Parametric Curves

Parametric equations \(x = f(t),\; y = g(t)\) trace a curve as the parameter \(t\) varies — describing paths with loops and vertical segments that no function \(y = f(x)\) can draw, with direction of travel given by increasing \(t\).

By the end you'll be able to eliminate the parameter to find the Cartesian shape of a parametric curve — keeping only the piece the parameter's domain allows — and read off the direction of travel from the sign of \(dx/dt\).

Predict: increase t — which way along the curve does the dot move? For the parabola arc below, will it travel left-to-right or right-to-left? Slide t to check, then try the Loop and Vertical segment presets: watch the dot revisit a point or climb straight up — things a graph \(y = f(x)\) can never do.

t = −1.00 · (x, y) = (0.00, 3.00) · dx/dt = 1.00 > 0 → moving right

x = t + 1, y = t² − 2t — t from −1 to 3
traced so far (increasing t) full curve start point

A parametric curve \(x = f(t),\; y = g(t)\) is a path traced by a moving point. Eliminating \(t\) recovers a Cartesian equation — often only a piece of it — while the orientation comes from increasing \(t\) and is invisible in the Cartesian equation alone.

Visual

Picture the demo above: the same picture can hide different motions. The curve \(x = \cos t,\; y = \sin t\) starts at \((1,0)\) and sweeps the unit circle counterclockwise as \(t\) runs from \(0\) to \(2\pi\); swap to \(x = \sin t,\; y = \cos t\) and you get the identical circle traced clockwise from \((0,1)\). The parametrization carries more information than the shape — starting point and direction — which is why every parametric sketch gets arrows. And some shapes only exist parametrically: \(x = 3,\; y = t\) draws a vertical segment, and \(x = t^2 - 2,\; y = t^3 - 3t\) passes through \((1, 0)\) twice (at \(t = \pm\sqrt{3}\)), closing a loop. Both fail the vertical line test, so neither is a graph \(y = f(x)\).

Applied

Real paths come with a clock attached. A thrown ball's trajectory, a planet's orbit, the path traced by a point on a rolling wheel — each is naturally "where is it at time \(t\)?", which is exactly \(x = f(t),\; y = g(t)\). The ball's arc loops back down through heights it already visited; the wheel-point's path touches the ground with a vertical approach; the orbit closes on itself. None of these is a function \(y = f(x)\), but all are effortless parametrically — and the physics (which way, how fast, where at what time) lives in the parametrization, not the picture.

Worked example

Eliminate the parameter for \(x = t + 1,\; y = t^2 - 2t\), \(t \in [-1, 3]\), and describe the direction as \(t\) increases. Solve the simpler equation for \(t\): \(t = x - 1\). Substitute: \(y = (x-1)^2 - 2(x-1) = x^2 - 4x + 3\) — a parabola. But only a piece: \(t \in [-1,3]\) forces \(x = t + 1 \in [0, 4]\). Endpoints: at \(t = -1\), \((x,y) = (0,3)\); at \(t = 3\), \((x,y) = (4,3)\). Direction: \(dx/dt = 1 > 0\) throughout, so the dot moves left-to-right, from \((0,3)\) down through the vertex and back up to \((4,3)\).

Your turn

Try \(x = 2t,\; y = 4t + 1\), \(t \in [0, 3]\). Solve for the parameter: \(t = x/2\). Substitute: \(y = 4(x/2) + 1 = \) ____. Endpoints: \(t=0\) gives \((0, 1)\) and \(t=3\) gives \((6, 13)\). Direction: \(dx/dt = \) ____, so the segment is traced ____-to-____.

Reveal the answer

\(y = 2x + 1\) — a line segment, but only the piece with \(0 \le x \le 6\). \(dx/dt = 2 > 0\), so \(x\) increases with \(t\): the segment is traced left-to-right, from \((0,1)\) up to \((6,13)\). Both answers came from the same two moves as the worked example: substitute to get the shape, check the sign of \(dx/dt\) for the direction.

More info — why eliminating the parameter loses information

Eliminating \(t\) is a projection: you throw away the clock. Two things vanish. First, the domain: in the worked example the Cartesian equation \(y = x^2 - 4x + 3\) describes an infinite parabola, but the parameter interval \(t \in [-1,3]\) only produces \(0 \le x \le 4\) — always translate the \(t\)-interval into an \(x\)-interval before sketching. Second, the orientation: \(y = x^2 - 4x + 3\) says nothing about which end the dot starts from — that is recovered from \(dx/dt\) (and \(dy/dt\)), or by plotting a couple of \(t\)-values in order. Paul's Online Math Notes (linked under Dive deeper) works several examples where the eliminated equation is a whole curve but the parametric one traces only part of it, or retraces it repeatedly.

Check your understanding

Question 1 of 4

Eliminate the parameter for \(x = 3t,\; y = 9t^2\), \(t \in [0, 2]\). Which description is correct?

Question 2 of 4

The curve \(x = t^2 - 2,\; y = t^3 - 3t\) passes through the point \((1, 0)\) at both \(t = \sqrt{3}\) and \(t = -\sqrt{3}\). What does this tell you about the curve?

Question 3 of 4

As \(t\) increases from \(0\) to \(2\pi\), which way does \(x = \cos t,\; y = \sin t\) travel around the unit circle, and where does it start?

Question 4 of 4

For \(x = 1 - t,\; y = t^2\), \(t \in [0, 2]\): eliminate the parameter AND give the direction of travel as \(t\) increases.

Recap

  • \(x = f(t),\; y = g(t)\) traces a curve as \(t\) varies — including loops, self-intersections, and vertical segments that no function \(y = f(x)\) can draw.
  • Eliminate the parameter by solving one equation for \(t\) and substituting — but the result may be only a piece of the Cartesian curve: always convert the \(t\)-domain into an \(x\)-domain.
  • The direction of travel is set by increasing \(t\): \(dx/dt > 0\) means moving right, \(dx/dt < 0\) means moving left — extra information the Cartesian equation alone doesn't show.
  • The same shape admits many parametrizations differing in start point, direction, and speed — e.g. \(x=\cos t, y=\sin t\) (counterclockwise) vs. \(x=\sin t, y=\cos t\) (clockwise) for the unit circle.

Dive deeper

Sources

  • Parametric Curves