Parametric Equations

Parametric equations describe x and y separately using a third variable t — usually time. A regular equation tells you the shape of a path. Parametric equations tell you the shape AND how something moves along it — direction, speed, position at any moment.

The key ideaInstead of y = f(x), you have x = f(t) AND y = g(t). Both x and y depend on a third variable t. As t increases, you get a sequence of (x, y) points that trace a path — with direction built in.

x = f(t) y = g(t) t ∈ [a, b]

The point at time t is (f(t), g(t)). t is the parameter. As t increases, the point traces a path with direction.

Live animation — watch the point trace the path

t = 0.00 · x = — · y = —

Why parametric vs regular equations?Some curves aren't functions of x — a circle, for example, fails the vertical line test. Parametric equations describe any curve including ones that loop back on themselves or go left and right freely. They also encode direction: which way does the curve trace as t increases?

Building a t-table

x = t + 1, y = t² − 2, for t ∈ [−2, 2]

Choose t values in the interval, evaluate both equations

t	x = t + 1	y = t² − 2	Point (x, y)
−2	−1	2	(−1, 2)
−1	0	−1	(0, −1)
0	1	−2	(1, −2)
1	2	−1	(2, −1)
2	3	2	(3, 2)

Plot the points in order and draw arrows showing direction of increasing t

Curve traces left-to-right as t increases. Shape: parabola opening upward.

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Orientation — arrows are everything

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The orientation of a parametric curve is the direction it traces as t increases. Always mark this with arrows on your graph.

Two parametric equations can produce the same rectangular curve but with opposite orientations — they look identical on paper but are different parametrizations. The exam will ask about this.

Example: x = cos(t), y = sin(t) traces the unit circle counterclockwise. x = cos(−t), y = sin(−t) traces it clockwise. Same circle, opposite direction.

Vocabulary

Parameter

The variable t that both x and y depend on. Often represents time but not always.

Parametric equations

A pair x = f(t), y = g(t) that together describe a curve.

Orientation

The direction the curve traces as t increases. Shown with arrows on the graph.

t-table

A table of t, x(t), y(t) values used to plot parametric curves manually.

Eliminating the parameter

Solving for t in one equation and substituting into the other to get y = f(x).

Key patterns

x = t, y = t² → vertical parabola y = x²

x = t², y = t → horizontal parabola x = y²

x = cos(t), y = sin(t) → unit circle x²+y²=1

x = a·cos(t), y = b·sin(t) → ellipse x²/a²+y²/b²=1

Arrows show orientation — always draw them.

Eliminating the parameterConverting parametric equations to a rectangular equation (y in terms of x). Solve one equation for t, substitute into the other. Always carry over domain restrictions — the rectangular equation may look familiar but only covers part of it.

Method 1 — Solve and substitute

x = t + 2, y = t² — eliminate t

Solve x equation for t: x = t + 2 → t = x − 2

Substitute into y: y = t² = (x − 2)²

Check restrictions: if t ∈ [−2, 2], then x = t+2 ∈ [0, 4]. So restrict x to [0, 4].

y = (x − 2)², x ∈ [0, 4] — a parabola but only the portion for x between 0 and 4

x = 2t − 1, y = 3t + 4 — eliminate t (linear)

Solve for t from x: t = (x + 1)/2

Substitute: y = 3·(x+1)/2 + 4 = (3x+3)/2 + 4 = (3x + 11)/2

y = (3x + 11)/2 — a straight line

Method 2 — Trig identity (sin²t + cos²t = 1)

x = 3cos(t), y = 2sin(t) — eliminate t

Isolate trig functions: cos(t) = x/3, sin(t) = y/2

Use identity sin²t + cos²t = 1: (y/2)² + (x/3)² = 1

Rewrite: x²/9 + y²/4 = 1

x²/9 + y²/4 = 1 — an ellipse with a = 3, b = 2

x = cos(t), y = sin(t) — eliminate t

cos(t) = x, sin(t) = y

sin²t + cos²t = 1 → x² + y² = 1

x² + y² = 1 — the unit circle, traced counterclockwise

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The restriction trap — most common exam mistake

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When you eliminate t, the rectangular equation may look like a familiar full curve — but parametric equations usually only trace PART of it.

x = t², y = t for t ≥ 0 eliminates to x = y², but only for y ≥ 0. The full parabola goes both up and down; this version only traces the top half.

