Parametric curve/ellipsoid question

[kankerfist]

I am studying curves in 3d and I came across the folowing question:

A curve C has the following parameterization:
x = a(Sin[t])(Cos[d]), y = b(Sin[t])(Sin[d]), z = c(Cos[t])
t >= 0 and a,b,c,d are positive constants.

Show that C lies on the ellipsoid:
(x^2)/(a^2) + (y^2)/(b^2) +( z^2)/(c^2) = 1.

My question: Can I show this by simply plugging in the x,y, and z values
of the curve C into the x,y and z of the ellipsoid? I'm a little confused about
how to show that a curve lies on this ellipsoid. Any help would be appreciated!

 

[soroban]

Re: Parametric curve and an ellipsoid question

Hello, kankerfist!

A curve \(\displaystyle C\) has the following parameterization: \(\displaystyle \:\begin{array}{ccc}x\:=\a\cdot\cos d)\sin\theta \\y\:=\b\cdot\sin d)\sin\theta \\z\:=\:c\cdot\cos\theta\end{array}\)

\(\displaystyle \;\;\)where: \(\displaystyle \,t\,\geq\,0\) and \(\displaystyle a,b,c,d\) are positive constants.

Show that \(\displaystyle C\) lies on the ellipsoid: \(\displaystyle \L\,\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,+\,\frac{z^2}{c^2}\;=\;1\)

To do what you suggested, we'd have to know that it is an ellipsoid.

The proper way is to eliminate the parameter.


The three equations are: \(\displaystyle \L\:\begin{array}{ccc}\frac{x}{a}\,=\,\cos d\cdot\sin\theta \\ \frac{y}{b}\,=\,\sin d\cdot\sin\theta \\ \frac{z}{c}\,=\,\cos\theta\end{array}\)


Square the equations: \(\displaystyle \L\:\begin{array}{ccc}\frac{x^2}{a^2}\,=\,\cos^2d\cdot\sin^2\theta \\ \frac{y^2}{b^2}\,=\,\sin^2d\cdot\sin^2\theta \\ \frac{z^2}{c^2}\,=\,\cos^2\theta \end{array}\)


Add: \(\displaystyle \L\:\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,+\,\frac{z^2}{c^2}\;=\;\cos^2d\cdot\sin^2\theta\,+\,\sin^2d\cdot\sin^2\theta\,+\,\cos^2\theta\)

\(\displaystyle \L\;\;\;=\;\left(\cos^2d\,+\,\sin^2d)\sin^2\theta\,+\,\cos^2\theta \;= \;\sin^2\theta\,+\,\cos^2\theta\;=\;1\)


Therefore: \(\displaystyle \L\,\frac{x^2}{a^2}\,+\,\frac{y^2}{b^2}\,+\,\frac{z^2}{c^2}\;=\;1\;\;\) . . . ta-DAA!

 

[kankerfist]

That helped a lot, thanks!