---
title: Parametric Curves and Their Calculus
module: Parametric Equations and Polar Coordinates
moduleNumber: 8
lessonNumber: 1
order: 801
summary: >
  A parametric curve gives x and y separately as functions of a third variable,
  recording not only a path but the direction and timing with which it is traced.
  Eliminating the parameter recovers a Cartesian equation; the slope, area,
  arc-length, and surface-area formulas run directly on the parameter, with the
  cycloid and astroid as worked examples.
topics: [Parametric Equations and Polar Coordinates]
sources:
  - book: Stewart
    ref: "Ch. 10 — Parametric Equations and Polar Coordinates; §10.1 Curves Defined by Parametric Equations, §10.2 Calculus with Parametric Curves"
---

A curve in the plane need not be the graph of a function. A particle can loop
back over itself, cross its own path, or trace a closed circuit, and no equation
$y = F(x)$ can describe such a path because it fails the Vertical Line Test. To
address this, let both coordinates depend on a common third variable instead of
making $y$ depend on $x$.

> **Definition (Parametric curve).** Given functions $f$ and $g$ on an interval,
> the equations
> $$
> x = f(t), \qquad y = g(t)
> $$
> are **parametric equations** and $t$ is the **parameter**. As $t$ varies, the
> point $(x, y) = (f(t), g(t))$ traces a **parametric curve** $C$. When $t$
> ranges over $[a, b]$, the curve has **initial point** $(f(a), g(a))$ and
> **terminal point** $(f(b), g(b))$.

The parameter carries information that a bare curve discards. It fixes an
**orientation** (the direction of increasing $t$) and a schedule (which point is
occupied when). If $t$ is time, $(f(t), g(t))$ is the position of a moving
particle, and the same set of points can be traced by different parametric
equations at different speeds and in different directions.

## Tracing and orientation

Consider $x = t^2 - 2t$, $y = t + 1$. Tabulating a few values of $t$ and plotting
the points reveals a curve that opens to the right, traced upward as $t$
increases.

$$
% caption: The curve $x = t^2-2t$, $y = t+1$ traced for $t \in [-2, 4]$; arrows
% mark the direction of increasing $t$, and the labelled points are equally
% spaced in $t$ but not in distance.
\begin{tikzpicture}[>=stealth, scale=0.62, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-2,0) -- (9,0) node[right, font=\footnotesize] {$x$};
  \draw[->, black] (0,-2) -- (0,6.4) node[above, font=\footnotesize] {$y$};
  \draw[acc, very thick, ->] plot[domain=-2:0, samples=30] ({\x*\x-2*\x},{\x+1});
  \draw[acc, very thick, ->] plot[domain=0:2, samples=30] ({\x*\x-2*\x},{\x+1});
  \draw[acc, very thick, ->] plot[domain=2:4, samples=30] ({\x*\x-2*\x},{\x+1});
  \fill[black] (8,-1) circle (2.5pt) node[right, font=\footnotesize] {t=-2};
  \fill[black] (0,1) circle (2.5pt) node[left, font=\footnotesize] {$t=0$};
  \fill[black] (-1,2) circle (2.5pt) node[left, font=\footnotesize] {$t=1$};
  \fill[black] (0,3) circle (2.5pt) node[left, font=\footnotesize] {$t=2$};
  \fill[black] (8,5) circle (2.5pt) node[right, font=\footnotesize] {$t=4$};
\end{tikzpicture}
$$

To confirm the shape, eliminate the parameter. From $y = t + 1$ we get
$t = y - 1$, and substituting into the first equation,

$$
x = (y-1)^2 - 2(y-1) = y^2 - 4y + 3,
$$

a parabola opening rightward. Eliminating $t$ produces a Cartesian equation for
the underlying set of points, but it discards the orientation and the timing.
Those are retained only in the parametric form.

