Developable surface
Envelope of one parameter family of plane is called developable surface. It is given by the equation
\( \vec{r}.\vec{N}=p \)
where \(\vec{N} \) and \(p \) are function of real parameter \( u\)
The equation may be put in the form
\(f( u ): \vec{r}.\vec{N}( u )-p( u )=0 \)
Theorem 1
Show that tangent plane is same at all points of generators of developable surface.
Proof
Let S be a developable surface, then its equation is
\(R=\vec{r}(s)+v\vec{t}( s ) \) (i)v is real parameter
Differentiating (i) w. r. to. s and v respectively, we get
\( R_1=\frac{\partial R}{\partial s} \)
or
\( R_1= \vec{t}+v\kappa \vec{n} \)
And
\( R_2=\frac{\partial R}{\partial v} \)
or
\( R_2=\vec{t} \)
Now
\({R_1}\times {R_2}= ( \vec{t}+v\kappa \vec{n} )\times \vec{t} \)
or
\({R_1}\times {R_2}=-v\kappa \vec{b} \)
or
\( H\vec{N}=-v\kappa \vec{b} \)
Taking magnitude, we get
\( H=v \kappa \)
Substituting value of H, we get
\( \vec{N}=-\vec{b} \)
Now, equation of tangent plane to the developable surface is
\( ( R-\vec{r} ).\vec{N}=0 \)
or
\(( R-\vec{r} ).\vec{b}=0 \)
Since \(\vec{b}\) is function of s only, tangent plane is same at all points of generators of the developable surface.
Theorem 2
Show that osculating plane to the edge of regression is tangent plane to the developable surface
Proof
Let S be a developable surface, then its equation is
\(R=\vec{r}(s)+v\vec{t}( s ) \) (i)v is real parameter
Then, we get
\( \vec{N}=-\vec{b} \) (A)
Now, equation of tangent plane to the developable surface is
\( ( R-\vec{r} ).\vec{N}=0 \) (B)
Also, equation of osculating plane to the edge of regression is
\( (R-\vec{r}).\vec{b}=0 \) (C)
Thus, from (A) ,(B) and (C), it shows that osculating plane to the edge of regression is tangent plane to the developable surface.
Developable of a space curve
Envelope of one parameter family of plane is called developable surface. Since, principal planes consists only single parameter, s, therefore their envelopes are developable surface. These developable surface are respectively called: Osculating developable, Polar developable and Rectifying developable
Theorem 1
The generators of osculating developable of a space curve are tangent to the curve and the edge of regression of osculating developable of a space curve is the curve itself.
Proof
Let \( \vec{r}=\vec{r}( s ) \) be a space curve and P be a point with
\(\vec{OP}=\vec{r} \)
Then family of equation of osculating plane is
F: \( ( R-\vec{r} ).\vec{b}=0 \)
(i)
By successive differentiation of (i) we get
\( \partial F =0 \)
or
\( ( R-\vec{r} ).( -\tau \vec{n} )-\vec{t}.\vec{b}=0 \)
or
\(( R-\vec{r} ).\vec{n}=0 \)
(ii)
Next
\(\partial ^2 F = 0\)
or
\( ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )+\vec{t}.\vec{n}=0 \)
or
\(( R-\vec{r} ).\vec{t}=0 \)
(iii)
Now, characteristic line (generator) of osculating developable is
\( F=0;\partial F=0 \)
or
\(( R-\vec{r} ).\vec{b}=0 ; ( R-\vec{r} ).\vec{n}=0\)
The first is osculating plane, second is rectifying plane, so intersection of osculating plane and rectifying plane is tangent
Hence, generators of osculating developable of a space curve are tangent to the curve
Next, characteristic point of osculating developable is
\( F=0;\partial F=0, \partial^2 F =0\)
or
\( ( R-\vec{r} ).\vec{b}=0, ( R-\vec{r} ).\vec{n}=0, ( R-\vec{r} ).\vec{t}=0 \)
The first is osculating plane, second is rectifying plane, and third is normal plane
Hence, intersection of osculating plane, rectifying plane, and normal plane is point on the curve
Thus, edge of regression of osculating developable is the curve itself
Theorem 2
The generator of polar developable passes through the center of osculating circle and the edge of regression of polar developable of a space curve is the locus of center of osculating sphere.
Proof
Let \( \vec{r}=\vec{r}( s ) \) be a space curve and P be a point with
\(\vec{OP}=\vec{r} \)
Then family of equation of normal plane is
F: \( ( R-\vec{r} ).\vec{t}=0 \) (i)
By successive differentiation of (i) we get
\( \partial F =0 \)
\( \partial F : ( R-\vec{r} ) \cdot ( \kappa \vec{n} )- \vec{t} \cdot \vec{t}=0 \)
or
\( ( R-\vec{r} ) \cdot \vec{n}=\rho \) (ii)
Next
\( \partial ^2 F =0\)
or
\( ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )+\vec{t}.\vec{n}=\rho' \)
or
\(( R-\vec{r} ).\vec{b}=\sigma \rho '\) (iii)
Now, generators of polar developable is
\(F=0;\partial F=0 \)
\(( R-\vec{r} ).\vec{t}=0, R-\vec{r} ).\vec{n}=\rho \)
\(( R-\vec{r} ).\vec{t}=0 \) and \(( R-\vec{r}-\rho \vec{n} ).\vec{n}=0 \)
It shows that generator of polar developable passes through the center of circle of curvature.
