Find parametric equations for the rectangle in the accompanying figure, assuming that the rectangle is traced counterclockwise as t varies from 0 to 1, starting at (1/2, 1/2)

when
t = 0. [Hint: Represent the rectangle piecewise, letting t vary from 0 to 14
for the first edge, from 14
to 12 for the second
edge, and so forth.]

Question

Find parametric equations for the rectangle in the accompanying figure, assuming that the rectangle is traced counterclockwise as t varies from 0 to 1, starting at (1/2, 1/2)

when
t = 0. [Hint: Represent the rectangle piecewise, letting t vary from 0 to 14
for the first edge, from 14
to 12 for the second
edge, and so forth.]

anonymous · Answer

What's the rectangle look like?

anonymous · Answer

attachment

anonymous · Answer

Here it is wio^

anonymous · Answer

You can use a piece-wise function, right?

anonymous · Answer

Yes @wio

anonymous · Answer

okay, so as $t$ goes from $0$ to $1/4$, we need $(1/2,1/2)$ to go to $(-1/2,1/2)$

anonymous · Answer

How would I know where is 1/4?

anonymous · Answer

We can leave the $y$ component alone... it's going to be $(x(t),1/2)$

anonymous · Answer

Th reason I chose $1/4$ is because we want nice intervals... $(0,1/4),(1/4,1/2),(1/2,3/4),(3/4,1)$

anonymous · Answer

Yes, but where is 1/4 on the rectangle, how would I know that?

anonymous · Answer

And why to leave the y component?

anonymous · Answer

We know $x(t)$ will be a line so....$$
x(t) = mt+b
$$And we use the two points [$x(0) = 1/2, x(1/4)=-1/2$] we have to solve for $m,b$. $$ \begin{array}{rcl}
(1/2) &=& (0)m+b\
(-1/2) &=& (1/4)m+b
\end{array}$$

anonymous · Answer

Because when going from $(1/2,1/2)$ to $(-1/2,1/2)$ clearly $y$ stays the same, constant at $1/2$

anonymous · Answer

Got these step

anonymous · Answer

Can you solve for $x(t)$ then?