Question

\( x = t,y = \frac{1}{{1 + {t^2}}},z = {t^2}\)

Short answer

The parametric equations \( x = t,y = \frac{1}{{1 + {t^2}}},\) and \(z = {t^2}\)matches with the graph \(V\).

Step-by-step solution

Parametric Equations.

Consider the parametric equations \(x = t,y = \frac{1}{{1 + {t^2}}}\), and \(z = {t^2}\).

Substitute \(x = t\) in the equation \(z = {t^2}\) as \(z = {x^2}\).

The equation \(z = {x^2}\) depicts that the curve lies on the parabolic and is parallel to the \(y\)-plane. According to the domain, the intervals are \(0 < y \le 1\) and \(z \ge 0\). The curve passes through the point \((0,1,0)\). When \(t\) is zero, \(y\) tends to one, and \(z\) tends to infinity as \(t\) tends to \( \pm \infty \).

Thus, the parametric equations \(x = t,y = \frac{1}{{1 + {t^2}}}\), and \(z = {t^2}\) matches with the graph \(V\).