**Different parametrizations, same points.** The equations
$x = \cos t$, $y = \sin t$ with $0 \le t \le 2\pi$ satisfy $x^2 + y^2 = 1$ and
trace the unit circle once counterclockwise from $(1,0)$. The equations
$x = \sin 2t$, $y = \cos 2t$ satisfy the same Cartesian equation but start at
$(0,1)$ and trace the circle twice clockwise. A curve is a set of points; a
parametric curve is that set together with a way of moving along it.

| Representation | Kind of object | Retains orientation? | Retains speed? |
| --- | --- | --- | --- |
| $y = F(x)$ | function graph | no | no |
| $F(x, y) = 0$ | implicit curve (point set) | no | no |
| $x = f(t)$, $y = g(t)$ | parametric curve | yes | yes |

Any function graph $y = F(x)$ is trivially parametric with $x = t$, $y = F(t)$,
and any curve $x = G(y)$ becomes $x = G(t)$, $y = t$. Parametric form is strictly
more general.

## The cycloid

Parametric form is essential when a curve's natural description is kinematic.
Roll a circle of radius $r$ along the $x$-axis and track a
fixed point $P$ on its rim, starting at the origin. The path $P$ traces is a
**cycloid**.

Take the angle of rotation $\theta$ as the parameter, with $\theta = 0$ when $P$
is at the origin. After the circle has turned through $\theta$ radians it has
rolled a distance equal to the arc it laid down, so its center sits at
$(r\theta, r)$. Measuring $P$'s displacement from that center gives its
coordinates.

$$
% caption: One arch of the cycloid $x = r(\theta - \sin\theta)$,
% $y = r(1-\cos\theta)$; the point $P$ rides the rim of a circle of radius $r$
% rolling along the axis, and the center advances to $(r\theta, r)$.
\begin{tikzpicture}[>=stealth, scale=1.05, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (-0.3,0) -- (7.0,0) node[right, font=\footnotesize] {ground};
  \draw[acc, very thick] plot[domain=0:6.2832, samples=110] ({\x - sin(deg(\x))}, {1 - cos(deg(\x))});
  \def\tp{4.0}
  \draw[black] (\tp,1) circle (1);
  \fill[black] (\tp,1) circle (1.4pt) node[below left, font=\footnotesize, black] {$C$};
  \coordinate (P) at ({\tp - sin(deg(\tp))}, {1 - cos(deg(\tp))});
  \fill[acc] (P) circle (2.6pt);
  \node[acc, anchor=east, font=\footnotesize] at (P) {$P$};
  \draw[black, dashed] (\tp,1) -- (P);
  \fill[black] (\tp,0) circle (1.6pt);
  \node[black, anchor=north, font=\footnotesize] at (\tp,0) {contact};
\end{tikzpicture}
$$

The horizontal shift is the rolled distance $r\theta$ minus the horizontal offset
$r\sin\theta$ of $P$ from the center; the height is $r$ minus the vertical offset
$r\cos\theta$:

$$
x = r(\theta - \sin\theta), \qquad y = r(1 - \cos\theta), \qquad \theta \in \mathbb{R}.
$$

One arch corresponds to $0 \le \theta \le 2\pi$. Eliminating $\theta$ here yields
a Cartesian equation too complicated to be useful, so the parametric description
is the one used in practice.

## Tangents

The slope of a parametric curve follows from the chain rule. Writing $y$ as a
function of $x$ along the curve,[^chain]

$$
\frac{\d y}{\d t} = \frac{\d y}{\d x}\cdot\frac{\d x}{\d t},
$$

and solving gives the slope directly in terms of the parameter derivatives.

> **Theorem (Slope of a parametric curve).** If $x = f(t)$, $y = g(t)$ are
> differentiable and $\dfrac{\d x}{\d t} \ne 0$, then
> $$
> \frac{\d y}{\d x} = \frac{\,\d y/\d t\,}{\,\d x/\d t\,}.
> $$
> The curve has a **horizontal tangent** where $\d y/\d t = 0$ (and $\d x/\d t \ne 0$)
> and a **vertical tangent** where $\d x/\d t = 0$ (and $\d y/\d t \ne 0$).

