Parametric Equations
A parametric curve describes a path by giving both coordinates as functions of a shared parameter $t$:
$$x = f(t), \qquad y = g(t)$$
As $t$ moves through an interval, the point $(x, y) = (f(t), g(t))$ traces a path in the plane. The parameter $t$ often represents time — you can think of the curve as the trajectory of a moving point.
Why Not Just Use $y = f(x)$?
A function $y = f(x)$ assigns exactly one $y$ value to each $x$. That works for a parabola or a sine wave, but it fails for any curve that doubles back on itself. A vertical line $x = 1$ violates it. A circle violates it. The figure-eight of a Lissajous curve violates it.
Parametric equations have no such restriction. The point $(f(t), g(t))$ can move in any direction, revisit the same $x$ coordinates, cross itself, trace loops, and spiral outward. The only rule is that both $f$ and $g$ must be well-defined functions of $t$.
A circle is the simplest demonstration. In Cartesian form, a unit circle requires two equations:
$$y = \sqrt{1 - x^2} \qquad \text{(upper half)}$$ $$y = -\sqrt{1 - x^2} \qquad \text{(lower half)}$$
In parametric form, the full circle is just:
$$x = \cos(t), \qquad y = \sin(t), \qquad t \in [0, 2\pi]$$
One clean pair of equations traces the entire circle exactly once.
The Unit Circle
The parametric circle works because $\cos^2(t) + \sin^2(t) = 1$ for all $t$. Substituting $x = \cos(t)$ and $y = \sin(t)$ gives $x^2 + y^2 = 1$, confirming the point always lies on the unit circle. The parameter $t$ is the angle measured counterclockwise from the positive $x$-axis.
Scaling by constants $a$ and $b$ gives an ellipse:
$$x = a\cos(t), \qquad y = b\sin(t)$$
With $a = 2$ and $b = 1$, the ellipse is twice as wide as it is tall. This is far more natural parametrically than the implicit form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
The Astroid
The astroid is the curve traced by a point on the rim of a small circle rolling inside a larger circle. Its parametric equations are:
$$x = \cos^3(t), \qquad y = \sin^3(t)$$
As $t$ sweeps from $0$ to $2\pi$, the point traces four sharp cusps — one on each axis — connected by four inward-curving arcs.
The implicit form of the astroid exists — it is $x^{2/3} + y^{2/3} = 1$ — but writing it down hides the structure. The parametric form makes it transparent: the cusps occur when $\cos(t)$ or $\sin(t)$ equals $\pm 1$, i.e. at $t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$. At those points one of the derivatives is zero, and the curve changes direction abruptly.
The total arc length of the astroid is $6a$ where $a$ is the radius of the outer circle — in our case $a = 1$, so the perimeter is exactly $6$. This is easy to derive using parametric arc length:
$$L = \int_0^{2\pi} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}, dt$$
Lissajous Figures
A Lissajous figure uses different frequencies for $x$ and $y$:
$$x = \sin(at), \qquad y = \sin(bt)$$
The ratio $a:b$ controls the shape. When $a = b$, the curve is a line or ellipse. When $a \neq b$, the curve weaves back and forth across itself, forming a mesh of loops.
Ratio 3:2
With $a = 3$ and $b = 2$, the curve closes after one full pass through $t \in [0, 2\pi]$. It has $a - 1 = 2$ inner loops in the horizontal direction and $b - 1 = 1$ inner loop in the vertical direction.
Ratio 4:3
The 4:3 ratio produces a denser figure with 3 horizontal loops and 2 vertical. A useful rule: a Lissajous figure with frequency ratio $a:b$ in lowest terms has $b$ tangencies on the left edge and $a$ tangencies on the top edge. Counting these tangencies is how oscilloscope operators read off frequency ratios.
Roulettes: Hypotrochoids
A hypotrochoid is traced by a point attached to a small circle rolling inside a larger one — the same mechanism as a spirograph toy. With an outer radius of $R = 3$ and inner radius $r = 1$, with the tracing point at distance $d = 1$ from the inner center:
$$x = (R - r)\cos(t) + d\cos!\left(\tfrac{R-r}{r},t\right)$$ $$y = (R - r)\sin(t) - d\sin!\left(\tfrac{R-r}{r},t\right)$$
For $R = 3$, $r = d = 1$:
$$x = 3\cos(t) + \cos(3t), \qquad y = 3\sin(t) - \sin(3t)$$
The curve has three-fold symmetry and three outer loops. The shape depends on the ratio $R/r$: when this ratio is rational, the curve closes; when irrational, it fills an annular region indefinitely.
The Cycloid
The cycloid is traced by a point on the rim of a circle of radius $r$ rolling along a straight line. For $r = 1$:
$$x = t - \sin(t), \qquad y = 1 - \cos(t)$$
As $t$ increases from $0$ to $2\pi$, the circle completes one full revolution and the point traces a single arch: it starts at the origin, rises to a maximum height of $2$ at $t = \pi$, and returns to the line at $(2\pi, 0)$.
The cycloid has two famous properties. First, it is the brachistochrone — the curve of fastest descent under gravity between two points not directly above each other, as proved by Johann Bernoulli in 1696. Second, it is the tautochrone — regardless of where on the arch a ball is released, it reaches the bottom in the same amount of time. Both properties depend on the particular shape of the cusp, which is what parametric equations capture naturally.
The Derivative
For a parametric curve $(f(t), g(t))$, the slope of the tangent line is given by the chain rule:
$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$$
This is valid wherever $f'(t) \neq 0$. When $f'(t) = 0$ and $g'(t) \neq 0$, the tangent is vertical. When both derivatives vanish simultaneously, the point is a cusp — the curve meets itself with an abrupt change of direction, as at the tips of the astroid and the cusps of the cycloid.
Conclusion
Parametric equations separate the description of a curve from the constraint that $y$ be a function of $x$. By treating $x$ and $y$ symmetrically as outputs of a shared parameter, they make naturally dynamic objects — trajectories, rolling curves, oscillating signals — easy to write down and analyze. The circle, the astroid, Lissajous figures, and roulettes are all parametric curves that resist a clean Cartesian description. The change in perspective is small, but the range of shapes it unlocks is vast.