Answer
(a) The parametrizations of the two parts of $C$ corresponding to $x \ge 0$ and $x \le 0$ are
${{\bf{r}}_1}\left( t \right) = \left( {t\sqrt {1 - {t^2}} ,{t^2},t} \right)$,
${{\bf{r}}_2}\left( t \right) = \left( { - t\sqrt {1 - {t^2}} ,{t^2},t} \right)$,
for $ - 1 \le t \le 1$.
(b) The projection of the curves onto the $xy$-plane is a circle of radius $\frac{1}{2}$ with center $\left( {x,y} \right) = \left( {0,\frac{1}{2}} \right)$.
(c) The parametrizations of $C$ satisfy the equation:
${x^2} + {\left( {y - 1} \right)^2} + {z^2} = 1$
Hence, $C$ lies on the sphere of radius $1$ with its center $\left( {0,1,0} \right)$.
Work Step by Step
(a) We have
${x^2} + {y^2} = {z^2}$,
$y = {z^2}$
We solve for $x$ in terms of $z$
The first equation can be written ${x^2} = {z^2} - {y^2}$.
From the second equation we obtain $y = {z^2}$. Thus, we can solve for $x$ by substituting $y$:
${x^2} = {z^2} - {z^4} = {z^2}\left( {1 - {z^2}} \right)$
$x = \pm z\sqrt {1 - {z^2}} $
Using $t=z$ as the parameter, we get $x = \pm t\sqrt {1 - {t^2}} $ and $y = {t^2}$. Therefore, we obtain two vector-valued functions to parametrize the entire curve $C$, which is the intersection of the two surfaces:
${{\bf{r}}_1}\left( t \right) = \left( {t\sqrt {1 - {t^2}} ,{t^2},t} \right)$,
${{\bf{r}}_2}\left( t \right) = \left( { - t\sqrt {1 - {t^2}} ,{t^2},t} \right)$,
for $ - 1 \le t \le 1$.
Notice that the two signs of the square root correspond to the two parts of $C$ for $x \ge 0$ and $x \le 0$.
(b) From part (a) we obtain two parametrizations of the entire curve $C$ given by:
${{\bf{r}}_1}\left( t \right) = \left( {t\sqrt {1 - {t^2}} ,{t^2},t} \right)$,
${{\bf{r}}_2}\left( t \right) = \left( { - t\sqrt {1 - {t^2}} ,{t^2},t} \right)$
The projection of the curves onto the $xy$-plane is obtained by setting the $z$-components to zero. Thus, ${{\bf{r}}_1}\left( t \right)$ and ${{\bf{r}}_2}\left( t \right)$ become $\left( {t\sqrt {1 - {t^2}} ,{t^2},0} \right)$ and $\left( { - t\sqrt {1 - {t^2}} ,{t^2},0} \right)$, respectively.
So, we have the coordinate functions:
$x = t\sqrt {1 - {t^2}} $, ${\ \ }$ $y = {t^2}$, ${\ \ }$ for $x \ge 0$
$x = - t\sqrt {1 - {t^2}} $, ${\ \ }$ $y = {t^2}$, ${\ \ }$ for $x \le 0$
It follows that
${x^2} = {t^2}\left( {1 - {t^2}} \right) = y\left( {1 - y} \right)$
${x^2} + {y^2} - y = 0$
${x^2} + {\left( {y - \frac{1}{2}} \right)^2} - \frac{1}{4} = 0$
${x^2} + {\left( {y - \frac{1}{2}} \right)^2} = \frac{1}{4}$
Notice that this is the equation of a circle of radius $\frac{1}{2}$ centered at $\left( {x,y} \right) = \left( {0,\frac{1}{2}} \right)$.
Hence, the projection of the curves onto the $xy$-plane is a circle of radius $\frac{1}{2}$ with center $\left( {x,y} \right) = \left( {0,\frac{1}{2}} \right)$.
(c) To show that $C$ lies on the sphere of radius $1$ with its center $\left( {0,1,0} \right)$, we must show that the curves parametrized by ${{\bf{r}}_{1,2}}\left( t \right) = \left( { \pm t\sqrt {1 - {t^2}} ,{t^2},t} \right)$ for $ - 1 \le t \le 1$ satisfy the equation ${x^2} + {\left( {y - 1} \right)^2} + {z^2} = 1$.
Since ${x^2} = {t^2}\left( {1 - {t^2}} \right)$, $y = {t^2}$ and $z=t$, we get
${x^2} + {\left( {y - 1} \right)^2} + {z^2} = {t^2}\left( {1 - {t^2}} \right) + {\left( {{t^2} - 1} \right)^2} + {t^2}$
${x^2} + {\left( {y - 1} \right)^2} + {z^2} = {t^2} - {t^4} + {t^4} - 2{t^2} + 1 + {t^2}$
${x^2} + {\left( {y - 1} \right)^2} + {z^2} = 1$
Hence, $C$ lies on the sphere of radius $1$ with its center $\left( {0,1,0} \right)$.
