Parametric Functions — AHL 2.10 Parametric functions and graphs (HL) - IB Mathematics Analysis and Approaches

SECTION 1

Historical Context & Motivation

For centuries mathematicians described curves using a single equation relating x and y directly, like y = x² for a parabola. This works well for many curves, but some shapes — spirals, cycloids, and the paths of planets — are difficult or impossible to capture with a single y = f(x) equation. The idea of introducing a third variable, called a parameter, to describe both coordinates independently opened up an entirely new way of thinking about curves. Instead of asking "what is y when x is 3?", we ask "where is the point when the parameter is 3?"

1637 • Descartes & Coordinate Geometry

René Descartes published La Géométrie, uniting algebra and geometry on the coordinate plane and making y = f(x) the standard way to describe curves.

1696 • The Brachistochrone Problem

Johann Bernoulli posed the challenge of finding the fastest-descent curve. The solution, a cycloid, is most naturally described using parametric equations — showing that some curves need a parameter.

1748 • Euler's Parametric Approach

Leonhard Euler systematically used parameters to describe motion along curves, linking the parameter to time and laying groundwork for modern parametric analysis.

1800s • Applications in Physics & Engineering

Parametric descriptions became essential for projectile motion, planetary orbits, and later for computer graphics, where curves are rendered parametrically in real time.

The central question this topic addresses is: how can we describe curves that fail the vertical-line test, or curves where position depends on time, using a pair of functions that work together? Parametric functions provide the answer, and they are a core tool in the IB HL curriculum.

SECTION 2

Core Principles & Definitions

A parametric function defines a curve by expressing both x and y as separate functions of an independent variable, usually called t. As t changes, the point (x(t), y(t)) traces out a path on the coordinate plane. The variable t is the parameter, and it often represents time, an angle, or any other quantity that drives the motion.

The Parameter t

An independent variable that controls both x and y. It does not appear on the final graph; it acts behind the scenes to generate the curve.

Component Functions

x = f(t) and y = g(t) are the two equations that together define every point on the curve. Each is a function in its own right.

Direction & Orientation

As t increases, the point moves along the curve in a specific direction. This orientation distinguishes parametric curves from static Cartesian equations.

Eliminating the Parameter

Sometimes you can solve one equation for t and substitute into the other, converting back to a Cartesian equation y = f(x). This is not always possible or desirable.

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KEY TAKEAWAY

Think of parametric equations like GPS coordinates being broadcast over time. Your phone's x-position and y-position are each a separate function of time. At any instant t, the pair (x(t), y(t)) tells you exactly where you are. If you plot all those points, you get the path you walked — that path is the parametric curve.

SECTION 3

Visualising Parametric Curves

The diagram below shows how the parametric equations x = 2cos(t) and y = 2sin(t) generate a circle. On the left, you can see how x and y each vary as functions of t. On the right, the resulting curve is plotted on the xy-plane with arrows indicating the direction of travel as t increases from 0 to 2π.

Left: the cyan curve is x(t) = 2cos(t) and the pink curve is y(t) = 2sin(t), both plotted against t. Right: the resulting circle on the xy-plane with direction of travel (counter-clockwise) shown by the arrow. Yellow dots mark key parameter values.

Notice that neither the cyan curve nor the pink curve alone is a circle. It is only when we combine x(t) and y(t) into ordered pairs that the circular shape emerges. This illustrates the power of parametric representation: individual component functions can be simple waves, yet together they generate complex and beautiful curves.

SECTION 4

Mathematical Framework

The formal definition is straightforward. Given two functions f and g, the parametric equations x = f(t) and y = g(t) define a curve C in the plane. The variable t ranges over some interval, often [0, 2π] for closed curves or all real numbers for open curves.

PARAMETRIC DEFINITION

x = f(t), y = g(t), t ∈ [a, b]

f(t) controls the horizontal position, g(t) controls the vertical position, and t is the parameter that drives both.

