Let $C$ be the line segment from $(0,0)$ to $(2,2)$, and let $f(x,y)=x^2+y$.
Write down a vector equation $r(t)$ of the line segment, that is, find a parametrization of $C$.
The answer given is $r(t)=\langle t,t\rangle,0\le t\le 2$.
My Question: How did they get $\boldsymbol{r(t)=\langle t,t\rangle}$?
On
A (non-uniquely) parametrized straight line segment that goes from a startpoint $(x_0,y_0)$ to an endpoint $(x_1,y_1)$ is given in vector form as $$\langle x,y\rangle = \langle x_0,y_0\rangle +\lambda \langle x_1-x_0,y_1-y_0\rangle, \qquad 0 \leq \lambda \leq 1.$$ After plugging in your numbers, you end up with $\langle x,y\rangle = \lambda \langle 2,2\rangle = \langle 2\lambda,2\lambda\rangle$. Letting $t:=2\lambda$ gives the desired parametrization, but that last step is more of a personal choice.
The line going from $(0, 0)$ to $(2, 2)$ has equation $y = x$, so a parametrization could be
\begin{eqnarray} x(t) &=& t \\ y(t) &=& x(t) = t \end{eqnarray}
For $0 \leq t \leq 2$ a parametrization of $C$ is
$$ {\bf r}(t) = (t, t)~~~\mbox{for}~~ 0\leq t \leq 2 $$