Stewart Chapter 8 · Parametric Equations

Parametric Equations — Precalculus Complete Guide

Parametric form, eliminating the parameter by three methods, graphing with orientation arrows, cycloids, Lissajous figures, projectile motion, and parametric-to-polar conversion — with full worked examples and exam tips.

Definition and notationEliminating the parameterDirection and orientationCircles and ellipsesCycloidLine segmentsConic sectionsProjectile motionParticle motionParametric to polarLissajous figuresRectangular vs parametric vs polar

What Are Parametric Equations?

A parametric curve is defined by expressing both coordinates as functions of a separate variable called the parameter. In precalculus, the parameter is almost always called t and often represents time.

Standard Parametric Form

x = f(t)
y = g(t)
where t is the parameter (often representing time)

As the parameter t increases through its range (called the parameter interval), the point (x, y) = (f(t), g(t)) traces a path in the plane. The resulting set of points is called the parametric curve or plane curve.

Rectangular
y = f(x)
No direction, no crossing
Parametric
x = f(t)
y = g(t)
Direction + self-crossing allowed
Polar
r = f(θ)
Best for symmetric curves

Why Use Parametric Equations?

Direction of travel: Parametric form encodes which way the curve is traced as t increases. Rectangular form cannot distinguish counterclockwise from clockwise motion.
Self-crossing curves: Curves like the cycloid and Lissajous figures cross themselves and cannot be written as y = f(x) at all.
Physics modeling: Projectile motion naturally separates into horizontal x(t) and vertical y(t) components, both functions of time.

Plotting a Parametric Curve Step by Step

To graph the parametric curve x = t squared minus 1, y = 2t for t from negative 2 to 2, make a table of values by substituting several t values.

tx = t² − 1y = 2tPoint (x, y)
-23-4(3, -4)
-10-2(0, -2)
0-10(-1, 0)
102(0, 2)
234(3, 4)

Eliminating the parameter: From y = 2t we get t = y/2. Substituting into x = t squared minus 1 gives x = (y/2) squared minus 1, or x = y squared over 4 minus 1. This is a parabola opening rightward with vertex at (negative 1, 0). The arrows on the curve point downward-to-upward as t increases from negative 2 to 2.

Orientation — Direction of Motion

Orientation is the direction in which the curve is traced as the parameter t increases. On exam graphs, orientation is shown with arrows pointing in the direction of increasing t.

Finding orientation

Substitute two or three increasing t values, plot the points in order, and place arrows showing the sequence.

x = cos t, y = sin t
t = 0: (1, 0)
t = π/2: (0, 1)
t = π: (-1, 0)
t = 3π/2: (0, -1)
Counterclockwise circle

Reversing orientation

Replace t with negative t to reverse the direction of travel on the same curve.

x = cos(-t) = cos t
y = sin(-t) = -sin t
t = 0: (1, 0)
t = π/2: (0, -1)
t = π: (-1, 0)
Clockwise circle (same shape)

Exam Trap

Two parametric pairs can produce the exact same rectangular equation but trace the curve in opposite directions. If an exam asks you to identify orientation, always work from the parametric form by testing increasing t values — never assume from the rectangular equation.

Common Parametric Forms to Memorize

CurveParametric Form
Line segmentx = x₁ + (x₂−x₁)t, y = y₁ + (y₂−y₁)t, 0 ≤ t ≤ 1
Circle (center h, k)x = h + r cos t, y = k + r sin t, 0 ≤ t ≤ 2π
Ellipsex = h + a cos t, y = k + b sin t
Parabolax = at², y = 2at
Hyperbolax = a sec t, y = b tan t
Cycloidx = r(t − sin t), y = r(1 − cos t)
Projectile motionx = v₀ cosθ · t, y = −½gt² + v₀ sinθ · t

Eliminating the Parameter — Four Worked Examples

Eliminating the parameter converts the parametric pair into a single rectangular equation in x and y. Note that the resulting rectangular form may require a domain restriction to match the original parametric curve exactly.

