7.1 Parametric Equations

Big idea

An equation of the form $y = f(x)$ can only describe a curve that passes the vertical line test. That rules out the circle, the ellipse, every closed loop, and every path that doubles back on itself. Since those are exactly the curves that show up when something moves - a planet, a wheel rim, a projectile - the single-equation description is the wrong tool for a large part of the subject.

The fix is to stop treating one coordinate as a function of the other and make both coordinates functions of a third quantity. Call it $t$ and think of it as time. At each instant the pair $(x(t), y(t))$ names a point, and as $t$ runs through an interval the point traces a path. Nothing forbids the path from crossing itself or returning to where it started, because the constraint that killed the circle - one output per input - no longer applies.

A parametrization says more than the curve it draws. Two different parametrizations can trace exactly the same set of points while disagreeing about direction, speed, and how many times the set gets covered. The Cartesian equation records only the set. That extra information is not clutter: the direction of travel is what makes a tangent vector point one way rather than the other, and the speed is what makes an arc length integral come out right. When you eliminate the parameter you gain a familiar equation and you lose all of it.

So the two skills of this section pull in opposite directions and you need both. Eliminating the parameter turns an unfamiliar pair of equations into a curve you recognize, which is how you find out what you are looking at. Building a parametrization turns a curve you recognize into something you can do calculus on, which is how you get slopes, lengths and areas out of it. Keep the two moves distinct and remember that the second one is not simply undoing the first.

Decoder

A parametric curve is a pair of functions giving the two coordinates in terms of a common parameter, and it carries a direction of travel that the equation relating $x$ and $y$ does not.

The phrase common parameter is the whole construction. The two coordinate functions are independent of each other but share an input, and that shared input is what links them into a single moving point instead of two unrelated graphs. Change the shared input and both coordinates respond at once.

The phrase direction of travel is what you lose on elimination. Reading the parametrization at increasing values of $t$ gives an ordering of the points, drawn as an arrow on the curve and called its orientation. Reversing the sign of the parameter reverses the arrow while leaving every point in place, so the same Cartesian equation can correspond to two curves that differ in the one respect you care about.

The classic mistake is eliminating the parameter and forgetting the domain. If $t$ ranges over an interval rather than the whole line, or if the coordinate functions have restricted ranges, the resulting curve is only a piece of the Cartesian graph. Substituting away $t$ produces an equation that is true on the curve but usually describes more points than the curve contains. After every elimination, ask what values $x$ and $y$ can actually take and cut the graph down to that.

Definitions and results

Parametric equations. A plane curve is given parametrically by $x = x(t)$ and $y = y(t)$ for $t$ in an interval $I$. Each $t$ produces one point $(x(t), y(t))$, and the curve is the set of all such points. The variable $t$ is the parameter and $I$ is the parameter interval.

Orientation. The curve inherits a direction from increasing $t$. Marking that direction with an arrow is part of a complete description, and two parametrizations tracing the same points in opposite orders are different curves for the purposes of calculus.

Endpoints. If $I = [a,b]$ the curve has an initial point $(x(a), y(a))$ and a terminal point $(x(b), y(b))$. When the two coincide the curve is closed.

Eliminating the parameter. Solve one equation for $t$ and substitute into the other, or combine the two using an identity. The result is an equation in $x$ and $y$ satisfied by every point of the curve. It may be satisfied by other points too, so restrict it using the ranges of $x(t)$ and $y(t)$ on $I$.

Every graph is a curve. Given $y = f(x)$, the parametrization $x = t$, $y = f(t)$ traces the graph left to right. Parametrizing is strictly more general than graphing, never less.

Lines. The segment from $(x_0,y_0)$ to $(x_1,y_1)$ is $x = x_0 + (x_1 - x_0)t$, $y = y_0 + (y_1-y_0)t$ for $0 \le t \le 1$, arriving at $t = 1$.

Circles and ellipses. The circle of radius $r$ centred at $(h,k)$ is $x = h + r\cos t$, $y = k + r\sin t$ for $0 \le t \le 2\pi$, traversed counterclockwise from the rightmost point. Replacing the two radii independently, $x = h + a\cos t$ and $y = k + b\sin t$ gives the ellipse with semi-axes $a$ and $b$, since $\cos^2 t + \sin^2 t = 1$ turns into the standard equation.

