Ultimate Guide to Vector Equations of Curves and Spheres

Introduction

We have seen how vector equations can describe perfectly straight lines using $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ . But the physical world is rarely straight. How do we describe the circular orbit of a satellite, or a complex 3D spiral, using vectors? In HSC Mathematics Extension 2, we extend our vector knowledge to describe Parametric Curves, Circles, and Spheres.

Executive Summary

This guide covers curved vector geometry:

Parametric Curves: Expressing curves as $\mathbf{r}(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}$ .
Circles in 2D: Deriving and using the vector equation $|\mathbf{r} - \mathbf{c}| = R$ .
Spheres in 3D: Extending the circle equation to describe a 3D sphere.
Cartesian Conversion: Converting vector curve equations back into standard $y=f(x)$ Cartesian algebra.

What is this about?

When defining a straight line, the parameter $\lambda$ multiplied a constant direction vector. To define a curve, the components of the position vector themselves must change according to a parameter (often $t$ for time or $\theta$ for angle).

Furthermore, the geometric definition of a circle is "the set of all points that are a fixed distance from a centre point". Because vectors are incredibly good at describing distances (using the magnitude/absolute value signs $|\mathbf{v}|$ ), the vector equation of a circle perfectly translates this English definition into mathematics. This exact same logic extends flawlessly into 3-dimensional spheres.

Main Content

1. Vector Equations of Circles

Let $\mathbf{c}$ be the position vector of the centre of a circle, and $R$ be its radius. Let $\mathbf{r}$ be the position vector of any point on the circumference of the circle.

The vector pointing from the centre to the edge is $\mathbf{r} - \mathbf{c}$ . Because every point on the edge is exactly distance $R$ away from the centre, the magnitude of this vector must be $R$ .

Therefore, the vector equation of a circle is: $|\mathbf{r} - \mathbf{c}| = R$

If the circle is centred at the origin $(0,0)$ , then $\mathbf{c} = \mathbf{0}$ , and the equation simplifies to: $|\mathbf{r}| = R$

Alternatively, using parametric equations with angle $\theta$ , a circle at the origin can be written as: $\mathbf{r}(\theta) = \begin{pmatrix} R\cos\theta \\ R\sin\theta \end{pmatrix}$

2. Vector Equations of Spheres

One of the most beautiful aspects of vector mathematics is how easily it scales to higher dimensions. The geometric definition of a sphere in 3D space is identical to a circle: "the set of all points in 3D space a fixed distance from a centre point".

Therefore, if $\mathbf{r}$ is a 3D vector $\begin{pmatrix} x\\y\\z \end{pmatrix}$ and $\mathbf{c}$ is a 3D centre $\begin{pmatrix} c_1\\c_2\\c_3 \end{pmatrix}$ , the equation of the sphere is exactly the same! $|\mathbf{r} - \mathbf{c}| = R$

3. Parametric Curves and Cartesian Conversion

A curve can be defined generally as $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$ . To find the Cartesian equation (an equation with only $x$ and $y$ , no $t$ ), you must:

Write the two parametric equations: $x = x(t)$ and $y = y(t)$ .
Use algebra (or trigonometric identities) to eliminate the parameter $t$ .

Example: Find the Cartesian equation for the curve $\mathbf{r}(t) = \begin{pmatrix} 2t \\ t^2 - 1 \end{pmatrix}$ .

$x = 2t \implies t = \frac{x}{2}$
$y = t^2 - 1$ Substitute (1) into (2): $y = \left(\frac{x}{2}\right)^2 - 1$ $y = \frac{x^2}{4} - 1$ (This is a parabola!)

Simple Worked Example

Question: A sphere has the vector equation $|\mathbf{r} - \mathbf{c}| = 5$ , where $\mathbf{c} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ . Find the Cartesian equation of this sphere.

Solution: Step 1: Set up the vector expression inside the magnitude. Let $\mathbf{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ . $\mathbf{r} - \mathbf{c} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} = \begin{pmatrix} x - 2 \\ y + 1 \\ z - 3 \end{pmatrix}$

Step 2: Apply the magnitude formula. The magnitude of a 3D vector $\begin{pmatrix} a\\b\\c \end{pmatrix}$ is $\sqrt{a^2 + b^2 + c^2}$ . $|\mathbf{r} - \mathbf{c}| = \sqrt{(x - 2)^2 + (y + 1)^2 + (z - 3)^2}$

Step 3: Equate to the radius and square both sides. We are given that the magnitude equals $5$ . $\sqrt{(x - 2)^2 + (y + 1)^2 + (z - 3)^2} = 5$ Square both sides: $(x - 2)^2 + (y + 1)^2 + (z - 3)^2 = 25$

Final Answer: The Cartesian equation of the sphere is $(x - 2)^2 + (y + 1)^2 + (z - 3)^2 = 25$ .

mini-FAQ page

Q: If $|\mathbf{r} - \mathbf{c}| = R$ is the equation for both a circle and a sphere, how do I know which one it is? A: You look at the dimension of the vectors $\mathbf{r}$ and $\mathbf{c}$ . If they have two components $\begin{pmatrix} x\\y \end{pmatrix}$ , it is a 2D circle. If they have three components $\begin{pmatrix} x\\y\\z \end{pmatrix}$ , it is a 3D sphere. The context of the question will always specify the dimension.

Q: How do I eliminate $t$ if the vector is $\mathbf{r}(\theta) = \begin{pmatrix} 3\cos\theta \\ 3\sin\theta \end{pmatrix}$ ? A: You use the Pythagorean identity! $x = 3\cos\theta \implies \frac{x}{3} = \cos\theta \implies \frac{x^2}{9} = \cos^2\theta$ $y = 3\sin\theta \implies \frac{y}{3} = \sin\theta \implies \frac{y^2}{9} = \sin^2\theta$ Add them together: $\frac{x^2}{9} + \frac{y^2}{9} = \cos^2\theta + \sin^2\theta = 1$ . $x^2 + y^2 = 9$ (A circle of radius 3).

Common mistakes to avoid

Forgetting to square the radius: When converting $|\mathbf{r} - \mathbf{c}| = R$ to Cartesian, the right-hand side of the Cartesian equation becomes $R^2$ , not $R$ . In the example above, the radius was 5, but the Cartesian equation equals 25.
Messing up the signs: The vector equation uses $\mathbf{r} - \mathbf{c}$ . If the centre is $(2, -1, 3)$ , the vector is $\mathbf{r} - \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ . This creates a $(y - (-1))$ which becomes $(y+1)$ in the Cartesian bracket. Watch your negative signs carefully!

Practice on Vu's Maths Hub

Converting between vector parameters and Cartesian curves is a common exam trick.

Practice eliminating difficult trigonometric parameters in the HSC Vectors Booklet.
Visualize 3D spheres and intersections at Vu's Maths Hub.

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