Reading the curve as a moving particle, $\d x/\d t$ and $\d y/\d t$ are the horizontal
and vertical velocities, and the slope is their ratio. The mnemonic is that the
$\d t$'s cancel.

$$
% caption: The tangent's slope is rise over run: a step along the tangent has
% horizontal leg $dx$ and vertical leg $dy$, and their ratio $dy/dx$ equals the
% velocity ratio $(dy/dt)/(dx/dt)$.
\begin{tikzpicture}[>=stealth, scale=1.0, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (2,1.6) .. controls (3.0,1.9) and (3.6,2.9) .. (4.2,3.7);
  \draw[black] (0.6,0.7) .. controls (1.2,1.0) and (1.6,1.35) .. (2,1.6);
  \coordinate (Pt) at (2,1.6);
  \fill[black] (Pt) circle (2pt);
  \draw[acc, very thick, ->] (Pt) -- (3.4,2.6) node[right, font=\footnotesize] {tangent};
  \draw[black, ->] (Pt) -- (3.4,1.6) node[midway, below, font=\footnotesize] {$dx$};
  \draw[black, ->] (3.4,1.6) -- (3.4,2.6) node[midway, right, font=\footnotesize] {$dy$};
  \draw[black, dashed] (Pt) -- (3.4,1.6);
  \node[anchor=east, font=\footnotesize] at (Pt) {slope $=\dfrac{dy}{dx}$};
\end{tikzpicture}
$$

> **Worked example.** Find the slope of the cycloid $x = r(\theta - \sin\theta)$,
> $y = r(1 - \cos\theta)$ at $\theta = \pi/3$, and locate its horizontal and
> vertical tangents.
>
> The parameter derivatives are $\d x/\d\theta = r(1 - \cos\theta)$ and
> $\d y/\d\theta = r\sin\theta$, so
>
> $$
> \frac{\d y}{\d x} = \frac{r\sin\theta}{r(1 - \cos\theta)} = \frac{\sin\theta}{1 - \cos\theta}.
> $$
>
> At $\theta = \pi/3$: $\sin\theta = \tfrac{\sqrt3}{2}$ and $1 - \cos\theta = \tfrac12$,
> so the slope is $(\sqrt3/2)/(1/2) = \sqrt3$. The tangent is horizontal where
> $\sin\theta = 0$ with $1 - \cos\theta \ne 0$, i.e. $\theta = (2n-1)\pi$, at the
> arch tops $\bigl((2n-1)\pi r,\, 2r\bigr)$. At $\theta = 2n\pi$ both derivatives
> vanish; l'Hospital's rule gives
>
> $$
> \lim_{\theta \to 2n\pi} \frac{\sin\theta}{1 - \cos\theta}
> = \lim_{\theta \to 2n\pi} \frac{\cos\theta}{\sin\theta} = \pm\infty,
> $$
>
> so the cusps at $\theta = 2n\pi$ carry vertical tangents.

**Second derivative.** Concavity needs $\d^2 y/\d x^2$, obtained by applying the same
slope rule to $\d y/\d x$ in place of $y$:

$$
\frac{\d^2 y}{\d x^2} = \frac{\dfrac{\d}{\d t}\!\left(\dfrac{\d y}{\d x}\right)}{\dfrac{\d x}{\d t}}.
$$

A common error is to write $\d^2 y/\d x^2 = (\d^2 y/\d t^2)/(\d^2 x/\d t^2)$. That is false;
the outer derivative acts on the whole slope quotient.

### A curve that crosses itself

A single parametric curve can pass through one point at two different times, and
the two passages carry different tangents.

