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Parametric equations

Parametric equations

Parametric equations

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Further kinematics

Vector methods with projectiles

Variable acceleration in one dimension

Differentiating vectors

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Friction with static particles

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Functions of time: Displacement, velocity, acceleration

Using differentiation to find velocity and acceleration

Maxima and minima problems

Using integration to find velocity and displacement

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Constant acceleration

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Velocity-time graphs

Constant acceleration: Deriving formulae

Constant acceleration formulae: SUVAT

Vertical motion under gravity

Modelling in mechanics

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Quantities and units in mechanics

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The normal distribution

The normal distribution: Finding probabilities

The inverse normal distribution function

The standard normal distribution

Finding the mean and standard deviation

Normal approximation to the binomial distribution

Hypothesis testing with the normal distribution

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Set notation

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Probability formulae

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Hypothesis testing I

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Vectors II

3D coordinates

Vectors in 3D

Solving geometric problems

Applications to mechanics: 3D vectors

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Locating roots

Iteration

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Newton-Raphson method

Numerical methods: Applications to modelling

Integration II

Integrating standard functions

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Differentiation II

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Differentiating exponentials and logarithmic functions

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Parametric differentiation

Implicit differentiation

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Connected rates of change

Trigonometry and modelling

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Solving trigonometric equations: Double angle formula

Simplifying a cos x + b sin x

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Solving equations with sec x, cosec x and cot x

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Sequences revision

Arithmetic sequences

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Sum to infinity

Sigma notation

Modelling with sequences and series

Binomial expansion II

Binomial expansion: (1+x)^n

Binomial expansion: (a+bx)^n

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Binomial expansion of compound expressions

Parametric equations

Parametric equations

Using trigonometric identities

Sketching parametric curves

Points of intersection of parametric curves

Modelling with parametric equations

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The modulus function

y = |f(x)| and y = f(|x|)

Combining transformations

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Algebraic division

Proof II

Proof by contradiction

Criticising proofs

Vectors I

Vectors

Representing vectors

Magnitude and direction of a vector

Position vectors and displacement vectors

Solving geometric problems

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Integration I

Integrating x^n

Indefinite integrals

Finding functions by integrating

Definite integrals

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Area between a curve and a line

Differentiation I

Finding the gradient of a curve

Differentiation from first principles

Differentiating x^n

Differentiating functions with multiple terms

Gradients, tangents and normals

Increasing and decreasing functions

Second order derivatives

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Exponential functions

y = e^x

Exponential modelling

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Trigonometric equations

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Binomial expansion I

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Using the discriminant

Direct and inverse proportion

Polynomials

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Proof I

Mathematical structure and arguments

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Proof by deduction

Proof by exhaustion

Explainer Video

Tutor: Alice

Summary

Parametric equations

In a nutshell

The coordinates of a point on a curve can be written as functions of a third variable. This is called the parameter and is often represented by ttt.



Parametric equations

Parametric equations may be used to define a curve, where x=p(t)x=p(t)x=p(t)​ and y=q(t)y=q(t)y=q(t). Each point on the curve will have coordinates (p(t),q(t))(p(t),q(t))(p(t),q(t)).​


Note: A given x−x-x−coordinate may correspond to more than one point on the curve.



Parametric and Cartesian equations

You can convert parametric equations into Cartesian equations using substitution.


Note: The range and domain of the cartesian function can be found using the range and domain of the parametric functions. If the parametric equations are given by x=p(t)x=p(t)x=p(t) and y=q(t)y=q(t)y=q(t) and the Cartesian equation has equation y=f(x)y=f(x)y=f(x), then:

  • The domain of f(x)f(x)f(x) will be the range of p(t)p(t)p(t).
  • The range of f(x)f(x)f(x) will be the range of q(t)q(t)q(t).


Example 1

A curve has parametric equations x=t+1x=t+1x=t+1 and y=t2−3,  −1<t<5y=t^2-3, \ \ -1\lt t\lt 5y=t2−3,  −1<t<5.

i) Find a Cartesian equation of the curve in the form y=f(x)y=f(x)y=f(x).

First, eliminate ttt:

​x=t+1x=t+1x=t+1​​

So, t=x−1t=x-1t=x−1.

y=t2−3y=t^2-3y=t2−3

This means y=(x−1)2−3‾\underline{y=(x-1)^2-3}y=(x−1)2−3​.


ii) Write down the range of f(x)f(x)f(x).

Find the range of y(t)=t2−3y(t)=t^2-3y(t)=t2−3.

This is a parabola, so it will have a minimum point of (0,−3)(0,-3)(0,−3), which occurs when t=0t=0t=0.

The maximum point in the given range is given by f(5)=52−3=22f(5)=5^2-3=22f(5)=52−3=22.


The range is −3<f(x)<22‾\underline{-3 \lt f(x) \lt 22}−3<f(x)<22​.


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FAQs - Frequently Asked Questions

How can the range of a cartesian equation be found using parametric equations?

The range will be given by the range of the parametric equation for the y-coordinates.

How can the domain of a cartesian equation be found using parametric equations?

The domain will be the range of the parametric equation for the x-coordinates.

How can parametric equations be converted into cartesian equations?

Using substitution.

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