Eliminating the Parameter

To convert parametric equations into a Cartesian equation, solve one equation for t and substitute into the other. For the circle example, use the Pythagorean identity: cos²(t) + sin²(t) = 1. Since x = 2cos(t) means cos(t) = x/2 and y = 2sin(t) means sin(t) = y/2, substitution gives the familiar Cartesian form.

ELIMINATION — CIRCLE EXAMPLE

(x/2)² + (y/2)² = 1 → x² + y² = 4

This confirms the curve is a circle of radius 2 centred at the origin. However, the Cartesian form loses the direction and speed information that the parametric form carries.

Gradient of a Parametric Curve

To find the slope dy/dx without eliminating the parameter, use the chain rule. Since dy/dx = (dy/dt) ÷ (dx/dt), you can differentiate each component function with respect to t and divide.

PARAMETRIC GRADIENT

dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0

When dx/dt = 0 and dy/dt ≠ 0, the tangent line is vertical. When both are zero, the point may be a cusp or self-intersection.

SECOND DERIVATIVE

d²y/dx² = (d/dt)(dy/dx) ÷ (dx/dt)

This formula lets you determine concavity of the parametric curve. First find dy/dx as a function of t, differentiate that with respect to t, then divide by dx/dt.

SECTION 5

Important Parametric Curves

Several families of curves appear frequently in the IB syllabus and beyond. Understanding these standard examples helps you recognise patterns and quickly set up parametric equations for new situations.

Curve	Parametric Equations	Cartesian Form
Circle	x = r cos(t), y = r sin(t)	x² + y² = r²
Ellipse	x = a cos(t), y = b sin(t)	x²/a² + y²/b² = 1
Line Segment	x = x₁ + t(x₂ − x₁), y = y₁ + t(y₂ − y₁), t ∈ [0,1]	y − y₁ = m(x − x₁)
Parabola	x = t, y = t²	y = x²
Cycloid	x = r(t − sin t), y = r(1 − cos t)	No simple Cartesian form

Common parametric curves and their Cartesian equivalents

Three standard parametric curves: a circle (cyan), an ellipse (violet), and a parabola (pink). Each is generated by a pair of parametric equations shown below the graph.

The ellipse is a generalisation of the circle: when a = b, the ellipse becomes a circle. The line segment is particularly useful in IB problems involving vectors, since the parametric form x = x₁ + t(x₂ − x₁) is equivalent to a vector equation of a line. The cycloid is famous for being the curve of fastest descent — a result that can only be elegantly expressed parametrically.

SECTION 6

Worked Example

Let us work through a complete IB-style problem. We are given the parametric equations x = 3cos(t) + 1 and y = 2sin(t) − 2 for 0 ≤ t ≤ 2π. We will sketch the curve, eliminate the parameter, and find the gradient at t = π/4.

Sketching, Eliminating, and Differentiating a Parametric Curve

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Step 1 — Identify the Shape

The equations x = 3cos(t) + 1 and y = 2sin(t) − 2 follow the pattern x = a cos(t) + h, y = b sin(t) + k. This is an ellipse centred at (h, k) = (1, −2) with semi-major axis a = 3 (horizontal) and semi-minor axis b = 2 (vertical).

Ellipse centred at (1, −2), a = 3, b = 2
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Step 2 — Build a Table of Key Values

Substitute key values of t to find reference points. At t = 0: x = 3(1) + 1 = 4, y = 2(0) − 2 = −2 → (4, −2). At t = π/2: x = 3(0) + 1 = 1, y = 2(1) − 2 = 0 → (1, 0). At t = π: x = 3(−1) + 1 = −2, y = 2(0) − 2 = −2 → (−2, −2). At t = 3π/2: x = 1, y = 2(−1) − 2 = −4 → (1, −4). These four points are the ends of the axes of the ellipse.

Key points: (4, −2), (1, 0), (−2, −2), (1, −4)
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Step 3 — Eliminate the Parameter

Isolate the trig functions: cos(t) = (x − 1)/3 and sin(t) = (y + 2)/2. Apply the Pythagorean identity cos²(t) + sin²(t) = 1.

