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Math · Calculus

Parametric Equations

Plot x(t) and y(t), with the tangent slope, arc length and where the tangents turn.

Use t as the parameter, with an explicit * for multiplication.
Where the tangent and the point are reported.
Point at t = 1
(0.5403, 0.84147)

the curve is closed

Slope dy/dx
-0.64209

dx/dt = -0.84147, dy/dt = 0.5403

Arc length
6.2832

over t from 0 to 6.283

The curve traced by x = cos(t) and y = sin(t) as t runs from 0 to 6.283, with the point at t = 1 marked.

x from -1.10equal scale on both axesto 1.10

dx/dt at t-0.841471zero means a vertical tangent
dy/dt at t0.540302zero means a horizontal tangent
d²y/dx²-1.678399negative, so the curve is concave down here
Horizontal tangents1.5708, 4.7124values of t where dy/dt = 0
Vertical tangents3.1416, 6.2832values of t where dx/dt = 0
x range-1 to 1over the t interval given
y range-1 to 1

The derivatives are central differences and the length is Simpson’s rule over 1,000 intervals — both accurate to about ten digits on smooth curves, and neither exact.

Classic curves — tap to load

Parametric equations give x and y each as a function of a third variable, usually t. That lets a curve double back, cross itself or close a loop — things y = f(x) cannot do. The slope is dy/dx = (dy/dt) ÷ (dx/dt).

What the parameter buys you

Writing y as a function of x forces one y for every x, which rules out most interesting curves. A circle fails immediately: above x = 0 there are two points. Splitting it into two half-circles works but is clumsy, and it hides the thing that matters — a circle is one curve, traced once.

Parametrising fixes this by describing motion rather than shape. As t advances, the point (x(t), y(t)) moves, and the path it traces is the curve. Nothing stops it from returning to an x it has already visited, so loops, crossings and closed curves come for free. The cycloid, the astroid and the Lissajous figures on this page all have short parametric forms and either ugly or impossible rectangular ones.

The slope comes from the chain rule

Since both coordinates depend on t, the chain rule gives dy/dx = (dy/dt) ÷ (dx/dt) wherever dx/dt is not zero. When dx/dt is zero and dy/dt is not, the curve is momentarily moving straight up: the tangent is vertical and the slope genuinely does not exist, which is reported rather than papered over. When both vanish at once the curve has a cusp — the cycloid does this where it touches the ground — and no tangent direction is defined at all.

dy/dx = (dy/dt) ÷ (dx/dt)     L = ∫ √((dx/dt)² + (dy/dt)²) dt

the length integral is the speed of the tracing point, integrated over t

  1. 1
    Differentiate each coordinate with respect to t. For x = cos t and y = sin t: dx/dt = −sin t and dy/dt = cos t.
  2. 2
    Divide to get the slope. dy/dx = cos t ÷ (−sin t) = −cot t. At t = 1 that is −0.642.
  3. 3
    Set dy/dt = 0 for horizontal tangents. cos t = 0 at t = π/2 and 3π/2 — the top and bottom of the circle.
  4. 4
    Set dx/dt = 0 for vertical ones. sin t = 0 at t = 0 and π, the left and right extremes, where dy/dx is undefined.
  5. 5
    Integrate the speed for the arc length. √(sin²t + cos²t) = 1, so the length over one full turn is 2π — the circumference of the unit circle, as it must be.

Classic curves and their lengths

Each length is a closed form the tool reproduces to about ten digits.

Curvex(t), y(t)Range of tArc length
Unit circlecos t, sin t0 to 2π2π ≈ 6.283185
Cycloid, one archt − sin t, 1 − cos t0 to 2πexactly 8
Astroid, whole curvecos³t, sin³t0 to 2πexactly 6
Parabola y = x²t, t²0 to 1≈ 1.478943
Line to (3, 4)3t, 4t0 to 1exactly 5

Eliminating the parameter, and when not to

Exam questions often ask you to remove t and recover a rectangular equation. For x = cos t, y = sin t, squaring and adding gives x² + y² = 1 immediately. It usually works by solving one equation for t and substituting, or by spotting an identity that links the two.

What the rectangular form loses is the direction and the timing. x² + y² = 1 is the circle as a set of points; the parametric form says it is traced anticlockwise, starting at (1, 0), once per 2π. For a projectile, x = 20t and y = 20t − 4.9t² eliminate to a parabola, but only the parametric version tells you the apex is reached at t = 20/9.8 ≈ 2.04 seconds. Where the question is about motion, eliminating the parameter throws away the answer.

A note on what this page computes numerically. The derivatives are central differences and the arc length is Simpson’s rule over a thousand intervals — accurate to roughly ten digits on a smooth curve, which is why the closed forms above reproduce exactly to the digits shown. They are not symbolic: a curve with a genuine singularity inside the interval will give a length that is wrong rather than one flagged as impossible, so keep the range on the smooth part.

What are parametric equations?
A pair x(t) and y(t) giving both coordinates in terms of a third variable. As t runs through its range the point (x(t), y(t)) traces the curve.
Why use them instead of y = f(x)?
Because y = f(x) allows only one y per x, which rules out circles, loops and any curve that doubles back. A parametrisation has no such restriction and also records the direction of travel.
How do I find dy/dx?
Divide: dy/dx = (dy/dt) ÷ (dx/dt). Both derivatives are with respect to the parameter, and the division is the chain rule.
What happens where dx/dt is zero?
The tangent is vertical and dy/dx does not exist. If dy/dt is zero at the same moment the curve has a cusp — as the cycloid does where it meets the ground — and there is no tangent direction at all.
How is arc length computed?
By integrating √((dx/dt)² + (dy/dt)²) over the range of t. That square root is the speed of the tracing point, so the integral is total distance travelled.
How do I eliminate the parameter?
Solve one equation for t and substitute into the other, or use an identity. For x = cos t and y = sin t, squaring and adding gives x² + y² = 1 in one step.
What is lost when the parameter is eliminated?
The direction, the starting point and the timing. A projectile’s path eliminates to a parabola, but only the parametric form says the apex comes at t ≈ 2.04 seconds.