Reparametrization. Substituting a new parameter for $t$ retraces the same points at a different rate. For instance $x = \cos(2t)$, $y = \sin(2t)$ on $[0,2\pi]$ covers the unit circle twice, and $x = \cos(-t)$, $y = \sin(-t)$ covers it once clockwise.

Slope of a parametric curve. Where the coordinate functions are differentiable and $dx/dt \neq 0$, the chain rule gives

$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $$

so the tangent direction comes from the two rates without ever solving for $y$ in terms of $x$.

Worked examples

A segment, and what the equation forgets

Consider $x = 1 + 2t$ and $y = 3 - t$ for $0 \le t \le 2$.

Solve the first equation for the parameter: $t = (x-1)/2$. Substitute into the second:

$$ y = 3 - \frac{x-1}{2}, \qquad \text{or} \qquad x + 2y = 7 $$

That is a full line, but the curve is not. As $t$ runs from $0$ to $2$, $x$ runs from $1$ to $5$, so the curve is the segment of $x + 2y = 7$ with $1 \le x \le 5$. It starts at $(1,3)$, ends at $(5,1)$, and is traversed left to right at constant speed.

Check the endpoints against the parametrization: at $t = 0$ you get $(1,3)$ and $1 + 6 = 7$; at $t = 2$ you get $(5,1)$ and $5 + 2 = 7$. Both lie on the line, and the equation alone would never have told you which part of it or in which direction.

An ellipse from a trigonometric identity

Consider $x = 3\cos t$ and $y = 2\sin t$ for $0 \le t \le 2\pi$.

Solving for $t$ is awkward, so use the Pythagorean identity instead. Isolate the trigonometric functions first, then square and add:

$$ \left(\frac{x}{3}\right)^2 + \left(\frac{y}{2}\right)^2 = \cos^2 t + \sin^2 t = 1 $$

so the curve is the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here the elimination loses nothing about which points are covered, since the parameter interval is a full period, but it still loses the orientation. At $t = 0$ the point is $(3,0)$; a little later, at $t = \pi/2$, it is $(0,2)$. The motion is counterclockwise, starting at the right-hand vertex.

The same graph traced badly

Consider $x = \sin t$ and $y = \sin^2 t$ for all real $t$.

Since $y = (\sin t)^2 = x^2$, every point lies on the parabola $y = x^2$. But $\sin t$ never leaves $[-1,1]$, so the curve is only the arc of the parabola with $-1 \le x \le 1$. Worse, as $t$ increases the point does not sweep across that arc once. It slides from $(0,0)$ out to $(1,1)$, back through the origin to $(-1,1)$, and returns, repeating forever with period $2\pi$.

The Cartesian equation $y = x^2$ says none of this. It does not know the curve is bounded, it does not know the motion oscillates, and it does not know the origin is visited infinitely often. This is the clearest case of the information a parametrization holds and an equation in $x$ and $y$ cannot.

Reading a slope off the two rates

Find the slope of the tangent to $x = 3\cos t$, $y = 2\sin t$ at $t = \pi/4$.

Differentiate each coordinate with respect to the parameter: $dx/dt = -3\sin t$ and $dy/dt = 2\cos t$. At $t = \pi/4$ both sine and cosine equal $\sqrt{2}/2$, so

$$ \frac{dy}{dx} = \frac{2\cos t}{-3\sin t} = \frac{\sqrt{2}}{-3\sqrt{2}/2} = -\frac{2}{3} $$

Check against implicit differentiation of the Cartesian form. From $\frac{x^2}{9} + \frac{y^2}{4} = 1$ you get $\frac{2x}{9} + \frac{y}{2}\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{4x}{9y}$. At $t = \pi/4$ the point is $x = 3\sqrt{2}/2$ and $y = \sqrt{2}$, giving $-\frac{4 \cdot 3\sqrt{2}/2}{9\sqrt{2}} = -\frac{6\sqrt{2}}{9\sqrt{2}} = -\frac{2}{3}$. The two routes agree, and the parametric one never required solving for $y$.

Practice

These drills mix the two directions. Some ask you to eliminate the parameter and identify the curve with its correct restrictions and orientation; others give a parametrization and ask for the tangent slope from the two coordinate rates.

Quiz

Five items on parametrizing curves, eliminating the parameter, orientation, and slopes from parametric equations.