Mathematics.

parametric and polar

Parametric Equations

Calculus II35 minDifficulty5 out of 10

You should know: functions, coordinate plane

Overview

A parametric curve describes x and y (and possibly z) each as separate functions of a third variable, the parameter t — typically thought of as time. Instead of y = f(x), you write x = x(t), y = y(t). This lets you describe curves that fail the vertical line test (circles, loops, cusps) and naturally encodes motion: position, velocity, and speed all fall out of differentiating the parametric functions.

Intuition

Imagine tracking a bug crawling on a table: at each moment t, it has an x-coordinate and a y-coordinate. Plotting (x(t), y(t)) for all t traces the bug's path. This is fundamentally different from y=f(x), which requires exactly one y for each x — a parametric curve can loop back, cross itself, or retrace, because the parameter t supplies the 'when', decoupling the path's shape from any functional restriction on x and y.

Interactive Graph

Trace a parametric curve as t increases

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Formal Definition

Definition

A parametric curve in the plane is a pair of functions of a parameter t:

x=x(t),y=y(t),t[α,β]x = x(t), \quad y = y(t), \quad t \in [\alpha, \beta]

Each point on the curve corresponds to one value of t

dydx=dy/dtdx/dt=y(t)x(t)\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)}

Chain rule applied to eliminate t, valid where x'(t) ≠ 0

Parametric slope formula
d2ydx2=ddt(dydx)dx/dt\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}

Second derivative requires differentiating dy/dx with respect to t again, then dividing by dx/dt

L=αβ(dxdt)2+(dydt)2dtL = \int_{\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt

Total distance traveled along the curve from t=α to t=β

Arc length

Notation

NotationMeaning
x(t), y(t)x(t),\ y(t)Coordinate functions of the parameter t
y(t)x(t)\frac{y'(t)}{x'(t)}Slope of the tangent line, computed without eliminating t

Derivation

The arc length formula follows from approximating the curve by straight segments and taking a limit, exactly as with Riemann sums:

Δsi(Δxi)2+(Δyi)2=(ΔxiΔt)2+(ΔyiΔt)2Δt\Delta s_i \approx \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2} = \sqrt{\left(\frac{\Delta x_i}{\Delta t}\right)^2+\left(\frac{\Delta y_i}{\Delta t}\right)^2}\, \Delta t

Distance formula on each small segment, rewritten in terms of Δt

L=limniΔsi=αβx(t)2+y(t)2dtL = \lim_{n\to\infty}\sum_i \Delta s_i = \int_\alpha^\beta \sqrt{x'(t)^2+y'(t)^2}\,dt

The Riemann sum converges to the arc length integral as Δt → 0

Properties

Non-uniqueness of parametrization

The same curve can be traced by infinitely many different (x(t),y(t))\text{The same curve can be traced by infinitely many different } (x(t),y(t))

Condition: e.g. (cos t, sin t) and (cos 2t, sin 2t) both trace the unit circle, at different speeds

Horizontal/vertical tangents

Horizontal tangent where y(t)=0, x(t)0; vertical tangent where x(t)=0, y(t)0\text{Horizontal tangent where } y'(t)=0,\ x'(t)\neq 0;\ \text{vertical tangent where } x'(t)=0,\ y'(t)\neq 0

Applications

Projectile motion is naturally parametric: x(t) = v₀cos(θ)t, y(t) = v₀sin(θ)t − ½gt², with t as actual time.

Worked Examples

  1. Differentiate each component with respect to t.

    x(t)=2t,y(t)=3t23x'(t) = 2t, \quad y'(t) = 3t^2 - 3
  2. Apply the parametric slope formula.

    dydx=3t232t\frac{dy}{dx} = \frac{3t^2-3}{2t}
  3. Substitute t = 1.

    dydxt=1=332=0\frac{dy}{dx}\Big|_{t=1} = \frac{3-3}{2} = 0

Answer: dy/dx = 0 at t=1 (a horizontal tangent)

Practice Problems

Difficulty 5/10

Find dy/dx for x = e^t, y = t·e^t at t = 0.

Difficulty 6/10

A particle moves with x(t) = t² − 4t, y(t) = 2t. Find the arc length traveled from t=0 to t=2.

Common Mistakes

Common Mistake

Computing d²y/dx² by simply differentiating dy/dx with respect to t (forgetting to divide by dx/dt again).

d²y/dx² = [d/dt(dy/dx)] / (dx/dt), NOT d/dt(dy/dx) alone — the chain rule must be applied a second time.

Common Mistake

Assuming arc length only requires ∫√(1+(dy/dx)²)dx as in the Cartesian case.

For parametric curves, use L=∫√(x'(t)²+y'(t)²)dt directly — converting to the Cartesian arc-length formula is often impossible or unnecessarily complicated when x(t) isn't invertible.

Historical Background

Parametric descriptions of curves arose naturally from mechanics — describing the path of a projectile or a point on a rolling wheel (the cycloid) required expressing x and y in terms of time rather than each other. Roberval, Pascal, and Huygens studied the cycloid parametrically in the 17th century, and Euler formalized the general parametric approach to curves in the 18th century as part of his systematic treatment of analytic geometry.

  1. 1630s-1650s

    Roberval, Torricelli, and Pascal study the cycloid using motion-based (parametric) descriptions

    Gilles de Roberval, Blaise Pascal

  2. 1748

    Euler systematizes parametric curves in Introductio in analysin infinitorum

    Leonhard Euler

Summary

  • Parametric equations describe x and y as separate functions of a parameter t, allowing curves that aren't graphs of functions.
  • Slope: dy/dx = y'(t)/x'(t) (chain rule), valid where x'(t) ≠ 0.
  • Second derivative requires re-dividing by dx/dt after differentiating dy/dx with respect to t.
  • Arc length: L = ∫√(x'(t)² + y'(t)²) dt over the parameter interval.
  • The same curve can have many different parametrizations tracing it at different speeds or directions.

References