Parametric and Polar Functions

Most functions describe y directly from x. Parametric and polar descriptions instead give a point by a changing parameter or by a distance and direction. They are especially useful for paths, circles, repeated loops, and models in which time or angle is the natural input. Familiarity with coordinate geometry, trigonometric ratios, and functions is helpful; the lessons on Functions and Trigonometry provide useful background.

Parametric Equations: One Input, Two Coordinates

A parametric curve gives both coordinates in terms of a parameter, often t:

x = f(t) and y = g(t).

For each allowed value of t, calculate an ordered pair (x, y) and plot it. The parameter does more than locate points: as it increases, it tells us the direction in which the curve is traced. A single Cartesian equation can hide that direction or can fail to describe a path that passes through the same point more than once.

Worked example: make a table and trace a line

Graph x = 2t − 1, y = t + 2 for −1 ≤ t ≤ 3.

t	x = 2t − 1	y = t + 2	Point
−1	−3	1	(−3, 1)
0	−1	2	(−1, 2)
1	1	3	(1, 3)
3	5	5	(5, 5)

The points lie on a line segment, traced from (−3, 1) toward (5, 5). The restriction on t matters: without it, the same rules would trace the entire line.

Eliminating the Parameter

Eliminating the parameter means combining the two parametric equations to obtain a relation involving only x and y. Solve one equation for t when possible, then substitute. Keep the stated parameter interval: it can restrict the resulting Cartesian curve.

Worked example: a parabola with only one branch

Eliminate t from x = t², y = 3t − 1, where t ≥ 0.

y = 3t − 1
y + 1 = 3t
t = (y + 1)/3

x = t² = [(y + 1)/3]²
9x = (y + 1)²

Because t ≥ 0, (y + 1)/3 ≥ 0, so y ≥ −1.

The equation 9x = (y + 1)² describes a sideways parabola, but the original parametrization traces only the part with y ≥ −1. For instance, t = 0 gives (0, −1), then increasing t moves up and to the right.

Worked example: recognize a circle

For x = 4 cos t, y = 4 sin t, use sin² t + cos² t = 1:

x² + y² = (4 cos t)² + (4 sin t)²
          = 16(cos²t + sin²t)
          = 16

This is the circle centred at the origin with radius 4. For 0 ≤ t ≤ 2π, it begins at (4, 0) and travels counterclockwise once. Here the parameter can be interpreted as an angle measured in radians.

Parameters as Motion

When t represents time, the pair (x(t), y(t)) gives position. Always attach units. A repeated position need not mean the object has stopped; it may have returned to an earlier location at a later time.

Worked example: locate a moving object

A robot's position in metres after t seconds is x = 2t + 1 and y = 10 − t, for 0 ≤ t ≤ 4.

At t = 0:  (x, y) = (1, 10)
At t = 4:  (x, y) = (2·4 + 1, 10 − 4) = (9, 6)

Eliminate t:
x = 2t + 1  gives  t = (x − 1)/2
y = 10 − (x − 1)/2
2y = 21 − x, so x + 2y = 21.

The robot follows the segment of x + 2y = 21 from (1, 10) to (9, 6), moving right and downward. Its horizontal change is 8 m and vertical change is −4 m over 4 s, so its average velocity is (2, −1) m/s.

Polar Coordinates

A polar coordinate is written (r, θ). The radius r is the directed distance from the pole (the origin); θ is measured from the positive x-axis, usually counterclockwise. Angles may be in degrees or radians, but use the same unit throughout a calculation.

The same point can have many polar names. Adding a full turn leaves the point unchanged: (r, θ) = (r, θ + 2πk) for any integer k. A negative radius points in the opposite direction: (−r, θ) = (r, θ + π).

Converting between polar and Cartesian form

Draw the right triangle from the origin to the point. Its horizontal and vertical components give

x = r cos θ,   y = r sin θ,   r² = x² + y².

For a Cartesian-to-polar conversion, determine the quadrant before choosing θ. The expression tan θ = y/x alone cannot distinguish angles that differ by π.

Worked example: polar to Cartesian

Convert (6, 150°) to Cartesian coordinates.

x = 6 cos 150° = 6(−√3/2) = −3√3
y = 6 sin 150° = 6(1/2) = 3

Cartesian point: (−3√3, 3)

Worked example: Cartesian to polar

Convert (−3, 3√3) to polar coordinates with 0 ≤ θ < 2π.

r = √[(-3)² + (3√3)²] = √(9 + 27) = 6

tan θ = y/x = (3√3)/(−3) = −√3.
The point is in Quadrant II, where the reference angle is π/3.
Therefore θ = 2π/3.

Polar coordinate: (6, 2π/3)

Basic Polar Curves

A polar equation gives r as a function of θ. To sketch one, make a small angle table, convert selected points if needed, and connect them in the order θ increases. Negative r values plot on the ray opposite the stated angle.