1

Method 1 — Solve for t and substitute (linear equation)

x = 3t − 2, y = t² + 1, t ∈ ℝ
  1. 1.Solve the linear equation for t: x = 3t − 2 → t = (x + 2)/3
  2. 2.Substitute into y = t² + 1:
  3. 3.y = ((x + 2)/3)² + 1
  4. 4.y = (x + 2)²/9 + 1
  5. 5.Result: y = (x + 2)²/9 + 1 (upward-opening parabola, vertex at (−2, 1))

Because t ranges over all reals, x ranges over all reals too — no domain restriction on the rectangular form.

2

Method 2 — Pythagorean identity (cos and sin)

x = 5 cos t, y = 3 sin t, 0 ≤ t ≤ 2π
  1. 1.Isolate each trig function: cos t = x/5, sin t = y/3
  2. 2.Apply the identity cos² t + sin² t = 1:
  3. 3.(x/5)² + (y/3)² = 1
  4. 4.x²/25 + y²/9 = 1
  5. 5.Result: an ellipse with a = 5 (horizontal semi-axis) and b = 3 (vertical semi-axis)

The full interval 0 to 2π traces the complete ellipse counterclockwise starting at (5, 0).

3

Method 3 — Pythagorean identity (sec and tan)

x = 2 sec t, y = 3 tan t, −π/2 < t < π/2
  1. 1.Isolate: sec t = x/2, tan t = y/3
  2. 2.Apply the identity sec² t − tan² t = 1:
  3. 3.(x/2)² − (y/3)² = 1
  4. 4.x²/4 − y²/9 = 1
  5. 5.Result: right branch only of a hyperbola (since sec t ≥ 1 means x ≥ 2)

The domain restriction t ∈ (−π/2, π/2) limits the curve to the right branch where x is positive.

4

Method 4 — Restricting the domain (line segment)

x = 1 + 2t, y = 3 − t, 0 ≤ t ≤ 4
  1. 1.Solve for t from x: t = (x − 1)/2
  2. 2.Substitute into y: y = 3 − (x − 1)/2 = (−x + 7)/2
  3. 3.Find x-range: t = 0 → x = 1; t = 4 → x = 9
  4. 4.Result: y = (−x + 7)/2 restricted to 1 ≤ x ≤ 9

This is a line segment, not an infinite line. The parameter restriction converts to a domain restriction in rectangular form.

The Cycloid

The cycloid is the curve traced by a fixed point on the rim of a circle as the circle rolls without slipping along a straight line. It is one of the most important curves in mathematics and has a parametric form that cannot be simplified to a manageable rectangular equation.

Cycloid Equations (circle of radius r)

x = r(t − sin t)
y = r(1 − cos t)
t = angle the circle has rotated (in radians). Cusps occur at t = 0, 2π, 4π, ... (point touches ground).

Key features

  • Cusps: The tracing point touches the ground at x = 0, 2πr, 4πr, ...
  • Arch height: Maximum y = 2r (diameter of rolling circle)
  • Arch width: One arch spans a horizontal distance of 2πr
  • No rectangular form: The curve self-intersects at cusps and cannot be expressed as y = f(x)

Historical importance

  • Brachistochrone: A bead slides fastest from point A to lower point B along an inverted cycloid arch, not a straight line
  • Tautochrone: A bead released from any point on the arch reaches the bottom in the same time
  • Arch area: Equals 3πr squared (three times the rolling circle area)

Checking a point on the cycloid (r = 1)

At t = π/2: x = 1(π/2 − sin(π/2)) = π/2 − 1 ≈ 0.571
y = 1(1 − cos(π/2)) = 1 − 0 = 1
Point: (π/2 − 1, 1) on the ascending part of the first arch

Line Segments in Parametric Form

Any line segment from point P1 = (x1, y1) to point P2 = (x2, y2) can be parameterized so that t = 0 gives the start and t = 1 gives the end.

Line Segment Formula

x = x&sub1; + (x&sub2; − x&sub1;)t
y = y&sub1; + (y&sub2; − y&sub1;)t
0 ≤ t ≤ 1

Worked example: segment from (2, 5) to (8, -1)

x = 2 + (8 − 2)t = 2 + 6t
y = 5 + (-1 − 5)t = 5 − 6t
0 ≤ t ≤ 1
Check: t = 0 gives (2, 5). t = 0.5 gives (5, 2) (midpoint). t = 1 gives (8, -1). Correct.

