Parametric Curves
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.1: “Parametric Equations” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-1-parametric-equations |
| Textbook used in class | Stewart, Calculus, Section 10.1: “Curves Defined by Parametric Equations” |
Opening Scenario
A particle moves along a curved path in the plane. Its $x$-coordinate at time $t$ is $x = \cos t$ and its $y$-coordinate is $y = \sin t$. If you eliminate $t$ (using $\cos^2 t + \sin^2 t = 1$), you find $x^2 + y^2 = 1$ -- the unit circle. But the parametric form carries extra information: as $t$ increases from $0$ to $2\pi$, the particle moves counterclockwise around the circle. The equation $x^2 + y^2 = 1$ alone does not tell you which direction or how fast.
Parametric equations describe not just the shape of a curve but also how a point moves along it.
Quick Reference
A parametric curve is the set of points $(x, y) = (f(t), g(t))$ as $t$ ranges over an interval $I$. The variable $t$ is the parameter; it does not appear in the final curve but controls the motion.
Plotting: Make a table of $(t, x, y)$ values; plot the resulting $(x,y)$ points; connect in order of increasing $t$ and mark the direction with an arrow.
Eliminating the parameter: Solve one equation for $t$ (if possible) and substitute into the other, obtaining a Cartesian equation in $x$ and $y$ alone. Always note any restrictions on $x$ or $y$ that arise from the parameter range.
Key Concepts
1. What a Parameter Does
In the equation $y = f(x)$, the independent variable is $x$ and the curve is drawn from left to right as $x$ increases. In parametric form, neither $x$ nor $y$ is the independent variable -- both depend on $t$. This allows curves that:
- Move right and left (backtrack in $x$),
- Cross themselves (a point visited twice corresponds to two $t$-values with the same $(x,y)$),
- Loop (closed curves, like circles).
2. Plotting: Build a Table
To plot $x = f(t)$, $y = g(t)$ for $t \in [a, b]$:
- Choose several $t$-values (endpoints and key intermediate values).
- Compute $x = f(t)$ and $y = g(t)$ at each $t$.
- Plot the points $(x, y)$ in the order of increasing $t$.
- Connect smoothly; add an arrow showing the direction of increasing $t$.
3. Eliminating the Parameter
Algebraic elimination: If $x = t^2$ and $y = t^3$, then $t = x^{1/2}$ (for $x \geq 0$), so $y = (x^{1/2})^3 = x^{3/2}$.
Trigonometric elimination: Use Pythagorean identities. If $x = a\cos t$, $y = b\sin t$, then $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = \cos^2 t + \sin^2 t = 1$.
Important: Elimination may extend the curve. Always check the range of $x$ and $y$ imposed by the parameter domain.
Worked Example
Sketch the curve $x = t^2 - 1$, $y = 2t$ for $-2 \leq t \leq 2$. Eliminate the parameter.
Table of values:
| $t$ | $x = t^2-1$ | $y = 2t$ |
|---|---|---|
| $-2$ | 3 | $-4$ |
| $-1$ | 0 | $-2$ |
| 0 | $-1$ | 0 |
| 1 | 0 | 2 |
| 2 | 3 | 4 |
The curve passes through $(3,-4)$, $(0,-2)$, $(-1,0)$, $(0,2)$, $(3,4)$ in that order.
Eliminate: From $y = 2t$: $t = y/2$. Substitute: $x = (y/2)^2 - 1 = \dfrac{y^2}{4} - 1$.
Rearranging: $y^2 = 4(x+1)$. This is a parabola opening to the right with vertex at $(-1, 0)$.
Range check: $t \in [-2,2]$ gives $y \in [-4,4]$, so only the portion with $-4 \leq y \leq 4$ is traced. The direction is upward (from $y = -4$ to $y = 4$) as $t$ increases.
Boxed answer: Cartesian equation $y^2 = 4(x+1)$, traced from $(3,-4)$ to $(3,4)$ upward as $t$ goes from $-2$ to $2$.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Forgetting direction | Drawing the curve without an arrow | The arrow shows which way $(x(t),y(t))$ moves as $t$ increases; this is part of the answer |
| Ignoring parameter range after eliminating | Claiming the full parabola $y^2 = 4(x+1)$ for all $y$ | If $t \in [-2,2]$, then $y = 2t \in [-4,4]$; restrict the Cartesian curve to that range |
| Confusing the parameter with $x$ | Writing $t = x$ and concluding $y = 2x$ | Read the parametric equations: $x$ and $t$ are different; $t = y/2$ here, not $t = x$ |
Common Misconceptions
the parameter $t$ is plotted on one of the axes.
This is the input-output-confusion error. The parameter $t$ is an independent variable that drives both $x$ and $y$; it does not appear on the Cartesian plane at all. The curve is the set of points $(x(t), y(t))$ in the $xy$-plane. Treating $t$ as though it were $x$ produces a graph of $y$ against $t$, not the parametric curve.
eliminating the parameter gives the complete curve.
This is the iconic-graph error. The Cartesian equation obtained by eliminating $t$ may describe a larger set of points than the parametric curve traces. The parameter range restricts which portion of the Cartesian curve is actually swept out. For $x = t^2-1$, $y = 2t$, $-2 \le t \le 2$, the full Cartesian parabola $y^2 = 4(x+1)$ extends to all $y$, but the parametric curve covers only $-4 \le y \le 4$. Always note the restrictions imposed by the domain of $t$.
Leveled Practice
Level 1 -- Table and Sketch
Problem 1. Sketch the curve $x = 1 + 3t$, $y = 2 - t^2$ for $-2 \leq t \leq 2$. Identify the Cartesian curve.
Show answer
From $x = 1+3t$: $t = (x-1)/3$. Substitute: $y = 2 - \left(\dfrac{x-1}{3}\right)^2 = 2 - \dfrac{(x-1)^2}{9}$.
A downward parabola with vertex at $(1, 2)$. Parameter range $t \in [-2,2]$ gives $x \in [-5, 7]$.
Direction: as $t$ increases, $x = 1+3t$ increases, so the curve is traced from left to right.
Level 2 -- Trigonometric Elimination
Problem 2. Identify and sketch the curve $x = 3\cos t$, $y = 2\sin t$, $0 \leq t \leq 2\pi$.
Show answer
$\dfrac{x^2}{9} + \dfrac{y^2}{4} = \cos^2 t + \sin^2 t = 1$.
The curve is the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$.
At $t = 0$: $(3, 0)$. At $t = \pi/2$: $(0, 2)$. At $t = \pi$: $(-3, 0)$. At $t = 3\pi/2$: $(0, -2)$. The curve is traced counterclockwise.
Mastery Checklist
Mental Model
A parametric curve is like a movie of a moving dot: at each frame $t$, the dot is at position $(x(t), y(t))$. The Cartesian equation is the set of all positions the dot ever occupies -- it is the “shadow” of the movie, showing the path but not the motion. Two different movies can produce the same shadow: $(\cos t, \sin t)$ and $(\cos 2t, \sin 2t)$ both trace the unit circle, but the second one goes around twice as fast.
Connections
Looking back
- Trigonometric identities: Pythagorean identities eliminate trigonometric parameters.
Looking ahead
- Direction and orientation (Section 10.1): The arrow on the curve is formalized.
- Parametric derivatives (Section 10.2): The slope $dy/dx$ is computed from the parameter $t$.