Lines and circles

The equation r = a is a circle of radius |a| centred at the pole. Equations of the form r = 2a cos θ or r = 2a sin θ are circles through the pole.

Worked example: convert a polar circle

Convert r = 8 cos θ to Cartesian form.

r = 8 cos θ
r² = 8r cos θ
x² + y² = 8x
x² − 8x + y² = 0
(x − 4)² + y² = 16

The curve is a circle with centre (4, 0) and radius 4. Multiplying by r in the second line is valid here, including the pole: the final equation confirms that (0, 0) lies on the circle.

Rose curves and symmetry

Equations such as r = a cos(nθ) or r = a sin(nθ) form petal-shaped rose curves. If n is odd, there are n petals; if n is even, there are 2n petals. Each petal reaches radius |a|.

Useful symmetry tests are substitutions: if replacing θ by −θ leaves an equation unchanged, the curve is symmetric about the polar axis (the x-axis). If replacing θ by π − θ leaves it unchanged, it is symmetric about the vertical axis. If replacing r by −r, or θ by θ + π, leaves it unchanged, it is symmetric about the pole. These tests are evidence, not a substitute for plotting points.

Worked example: count and orient petals

For r = 3 sin(2θ), n = 2 is even, so the graph has 2(2) = 4 petals, each of length 3. A maximum occurs when sin(2θ) = 1:

2θ = π/2, so θ = π/4.

One petal points at π/4.
The other petals repeat every π/2:
π/4, 3π/4, 5π/4, 7π/4.

Practical Modelling

Choose coordinates that match the situation. Parametric equations are natural when time controls both horizontal and vertical position. Polar equations are natural when a quantity depends on direction from a fixed centre, such as the reach of a rotating sensor or a circular antenna pattern.

Worked example: a rotating sprinkler

A sprinkler reaches 12 m in every direction except that a wall blocks all directions from 120° through 180°. Its wet boundary can be modelled by r = 12 for 0° ≤ θ < 120° and 180° < θ ≤ 360°. The missing angular interval is essential: r = 12 alone would incorrectly claim a full circle.

Common Mistakes

• Dropping a parameter restriction. Elimination may produce a larger curve than the parametric path. State the corresponding restriction in x, y, or both.
• Forgetting direction. A Cartesian equation identifies a locus; it does not show where a moving object starts or which way it travels.
• Using the wrong quadrant for an inverse tangent. Check the signs of x and y after finding a reference angle.
• Treating a negative radius as impossible. Plot (−2, 30°) as (2, 210°), not 2 units along the 30° ray.
• Mixing degrees and radians. In particular, calculator trigonometric mode must agree with the angle unit in the problem.

Practice Set

1. Make a table for x = t − 2, y = 2t + 1 at t = 0, 1, 2. State the direction of travel as t increases.
2. Eliminate t from x = t + 4 and y = 2t − 1.
3. Eliminate t from x = 5 cos t, y = 5 sin t. Identify the curve.
4. A particle has x = 3t, y = 8 − 2t, 0 ≤ t ≤ 2. Give its starting point, ending point, and Cartesian path.
5. Convert the polar coordinate (4, 225°) to Cartesian form.
6. Convert the Cartesian point (0, −7) to polar form with 0 ≤ θ < 2π.
7. Convert r = 6 sin θ to Cartesian form and identify its centre and radius.
8. How many petals does r = 2 cos(3θ) have, and what is the length of each petal?
9. Give an equivalent polar coordinate with positive radius for (−5, π/6).
10. Does r = 4 cos θ have symmetry about the polar axis? Verify using a substitution.

Answer Checks

1. The points are (−2, 1), (−1, 3), and (0, 5). The path moves right and upward.
2. From x = t + 4, t = x − 4. Thus y = 2(x − 4) − 1 = 2x − 9.
3. x² + y² = 25(cos² t + sin² t) = 25. It is a circle centred at the origin with radius 5.
4. At t = 0 the point is (0, 8); at t = 2 it is (6, 4). Since t = x/3, y = 8 − 2x/3, or 2x + 3y = 24, restricted to that segment.
5. x = 4 cos 225° = −2√2 and y = 4 sin 225° = −2√2.
6. The radius is 7 and the point lies on the negative y-axis, so one answer is (7, 3π/2).
7. r² = 6r sin θ gives x² + y² = 6y, so x² + (y − 3)² = 9. The centre is (0, 3) and the radius is 3.
8. Because 3 is odd, there are 3 petals. Each reaches radius 2.
9. Add π to the angle: (5, 7π/6).
10. Yes. Replacing θ by −θ gives r = 4 cos(−θ) = 4 cos θ, the unchanged equation.