A particle in the $xy$-plane starts at the point $(1, 1)$ and moves along the curve $y = x^2$ to the point $(−1, 1)$ and then returns along the same curve to the point $(1, 1)$. Use a single vector-valued function to represent the path of the particle over a single interval (for the parameter). Don’t forget to give the interval for the parameter.
From what I understand, the x value starts at $1$, decreases to $-1$, and increases to $1$. One function that does this is cosine. So if I set $x = \cos(t)$, then $y = \cos(t)^{2}$ (due to $y = x^2$). Further, the interval for the parameter is $[0, \pi]$. So then, the vector-valued function is $r(t) = \langle \cos(t), \cos(t)^2 \rangle$.
Is my understanding correct? I am kind of hesitant because of how straightforward this question was, and I'm worried that I'm missing something or doing something wrong.
You’re right except for the interval. You need the point to return to $(1,1)$, so the parameter should continue to increase to $2\pi$. The interval is $[0,2\pi]$.