#### Ruled Surface

A surface generated by motion of straight line along a given curve is called ruled surface.

Here, straight line is called generator of ruled surface. The given curve is called directrix of ruled surface.

For example, a cone is formed by keeping one point of a line fixed and moving the line along a circle.

Thus, cone is a ruled surface.

Other examples of ruled surface are cylinder, right conoid and helicoid.

#### Equation of Ruled Surface

Let S:\( \vec{r}=\vec{r}( s ) \) be a directrix of a ruled surface and P be a point on it such that

\( \vec{OP}=\vec{r} \)

If \( \vec{g}( s ) \) be unit vector along generator and \( R \) be position of arbitrary point T on it.

Then equation of ruled surface S is

\( R=\vec{r}( s )+v\vec{g}( s ) \) (i)where \( v \) is real parameter

#### Types of Ruled Surface

There are two types of ruled surface. One is developable and other is skew.

A ruled surface whose consecutive generators do intersect is called developable surface.

For example, cone is developable surface.

A ruled surface whose consecutive generators do not intersect is called skew surface.

For example, cylinder is skew surface.

#### Theorem

Show that necessary and sufficient condition for a ruled surface to be a developable is \([ \vec{t},\vec{g},\vec{g}' ]=0 \).

Proof

Let S be a ruled surface, then its equation is

\(R=\vec{r}( s )+v\vec{g}( s ) \) (i)\(v \) is real parameter

Differentiating (i) w. r. to. s and v respectively, we get

\({R_1} =\frac{\partial R}{\partial s}= \vec{t}+v \vec{g}'\)

\({R_2}=\frac{\partial R}{\partial v}=\vec{g} \)

\({R_{11}}=\vec{t}'+v\vec{g}''\)

\({R_{12}}=\vec{g}'\)

\({R_{22}}=0 \)

Here

\([ {R_1},{R_2},{R_{12}} ]=HM\)

or
\([ \vec{t}+v\vec{g}',\vec{g},\vec{g}' ] HM \)

or
\( [ \vec{t},\vec{g},\vec{g}' ]=HM\)

Next

\([ {R_1},{R_2},{R_{22}} ]=HN\)

or
\([ \vec{t}+v\vec{g}',\vec{g},0 ]=HN \)

or
\(0=HN \)

or
\(N=0 \)

Now, Gaussian curvature of the surface is

\(K =\frac{LN-M^2}{H^2}\)

or
\( K= \frac{-M^2}{H^2} \)

or
\( K=-\frac{[ \vec{t},\vec{g},\vec{g}' ]^2}{H^4} \)

Since, necessary and sufficient condition for a surface to be developable surface is

\(K=0 \)

So, necessary and sufficient condition for a ruled surface to be a developable surface is

\( \frac{[ \vec{t},\vec{g},\vec{g}' ]^2}{H^4}=0 \)

or
\( [\vec{t},\vec{g},\vec{g}' ]=0 \)

#### Theorem 2

Show that necessary and sufficient condition for a ruled surface to be skew is \([ \vec{t},\vec{g},\vec{g}' ] \ne 0 \).

Proof

Let S be a ruled surface, then its equation is

\(R=\vec{r}( s )+v\vec{g}( s ) \) (i)\(v \) is real parameter

Differentiating (i) w. r. to. s and v respectively, we get

\({R_1} =\frac{\partial R}{\partial s}= \vec{t}+v \vec{g}'\)

\({R_2}=\frac{\partial R}{\partial v}=\vec{g} \)

\({R_{11}}=\vec{t}'+v\vec{g}''\)

\({R_{12}}=\vec{g}'\)

\({R_{22}}=0 \)

Here

\([ {R_1},{R_2},{R_{12}} ]=HM\)

or
\([ \vec{t}+v\vec{g}',\vec{g},\vec{g}' ] HM \)

or
\( [ \vec{t},\vec{g},\vec{g}' ]=HM\)

