Calculus on Curves

Given any $\mathbf{v} = \begin{bmatrix} v_1 \\ \vdots \\ v_n \end{bmatrix} \in \mathbb{R}^n$ , denote $\displaystyle \mathbf{v} \equiv \sum_{i=1}^n v_i \mathbf{e}_i.$ Given sufficiently differentiable functions $f_i : \mathbb{R} \to \mathbb{R}$ , denote

$\displaystyle \mathbf{f}(t) := \begin{bmatrix} f_1(t) \\ \vdots \\ f_n(t) \end{bmatrix} \equiv \sum_{i=1}^n f_i(t)\, \mathbf{e}_i \equiv \left( \sum_{i=1}^n f_i \mathbf{e}_i \right)(t).$

That is, $\displaystyle \mathbf{f} = \sum_{i=1}^n f_i : \mathbb{R} \to \mathbb{R}^n$ . We call $\mathbf{f}$ a curve.

Definition 1. Define differentiation on vector-valued functions by linearity:

$\displaystyle \mathbf{f}' := \sum_{i=1}^n f_i'\, \mathbf{e}_i, \quad \frac{\mathrm{d}}{\mathrm{d}t}(\mathbf{f}(t)) := \mathbf{f}'(t).$

In what follows, let bold-face letters (e.g. $\mathbf{f}, \mathbf{g}, \mathbf{r}$ ) denote curves $\mathbb{R} \to \mathbb{R}^n$ .

Problem 1. Verify the following differentiation properties for curves.

Linearity: $(\mathbf{f} + \mathbf{g})' = \mathbf{f}' + \mathbf{g}'$ and $(\alpha\mathbf{f})' = \alpha\mathbf{f}'$ for any $\alpha \in \mathbb{R}$ .
Scalar product: $(\phi\mathbf{f})' = \phi'\mathbf{f} + \phi\mathbf{f}'$ for any differentiable $\phi : \mathbb{R} \to \mathbb{R}$ .
Dot product: $(\mathbf{f} \cdot \mathbf{g})' = \mathbf{f}' \cdot \mathbf{g} + \mathbf{f} \cdot \mathbf{g}'$ .
Cross product for n = 3: $(\mathbf{f} \times \mathbf{g})' = \mathbf{f}' \times \mathbf{g} + \mathbf{f} \times \mathbf{g}'$ .

(Click for Solution)

Solution. By Definition 1,

$\displaystyle \begin{aligned} (\mathbf{f}+\mathbf{g})' &= \sum_{i=1}^n (f_i+g_i)' \mathbf{e}_i \\ &= \sum_{i=1}^n (f_i'+g_i') \mathbf{e}_i \\ &= \sum_{i=1}^n (f_i'\mathbf{e}_i+g_i'\mathbf{e}_i) \\ &= \sum_{i=1}^n f_i' \mathbf{e}_i + \sum_{i=1}^n g_i' \mathbf{e}_i = \mathbf{f}'+\mathbf{g}'. \end{aligned}$

The scalar case $(\alpha\mathbf{f})' = \alpha\mathbf{f}'$ follows similarly.

By Definition 1, $(\phi\mathbf{f})_i = \phi f_i$ . By the usual product rule,

$\begin{aligned} (\phi\mathbf{f})' &= \sum_{i=1}^n (\phi f_i)' \mathbf{e}_i \\ &= \sum_{i=1}^n (\phi'f_i + \phi f_i') \mathbf{e}_i \\ &= \sum_{i=1}^n \phi'f_i \mathbf{e}_i + \sum_{i=1}^n \phi f_i' \mathbf{e}_i \\ &= \phi'\sum_{i=1}^n f_i \mathbf{e}_i + \phi \sum_{i=1}^n f_i' \mathbf{e}_i \\ &= \phi'\mathbf{f} + \phi\mathbf{f}'. \end{aligned}$

By the usual dot product,

$\begin{aligned} \mathbf{f}\cdot\mathbf{g} &= \left( \sum_{i=1}^n f_i \mathbf{e}_i \right) \cdot \left( \sum_{j=1}^n g_j \mathbf{e}_j \right) \\ &= \sum_{i=1}^n \sum_{j=1}^n (f_i g_j)\, \mathbf e_i \cdot \mathbf e_j \\ &= \sum_{i=1}^n f_i g_i.\end{aligned}$

The usual product rule gives

$\displaystyle \begin{aligned} (\mathbf{f}\cdot\mathbf{g})' &= \left( \sum_{i=1}^n f_i g_i \right)' \\ &= \sum_{i=1}^n (f_i g_i)' \\ &= \sum_{i=1}^n (f_i' g_i + f_i g_i') \\ &= \sum_{i=1}^n f_i' g_i + \sum_{i=1}^n f_i g_i' \\ &= \mathbf{f}'\cdot\mathbf{g} + \mathbf{f}\cdot\mathbf{g}'. \end{aligned}$