$$
% caption: The curve $x = t^2$, $y = t^3-3t$ crosses itself at $(3,0)$, reached at
% $t = \sqrt3$ and $t = -\sqrt3$ with tangent slopes $+\sqrt3$ and $-\sqrt3$;
% horizontal tangents sit at the dots on the line $x=1$ and a vertical tangent at
% the origin.
\begin{tikzpicture}[>=stealth, scale=0.85, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (-0.4,0) -- (4.6,0) node[right, font=\footnotesize] {$x$};
  \draw[->, black] (0,-3.0) -- (0,3.3) node[above, font=\footnotesize] {$y$};
  \draw[acc, very thick] plot[domain=-2.05:2.05, samples=140] ({\x*\x},{\x*\x*\x-3*\x});
  \draw[black, dashed] (2.1,-1.386) -- (3.9,1.386);
  \draw[black, dashed] (2.1,1.386) -- (3.9,-1.386);
  \fill[black] (3,0) circle (2.2pt);
  \node[font=\footnotesize, anchor=north] at (3,-0.5) {(3,0)};
  \fill[acc] (1,2) circle (1.8pt) node[left, font=\footnotesize] {(1,2)};
  \fill[acc] (1,-2) circle (1.8pt);
  \draw[black] (0.55,2) -- (1.45,2);
  \draw[black] (0.55,-2) -- (1.45,-2);
  \draw[black] (0,-0.6) -- (0,0.6);
  \fill[black] (0,0) circle (1.8pt);
\end{tikzpicture}
$$

> **Worked example.** For $C : x = t^2$, $y = t^3 - 3t$, find the tangents at the
> self-crossing, the horizontal and vertical tangents, and the concavity.
>
> _Self-crossing._ Since $y = t(t^2 - 3) = 0$ at $t = 0$ and $t = \pm\sqrt3$, and
> $x = 3$ at $t = \pm\sqrt3$, the point $(3, 0)$ is reached by two parameter
> values, so $C$ crosses itself there. The slope is
>
> $$
> \frac{\d y}{\d x} = \frac{3t^2 - 3}{2t} = \frac{3}{2}\!\left(t - \frac1t\right),
> $$
>
> equal to $\pm\sqrt3$ at $t = \pm\sqrt3$. The two tangents at $(3,0)$ are
> $y = \sqrt3\,(x - 3)$ and $y = -\sqrt3\,(x - 3)$.
>
> _Horizontal and vertical tangents._ $\d y/\d t = 3t^2 - 3 = 0$ at $t = \pm1$
> gives horizontal tangents at $(1, -2)$ and $(1, 2)$; $\d x/\d t = 2t = 0$ at
> $t = 0$ gives a vertical tangent at $(0, 0)$.
>
> _Concavity._
>
> $$
> \frac{\d^2 y}{\d x^2}
> = \frac{\dfrac{\d}{\d t}\!\left[\tfrac{3}{2}\!\left(t - t^{-1}\right)\right]}{2t}
> = \frac{\tfrac32\!\left(1 + t^{-2}\right)}{2t}
> = \frac{3\,(t^2 + 1)}{4t^3},
> $$
>
> positive for $t > 0$ (concave up) and negative for $t < 0$ (concave down).

## Areas

The area under a curve traced by $x = f(t)$, $y = g(t)$ as $t$ runs from $\alpha$
to $\beta$ comes from the substitution rule applied to $A = \int_a^b y\,\d x$ with
$\d x = f'(t)\,\d t$:[^subst]

$$
A = \int_a^b y\,\d x = \int_\alpha^\beta g(t)\,f'(t)\,\d t,
$$

where the limits $\alpha, \beta$ are the parameter values giving $x = a$ and
$x = b$. For the area under one arch of the cycloid, $x = r(\theta - \sin\theta)$
gives $\d x = r(1 - \cos\theta)\,\d\theta$, and $\theta$ runs from $0$ to $2\pi$:

$$
A = \int_0^{2\pi} r(1-\cos\theta)\cdot r(1-\cos\theta)\,\d\theta
= r^2 \int_0^{2\pi} (1 - \cos\theta)^2\,\d\theta.
$$

Expanding $(1 - \cos\theta)^2 = 1 - 2\cos\theta + \tfrac12(1 + \cos 2\theta)$ and
integrating over a full period leaves only the constant terms $1 + \tfrac12$:

$$
A = r^2\!\left(\frac32\cdot 2\pi\right) = 3\pi r^2.
$$

The area under one arch is exactly three times the area of the rolling circle.

