Question

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

Short answer

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

Step-by-step solution

Step 1:

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

Square and add the parametric equations of\(x = cost\)and\(y = sint\).

That is, calculate the value of the expression \({x^2} + {y^2}\).

\({x^2} + {y^2} = {(cost)^2} + {(sint)^2}\)

\( = co{s^2}t + si{n^2}t\)

\( = 1\)

The equation of the curve \({x^2} + {y^2} = 1\) forms a circular cylinder along the \(z - \)axis.\(z\)

The point \(t = 0\) gives the value \(z = 1\), hence the range of \(z\) is \(0 < z \le 1\). The location of points \((x,y,z)\) are on the curve which is above the point \((x,y,0)\) and that moves in a counter clockwise direction around the unit circle on the \(xy - \)plane and the value of \(t\) increases as the \(z\) value tends to zero and the \(t\) value tends to infinity.

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