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 \(\normalsize{xy}\) 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:
\[\Large{p(t)=[t+1, 2t3]}.\]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.
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 \(\normalsize{p(t)=[t+1, 2t3]}\) we have
\[\Large{x=t+1}\] \[\Large{y=2t3}\]so that \(\normalsize{y=2(x1)3}\) or \(\normalsize{y=2x5}\). Please check that this is the same line – with a different equation.
Q1 (E): Find a Cartesian equation for the parametric line \({\normalsize p(s) = [2s+1, 4s3]}\).
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 \(\normalsize{y=2+4(x3)=4x10}\) so the equation is \(\normalsize{y=4x10}\).
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 \(\normalsize{y=2x5}\).
A2. Another parameterisation is \(p(t) = [3t, 9t+2]\).
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