Useful trick: To write only part of a line (not from 0 to 1), use t ranging over any interval you choose. For a segment traversed twice as slowly, use 0 ≤ t ≤ 2 with x = x1 + (x2 minus x1) times t/2.

Parametric Form of Conic Sections

All conic sections — circles, ellipses, parabolas, and hyperbolas — have natural parametric representations. These are especially useful for graphing and for calculus applications.

Circle

x = h + r cos t
y = k + r sin t
0 ≤ t ≤ 2π

Center (h, k), radius r. Counterclockwise orientation. Eliminate the parameter using cos squared t plus sin squared t = 1 to get (x minus h) squared plus (y minus k) squared = r squared.

Ellipse

x = h + a cos t
y = k + b sin t
0 ≤ t ≤ 2π

Center (h, k), horizontal semi-axis a, vertical semi-axis b. When a = b this is a circle. Eliminate parameter to get standard ellipse form.

Parabola (horizontal axis)

x = at squared
y = 2at
t ∈ &reals;

From y = 2at, get t = y divided by (2a). Substitute to get x = a times (y squared divided by 4a squared) = y squared divided by (4a). So y squared = 4ax. Focus at (a, 0).

Hyperbola

x = a sec t
y = b tan t
t ≠ ±π/2

Uses identity sec squared t minus tan squared t = 1. Eliminates to x squared over a squared minus y squared over b squared = 1. Different t-intervals give different branches.

Projectile Motion — The Classic Application

Projectile motion separates naturally into independent horizontal and vertical components, making parametric equations the ideal model. Gravity acts only on the vertical component.

Standard Projectile Equations

x(t) = v&sub0; cos θ · t + x&sub0;
y(t) = −½g t² + v&sub0; sin θ · t + y&sub0;
v&sub0; = initial speed (ft/s or m/s)
θ = launch angle above horizontal
g = 32 ft/s² in feet, 9.8 m/s² in meters
x&sub0;, y&sub0; = initial position (often both 0)

Worked Example — Full Solution

A projectile is launched from the ground at 80 ft/s at 45 degrees. Find: (a) maximum height, (b) time of flight, (c) horizontal range.

Setup
v₀ = 80 ft/s, θ = 45°, launched from ground. x = 80 cos(45°) · t = 40√2 · t, y = −16t² + 80 sin(45°) · t = −16t² + 40√2 · t
Max height
dy/dt = −32t + 40√2 = 0 → t = 40√2/32 = 5√2/4 ≈ 1.77 s. y(5√2/4) = −16(5√2/4)² + 40√2(5√2/4) = −16(50/16) + 200/4 = −50 + 50 = 50 ft
Time of flight
y = 0: −16t² + 40√2 t = 0 → t(−16t + 40√2) = 0 → t = 5√2/2 ≈ 3.54 s (nonzero root)
Range
x(5√2/2) = 40√2 · (5√2/2) = 40√2 · 5√2/2 = 200 ft
50 ft
Maximum height
at t ≈ 1.77 s
3.54 s
Time of flight
when y returns to 0
200 ft
Horizontal range
maximum at 45°

Key Formulas for Projectile Motion

Max height time
t = v&sub0; sin θ / g
Time of flight
T = 2v&sub0; sin θ / g
Range
R = v&sub0;² sin(2θ) / g

Parametric Equations and Particle Motion

When parametric equations model a particle moving in the plane, t represents time and (x(t), y(t)) is the position at time t. We can analyze speed, direction, and path from the parametric form.

Position, velocity, speed

Position: (x(t), y(t))
Velocity: (dx/dt, dy/dt)
Speed: √((dx/dt)² + (dy/dt)²)
Direction: arctan(dy/dt &div; dx/dt)

Worked example

x = 3 cos t, y = 2 sin t
dx/dt = -3 sin t
dy/dt = 2 cos t
Speed = √(9 sin²t + 4 cos²t)
Max speed at t = π/2: √9 = 3

Common Particle Motion Questions

  • When does the particle move right? When dx/dt is positive.
  • When does the particle move up? When dy/dt is positive.
  • When does the particle stop? When both dx/dt = 0 and dy/dt = 0.
  • How many times does the particle pass through a point? Solve x(t) = a and y(t) = b simultaneously.

