Double family of curves

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Double family of curves

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface then the quadratic differential equation
\( Pdu^2+2Qdudv+Rdv^2=0\)
or\( P \left( \frac{du}{dv}\right)^2+2Q\left( \frac{du}{dv}\right)+R=0 \)
where P,Q,R are continuous function of u,v and do not vanish together, is called double family of curves on the surface.

Condition for orthogonality of double family

Let S: \(\vec{r}=\vec{r}(u,v) \) be a surface, and
\( Pdu^2+2Qdudv+Rdv^2=0\)
or\( P \left( \frac{du}{dv}\right)+2Q\left( \frac{du}{dv}\right)+R=0 \)
be double family of curves.
If \(\left( \frac{\lambda}{\mu}, \frac{\lambda '}{\mu '}\right) \) be the roots of the curves, then.
\( \frac{\lambda}{\mu}+ \frac{\lambda '}{\mu '}= \frac{-2Q}{P}\) and \( \frac{\lambda}{\mu}. \frac{\lambda '}{\mu '}= \frac{R}{P}\)
Now, the curves are orthogonal if
\( E \frac{\lambda}{\mu}. \frac{\lambda '}{\mu '}+ F\left( \frac{\lambda}{\mu}+\frac{\lambda '}{\mu '}\right)+G=0\)
or\( E \left ( \frac{R}{P} \right ) + F \left (\frac{-2Q}{P} \right )+G=0\)
or\( ER-2QF+GP=0\)

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Theorem 1

The necessary and sufficient condition for parametric curves to be orthogonal is F=0
Proof
Let \(S: \vec{r}=\vec{r}(u,v)\) be a surface with parametric curves u =constant and v =constant on it.
Then, differential equation of parametric curves is
dudv=0 (i)
Also, differential equation of double family of curves is
\(P{du}^2+2Qdudv+R{dv}^2=0\) (ii)
Comparing (i) and (ii) we get
P=0, Q≠0 and R=0
Now, necessary and sufficient condition for parametric curves to be orthogonal is
ER-2QF+GP=0
or E.0-2QF+G.0=0
or F=0

Theorem 2

Show that parametric curves form an orthogonal system on a sphere x=asinu cosv,y=asinu sinv,z=acosu.
Solution
The sphere is
x=asinucosv,y=asinusinv,z=acosu
or \( \vec{r}=(a \sin u \cos v, a \sin u \sin v,a \cos u \) (i)
By successive differentiation w. r. to. u and v, we get
\( \vec{r}_1=(a \cos u \cos v,a \cos u \sin v,-a \sin u \)
\( \vec{r}_2=(-a \sin u \sin v,a \sin u \cos v,0 \)
Now, the fundamental coefficient are
\( F=\vec{r}_1.\vec{r}_2=0\)
Hence, the parametric curves on a sphere form an orthogonal system.

Theorem 3

Prove that, if θ is angle between two directions of \(Pdu^2+2Qdudv+Rdv^2=0\) then \( \tan \theta =\frac{2H \sqrt{Q^2-PR}}{ER-2FQ+GP} \)
Solution
The double family of curves is
\(Pdu^2+2Qdudv+Rdv^2=0\)
or \(P \left ( \frac{du}{dv} \right )^2+2Q \frac{du}{dv}+R=0\)
Let \( \left ( \frac{l}{m} ,\frac{l'}{m'} \right ) \) be the roots, then
\( \frac{l}{m}+\frac{l'}{m'}=\frac{-2Q}{P}\) and \( \frac{l}{m}.\frac{l'}{m'}=\frac{R}{P}\)
Hence
\( \cos \theta =Ell'+F(lm'+l'm)+Gmm'\)
or \( \cos \theta =E \left( \frac{l}{m}.\frac{l'}{m'}\right )+F \left( \frac{l}{m}+\frac{l'}{m'}\right )+G\)
or \( \cos \theta =E \left( \frac{R}{P} \right ) +F \left( \frac{-2Q}{P} \right )+G\)
or \( \cos \theta =\frac{ER-2FQ+GP}{P} \) (A)
Again
\( \sin \theta =H(lm'-l'm)\)
or \( \sin \theta =H\left( \frac{l}{m}-\frac{l'}{m'}\right )\)
or \( \sin \theta =H \sqrt{\left( \frac{l}{m}+\frac{l'}{m'}\right )^2-4 \left( \frac{l}{m}.\frac{l'}{m'}\right ) } \)
or \( \sin \theta =H \sqrt{\left( \frac{-2Q}{P} \right )^2-4 \left( \frac{R}{P} \right ) } \)
or \( \sin \theta =\frac{2H \sqrt{Q^2-PR}}{P} \) (B)
Thus, using (A) and (B), we get
\( \tan \theta =\frac{2H \sqrt{Q^2-PR}}{ER-2FQ+GP} \)
This completes the solution