\) and \(\mathbf{r}''(t) = (-\cos t, -\sin t, 0)\). The cross product is

\[\mathbf{r}' \times \mathbf{r}'' = (\sin t,\ -\cos t,\ 1), \qquad \lVert \mathbf{r}' \times \mathbf{r}''\rVert = \sqrt{2}\]

and \(\lVert \mathbf{r}'\rVert = \sqrt{2}\), so the curvature is constant at \(\kappa = \sqrt{2}/(\sqrt{2})^3 = 1/2\). Since the speed \(v = \sqrt{2}\) is constant, by Proposition 9 (Decomposition of Acceleration) the acceleration has only a normal component and its magnitude is \(\kappa v^2 = (1/2)\cdot 2 = 1\), which indeed matches \(\lVert \mathbf{r}''(t)\rVert = \lVert(-\cos t, -\sin t, 0)\rVert = 1\).

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References

[Ste] J. Stewart, Calculus, 8th ed., Cengage Learning, 2016.
[Kim] 김홍종, 미적분학 1·2, 제3개정판, 서울대학교출판문화원, 2020.