Always ask: what values does x take as t varies over its interval? Those are your x-restrictions on the rectangular equation.

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Connection to conics — the big picture

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Every conic section has a parametric form:

Circle: x = r·cos(t), y = r·sin(t)
Ellipse: x = a·cos(t), y = b·sin(t)
Parabola: x = t, y = t² (or x = t², y = t)
Hyperbola: x = a·sec(t), y = b·tan(t)

This is why parametric equations and conics appear in the same chapter — they're two ways of describing the same curves.

Elimination methods

Solve & substitute

Isolate t in one equation, plug into the other. Works when t appears linearly or as a power.

Trig identity

When x and y involve sin and cos of the same t, use sin²t + cos²t = 1. Isolate each trig function first.

Domain restriction

After eliminating t, restrict x to the values it actually takes over the t interval.

Exam reminders

Always carry restrictions from t to x when eliminating.

sin²t + cos²t = 1 is the key for circular/elliptic parametrics.

The rectangular form may be a full parabola; the parametric version might only be half.

Check: does the rectangular curve match the parametric by plugging in a few t values?

Common parametric shapesKnow these cold. On the exam you'll likely see one of these or a variation. The key is recognizing the form and knowing how the orientation works.

The four most common forms

Line: x = at + h, y = bt + k

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Linear in t → straight line. Slope = b/a. Point at t = 0: (h, k). Direction: moving in direction (a, b) as t increases.

Eliminate: y − k = (b/a)(x − h) — slope-intercept form

Parabola: x = t, y = at² + bt + c (or x and y swapped)

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x = t → just substitute: y = at² + bt + c becomes y = ax² + bx + c (vertical parabola)

→

y = t, x = at² → x = ay² (horizontal parabola)

The squared variable determines which way it opens

Circle/Ellipse: x = a·cos(t), y = b·sin(t)

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a = b: circle of radius a, centered at origin

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a ≠ b: ellipse with semi-axes a (horizontal) and b (vertical)

→

Orientation: counterclockwise as t goes from 0 to 2π

x²/a² + y²/b² = 1 — recognizable from conic sections chapter

Reversed orientation: x = a·cos(−t), y = b·sin(−t)

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cos(−t) = cos(t) (even), sin(−t) = −sin(t) (odd)

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Same ellipse/circle shape, but traced clockwise instead

Same rectangular equation x²/a² + y²/b² = 1, opposite orientation

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How to identify shape from parametric form

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Both linear in t? → Line
One linear, one quadratic? → Parabola (squared one determines direction)
Both trig (cos/sin)? → Circle or ellipse
sec and tan? → Hyperbola

Quick check: eliminate t and see what you get. The rectangular form is the shape without direction.

Shape cheat sheet

x=at+h, y=bt+k

Line, slope b/a, through (h,k) at t=0

x=t, y=t²

Vertical parabola y=x², left to right orientation

x=t², y=t

Horizontal parabola x=y², bottom to top orientation

x=r·cos t, y=r·sin t

Circle radius r, CCW orientation

x=a·cos t, y=b·sin t

Ellipse x²/a²+y²/b²=1, CCW orientation

x=a·sec t, y=b·tan t

Hyperbola x²/a²−y²/b²=1

Orientation rules

t increases left→right for x=t forms.

cos/sin traces counterclockwise (CCW) as t: 0→2π.

Negative t or reversed cos/sin → clockwise (CW).

Same shape + different speed = different parametrization.

Projectile motionThe biggest real-world application of parametric equations. Horizontal and vertical motion are independent — gravity only affects vertical. x and y each get their own equation, both depending on time t.

x(t) = (v₀ cos θ) t y(t) = −½gt² + (v₀ sin θ)t + h₀

v₀ = initial speed · θ = launch angle · g = 32 ft/s² or 9.8 m/s² · h₀ = initial height

Interactive projectile launcher

Launch angle θ45°

Initial speed v₀60 ft/s

Initial height h₀0 ft

Max height

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Range (distance)

—

Time in air

—

Equations

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45° gives maximum range on flat ground · Try different angles to see why

Solving projectile problems — the three questions

A ball is launched at 80 ft/s at 30°, from ground level. Find: time in air, range, max height.