For $n=3$ , recall that the cross product is defined by

$\displaystyle \mathbf{f}\times\mathbf{g} := \det(\mathbf f, \mathbf g, \cdot)= \begin{bmatrix} f_2 g_3 - f_3 g_2 \\ f_3 g_1 - f_1 g_3 \\ f_1 g_2 - f_2 g_1 \end{bmatrix}.$

By the usual product rule,

$\begin{aligned} (f_2 g_3 - f_3 g_2)' &= (f_2 g_3)' - (f_3 g_2)' \\ &= (f_2' g_3 + f_2 g_3') - (f_3' g_2 + f_3 g_2') \\ &= (f_2'g_3 - f_3'g_2) + (f_2 g_3' - f_3 g_2'). \end{aligned}$

Similar for the other entries,

$\begin{aligned} (\mathbf{f}\times\mathbf{g})' &= \begin{bmatrix} f_2 g_3 - f_3 g_2 \\ f_3 g_1 - f_1 g_3 \\ f_1 g_2 - f_2 g_1 \end{bmatrix}' \\ &= \begin{bmatrix} ( f_2 g_3 - f_3 g_2 )' \\ ( f_3 g_1 - f_1 g_3 )' \\ ( f_1 g_2 - f_2 g_1 )' \end{bmatrix} \\ &= \begin{bmatrix} (f_2'g_3 - f_3'g_2) + (f_2 g_3' - f_3 g_2') \\ (f_3'g_1 - f_1'g_3) + (f_3 g_1' - f_1 g_3') \\ (f_1'g_2 - f_2'g_1) + (f_1 g_2' - f_2 g_1') \end{bmatrix} \\ &= \begin{bmatrix} f_2'g_3 - f_3'g_2 \\ f_3'g_1 - f_1'g_3 \\ f_1'g_2 - f_2'g_1 \end{bmatrix} + \begin{bmatrix} f_2 g_3' - f_3 g_2' \\ f_3 g_1' - f_1 g_3' \\ f_1 g_2' - f_2 g_1' \end{bmatrix} \\ &= \mathbf f' \times \mathbf g + \mathbf f \times \mathbf g'. \end{aligned}$

Problem 2. Given a differentiable function $\phi : \mathbb{R} \to \mathbb{R}$ , show that for any $t \in \mathbb{R}$ ,

$\displaystyle ( \mathbf{f} \circ \phi )'(t) = \phi'(t)(\mathbf f' \circ \phi)(t) .$

This result is the all-too-familiar chain rule, extended to curves.

(Click for Solution)

Solution. By Definition 1, $(\mathbf f \circ \phi)_i = f_i \circ \phi$ differentiating component-wise and applying the usual chain rule,

$\begin{aligned} ( \mathbf{f} \circ \phi )'(t) &= \sum_{i=1}^n (f_i \circ \phi)'(t) \mathbf{e}_i \\ &= \sum_{i=1}^n \phi'(t) (f_i' \circ \phi)(t) \mathbf{e}_i \\ &= \phi'(t) \sum_{i=1}^n (f_i' \circ \phi)(t) \mathbf{e}_i \\ &= \phi'(t)(\mathbf f' \circ \phi)(t) . \end{aligned}$

Problem 3. Suppose $|\mathbf{r}|$ is constant. Show that $\mathbf{r}(t) \perp \mathbf{r}'(t)$ for any $t$ .

(Click for Solution)

Solution. Since $|\mathbf{r}|$ is constant, so is $|\mathbf{r}|^2 = \mathbf{r}\cdot\mathbf{r}$ . Differentiating via the dot product from Problem 1,

$\displaystyle 0 = (\mathbf{r}\cdot\mathbf{r})' = \mathbf{r}\cdot\mathbf{r}' + \mathbf{r}' \cdot\mathbf{r} = 2 \mathbf r \cdot \mathbf r',$

so for any $t$ , $\mathbf{r}(t)\cdot\mathbf{r}'(t) = 0$ , that is, $\mathbf{r}(t) \perp \mathbf{r}'(t)$ .

Suppose furthermore that the components for all curves are continuous on $[a,b]$ .

Definition 2. Define integration on vector-valued functions by linearity:

$\displaystyle \int_a^b \mathbf{f}(t)\, \mathrm{d}t := \sum_{i=1}^n \left( \int_a^b f_i(t)\, \mathrm{d}t \right) \mathbf{e}_i.$

Problem 4. Verify the following integration properties for curves.

