← M408M Week 1: Parametric Equations & Curves
Multivariable Calculus Week 1 • Session 2

Curves Defined by Parametric Equations

1
What is a Parametric Curve?
2
Step-by-Step Sketching Animation
3
Parameter Elimination & Exercises
Concept & Motivation 02 / 10

What is a Parametric Curve?

A dynamic description of points in the plane $(x, y) = (f(t), g(t))$
💡 Core Concept: Motion in Time

A parametric curve is a set of points of the form:

$$(x, y) = (f(t), g(t))$$

where $f$ and $g$ are functions of a variable $t$, called the parameter (often representing time).

  • As $t$ increases over an interval $[a, b]$, the point $(x(t), y(t))$ traces out a curve.
  • Orientation: The curve has a natural direction of motion indicating increasing $t$.
⚠️ Why Cartesian $y = f(x)$ Falls Short
  • Vertical Line Test: Standard Cartesian functions $y = f(x)$ can only assign one $y$ to each $x$.
  • Cannot naturally describe closed loops, circles, self-intersecting figures, or spirals without splitting equations.
  • No velocity or direction: Cartesian graphs show the path, but cannot describe how fast or in what direction an object moves.
Sketching Technique 03 / 10

Method 1: Point Plotting & Orientation

Interactive Sketching Animation: $x = t^2 - 2t, \quad y = t + 1 \quad (-2 \le t \le 4)$
Step 0 / 8
Ready to Plot
Choose sample values of $t \in [-2, 4]$. Click Next or Auto Play to see how the curve is plotted point-by-point with orientation.
$t$ $x = t^2 - 2t$ $y = t + 1$ Point $(x, y)$
$-2$ $8$ $-1$ $(8, -1)$ (Initial)
$-1$ $3$ $0$ $(3, 0)$
$0$ $0$ $1$ $(0, 1)$
$1$ $-1$ $2$ $(-1, 2)$ (Vertex)
$2$ $0$ $3$ $(0, 3)$
$3$ $3$ $4$ $(3, 4)$
$4$ $8$ $5$ $(8, 5)$ (Terminal)
Algebraic Technique 04 / 10

Method 2: Eliminating the Parameter

Finding a direct Cartesian relationship $F(x, y) = 0$ by eliminating $t$
Worked Example: Parabola

Given the parametric equations:

$$x = t^2 - 2t, \qquad y = t + 1 \qquad (-2 \le t \le 4)$$
1. Solve for $t$ in terms of $y$: $$y = t + 1 \implies t = y - 1$$
2. Substitute into equation for $x$: $$x = (y - 1)^2 - 2(y - 1)$$ $$x = y^2 - 2y + 1 - 2y + 2$$
$$x = y^2 - 4y + 3 = (y - 2)^2 - 1$$
3. Identify the Cartesian geometry: A rightward-opening parabola with vertex at $(-1, 2)$.
💡 What is a Cartesian Equation?

A Cartesian Equation is an equation containing only the variables in the Cartesian coordinate system (in a 2D plane, only $x$ and $y$). To find the Cartesian equation from $(x(t), y(t))$, we eliminate the parameter $t$ so that only $x$ and $y$ remain.

⚠️ The Domain & Range Trap!

Eliminating $t$ yields the entire Cartesian curve, but the parametric curve may only trace a piece of it!

Example: $x = \sin\theta, \qquad y = \cos^2\theta$
Identity $\sin^2\theta + \cos^2\theta = 1 \implies y = 1 - x^2$.
However: since $-1 \le \sin\theta \le 1$, the parametric curve is only the portion of the parabola for $x \in [-1, 1]$ and $y \in [0, 1]$!
Interactive Laboratory 05 / 10

Interactive Parametric Curve Visualizer

Explore real-time curve tracing, parameter sweep, velocity vectors, and presets
$t$: 0.00
Point: (0.00, 0.00)
Speed $|\vec{v}(t)|$: 0.00
Cartesian: x = y² - 4y + 3
$t$ $x$ $y$ $(x, y)$
Active Learning 06 / 10

In-Class Practice Problems

Work on these three problems individually or in small groups before we discuss solutions
1 Problem 1: Cartesian Equation

Given the parametric equations:

$$x = \sin t, \quad y = \cos(2t)$$

(a) Eliminate the parameter $t$ to write this curve in Cartesian form: $$y = f(x)$$

(b) Identify the restricted domain of $x$ and describe the path traced out.