((x − 1)/3)² + ((y + 2)/2)² = 1 → (x − 1)²/9 + (y + 2)²/4 = 1
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Step 4 — Find the Gradient at t = π/4

Differentiate: dx/dt = −3sin(t) and dy/dt = 2cos(t). Therefore dy/dx = (2cos(t)) / (−3sin(t)) = −2cos(t) / (3sin(t)). At t = π/4: cos(π/4) = sin(π/4) = √2/2, so dy/dx = −2(√2/2) / (3(√2/2)) = −2/3.

dy/dx at t = π/4 is −2/3
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Step 5 — Interpret the Result

The slope −2/3 tells us that at the point corresponding to t = π/4, which is approximately (3.12, −0.59), the curve is falling to the right. The curve is traversed counter-clockwise, consistent with the standard parametric circle/ellipse orientation.

The tangent line slopes downward at a rate of 2 units vertically for every 3 units horizontally.

SECTION 7

Parametric vs Cartesian: Strengths & Limitations

You might wonder: if we can often convert back to Cartesian form, why bother with parametric equations at all? The two representations have distinct advantages depending on the situation.

Feature	Parametric Form	Cartesian Form
Vertical-line test	Not required — can describe any curve, even those that double back	y = f(x) must pass the test (one y per x)
Direction / motion	Built in — as t increases, you see which way the curve is traced	No inherent direction information
Complex curves	Handles spirals, cycloids, Lissajous figures naturally	Many complex curves have no simple Cartesian equation
Simplicity	Requires two equations	A single equation — often easier for basic curves
Finding intersections	More complex — must check if curves pass through the same point at the same or different t values	Straightforward — solve simultaneous equations

Comparison of parametric and Cartesian representations

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KEY TAKEAWAY

Parametric equations are like giving someone turn-by-turn driving directions ("at time t, go here"), while a Cartesian equation is like handing them a map of the entire road. Directions carry timing and direction; the map shows the whole shape but loses the journey. Use whichever form best fits the problem you are solving.

SECTION 8

Connections to Advanced Topics

Parametric functions in AHL 2.10 connect directly to several advanced areas you will encounter later in the IB and in university mathematics. Understanding these links helps you see why mastering the basics now pays off.

AHL 2.10 Concept	Advanced Extension
Parametric equations x = f(t), y = g(t)	Vector-valued functions r(t) = ⟨f(t), g(t), h(t)⟩ in 3D (multivariable calculus)
Gradient dy/dx = (dy/dt)/(dx/dt)	Arc length: L = ∫√((dx/dt)² + (dy/dt)²) dt (calculus option)
Eliminating the parameter	Polar coordinates r = f(θ) — another alternative coordinate system
Direction of traversal	Velocity and acceleration vectors in physics and kinematics

How AHL 2.10 concepts extend into more advanced mathematics

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IB Exam Tip

In Paper 1 (no calculator), you may be asked to sketch a parametric curve from a table of values or eliminate the parameter algebraically. In Paper 2, expect GDC-based questions where you graph parametric equations and use your calculator to find intersection points or gradients at specific values of t. Practice both approaches.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A curve is defined by x = t² and y = t³. Does this curve pass the vertical-line test? Explain why parametric representation is useful here.

Show answer

The curve does not pass the vertical-line test because for each positive x value, there are two corresponding y values (one from positive t, one from negative t). For example, when t = 2, the point is (4, 8), but when t = −2, the point is (4, −8). A Cartesian function y = f(x) cannot capture both branches simultaneously. Parametric representation handles this naturally because each value of t gives exactly one point, allowing the curve to fold back on itself without any issue.

Question 2: BASIC CALCULATION

A curve is given by x = 4cos(t) and y = 4sin(t). Eliminate the parameter to find the Cartesian equation, and state the radius and centre of the curve.

Show answer

From the equations, cos(t) = x/4 and sin(t) = y/4. Applying the Pythagorean identity cos²(t) + sin²(t) = 1 gives (x/4)² + (y/4)² = 1, which simplifies to x² + y² = 16. This is a circle centred at the origin (0, 0) with radius 4. The curve is traversed counter-clockwise as t increases from 0 to 2π, starting at the point (4, 0).

