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A parametric surface
Bézier surface example

Parametric forms for lines and vectors

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. This plays a similar role to the notion of slope, which is the ratio of the relative displacement of the \(\normalsize x\) and \(\normalsize y\)-directions or “rise over run”. Using vectors allows us to be more precise about the notion of direction of a line.

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)

Now imagine that we would like to traverse the same line, only faster. To move through the same line at double the speed, we replace \(\normalsize{t}\) with \(\normalsize{2t}\):

\[\Large{[3,2]+2t(1,4)=[3,2]+t(2,8)}.\]

Notice that the direction vector has doubled, but the Cartesian equation of this line remains unchanged. We can also move backwards along the line by relpacing \(\normalsize{t}\) with \(\normalsize{-t}\):

\[\Large{[3,2]-t(1,4)=[3,2]+t(-1,-4)}.\]

Now the direction vector has been negated, but again the Cartesian equation remains unchanged.

So we see that the parameterisation of a curve describes more than the Cartesian equation, a parameterisation describes how we traverse the curve: fast or slow, forwards or backwards.

Answers

A1. This is also \(\normalsize{y=2x-5}\).

A2. Another parameterisation is \(p(t) = [3t, 9t+2]\).

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