💡 Hint: Use $\cos(2t) = \cos^2 t - \sin^2 t$ and Pythagorean identity.
2 Problem 2: Coordinate Swap & Symmetry

Consider the two parametric curves on $0 \le t \le \pi$:

$$\begin{aligned} \text{Curve } 1&: \quad x_1(t) = t, \quad y_1(t) = \sin t \\ \text{Curve } 2&: \quad x_2(t) = \sin t, \quad y_2(t) = t \end{aligned}$$

(a) Find the Cartesian equations of both curves and explain the geometric difference between their graphs.

(b) Generalization: For any curve $x = t, y = f(t)$, what does the graph of the swapped curve $x = f(t), y = t$ look like?

💡 Hint: What geometric transformation in the plane does swapping $(x, y) \longleftrightarrow (y, x)$ correspond to?
3 Problem 3: The "Same Equation" Trap

Compare these three curves for $-\infty < t < \infty$:

$$\begin{aligned} C_1&: x = t, \quad y = t^2 \\ C_2&: x = \sin t, \quad y = \sin^2 t \\ C_3&: x = e^t, \quad y = e^{2t} \end{aligned}$$

(a) Show that eliminating $t$ gives the identical equation $y = x^2$ for all three.

(b) Explain how their domains, sketched paths, and particle motions actually differ.

💡 Hint: Analyze the specific ranges of $x(t)$ and $y(t)$ for each parameter domain.
Solution: Problem 1 07 / 10

Exercise 1: Cartesian Equation

Eliminating the parameter using trigonometric double-angle formulas
Problem: Given the parametric equations $$x = \sin t, \quad y = \cos(2t)$$ write this equation in the form $y = f(x)$.
Hint: Use the formula $\cos(2\theta) = \cos^2\theta - \sin^2\theta$.
1 Step 1: Express $\cos(2t)$ in Terms of $\sin t$

Using the given double-angle formula and the Pythagorean identity $\cos^2 t + \sin^2 t = 1 \implies \cos^2 t = 1 - \sin^2 t$: $$\cos(2t) = \cos^2 t - \sin^2 t = (1 - \sin^2 t) - \sin^2 t = 1 - 2\sin^2 t$$

2 Step 2: Substitute $x = \sin t$ into the Equation

Substitute $x = \sin t$ directly into $y = 1 - 2\sin^2 t$: $$y = 1 - 2(\sin t)^2 \implies \mathbf{y = 1 - 2x^2}$$ Therefore, in the required form $y = f(x)$, we have: $$\mathbf{f(x) = 1 - 2x^2}$$

3 Step 3: Domain & Geometric Interpretation

• Domain Restriction: Because $x = \sin t$, the domain is restricted to: $$-1 \le x \le 1$$ • Range: As $x \in [-1, 1]$, $y = 1 - 2x^2$ ranges from $-1$ (when $x = \pm 1$) to $1$ (vertex at $x = 0$).
• Motion: The curve is the parabolic arc opening downward from $(-1, -1)$ to $(1, -1)$ with vertex at $(0, 1)$. As $t$ increases, a particle oscillates back and forth along this arc.

Solution: Problem 2 08 / 10

Exercise 2: Swapped Coordinates & Reflection Across $y = x$

Analyzing the geometric effect of interchanging $x(t)$ and $y(t)$
Problem: Consider these two sets of parametric equations: $$\text{Curve } 1: \quad x(t) = t, \quad y(t) = \sin t \quad (0 \le t \le \pi)$$ $$\text{Curve } 2: \quad x(t) = \sin t, \quad y(t) = t \quad (0 \le t \le \pi)$$ 1. What is the difference between their associated curves?
2. Generalization: Given any curve $x(t) = t, \; y(t) = f(t)$, what does the graph of $x(t) = f(t), \; y(t) = t$ look like?
1 Step 1: Compare the Cartesian Equations

• Curve 1: Since $x = t$, substituting gives $\mathbf{y = \sin x}$ for $x \in [0, \pi]$ (horizontal arch peaking at $(\pi/2, 1)$).

