# Converting from cartesian to vector valued functions?

Find vector-valued functions forming the boundaries of the region in the figure. Intervals are given for the parameter of each function.

The figure is just the parabola $y=x^2$ bounded by the lines $y = 4$ and $x=0$ . The question basically asked for $r_1, r_2, r_3$ where

$a) (y=x^2)$, $0 \le t \le 2$, $r_1(t)=?$

$b) (y=4)$, $0 \le t \le 2, r_2(t)=?$

$c) (x=0)$, $0 \le t \le 4, r_3(t)=?$

For $a$ I got $i+t^2j$ which was correct, but for $b$ and $c$ I for $4j$ and $0$ respectively, and both of the are wrong.

## 1 Answer

You have a function $\mathbb{R}\{t\}\to\mathbb{R}^2$ so that the tip of the vectors (when centred at the origin) will draw the boundary's lines. Then $y = 4$ will yield for each $t$ a vector whose tip is on the line mentioned. This means that the function must have the form $g(t)\hat{\boldsymbol{\imath}} + 4\hat{\boldsymbol{\jmath}}$. If you just sum the part of $\hat{\boldsymbol{\jmath}}$ then the function will draw just a point all the time. $g(t)$ here is obviously $t$.

Now you should be able to do c) easily.

• You said $g(t)$ is $t$ so I tried $ti+4j$ but the answer was still wrong :/
– Ovi
Sep 21, 2015 at 14:57
• @Ovi are you convinced that $g(t)$ is t? I am convinced. Try drawing it. Sep 21, 2015 at 15:01