Question 3: INTERMEDIATE

Given x = 2t + 1 and y = t² − 3, find dy/dx in terms of t. Then determine the exact coordinates of the point where the tangent line is horizontal.

Show answer

Differentiate: dx/dt = 2 and dy/dt = 2t. Therefore dy/dx = (2t)/2 = t. A horizontal tangent occurs when dy/dx = 0, so t = 0. Substituting t = 0 into the original equations gives x = 2(0) + 1 = 1 and y = (0)² − 3 = −3. The point with a horizontal tangent is (1, −3). To verify, we can eliminate the parameter: from x = 2t + 1, t = (x − 1)/2, so y = ((x − 1)/2)² − 3 = (x − 1)²/4 − 3, which is a parabola with vertex at (1, −3), confirming our answer.

Question 4: APPLIED

A Ferris wheel has its centre 15 metres above the ground and a radius of 12 metres. It completes one revolution every 40 seconds, and riders board at the lowest point. Write parametric equations for a rider's position (x, y) where x is horizontal displacement from the centre and y is height above the ground, using t in seconds. Where is the rider after 10 seconds?

Show answer

The wheel completes one revolution in 40 seconds, so the angular speed is 2π/40 = π/20 radians per second. Since the rider starts at the bottom, the angle measured from the bottom is θ = (π/20)t. Using the convention that the bottom corresponds to angle −π/2 on a standard circle, the parametric equations are x = 12cos(πt/20 − π/2) = 12sin(πt/20) and y = 15 + 12sin(πt/20 − π/2) = 15 − 12cos(πt/20). After 10 seconds: x = 12sin(π/2) = 12 metres and y = 15 − 12cos(π/2) = 15 − 0 = 15 metres. So after 10 seconds, the rider is 12 metres to the right and 15 metres above the ground — exactly at the centre height, which makes sense because 10 seconds is one-quarter of the revolution.

Question 5: CRITICAL THINKING

Two curves are defined parametrically: Curve A has x = cos(t), y = sin(2t), and Curve B has x = cos(s), y = sin(2s) where the parameter for B is s = π − t. Show that both curves produce the same set of points on the xy-plane, but describe the difference in how each curve is traced. What does this reveal about the relationship between a parametric curve and its underlying point set?

Show answer

For Curve B, substitute s = π − t:

x_B

= cos(π − t) = −cos(t) and

y_B

= sin(2(π − t)) = sin(2π − 2t) = −sin(2t). Now consider the full range 0 ≤ t ≤ 2π. Curve A traces the figure-eight Lissajous pattern in one direction, while Curve B traces the same figure but reflected through the origin. However, as t runs from 0 to 2π, both parametrisations cover the exact same set of (x, y) points because the Lissajous figure possesses this symmetry — every point (cos(t), sin(2t)) has a corresponding point (−cos(t), −sin(2t)) that also lies on the curve. This demonstrates a fundamental idea: a parametric curve carries more information than its point set. Two different parametrisations can generate identical images but with different orientations, speeds, and starting points. The parametric form encodes the journey, not just the destination.

SUMMARY

Lesson Summary

Summary

Parametric functions describe curves by expressing both x and y as separate functions of a parameter t. This approach captures curves that fail the vertical-line test, encodes direction of traversal, and naturally models real-world motion. Standard examples include circles (x = r cos t, y = r sin t), ellipses (x = a cos t, y = b sin t), and line segments. You can convert to Cartesian form by eliminating the parameter, though this may lose direction and timing information.

The gradient of a parametric curve is found using dy/dx = (dy/dt) ÷ (dx/dt), which applies the chain rule without needing to eliminate t first. Horizontal tangents occur when dy/dt = 0 (and dx/dt ≠ 0), and vertical tangents occur when dx/dt = 0 (and dy/dt ≠ 0). These tools connect directly to vector-valued functions and arc length in more advanced courses. Master the interplay between parametric and Cartesian forms, and you will have a powerful toolkit for describing any curve you encounter.