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Part 14 - Curves

LMA017 Date: September 27, 2023 4 min read

Introduction

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

Definition 1 (Curve and parameterization)

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

Equivalently, we can say that:

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

We call this the parameterization of the curve. Where tt is the parameter.

Example 1

Given the function y=xy = x. In the interval, 0≤t≤10 \leq t \leq 1.

We can parameterize and say:

r⃗(t)=⟨t,t⟩ ∣ 0≤t≤1\vec{r}(t) = \langle t, t \rangle \ | \ 0 \leq t \leq 1

But we could also say that:

r⃗(t)=⟨t2,t2⟩ ∣ 0≤t≤1\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”:

r⃗(t)=⟨1−t,1−t⟩ ∣ 0≤t≤1\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, x2+y2=1x^2 + y^2 = 1.

r⃗(t)=⟨cos(t),sin(t)⟩ ∣ 0≤t≤2π\vec{r}(t) = \langle cos(t), sin(t) \rangle \ | \ 0 \leq t \leq 2\pi

In general:

r⃗(t)=⟨t,f(t)⟩ ∣ a≤t≤b\vec{r}(t) = \langle t, f(t) \rangle \ | \ a \leq t \leq b
Definition 2 (Tangent vector)

If a curve, CC, is given by a parameterization:

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

Then:

r⃗ ′(t)=⟨f′(t),g′(t),h′(t)\vec{r}\ ^\prime(t) = \langle f^\prime(t), g^\prime(t), h^\prime(t)

is the tangent vector.

Definition 3 (Unit tangent vector)T⃗(t)=r⃗ ′(t)∣r⃗ ′(t)∣\vec{T}(t) = \dfrac{\vec{r}\ ^\prime(t)}{|\vec{r}\ ^\prime(t)|}

is the unit tangent vector.

Example 2

For the curve r⃗(t)=(t)i⃗+(2−t)j⃗\vec{r}(t) = \sqrt(t) \vec{i} + (2 - t)\vec{j}.

Find T⃗(t)\vec{T}(t) at t=1t = 1.

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

r⃗(t)=⟨t,2−t⟩\vec{r}(t) = \langle \sqrt{t}, 2 - t \rangler⃗ ′(t)=⟨12t,−1⟩\vec{r}\ ^\prime (t) = \langle \dfrac{1}{2\sqrt{t}}, -1 \rangler⃗ ′(1)=⟨12,−1⟩\vec{r}\ ^\prime (1) = \langle \dfrac{1}{2}, -1 \rangle∣r⃗ ′(1)∣=…=52|\vec{r}\ ^\prime (1)| = \ldots = \dfrac{\sqrt{5}}{2}T⃗(1)=r⃗ ′(1)∣r⃗ ′(1)∣\vec{T}(1) = \dfrac{\vec{r}\ ^\prime(1)}{|\vec{r}\ ^\prime(1)|}T⃗(1)=⟨12,−1⟩52\vec{T}(1) = \dfrac{\langle \dfrac{1}{2}, -1 \rangle}{\dfrac{\sqrt{5}}{2}}T⃗(1)=⟨15,−25⟩\vec{T}(1) = \langle \dfrac{1}{\sqrt{5}}, - \dfrac{2}{\sqrt{5}} \rangle

Parameterization over line

Given a line that passes through the point (x0,y0,z0)(x_0, y_0, z_0), with a direction of the vector, v⃗=⟨a,b,c⟩\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:

r⃗(t)=r⃗(t0)+tv⃗\vec{r}(t) = \vec{r}(t_0) + t\vec{v} (x(t),y(t),z(t))=(x0,y0,z0)+t(a,b,c)(x(t), y(t), z(t)) = (x_0, y_0, z_0) + t(a, b, c)

This means that:

x(t)=x0+tax(t) = x_0 + ta y(t)=y0+tby(t) = y_0 + tb z(t)=z0+tcz(t) = z_0 + tc
Example 3

Find the parameterization equation to the tangent line to the helix, x=2cos(t),y=sin(t),z=tx = 2cos(t), y = sin(t), z = t. At point (0,1,π2)(0, 1, \dfrac{\pi}{2}).

From this we easily see that t=π2t = \dfrac{\pi}{2}.

The tangent line passes through T⃗(π2)\vec{T}(\dfrac{\pi}{2}).

Let’s find this.

r⃗(t)=⟨2cos(t),sin(t),t⟩\vec{r}(t) = \langle 2 cos(t), sin(t), t \rangler⃗ ′(t)=⟨−2sin(t),cos(t),1⟩\vec{r}\ ^\prime(t) = \langle -2 sin(t), cos(t), 1 \rangler⃗ ′(π2)=⟨−2,0,1⟩\vec{r}\ ^\prime(\dfrac{\pi}{2}) = \langle -2, 0, 1 \rangle∣r⃗ ′(π2)∣=(−2)2+02+12=5|\vec{r}\ ^\prime(\dfrac{\pi}{2})| = \sqrt{(-2)^2 + 0^2 + 1^2} = \sqrt{5}T⃗(π2)=⟨−25,0,15⟩\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⋅−25=−2t5x(t) = 0 + t \cdot -\dfrac{2}{\sqrt{5}} = \boxed{-\dfrac{2t}{\sqrt{5}}}y(t)=1+t⋅0=1y(t) = 1 + t \cdot 0 = \boxed{1}z(t)=π2+t15=π2+t5z(t) = \dfrac{\pi}{2} + t \dfrac{1}{\sqrt{5}} = \boxed{\dfrac{\pi}{2} + \dfrac{t}{\sqrt{5}}}