Question

\(x = cost,y = sint,z = cos2t\)

Short answer

The parametric equations \(x = cost,y = sint\), and \(z = cos2t\) matches with the graph I.

Step-by-step solution

Step 1:

Consider the parametric equations\(x = cost,y = sint\), and\(z = cos2t\).

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 circle with radius \(1\) in the \(xy\) plane.\(\)

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

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

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

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

\( = cos2t\)

Substitute \(z = cos2t\) in the above computed equation as \({x^2} - {y^2} = z\).

The equation of the curve \({x^2} - {y^2} = z\) indicates that one parabola is facing up toward \(zx\) plan and other facing down toward \(zy\) plan.

The range of the cosine function is \(( - 1,1)\). Hence the range of graph \(z = cos2t\) is \( - 1 \le z \le 1\) which oscillates between \( - 1\) and \(1\).

Thus, the parametric equations \(x = cost,y = sint\), and \(z = cos2t\) matches with the graph I.