• Curve 2: Since $y = t$, substituting gives $\mathbf{x = \sin y}$ for $y \in [0, \pi]$ (vertical arch bowing rightward to $(1, \pi/2)$).

2 Step 2: Geometric Transformation ($y = x$ Reflection)

Notice that every point $(x_1, y_1) = (t, \sin t)$ on Curve 1 corresponds directly to $(x_2, y_2) = (\sin t, t)$ on Curve 2.

In Cartesian geometry, the coordinate swap $\mathbf{(x, y) \longleftrightarrow (y, x)}$ is a reflection across the diagonal line $y = x$.

Therefore, Curve 2 is the exact reflection of Curve 1 across $y = x$!

3 Step 3: Generalization for Any Function $f(t)$

For any $C_1: x = t, y = f(t) \implies y = f(x)$, the swapped curve $C_2: x = f(t), y = t \implies x = f(y)$ is always the reflection of $y = f(x)$ across the line $y = x$ (the geometric inverse relation).

Curve 1: $(t, \sin t)$ Curve 2: $(\sin t, t)$ Line $y = x$
$t$: 1.00
$P_1$: (1.00, 0.84) $P_2$: (0.84, 1.00)
Solution: Problem 3 09 / 10

Exercise 3: The "Same Cartesian Equation" Trap

Why identical Cartesian equations can represent completely different curves
Problem: Compare the geometric curves and motions described by: $$\begin{aligned} C_1&: x = t, \; y = t^2 \quad (-\infty < t < \infty) \\ C_2&: x = \sin t, \; y = \sin^2 t \quad (-\infty < t < \infty) \\ C_3&: x = e^t, \; y = e^{2t} \quad (-\infty < t < \infty) \end{aligned}$$
1 Cartesian Analysis: What do all three share?

In all three cases, substituting $x$ into $y$ yields: $$\mathbf{y = x^2}$$ However, their domains and physical motions are completely different!

2 Curve 1 vs. Curve 2 vs. Curve 3 Breakdown
$C_1: (t, t^2)$: $x \in (-\infty, \infty)$ — traces the entire parabola once from left to right.
$C_2: (\sin t, \sin^2 t)$: $x \in [-1, 1], y \in [0, 1]$ — oscillates endlessly back and forth along this segment!
$C_3: (e^t, e^{2t})$: $x > 0, y > 0$ — traces only the right branch (excluding origin $(0,0)$!).
Select Display Curve:
C₁: Full Parabola C₂: [-1, 1] Segment C₃: Right Half (x > 0)
All 3 Curves: Same Cartesian Equation $y = x^2$, Different Domains & Motions!
$t$: 0.50
Summary & Review 10 / 10

Summary & Key Takeaways

Recap of the core concept and essential curve sketching techniques
💡 The Parametric Curve (Core Concept)
A set of points $(x, y)$ in the Cartesian plane traced out continuously as the parameter $t$ varies over an interval $I$:
$$(x, y) = \bigl(f(t),\, g(t)\bigr), \quad t \in I$$
🛠️ Methods & Techniques
📊 1. Point Plotting
Construct a table of values $(t, x, y)$ to plot key points sequentially and determine the curve's orientation (direction of motion).
🔄 2. Eliminating Parameter
Solve for $t$ algebraically or use trigonometric identities (e.g. $\cos^2 t + \sin^2 t = 1$) to convert to a Cartesian equation $y = f(x)$.
⚠️ 3. Domain & Range Check
Always verify the bounds on $x(t)$ and $y(t)$ when eliminating $t$ to ensure only the valid portion of the Cartesian curve is drawn.