#### 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\( ER-2QF+GP=0\)

#### 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

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