Ruled Surface

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Ruled Surface

A surface generated by motion of straight line along a given curve is called ruled surface, in which, striaght line is called generator and given curve is called directrix of the 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.

  
0,0
az = 1.00
el = 0.30
 rotate = 1.00

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

  
0,0
az = 1.00
el = 0.30
 rotate = 1.00

  
0,0
az = 1.00
el = 0.30
 rotate = 1.00

Equation of Ruled Surface

Let S:$\overrightarrow{r}=\overrightarrow{r}(s)$ be a directrix of a ruled surface and P be a point on it with
$\overrightarrow{OP}=\overrightarrow{r}$
If $\overrightarrow{g}(s)$ be unit vector along generator and $R$ be arbitrary point T on the generator, then equation of ruled surface S is
$R=\overrightarrow{r}(s)+v\overrightarrow{g}(s)$ (i)where $v$ is parameter

Types of Ruled Surface

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

1. A ruled surface whose consecutive generators do intersect is called developable surface.
   For example, cone is developable surface.
   
2. 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 $[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }]=0$.
Proof
Let S be a ruled surface, then its equation is
$R=\overrightarrow{r}(s)+v\overrightarrow{g}(s)$ (i)$v$ is real parameter
Differentiating (i) w. r. to. s and v respectively, we get
$R_1=\frac{\partial R}{\partial s}=\overrightarrow{t}+v{\overrightarrow{g}}^{\prime }$
$R_2=\frac{\partial R}{\partial v}=\overrightarrow{g}$
$R_{11}={\overrightarrow{t}}^{\prime }+v{\overrightarrow{g}}^{″}$
$R_{12}={\overrightarrow{g}}^{\prime }$
$R_{22}=0$
Here
$[R_1,R_2,R_{12}]=HM$
or $[\overrightarrow{t}+v{\overrightarrow{g}}^{\prime },\overrightarrow{g},{\overrightarrow{g}}^{\prime }]HM$
or $[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }]=HM$
Next
$[R_1,R_2,R_{22}]=HN$
or $[\overrightarrow{t}+v{\overrightarrow{g}}^{\prime },\overrightarrow{g},0]=HN$
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 $K=-\frac{[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }{]}^2}{H^4}$
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
$\frac{[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }{]}^2}{H^4}=0$
or $[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }]=0$

Theorem 2

Show that necessary and sufficient condition for a ruled surface to be skew is $[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }]\ne 0$.
Proof
Let S be a ruled surface, then
$K=-\frac{[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }{]}^2}{H^4}$
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
$\frac{[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }{]}^2}{H^4}\ne 0$
or $[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }]\ne 0$

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
$\overrightarrow{r}=(\alpha ,\beta ,0)$ and $\overrightarrow{g}=(a,b,1)$
Thus
$\overrightarrow{t}=({\alpha }^{\prime },{\beta }^{\prime },0)$ and ${\overrightarrow{g}}^{\prime }=(a^{\prime },b^{\prime },0)$
So
$\overrightarrow{g}\times {\overrightarrow{g}}^{\prime }=(a,b,1)\times (a^{\prime },b^{\prime },0)$
or $\overrightarrow{g}\times {\overrightarrow{g}}^{\prime }=(-b^{\prime },a^{\prime },a{b}^{\prime }-a^{\prime }b)$
Now, condition that the line generates a skew surface is
$[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }]\ne 0$
or $\overrightarrow{t}.(\overrightarrow{g}\times {\overrightarrow{g}}^{\prime })\ne 0$
or $({\alpha }^{\prime },{\beta }^{\prime },0).(-b^{\prime },a^{\prime },a{b}^{\prime }-a^{\prime }b)\ne 0$
or $a^{\prime }{\beta }^{\prime }-b^{\prime }{\alpha }^{\prime }\ne 0$

Example

Show that a line $x=3t^{2}z+2t(1-3t^2),y=-2tz+t^2(3+4t^2)$ generates a skew surface.
Solution
Given equation of line is
$x=3t^{2}z+2t(1-3t^2),y=-2tz+t^2(3+4t^2)$
or $x-2t(1-3t^2)=3t^{2}z,y-t^2(3+4t^2)=-2tz$
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
$\overrightarrow{r}=\{2t(1-3t^2),t^2(3+4t^2),0\}$ and $\overrightarrow{g}=(3t^2,-2t,1)$
Thus
$\overrightarrow{t}=(2-18t^2,6t+16t^3,0)$ and ${\overrightarrow{g}}^{\prime }=(6t,-2,0)$
So
$\overrightarrow{g}\times {\overrightarrow{g}}^{\prime }=(3t^2,-2t,1)\times (6t,-2,0)$
or $\overrightarrow{g}\times {\overrightarrow{g}}^{\prime }=(-2,6t,18t^2)$
Now, we have
$[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }]=\overrightarrow{t}.(\overrightarrow{g}\times {\overrightarrow{g}}^{\prime })$
or $[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }]=(2-18t^2,6t+16t^3,0).(-2,6t,18t^2)$
or $[\overrightarrow{t},\overrightarrow{g},{\overrightarrow{g}}^{\prime }]\ne 0$
Hence, given line generates a skew surface.

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