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Determine a parametric equation for A.

The problem gives me the following hint, but I don't see it:

The parametric equation can be written as:

$$\vec{OP}=b+xu+yv, x\in [a,b], y\in [c,d]$$

Steve
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  • Your equation doesn't make much sense.... and what it $P$ – Doug M Nov 29 '16 at 04:37
  • @Bye_World I know what the answer is, but I don't understand it. – Steve Nov 29 '16 at 04:47
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    The trick here is that $P$ represents any point in $A$. You are trying to write an equation in two parameters that "hits" every point in $A$. Notice that you already have a vector $\mathbf{b}$ that points to $B$. But then you are given vectors $\mathbf{v}$ and $\mathbf{u}$, and furthermore $\mathbf{b}+\mathbf{v}$ is the vector that points to the tip of the $\mathbf{v}$ vector shown there in the diagram. Does that makes sense? – Nico Nov 29 '16 at 05:19

1 Answers1

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The vector $\vec b$ points to $B$. The vector $\vec b + \vec v$ points to point one unit up on that grid. So what about the vector $\vec b + t\vec v$ for some $t\in[0,1]$? It would be somewhere in between $\vec b$ and $\vec b + \vec v$. So the set of all points $\vec{OP}(t) =\vec b + t\vec v$ for $t\in[0,1]$ would be the line segment along the grid from point $B$ to the point one unit up.

Like $\vec{OP}(s) = \vec b + s\vec u$ for $s\in[0,1]$ would be the line segment starting from $B$ and going one unit right.

So what if you wanted to end up somewhere within the parallelogram with sides $\vec{OP}(t) =\vec b + t\vec v$ for $t\in[0,1]$ and $\vec{OP}(s) = \vec b + s\vec u$ for $s\in[0,1]$. Well you'd want to go some distance in the $\vec v$ direction and some distance in the $\vec u$ direction. It would be $vec{OP}(s,t) = \vec b + s\vec u + t\vec v$ for $s,t\in[0,1]$. For any choice of $(s,t)\in [0,1]^2$, you'll end up a little bit over to the right and a little bit up -- but within (including on the boundary) the parallelogram you see on the grid with point $B$ at its lower left corner.

But that's not the parallelogram we want to parametrize. We want to parametrize $A$. So first we need to shift over. Instead of starting $s$ and $t$ at zero, we could start them at negative numbers and then we'd be left and below the point $B$. Notice that the lower left corner of $A$ is $\vec b -2\vec u -\vec v$. So we'll want to start our intervals at $-2$ and $-1$ respectively. Then the upper right corner is $\vec b -\vec u + \vec v$. So we'll want to end our intervals at $-1$ and $+1$ respectively.

Well then we can immediately write down the answer:

$$\vec{OP}(s,t)=\vec b+s\vec u+t\vec v,\quad s\in [-2,-1], t\in [-1,1]$$

Make sure you understand why this works.