Can anyone tell me how are parametric equations derived based on the hyperboloid of 1 sheet formula shown in the image?
Clearly my request is beyond what the textbook question is asking here and I successfully completed the task of graphing it, please see the image below. it's a very pretty curve overlaying on the hyperboloid of 1 sheet. It's very interesting to me to understand how to generate a curve or any curve that would overlap on top of the a shape like this.
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Consider cylindrical coordinates $r, \theta, z$. The given hyperboloid has equation $(12r)^2-(5z)^2=10^2$, and is thus given by the revolution of the $r,z$ hyperbola around the $z$ axis. Now, concerning your question, on how to generate a general curve that lays on that surface it is clear that we can always put $$ \left\{ \begin{gathered} \theta = \theta (t) \hfill \\ z = z(\theta ) \hfill \\ r = \frac{1} {{12}}\sqrt {10^{\,2} + \left( {5z} \right)^{\,2} } = \frac{{10}} {{12}}\sqrt {1 + \left( {\frac{z} {2}} \right)^{\,2} } \hfill \\ x = r\cos \theta \hfill \\ y = r\sin \theta \hfill \\ \end{gathered} \right. $$ As for the example you gave, we can write $$ \left\{ \begin{gathered} x = 27/26\sin \left( {8t} \right) - 8/39\sin \left( {18t} \right) = 27/26\cos \left( {8t - \pi /2} \right) + 8/39\cos \left( {18t + \pi /2} \right) \hfill \\ y = - 27/26\cos \left( {8t} \right) + 8/39\cos \left( {18t} \right) = 27/26\sin \left( {8t - \pi /2} \right) + 8/39\sin \left( {18t + \pi /2} \right) \hfill \\ z = 144/65\sin \left( {5t} \right) = \frac{{144}} {{65}}\sqrt {\frac{{1 - \cos \left( {10t} \right)}} {2}} \hfill \\ \end{gathered} \right. $$ wherefrom we get that $P=(x,y)$ is the composition of two circular orbits, as depicted in this figure
In fact, regarding $r$ we have that: $$ \begin{gathered} r^{\,2} = x^{\,2} + y^{\,2} = \left( { - 27/26} \right)^{\,2} + \left( {8/39} \right)^{\,2} - 2\left( {27/26} \right)\left( {8/39} \right)\left( {\cos \left( {8t} \right)\cos \left( {18t} \right) + \sin \left( {8t} \right)\sin \left( {18t} \right)} \right) = \hfill \\ = \frac{1} {{13^{\,2} }}\left( {\frac{{3^{\,6} }} {{2^{\,2} }} + \frac{{2^{\,6} }} {{3^{\,2} }}} \right) - 2\frac{1} {{13^{\,2} }}\left( {\frac{{3^{\,3} }} {2}\frac{{2^{\,3} }} {3}} \right)\cos \left( {10t} \right) = \frac{1} {{13^{\,2} \cdot 2^{\,2} \cdot 3^{\,2} }}\left( {3^{\,8} + 2^{\,8} - 2 \cdot 3^{\,4} \cdot 2^{\,4} \cos \left( {10t} \right)} \right) = \hfill \\ = \frac{1} {{13^{\,2} 2^{\,2} 3^{\,2} }}\left( {\left( {3^{\,8} + 2^{\,8} } \right) - 2\left( {3^{\,4} 2^{\,4} } \right)\left( {\cos ^{\,2} \left( {5t} \right) - \sin ^{\,2} \left( {5t} \right)} \right)} \right) = \left( {\frac{5} {6}} \right)^{\,2} + \left( {\frac{{12}} {{13}}} \right)^{\,2} \sin ^{\,2} \left( {5t} \right) \hfill \\ \end{gathered} $$ and $$ \begin{gathered} r^{\,2} = \frac{{10^{\,2} + \left( {5z} \right)^{\,2} }} {{12^{\,2} }} = \frac{{100}} {{144}}\left( {1 + \left( {72/65\sin \left( {5t} \right)} \right)^{\,2} } \right) = \hfill \\ = \left( {\frac{5} {6}} \right)^{\,2} \left( {1 + \frac{1} {2}\left( {\frac{{72}} {{65}}} \right)^{\,2} \left( {1 - \cos \left( {10t} \right)} \right)} \right) = \frac{1} {{13^{\,2} \cdot 2^{\,2} \cdot 3^{\,2} }}\left( {3^{\,8} + 2^{\,8} - 2 \cdot 3^{\,4} \cdot 2^{\,4} \cos \left( {10t} \right)} \right) = \hfill \\ = \left( {\frac{5} {6}} \right)^{\,2} + \left( {\frac{{12}} {{13}}} \right)^{\,2} \sin ^{\,2} \left( {5t} \right) \hfill \\ \end{gathered} $$ so demonstrating that the point $P$ is on the given hyperboloid.
Now we can derive the value of $\theta (t)$ to complete the exemplification, and we can do it better geometrically from the figure above.
However, since it is not immediate, I will omit it herewith.