Example: Parameterize the line through $P=(1,0)$ and $Q=(3,2)$ so that $P$ and $Q$ correspond to $t=1$ and $t=3$, respectively.

Solution 1: This is more-or-less the approach that we took in class. We look at the vector formed by going from point $P$ to point $Q$, which is $\vec{PQ}=2\vec{i}+2\vec{j}$. We create a line parallel to this emanating from $P$ and we get something that looks like

\[\vec{r}(t)=P+t\vec{PQ}=\langle 1,0\rangle+t\langle 2,2\rangle\]

The only problem with this is that it doesn’t satisfy the condition that $P$ occurs when $t=1$ and $Q$ occurs when $t=3$. We have component functions of the form

\[x=1+2t\quad\text{and}\quad y=0+2t=2t.\]

We can perform our standard algebraic manipulations here, multiplying $t$ by a scalar and adding some scalar to $t$. We know that increasing $t$ by 1 means that both $x$ and $t$ should increase by 1, and so we replace $t$ by $\frac{1}{2}t$ in the above component functions. (Notice that this just makes us traverse along the path from $P$ to $Q$ more slowly.) We now have component functions of the form

\[x=1+t\quad\text{and}\quad y=t\]

If we replace $t$ now by $t-1$, we get something that actually works. (What we’re doing here is just picking the line up and moving it along itself.) We now have

\[x=1+(t-1)=t\quad\text{and}\quad y=t-1.\]

Solution 2: Another way of doing this is to start with the displacement vector $\vec{OP}$ (from the origin to $P$), which we find to be $1\vec{i}+0\vec{j}$, and again consider the displacement vector $\vec{PQ}$ as being added on. We want $t=1$ to be the point along the line where we are at $P$, and so we know that we’ll have something along the lines of


since $t=1$ causes this to simply output the vector corresponding to $P$. Now, we need to know how to scale the right side of the equation (i.e., to traverse the line from $P$ to $Q$ at the appropriate pace). If $a=1$, then plugging in $t=3$ gives $\vec{OP}+2\vec{PQ}$, which overshoots $Q$. We find that $a=2$ works. So, we have

\[\vec{r}(t)=\vec{OP}+\frac{t-1}{2}\vec{PQ}=\langle 1,0\rangle+\frac{t-1}{2}\langle 2,2\rangle=t\vec{i}+(t-1)\vec{j},\]

which is exactly what we expected.


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