Set up: x(t) = 80cos(30°)t = 80(√3/2)t = 40√3 t. y(t) = −16t² + 80sin(30°)t = −16t² + 40t. (g = 32, so ½g = 16)

Time in air: Solve y(t) = 0. −16t² + 40t = 0 → t(−16t + 40) = 0 → t = 0 or t = 40/16 = 2.5 sec

Range: x(2.5) = 40√3(2.5) = 100√3 ≈ 173.2 ft

Max height: Vertex of y(t). t_max = −b/(2a) = −40/(2·(−16)) = 40/32 = 1.25 sec. y(1.25) = −16(1.5625) + 40(1.25) = −25 + 50 = 25 ft

Time in air: 2.5 sec · Range: 100√3 ≈ 173.2 ft · Max height: 25 ft

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The three projectile questions and how to answer each

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Time in air: Set y(t) = 0, solve for t. Discard t = 0 (launch). The positive t is when it lands.

Range (horizontal distance): Plug the landing time into x(t).

Maximum height: The vertex of y(t). Use t = −b/(2a) to find when, then plug in to find height.

Height at a given distance: Set x(t) = distance, solve for t, plug t into y(t).

g values: 32 ft/s² or 9.8 m/s². The ½ in ½gt² makes it 16 ft/s² or 4.9 m/s² in the equation.

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Why 45° maximizes range (the math)

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Range = (v₀²/g) · sin(2θ). This is maximized when sin(2θ) = 1, which means 2θ = 90°, so θ = 45°.

Also: complementary angles give the same range. 30° and 60° produce the same horizontal distance — try it in the simulator above.

Higher angle → more time in air but less horizontal speed. Lower angle → more horizontal speed but less time. 45° is the perfect balance.

Projectile formulas

x(t) = (v₀ cos θ)t

Horizontal — constant speed, no gravity effect

y(t) = −½gt² + (v₀ sin θ)t + h₀

Vertical — gravity pulls down, initial upward velocity

g = 32 ft/s²

In feet. Makes ½g = 16.

g = 9.8 m/s²

In meters. Makes ½g = 4.9.

t_max height = −b/(2a)

The vertex formula on y(t). Time when vertical velocity = 0.

Exam reminders

Horizontal: NO gravity. Vertical: YES gravity (−½gt²).

Time in air: set y(t) = 0, take positive root.

Range: plug landing time into x(t).

Max height: vertex of y(t) at t = −b/(2a).

45° = max range. 30° and 60° give equal ranges.

h₀ ≠ 0: launched from a cliff/building. Set y(t) = 0 for landing, not just factoring.

Build your own parametric curve. Enter any expressions for x(t) and y(t) using the sliders, and watch the curve trace live with orientation arrows.

Parametric explorer

t range: 0 to2π

Parameter a2

Parameter b3

Arrows show orientation — direction of increasing t

Worked example — recognize the curve

Identify the curve: x = 4cos(t), y = 3sin(t), t ∈ [0, 2π]

Both involve cos and sin of same t → ellipse form

Isolate: cos(t) = x/4, sin(t) = y/3

Identity: (x/4)² + (y/3)² = 1 → x²/16 + y²/9 = 1

Orientation: at t=0: (4, 0). At t=π/2: (0, 3). Moving counterclockwise.

Ellipse: x²/16 + y²/9 = 1, a=4, b=3, traced counterclockwise starting at (4,0)

Explorer presets

Line

x=at, y=bt — slope b/a through origin

Parabola ↕

x=t, y=t² — vertical, left to right

Parabola ↔

x=t², y=t — horizontal, bottom to top

Circle

x=a·cos t, y=a·sin t — radius a, CCW

Ellipse

x=a·cos t, y=b·sin t — semi-axes a, b

Figure-8

x=a·sin t, y=b·sin(2t) — Lissajous curve

Spiral

x=t·cos t, y=t·sin t — Archimedean spiral

Quick recognition

Both linear in t → line

One linear, one quadratic → parabola

Both cos/sin same t → circle or ellipse

sin(t) and sin(2t) → figure-8 / Lissajous

t·cos(t), t·sin(t) → spiral