Question

Graph the curve with parametric equations

\(\begin{array}{l}x = \sqrt {1 - 0.25{\rm{co}}{{\rm{s}}^2}\left( {10t} \right)} {\rm{cos}}t\\y = \sqrt {1 - 0.25{\rm{co}}{{\rm{s}}^2}\left( {10t} \right)} {\rm{sin}}t\\z = 0.5{\rm{cos}}\left( {10t} \right)\end{array}\)

Explain the appearance of the graph by showing that it lies on a sphere.

Short answer

The graph looks as follows:

Step-by-step solution

Graphical representation

The graphical representation are:

Projection of planes and explanation

Given equation are:

\(\begin{array}{l}x = \sqrt {1 - 0.25{\rm{co}}{{\rm{s}}^2}\left( {10t} \right)} {\rm{cos}}t\\y = \sqrt {1 - 0.25{\rm{co}}{{\rm{s}}^2}\left( {10t} \right)} {\rm{sin}}t\\z = 0.5{\rm{cos}}\left( {10t} \right)\end{array}\)

The projection on the xy plane is like that of \(x = r\cos \;t\)and\(y = r\;\sin \;t\).

Which is a circle of radius r. Here \(r = \sqrt {1 - 0.25co{s^2}(10t)} \) so the projection is a circular shape that has a radius varying from \(1\;to\;0.75\). The curve appears to lie on a sphere of radius one centered at the origin. In which case the equation \({x^2} + {y^2} + {z^2} = {r^2}\) would be satisfied.

\(\begin{array}{l}{1^2} = {\left( {\sqrt {1 - 0.25co{s^2}(10t)} \cos t} \right)^2} + {\left( {\sqrt {1 - 0.25co{s^2}(10t)} {\rm{sin}}t} \right)^2} + {\left( {0.5{\rm{cos}}10t} \right)^2}\\1 = (1 - 0.25co{s^2}10t)co{s^2}t + (1 - 0.25co{s^2}10t)si{n^2}t + 0.25co{s^2}10t\\1 = 1 - 0.25co{s^2}10t + 0.25co{s^2}10t\\1 = 1\end{array}\)