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Parametric forms for lines and vectors

We consider parametric forms for lines, and compare them to Cartesian forms.
A parametric surface
© UNSW Australia 2015
In many situations, it is useful to have an alternative way of describing a curve besides having an equation for it in the \(\normalsize{x-y}\) plane. A parametric form for a line occurs when we consider a particle moving along it in a way that depends on a parameter \(\normalsize{t}\), which might be thought of as time. Thus both \(\normalsize{x}\) and \(\normalsize{y}\) become functions of \(\normalsize{t}\). The simplest parameterisation are linear ones.
In this step we
  • look at parameterisations of the simplest curves: lines
  • show how to go from parametric equations to Cartesian equations
  • introduce vectors.

Parameterisations of a line

A simple example of a parameterisation is a linear parameterisation:
\[\Large{p(t)=[t+1, 2t-3]}.\]
As \(\normalsize{t}\) varies, the point \(\normalsize{p(t)}\) moves along a line. When \(\normalsize{t=0}\), \(\normalsize{p(0)=[1,-3]}\), when \(\normalsize{t=1}\), \(\normalsize{p(1)=[2,-1]}\), and when \(\normalsize{t=2}\), \(\normalsize{p(2)=[3,1]}\). Clearly also fractional values of \(\normalsize{t}\) are allowed.
An animation showing the trajectory p(t)=[t+1, 2t-3]
How do we go from a parametric form for a line to one of the more usual Cartesian forms involving \(\normalsize{x}\) and \(\normalsize{y}\)? We eliminate the parameter \(\normalsize{t}\). For example for the parametric equation \(\normalsize{p(t)=[t+1, 2t-3]}\) we have
\[\Large{x=t+1}\] \[\Large{y=2t-3}\]
so that \(\normalsize{y=2(x-1)-3}\) or \(\normalsize{y=2x-5}\). Please check that this is the same line – with a different equation.
Graph of y=2x-5
Q1 (E): Find a Cartesian equation for the parametric line \({\normalsize p(s) = [2s+1, 4s-3]}\).

How to go from a Cartesian equation to a parametric form?

On the other hand, how can we go from a Cartesian equation of a line, say \(\normalsize{y=3x+2}\) to a parametric form? It is important to realise that although the Cartesian form is more or less unique, this is not at all the case for a parametric form.
One way is to introduce a parameter in a simple way, say by setting \(\normalsize{x=t}\). Then the original equation gives \(\normalsize{y=3x+2=3t+2}\). Putting these together gives the parameterisation \(\normalsize{p(t)=[t,3t+2]}\).
Q2 (E): Find another parameterisation of \(\normalsize y=3x+2\) by setting \(\normalsize x=3t\).

Using vectors to parameterise lines

The rest of this step is about vectors, for those who are familiar with them. A vector is a directed line segment, or geometrically a difference between two points in the plane. We will denote vectors by round brackets, such as \(\normalsize{v=(1,4)}\). This represents a relative displacement of \(\normalsize{1}\) to the right in the \(\normalsize x\)-direction and \(\normalsize{4}\) up in the \(\normalsize y\)-direction. We agree that vectors are not fixed in place, as they are only determined in a relative sense.
Suppose that we consider the line through the point \(\normalsize{[3,2]}\) which goes in the direction \(\normalsize{v=(1,4)}\). This line can be expressed as
\[\Large{[3,2]+t(1,4)=[3+t,2+4t]}\]
which is now in parametric form with parameter \(\normalsize{t}\).
What is the Cartesian equation of this line? We may eliminate \(\normalsize{t}\) from the equations \(\normalsize{x=3+t}\), \(\normalsize{y=2+4t}\) to get \(\normalsize{y=2+4(x-3)=4x-10}\) so the equation is \(\normalsize{y=4x-10}\).
Graph of y=4x-10 showing also the point [3,2] and the vector (1,4)

Answers

A1. This is also \(\normalsize{y=2x-5}\).
A2. Another parameterisation is \(p(t) = [3t, 9t+2]\).
© UNSW Australia 2015
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Maths for Humans: Linear, Quadratic & Inverse Relations

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