For any $\alpha,\beta\in\mathbb{R}$ , $\displaystyle \int_a^b (\alpha\mathbf{f}+\beta\mathbf{g})(t)\,\mathrm{d}t = \alpha\int_a^b\mathbf{f}(t)\,\mathrm{d}t + \beta\int_a^b\mathbf{g}(t)\,\mathrm{d}t$ .
For any $c\in[a,b]$ , $\displaystyle \int_a^b\mathbf{f}(t)\,\mathrm{d}t = \int_a^c\mathbf{f}(t)\,\mathrm{d}t + \int_c^b\mathbf{f}(t)\,\mathrm{d}t$ .
$\displaystyle \int_a^b\mathbf{f}'(t)\,\mathrm{d}t = \mathbf{f}(b) - \mathbf{f}(a)$ .

(Click for Solution)

Solution. Each property reduces to the scalar case via Definition 2. For example,

$\begin{aligned} \int_a^b(\alpha\mathbf{f}+\beta\mathbf{g})(t)\,\mathrm{d}t &= \sum_{i=1}^n \left(\int_a^b(\alpha f_i+\beta g_i)(t)\,\mathrm{d}t\right)\mathbf{e}_i \\ &= \sum_{i=1}^n \left(\int_a^b(\alpha f_i)(t)\,\mathrm{d}t + \int_a^b(\beta g_i)(t)\,\mathrm{d}t\right)\mathbf{e}_i \\ &= \sum_{i=1}^n \left( \alpha \int_a^b f_i(t)\,\mathrm{d}t + \beta \int_a^b g_i(t)\,\mathrm{d}t\right)\mathbf{e}_i \\ &= \sum_{i=1}^n \left( \alpha \int_a^b f_i(t)\,\mathrm{d}t \right)\mathbf{e}_i + \sum_{i=1}^n \left( \beta \int_a^b g_i(t)\,\mathrm{d}t\right)\mathbf{e}_i \\ &= \alpha \sum_{i=1}^n \left( \int_a^b f_i(t)\,\mathrm{d}t \right)\mathbf{e}_i + \beta \sum_{i=1}^n \left( \int_a^b g_i(t)\,\mathrm{d}t\right)\mathbf{e}_i \\ &= \alpha\int_a^b\mathbf{f}(t)\,\mathrm{d}t + \beta\int_a^b\mathbf{g}(t)\,\mathrm{d}t. \end{aligned}$

The other properties follow analogously from the scalar versions applied component-wise.

Definition 3. Define the arc length $L(\mathbf{r})$ of $\mathbf{r} : [a,b] \to \mathbb{R}^n$ by

$\displaystyle L(\mathbf{r}) := \int_a^b |\mathbf{r}'(t)|\, \mathrm{d}t,$

where $|\mathbf{v}| := \sqrt{\mathbf{v} \cdot \mathbf{v}}$ denotes the usual Euclidean norm on $\mathbb{R}^n$ . We remark that if $\mathbf r(t) = t\mathbf v$ for $t \in [0, 1]$ , $\mathbf r' = \mathbf v$ , so that

$\displaystyle L(\mathbf r) = \int_0^1 |\mathbf v|\, \mathrm dt = |\mathbf v|.$

Problem 5. Use Definition 3 to evaluate the length of the curve defined by the equation $y = x^2$ between the origin $(0,0)$ and the point $(1,1)$ .

(Click for Solution)

Solution. Parametrize the curve by $\mathbf{r}(t) = \begin{bmatrix} t \\ t^2 \end{bmatrix}$ , $t \in [0,1]$ . Then $\mathbf{r}'(t) = \begin{bmatrix} 1 \\ 2t \end{bmatrix}$ , so $|\mathbf{r}'(t)| = \sqrt{1+4t^2}$ , and by Definition 3,

$\displaystyle L(\mathbf{r}) = \int_0^1 \sqrt{1+4t^2}\,\mathrm{d}t.$

Substitute $2t = \tan\theta$ , so $2\,\mathrm{d}t = \sec^2\theta\,\mathrm{d}\theta$ and $\sqrt{1+4t^2} = \sec\theta$ :

$\begin{aligned} \int\sqrt{1+4t^2}\,\mathrm{d}t &= \frac{1}{2}\int\sec^3\theta\,\mathrm{d}\theta \\ &= \frac{1}{4}\left[\sec\theta\tan\theta + \ln| {\sec\theta+\tan\theta} |\right] + C \\ &= \frac{1}{4}[2t\sqrt{1+4t^2} + \ln(2t+\sqrt{1+4t^2})] + C. \end{aligned}$

Evaluating over $[0,1]$ ,

$\begin{aligned} L(\mathbf{r}) &= \frac{1}{4}[2t\sqrt{1+4t^2} + \ln(2t+\sqrt{1+4t^2}) ]_0^1 \\ &= \frac{1}{4} (2\sqrt{5} + \ln (2+\sqrt{5}) - (0 + \ln 1)) \\ &= \frac{1}{4} (2\sqrt{5} + \ln (2+\sqrt{5})). \end{aligned}$

—Joel Kindiak, 18 Apr 26, 2000H

KindiakMath

Calculus on Curves

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Calculus on Curves

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