Converting Parametric Equations to Polar Form

To convert from parametric to polar, use the relationships r squared = x squared plus y squared and theta = arctan(y over x), along with x = r cos theta and y = r sin theta. The typical path is: eliminate t to get rectangular form, then convert rectangular to polar.

Example 1 — Circle

Parametric: x = 3 cos t, y = 3 sin t
Rectangular: x² + y² = 9
Polar: r = 3

Example 2 — Line

Parametric: x = t, y = 2t
Rectangular: y = 2x
Polar sub: r sin θ = 2r cos θ
Polar: tan θ = 2, so θ = arctan(2)

Example 3 — Parabola

Parametric: x = t², y = 2t (with a = 1)
Rectangular: y² = 4x
Polar sub: r² sin² θ = 4r cos θ
Polar: r = 4 cos θ / sin² θ

Quick conversion reference: x = r cos θ, y = r sin θ, r² = x² + y², tan θ = y/x. Always check the quadrant when computing theta = arctan(y/x) — the arctangent only gives the correct angle in quadrants I and IV without adjustment.

Lissajous Figures

Lissajous figures (also called Lissajous curves or Bowditch curves) are the parametric curves described by:

Lissajous General Form

x = A sin(at + δ)
y = B sin(bt)
a and b are frequencies, δ is the phase difference, A and B are amplitudes. The shape depends critically on the ratio a : b.
Ratio a : bEquations (A = B = 1)Shape
1 : 1, δ = 0x = sin t, y = sin tStraight line (diagonal)
1 : 1, δ = π/2x = sin(t + π/2), y = sin tCircle (if A = B) or ellipse
2 : 1x = sin(2t), y = sin tFigure-8 / parabolic arc
3 : 2x = sin(3t), y = sin(2t)Three-lobed closed curve
3 : 1x = sin(3t), y = sin tTwo-lobed figure

When do Lissajous figures close?

A Lissajous figure is a closed curve if and only if the ratio a/b is a rational number. When a/b = p/q in lowest terms, the curve closes after q cycles of y (or p cycles of x). When a/b is irrational, the curve never closes and eventually fills a rectangle.

Real-world applications

  • Oscilloscopes use Lissajous figures to measure frequency ratios
  • Guitar tuners display Lissajous patterns on analog screens
  • Clock-face designs (the Spirograph toy)
  • Identifying phase shifts in AC circuits

Rectangular vs Parametric vs Polar — When to Use Each

PropertyRectangular y = f(x)Parametric x=f(t), y=g(t)Polar r = f(θ)
Shows directionNoYes (orientation)Partial
Handles self-crossingNoYesYes
Best forAlgebra, most functionsMotion, physics, complex curvesRotationally symmetric curves
Calculus (slope)dy/dx directly(dy/dt) / (dx/dt)More complex formula
Area formula∫ f(x) dx∫ y(t) x'(t) dt½ ∫ r² dθ
Example curvesPolynomials, trig functionsCycloid, projectile, LissajousRoses, limacons, cardioids

Frequently Asked Questions

What are parametric equations and how do they differ from rectangular equations?

Parametric equations express both x and y as separate functions of a third variable called the parameter, usually t: x = f(t) and y = g(t). A rectangular equation directly links x and y, like y = x squared. The key advantage of parametric form is that it carries direction information — the orientation of the curve as t increases — which rectangular form cannot represent. Two parametric pairs can produce the same rectangular graph but trace it in opposite directions. Parametric form is also essential for curves that cross themselves, such as cycloids and Lissajous figures, which cannot be written as a single y = f(x).

How do you eliminate the parameter from parametric equations?