## Arc length

For a curve $y = F(x)$ the length is $\int \sqrt{1 + (\d y/\d x)^2}\,\d x$. Substituting
the parametric slope and $\d x = (\d x/\d t)\,\d t$ converts this to an integral over
$t$; the same formula holds for any traced-once curve, proved directly by
polygonal approximation.

$$
% caption: Arc length as a limit of polygon lengths: sample points
% $P_0, P_1, \dots, P_n$ on the curve at successive parameter values, joined by
% chords whose total length approaches the length of $C$ as the mesh shrinks.
\begin{tikzpicture}[>=stealth, scale=1.0, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[acc, very thick] (0,0.4) .. controls (1.4,1.9) and (3.0,-0.4) .. (5.2,1.5);
  \coordinate (p0) at (0,0.4);
  \coordinate (p1) at (0.87,0.91);
  \coordinate (p2) at (1.80,0.89);
  \coordinate (p3) at (2.82,0.72);
  \coordinate (p4) at (3.95,0.80);
  \coordinate (p5) at (5.2,1.5);
  \draw[black] (p0) -- (p1) -- (p2) -- (p3) -- (p4) -- (p5);
  \foreach \p in {p0, p1, p2, p3, p4, p5}
    \fill[black] (\p) circle (1.8pt);
  \node[anchor=north east, font=\footnotesize] at (p0) {$P_0$};
  \node[anchor=south, font=\footnotesize] at (p1) {$P_1$};
  \node[anchor=south, font=\footnotesize] at (p3) {$P_i$};
  \node[anchor=north west, font=\footnotesize] at (p5) {$P_n$};
\end{tikzpicture}
$$

> **Theorem (Parametric arc length).** If $x = f(t)$, $y = g(t)$ have continuous
> derivatives on $[\alpha, \beta]$ and $C$ is traversed exactly once as $t$
> increases from $\alpha$ to $\beta$, then the length of $C$ is
> $$
> L = \int_\alpha^\beta \sqrt{\left(\frac{\d x}{\d t}\right)^2 + \left(\frac{\d y}{\d t}\right)^2}\;\d t.
> $$

The traversed-once condition matters. For the circle $x = \sin 2t$,
$y = \cos 2t$ on $[0, 2\pi]$ the integral gives $4\pi$, twice the circumference,
because the point rounds the circle twice.

**Length of one cycloid arch.** With $\d x/\d\theta = r(1 - \cos\theta)$ and
$\d y/\d\theta = r\sin\theta$,

$$
\left(\frac{\d x}{\d\theta}\right)^2 + \left(\frac{\d y}{\d\theta}\right)^2
= r^2\bigl(1 - 2\cos\theta + \cos^2\theta + \sin^2\theta\bigr)
= 2r^2(1 - \cos\theta).
$$

The half-angle identity $1 - \cos\theta = 2\sin^2(\theta/2)$ turns the square
root into $2r\sin(\theta/2)$ on $0 \le \theta \le 2\pi$, so

$$
L = \int_0^{2\pi} 2r\sin\!\frac{\theta}{2}\,\d\theta
= 2r\Bigl[-2\cos\tfrac{\theta}{2}\Bigr]_0^{2\pi} = 8r.
$$

One arch has length $8r$, eight times the generating radius.

$$
% caption: The astroid $x = a\cos^3 t$, $y = a\sin^3 t$: a hypocycloid with four
% cusps on the axes, traced by a point on a circle of radius $a/4$ rolling inside
% a circle of radius $a$; its total length is $6a$.
\begin{tikzpicture}[>=stealth, scale=1.45, font=\small]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[black] (-1.35,0) -- (1.35,0) node[right, font=\footnotesize] {$x$};
  \draw[black] (0,-1.35) -- (0,1.35) node[above, font=\footnotesize] {$y$};
  \draw[acc, very thick] plot[domain=0:360, samples=180] ({(cos(\x))^3},{(sin(\x))^3});
  \foreach \pt in {(1,0),(0,1),(-1,0),(0,-1)} \fill[acc] \pt circle (1.3pt);
  \node[font=\footnotesize, anchor=south west] at (1,0.04) {cusp};
\end{tikzpicture}
$$

