Part 14 - Curves | rezarezvan.com

Introduction

In this part we’ll cover curves and how we can parameterize a curve.

Definition 1 (Curve and parameterization)

Let, $f(t)$ , $g(t)$ and $h(t)$ be continuous functions on some interval, $I$ . The connection of points $(x, y, z)$ , where $x = f(t)$ , $y = g(t)$ and $z = h(t)$ , is a curve.

Equivalently, we can say that:

\vec{r}(t) = \langle f(t), g(t), h(t) \rangle

We call this the parameterization of the curve. Where $t$ is the parameter.

Example 1

Given the function $y = x$ . In the interval, $0 \leq t \leq 1$ .

We can parameterize and say:

\vec{r}(t) = \langle t, t \rangle \quad \| \quad 0 \leq t \leq 1

But we could also say that:

\vec{r}(t) = \langle t^2, t^2 \rangle \quad \| \quad 0 \leq t \leq 1

So, there isn’t necessarily a unique parameterization for a function.

We can also change the “direction”:

\vec{r}(t) = \langle 1 - t, 1 - t \rangle \quad \| \quad 0 \leq t \leq 1

Let’s try for a more complex function. Given the unit circle, $x^2 + y^2 = 1$ .

\vec{r}(t) = \langle \cos(t), \sin(t) \rangle \quad \| \quad 0 \leq t \leq 2\pi

In general:

\vec{r}(t) = \langle t, f(t) \rangle \quad \| \quad a \leq t \leq b

Definition 2 (Tangent vector)

If a curve, $C$ , is given by a parameterization:

\vec{r}(t) = \langle f(t), g(t), h(t) \rangle

Then:

\vec{r}'(t) = \langle f^\prime(t), g^\prime(t), h^\prime(t) \rangle

is the tangent vector.

Definition 3 (Unit tangent vector)

\vec{T}(t) = \dfrac{\vec{r}'(t)}{|\vec{r}'(t)|}

is the unit tangent vector.

Example 2

For the curve $\vec{r}(t) = \sqrt{t} \vec{i} + (2 - t)\vec{j}$ .

Find $\vec{T}(t)$ at $t = 1$ .

Let’s rewrite $\vec{r}(t)$ in usual notation:

\vec{r}(t) = \langle \sqrt{t}, 2 - t \rangle

\vec{r}'(t) = \langle \dfrac{1}{2\sqrt{t}}, -1 \rangle

\vec{r}'(1) = \langle \dfrac{1}{2}, -1 \rangle

|\vec{r}'(1)| = \ldots = \dfrac{\sqrt{5}}{2}

\vec{T}(1) = \dfrac{\vec{r}'(1)}{|\vec{r}'(1)|}

\vec{T}(1) = \dfrac{\langle \dfrac{1}{2}, -1 \rangle}{\dfrac{\sqrt{5}}{2}}

\vec{T}(1) = \langle \dfrac{1}{\sqrt{5}}, - \dfrac{2}{\sqrt{5}} \rangle

Parameterization over line

Given a line that passes through the point $(x_0, y_0, z_0)$ , with a direction of the vector, $\vec{v} = \langle a, b, c \rangle$ .

If we want to parameterize this line, we can choose another point that this vector passes through as:

\vec{r}(t) = \vec{r}(t_0) + t\vec{v}

(x(t), y(t), z(t)) = (x_0, y_0, z_0) + t(a, b, c)

This means that:

x(t) = x_0 + ta

y(t) = y_0 + tb

z(t) = z_0 + tc

Example 3

Find the parameterization equation to the tangent line to the helix, $x = 2\cos(t), y = \sin(t), z = t$ . At point $(0, 1, \dfrac{\pi}{2})$ .

From this we easily see that $t = \dfrac{\pi}{2}$ .

The tangent line passes through $\vec{T}\left(\dfrac{\pi}{2}\right)$ .

Let’s find this.

\vec{r}(t) = \langle 2 \cos(t), \sin(t), t \rangle

\vec{r}'(t) = \langle -2 \sin(t), \cos(t), 1 \rangle

\vec{r}'\left(\dfrac{\pi}{2}\right) = \langle -2, 0, 1 \rangle

|\vec{r}'\left(\dfrac{\pi}{2}\right)| = \sqrt{(-2)^2 + 0^2 + 1^2} = \sqrt{5}

\vec{T}\left(\dfrac{\pi}{2}\right) = \langle - \dfrac{2}{\sqrt{5}}, 0, \dfrac{1}{\sqrt{5}} \rangle

Now let’s set this into our equation:

x(t) = 0 + t \cdot \left(-\dfrac{2}{\sqrt{5}}\right) = \boxed{-\dfrac{2t}{\sqrt{5}}}

y(t) = 1 + t \cdot 0 = \boxed{1}

z(t) = \dfrac{\pi}{2} + t \dfrac{1}{\sqrt{5}} = \boxed{\dfrac{\pi}{2} + \dfrac{t}{\sqrt{5}}}