Next

\([ {R_1},{R_2},{R_{22}} ]=HN\)

or
\([ \vec{t}+v\vec{g}',\vec{g},0 ]=HN \)

or
\(0=HN \)

or
\(N=0 \)

Now, Gaussian curvature of the surface is

\(K =\frac{LN-M^2}{H^2}\)

or
\( K= \frac{-M^2}{H^2} \)

or
\( K=-\frac{[ \vec{t},\vec{g},\vec{g}' ]^2}{H^4} \)

Since, necessary and sufficient condition for a surface to be developable surface is

\(K \ne 0 \)

So, necessary and sufficient condition for a ruled surface to be a developable surface is

\( \frac{[ \vec{t},\vec{g},\vec{g}' ]^2}{H^4}\ne 0\)

or
\( [\vec{t},\vec{g},\vec{g}' ]\ne 0 \)

#### Example

Find a condition that \( x=az+\alpha ,y=bz+\beta \) generates a skew surface where \( a,b,\alpha ,\beta \) are function of s.

Solution

Given the equation of line is

\(x=az+\alpha ,y=bz+\beta \)

or
\( \frac{x-\alpha }{a}=z,\frac{y-\beta }{b}=z\)

or
\(\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z}1\)

Here

\( \vec{r}=( \alpha ,\beta ,0 )\) and \(\vec{g}=( a,b,1 )\)

Thus

\( \vec{t}=( \alpha',\beta ',0 )\) and \(\vec{g}'=( a',b',0 ) \)

So

\( \vec{g}\times \vec{g}'=( a,b,1 )\times ( a',b',0 ) \)

or
\( \vec{g}\times \vec{g}'=( -{b}',{a}',a{b}'-a'b )\)

Now, condition that the line generates a skew surface is

\([ \vec{t},\vec{g},\vec{g}']\ne 0 \)

or
\(\vec{t}.( \vec{g}\times \vec{g}' )\ne 0\)

or
\(( \alpha',\beta ',0 ).( -b',a',ab'-a'b )\ne 0 \)

or
\(a' \beta '-b'\alpha ' \ne 0 \)

#### Example

Show that a line \( x=3t^2z+2t( 1-3t^2 ),y=-2tz+t^2( 3+4t^2)\) generates a skew surface.

Solution

Given equation of line is

\(x=3t^2z+2t( 1-3t^2 ),y=-2tz+t^2( 3+4t^2 )\)

or
\(x-2t( 1-3t^2 )=3t^2z,y-t^2( 3+4t^2 )=-2tz\)

or
\(\frac{x-2t( 1-3t^2 )}{3t^2}=z,\frac{y-t^2( 3+4t^2 )}{-2t}=z\)

or
\( \frac{x-2t( 1-3t^2 )}{3t^2}=\frac{y-t^2( 3+4t^2 )}{-2t}=\frac{z}{1} \)

Here

\(\vec{r}=\{ 2t( 1-3t^2 ),t^2( 3+4t^2 ),0 \}\) and \( \vec{g}=( 3t^2,-2t,1 )\)

Thus

\(\vec {t}=( 2-18t^2,6t+16t^3,0 )\) and \( \vec{g}'=( 6t,-2,0 )\)

So

\( \vec{g}\times \vec{g}'=( 3t^2,-2t,1 )\times ( 6t,-2,0 )\)

or
\( \vec{g}\times \vec{g}'=( -2,6t,18t^2 )\)

Now, we have

\([ \vec {t},\vec{g},\vec{g}' ]=\vec{t}.( \vec{g}\times \vec{g}' )\)

or
\([ \vec {t},\vec{g},\vec{g}' ] =( 2-18t^2,6t+16t^3,0 ).( -2,6t,18t^2 )\)

or
\([ \vec {t},\vec{g},\vec{g}' ]\ne 0 \)

Hence, given line generates a skew surface.

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