Part 14 - Curves

Introduction

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

Definition

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

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

We can parameterize and say: $$ \vec{r}(t) = \langle t, t \rangle \ | \ 0 \leq t \leq 1 $$

But we could also say that: $$ \vec{r}(t) = \langle t^2, t^2 \rangle \ | \ 0 \leq t \leq 1 $$

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

We can also change the “direction”: $$ \vec{r}(t) = \langle 1 - t, 1 - t \rangle \ | \ 0 \leq t \leq 1 $$

Let’s try for a more complex function. Given the unit circle, $x^2 + y^2 = 1$. $$ \vec{r}(t) = \langle cos(t), sin(t) \rangle \ | \ 0 \leq t \leq 2\pi $$

In general: $$ \vec{r}(t) = \langle t, f(t) \rangle \ | \ a \leq t \leq b $$

Definition

If a curve, $C$, is given by a parameterization: $$ \vec{r}(t) = \langle f(t), g(t), h(t) $$

Then: $$ \vec{r}\ ^\prime(t) = \langle f^\prime(t), g^\prime(t), h^\prime(t) $$

is the tangent vector.

Definition

$$ \vec{T}(t) = \dfrac{\vec{r}\ ^\prime(t)}{|\vec{r}\ ^\prime(t)|} $$

is the unit tangent vector.

Examples

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

$$ \vec{r}\ ^\prime (t) = \langle \dfrac{1}{2\sqrt{t}}, -1 \rangle $$

$$ \vec{r}\ ^\prime (1) = \langle \dfrac{1}{2}, -1 \rangle $$

$$ |\vec{r}\ ^\prime (1)| = \ldots = \dfrac{\sqrt{5}}{2} $$

$$ \vec{T}(1) = \dfrac{\vec{r}\ ^\prime(1)}{|\vec{r}\ ^\prime(1)|} $$

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

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

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

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

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

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

Let’s find this.

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

$$ \vec{r}\ ^\prime(t) = \langle -2 sin(t), cos(t), 1 \rangle $$

$$ \vec{r}\ ^\prime(\dfrac{\pi}{2}) = \langle -2, 0, 1 \rangle $$

$$ |\vec{r}\ ^\prime(\dfrac{\pi}{2})| = \sqrt{(-2)^2 + 0^2 + 1^2} = \sqrt{5} $$

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

Now let’s set this into our equation: $$ x(t) = 0 + t \cdot -\dfrac{2}{\sqrt{5}} = \boxed{-\dfrac{2t}{\sqrt{5}}} $$

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

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