> **Worked example.** Find the total length of the astroid $x = a\cos^3 t$,
> $y = a\sin^3 t$ ($a > 0$).
>
> Differentiating, $\d x/\d t = -3a\cos^2 t\,\sin t$ and
> $\d y/\d t = 3a\sin^2 t\,\cos t$, so
>
> $$
> \left(\frac{\d x}{\d t}\right)^2 + \left(\frac{\d y}{\d t}\right)^2
> = 9a^2\cos^2 t\,\sin^2 t\,(\cos^2 t + \sin^2 t) = 9a^2\sin^2 t\,\cos^2 t.
> $$
>
> The curve is symmetric across both axes, so its length is four times the
> first-quadrant arc $0 \le t \le \pi/2$, on which $\sin t\cos t \ge 0$ and the
> square root is $3a\sin t\cos t$:
>
> $$
> L = 4\int_0^{\pi/2} 3a\,\sin t\,\cos t\;\d t
> = 12a\left[\tfrac12\sin^2 t\right]_0^{\pi/2} = 6a.
> $$

## Surface area

Rotating a parametric curve about the $x$-axis and adapting the surface-of-
revolution formula gives, with $g(t) \ge 0$ and the curve traced once,

$$
S = \int_\alpha^\beta 2\pi\, g(t)\,\sqrt{\left(\frac{\d x}{\d t}\right)^2 + \left(\frac{\d y}{\d t}\right)^2}\;\d t
= \int 2\pi y\,\d s,
$$

where $\d s = \sqrt{(\d x/\d t)^2 + (\d y/\d t)^2}\;\d t$ is the arc-length element. For the
semicircle $x = r\cos t$, $y = r\sin t$, $0 \le t \le \pi$, the element is
$\d s = r\,\d t$, and rotation about the $x$-axis sweeps out a sphere:

$$
S = \int_0^\pi 2\pi (r\sin t)\, r\,\d t = 2\pi r^2\bigl[-\cos t\bigr]_0^\pi = 4\pi r^2,
$$

the standard surface area of a sphere of radius $r$.

## The parametric calculus, collected

Every formula above replaces $\d x$ by $(\d x/\d t)\,\d t$ and integrates over the
parameter. The single arc-length element $\d s$ recurs in length, surface area,
mass, and centroid computations.

| Quantity | Cartesian form | Parametric form |
| --- | --- | --- |
| Slope | $\dfrac{\d y}{\d x}$ | $\dfrac{\d y/\d t}{\d x/\d t}$ |
| Concavity | $\dfrac{\d^2 y}{\d x^2}$ | $\dfrac{(\d/\d t)(\d y/\d x)}{\d x/\d t}$ |
| Area | $\displaystyle\int_a^b y\,\d x$ | $\displaystyle\int_\alpha^\beta g(t)\,f'(t)\,\d t$ |
| Arc length | $\displaystyle\int_a^b \sqrt{1 + (y')^2}\,\d x$ | $\displaystyle\int_\alpha^\beta \sqrt{(x')^2 + (y')^2}\,\d t$ |
| Surface area ($x$-axis) | $\displaystyle\int 2\pi y\,\d s$ | $\displaystyle\int_\alpha^\beta 2\pi g(t)\,\sqrt{(x')^2 + (y')^2}\,\d t$ |

To study an unfamiliar parametric curve, a fixed sequence of steps recovers its
shape without ever eliminating the parameter.

```algorithm
caption: Analyze a parametric curve $x = f(t)$, $y = g(t)$
compute $dx/dt$ and $dy/dt$
mark horizontal tangents where $dy/dt = 0$ and $dx/dt \ne 0$
mark vertical tangents where $dx/dt = 0$ and $dy/dt \ne 0$
for each point reached by two parameter values do
  record a self-intersection with a tangent slope per value
compute $d^2y/dx^2$ and split the $t$-axis by its sign for concavity
trace the curve in the direction of increasing $t$
```

[^chain]: Stewart, §10.2 — the slope formula for parametric curves follows from
the [chain rule](/calculus/derivatives/differentiation-rules-and-the-chain-rule)
applied to $y$ as a function of $x$ along the curve.
[^subst]: Stewart, §10.2 — the parametric area formula is the
[substitution rule](/calculus/integrals/the-substitution-rule) applied to
$\int y\,\d x$; the arc-length and surface-area formulas adapt §8.1–§8.2.