Representing curves and points using alternative coordinate systems. Covers parametric equations: defining x(t) and y(t) to trace curves (lines, circles, projectile paths), eliminating the parameter to find Cartesian equations. Introduces polar coordinates (r, theta), converting between polar and Cartesian forms, and graphing basic polar curves.
High School Advanced • 9-12
Until now, you have described curves as y = f(x), where y is explicitly a function of x. But many curves cannot be expressed this way -- a full circle, for instance, fails the vertical line test. Parametric equations and polar coordinates offer two powerful alternatives. Parametric equations describe motion by letting both x and y depend on a third variable (time). Polar coordinates describe position by distance and direction rather than horizontal and vertical displacement. Together, they vastly expand what you can describe and analyze.
A parametric curve is defined by a pair of equations:
where t is the parameter (often representing time). As t varies, the point (x, y) traces out a curve in the plane. The parameter gives you something Cartesian equations do not: a sense of direction and speed along the curve.
| Curve | Parametric Equations | Parameter Range |
|---|---|---|
| Line through (x_0, y_0) with direction (a, b) | x = x_0 + at, y = y_0 + bt | all real t |
| Circle centered at origin, radius r | x = r cos(t), y = r sin(t) | 0 ≤ t < 2π |
| Ellipse with semi-axes a, b | x = a cos(t), y = b sin(t) | 0 ≤ t < 2π |
| Projectile (initial velocity v_0, angle α) | x = v_0 cos(α) t, y = v_0 sin(α) t - (1/2)g t^2 | t ≥ 0 |
To convert parametric equations to a single Cartesian equation, solve one parametric equation for t and substitute into the other.
For trigonometric parametrizations, use the Pythagorean identity sin^2(t) + cos^2(t) = 1 instead.
Polar coordinates describe a point by its distance from the origin (r) and angle from the positive x-axis (θ):
Unlike Cartesian coordinates, the same point can have multiple polar representations (by adding 2π to θ, or by using negative r with a shifted angle).
(Adjust θ for the correct quadrant.)
Some standard polar curves and their equations:
Problem: Eliminate the parameter from x = 2t + 1, y = 3t - 4.
Solution: Solve the first equation for t:
Substitute into the second equation:
Cartesian equation: y = (3/2)x - 11/2. This is a line with slope 3/2.
Problem: Eliminate the parameter from x = 5 cos(t), y = 5 sin(t).
Solution: From the equations: cos(t) = x/5 and sin(t) = y/5.
Use the Pythagorean identity:
This is a circle of radius 5 centered at the origin.
Problem: (a) Convert the polar point (4, π/3) to Cartesian. (b) Convert the Cartesian point (-3, 3) to polar.
Solution (a):
Cartesian point: (2, 2√3).
Solution (b):
Since the point is in Quadrant II (x < 0, y > 0), θ = π - π/4 = 3π/4.
Polar point: (3√2, 3π/4).
Parametric equations are indispensable in physics and computer graphics. Every animation, every simulated trajectory, every robot path is described parametrically. Polar coordinates simplify problems with circular or rotational symmetry -- radar systems, orbital mechanics, and antenna radiation patterns all use polar coordinates naturally. Mastering both systems gives you the flexibility to choose the best tool for each problem.
Problem 1: Eliminate the parameter from x = t^2, y = 2t + 1. What type of curve is this?
From y = 2t + 1: t = (y - 1)/2.
Substitute: x = ((y - 1)/2)^2 = (y - 1)^2/4.
Rewrite: 4x = (y - 1)^2, or equivalently x = (1/4)(y - 1)^2.
This is a parabola opening to the right with vertex at (0, 1).
Problem 2: Convert the polar point (6, 5π/6) to Cartesian coordinates.
x = 6 cos(5π/6) = 6 · (-√3/2) = -3√3
y = 6 sin(5π/6) = 6 · (1/2) = 3
Cartesian point: (-3√3, 3).
Problem 3: Convert the Cartesian equation x^2 + y^2 = 16 to polar form.
Since x^2 + y^2 = r^2 in polar coordinates:
r^2 = 16, so r = 4.
The polar equation is simply r = 4 (a circle of radius 4 centered at the origin).
Problem 4: A projectile is launched at 30 m/s at an angle of 60 degrees above horizontal (g = 10 m/s^2). Write parametric equations for its position. How long is it in the air, and how far does it travel horizontally?
x = 30 cos(60°) t = 30(1/2)t = 15t
y = 30 sin(60°) t - (1/2)(10)t^2 = 15√3 t - 5t^2
The projectile lands when y = 0: 15√3 t - 5t^2 = 0 → t(15√3 - 5t) = 0.
t = 0 (launch) or t = 3√3 seconds (landing).
Horizontal distance: x = 15 · 3√3 = 45√3 ≈ 77.9 meters.
Problem 5: Describe the polar curve r = 3(1 + cos(θ)). Identify the type of curve and find the value of r at θ = 0, π/2, π, and 3π/2.
This is a cardioid (heart-shaped curve) with a = 3.
θ = 0: r = 3(1 + cos(0)) = 3(1 + 1) = 6 (farthest from origin)
θ = π/2: r = 3(1 + cos(π/2)) = 3(1 + 0) = 3
θ = π: r = 3(1 + cos(π)) = 3(1 + (-1)) = 0 (curve passes through origin)
θ = 3π/2: r = 3(1 + cos(3π/2)) = 3(1 + 0) = 3
The cardioid is symmetric about the x-axis (the polar axis), extends from r = 0 to r = 6, and has its cusp at the origin when θ = π.
Parametric equations define curves by expressing x and y each as functions of a parameter t, capturing not just shape but direction and timing. Eliminating the parameter recovers a Cartesian equation but loses directional information. Polar coordinates (r, θ) describe points by distance and angle, offering an elegant framework for curves with rotational symmetry. Converting between polar and Cartesian uses x = r cos(θ), y = r sin(θ), r = √(x^2 + y^2), and θ = arctan(y/x) (with quadrant adjustment). These coordinate systems complement Cartesian coordinates and are essential tools in physics, engineering, and advanced mathematics.