There are three main methods. Method 1 (algebraic substitution): solve one equation for t, then substitute into the other. Works best when one equation is linear. Example: x = 2t minus 1 gives t = (x+1)/2, substitute into y = t squared plus 3 to get y = (x+1) squared over 4 plus 3. Method 2 (Pythagorean identity): when x = a times cos t and y = b times sin t, isolate the trig functions and apply cos squared t plus sin squared t = 1. This gives the ellipse equation x squared over a squared plus y squared over b squared = 1. Method 3 (other trig identities): when equations involve tan and sec, use sec squared t minus tan squared t = 1 to eliminate the parameter.

What is the cycloid and what are its parametric equations?

A cycloid is the curve traced by a point on the rim of a circle as the circle rolls along a straight line. Its parametric equations are x = r times (t minus sin t) and y = r times (1 minus cos t), where r is the radius of the rolling circle and t is the angle of rotation. The cycloid has cusps at every integer multiple of 2 pi (where the tracing point touches the ground) and arches in between. It cannot be expressed in a simple rectangular form, making parametric equations essential. The cycloid is historically famous as the solution to the brachistochrone problem — the path of fastest descent under gravity.

How do you find maximum height, time of flight, and range in projectile motion using parametric equations?

For a projectile launched from the origin at speed v0 and angle theta, the equations are x(t) = v0 times cos(theta) times t and y(t) = negative one-half times g times t squared plus v0 times sin(theta) times t. Maximum height occurs when the vertical velocity is zero: set dy/dt = negative g times t plus v0 sin theta = 0, solving t = v0 sin theta divided by g. Substitute back to find max height. Time of flight is found by setting y = 0 and solving for the nonzero t: t = 2 v0 sin theta divided by g. Range is then x at that time: R = v0 squared times sin(2 theta) divided by g. Maximum range occurs at theta = 45 degrees.

What are Lissajous figures in parametric equations?

Lissajous figures are parametric curves of the form x = A sin(a times t plus delta) and y = B sin(b times t), where a and b are frequencies and delta is a phase shift. The shape of the figure depends on the ratio a to b. When a to b = 1 to 1 and delta = pi over 2, the curve is an ellipse (or circle if A equals B). When a to b = 2 to 1, the figure forms a figure-8 or parabolic arc. When a to b = 3 to 2, a more complex closed loop appears. Lissajous figures are used in electronics to visualize phase relationships between two signals and appear on oscilloscopes.

How do you convert parametric equations to polar form?

To convert from parametric form x = f(t), y = g(t) to polar form r = h(theta), use the substitutions r = square root of x squared plus y squared and theta = arctan(y divided by x). In practice you usually eliminate t first to get rectangular form y = g(x) or F(x, y) = 0, then convert that to polar using x = r cos theta and y = r sin theta. For example, the parametric circle x = a cos t, y = a sin t gives rectangular x squared plus y squared = a squared, which converts to polar r = a.

Quick Reference — Exam Cheat Sheet

Eliminate parameter

  • Linear: solve for t, substitute
  • Cos/sin: use cos² + sin² = 1
  • Sec/tan: use sec² − tan² = 1
  • Keep domain restrictions!

Projectile formulas

  • x = v&sub0; cos θ · t
  • y = −½g t² + v&sub0; sin θ · t
  • Max height t = v&sub0; sin θ / g
  • Range = v&sub0;² sin 2θ / g

Orientation tips

  • Plot 3-4 increasing t values
  • Draw arrows between points
  • Negate t to reverse direction
  • Same shape, different orientation

Circle and ellipse

  • x = h + r cos t (circle)
  • y = k + r sin t (circle)
  • x = h + a cos t (ellipse)
  • y = k + b sin t (ellipse)

Cycloid

  • x = r(t − sin t)
  • y = r(1 − cos t)
  • Cusps at t = 0, 2π, 4π, ...
  • Max height = 2r (diameter)

Parametric to polar

  • x = r cos θ, y = r sin θ
  • r² = x² + y²
  • tan θ = y/x (check quadrant)
  • Eliminate t first, then convert

Practice Parametric Equations

Elimination, orientation, projectile motion, and Lissajous figures — hundreds of practice problems with full step-by-step solutions. Free to start, no account needed.

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