Next, edge of regression of polar
developable is locus of a point given by
\(F=0;\partial F=0;\partial ^2 F=0\)
or
\(( R-\vec{r} ).\vec{t}=0 , ( R-\vec{r} ).\vec{n}=\rho ; ( R-\vec{r} ).\vec{b}=\sigma \rho '\)
Since
\( R-\vec{r} \) is perpendicular to tangent
Thus, we write
\(R-\vec{r}=\lambda \vec{n}+\mu \vec{b}\) (A)
Taking dot product of (A) by \(\vec{n} \) we get
\(( R-\vec{r} ).\vec{n}=\lambda\)
or
\( \lambda =\rho \)from (ii)
Taking dot product of (A) by \(\vec{b}\) we get
\(( R-\vec{r} ).\vec{b}=\mu \)
or
\( \mu =\sigma \rho '\) from (iii)
Substituting the values of \( \lambda \) and \( \mu \) in (A), the edge of regression of polar developable is
\(R=\vec{r}+\rho \vec{n}+\sigma \rho '\vec{b}\)
This is the locus of center of osculating sphere.
Hence the theorem.
Theorem 3
The edge of regression of rectifying developable of a space curve has the equation of the form
\( R=\vec{r}+\kappa \frac{\tau \vec{t}+\kappa \vec{b}}{{\kappa }'\tau -\kappa {\tau }'}\)
Proof
Let \( C:\vec{r}=\vec{r}( s )\) be a space curve.
Then equation of rectifying plane is
\( F: ( R-\vec{r} ).\vec{n}=0 \) (i)
By successive differentiation (i), we get
\( \partial F=0 \)
or
\( ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )+\vec{t}.\vec{n}=0 \)
or
\( ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )=0 \)
(i)
Next
\( {{\partial }^{2}}F=0 \)
or
\( ( R-\vec{r} ).( {\tau }'\vec{b}-{\kappa }'\vec{t}-{\tau ^2}\vec{n}-\kappa ^2 \vec{n} )-\vec{t}.( \tau \vec{b}-\kappa \vec{t} )=0 \)
or
\( ( R-\vec{r} ).( \tau \vec{'b}-\kappa '\vec{t} )+\kappa =0 \) (iii)
Now, edge of regression of rectifying developable of a space curve is locus of point given by
\( F=0;\partial F=0;{{\partial }^{2}}F=0 \)
or
\( ( R-\vec{r} ).\vec{n}=0, ( R-\vec{r} ).( \tau \vec{b}-\kappa \vec{t} )=0, ( R-\vec{r} ).( \tau \vec{'b}-\kappa '\vec{t} )=-\kappa \)
From (i) and (ii) we see that
\( ( R-\vec{r} )=\lambda [ \vec{n}\times ( \tau \vec{b}-\kappa \vec{t} ) ] \)
or
\( ( R-\vec{r} )=\lambda ( \tau \vec{t}+\kappa \vec{b} ) \) (A)
Multiplying (iv) by we get \( ( \tau \vec{'b}-\kappa '\vec{t} ) \)
\( ( R-\vec{r} ).( \tau \vec{'b}-\kappa '\vec{t} )=\lambda ( \tau \vec{t}+\kappa \vec{b} )( \tau \vec{'b}-\kappa '\vec{t} )\)
or
\( -\kappa =\lambda ( \kappa {\tau }'-\kappa '\tau ) \)
or
\( \lambda =\frac{\kappa }{( \kappa '\tau -\kappa \tau ' )} \)
Substituting \( \lambda \) in (A) we get
\( ( R-\vec{r} )=\frac{\kappa }{( {\kappa }'\tau -\kappa {\tau }' )}( \tau \vec{t}+\kappa \vec{b} ) \)
or
\( R=\vec{r}+\kappa \frac{ \tau \vec{t}+\kappa \vec{b}} {\kappa '\tau -\kappa \tau '} \)
is the required equation.
Example
Prove that the rectifying developable of a curve is the polar developable of its involutes and conversely.
Proof
Let \( C \) be a given curve and \( C_1\) be its involutes then
\( \vec{r}=\vec{r}( s ) \) and \( {{\vec{r}}_{1}}=\vec{r}+( c-s )\vec{t} \)
Here
equation of involute \( C_1 \) is
\( \vec{r_1}= \vec{r} + (c-s) \vec{t} \) (i)
Differentiating (i) w. r. to. s, we get
\( \vec{t_1}\frac{ds_1}{ds}= \vec{t} + (c-s) \kappa \vec{n} + (-1)\vec{t} \)
or\( \vec{t_1}\frac{ds_1}{ds}= (c-s) \kappa \vec{n} \)
Taking magnitude, we get
\(\frac{ds_1}{ds}= (c-s) \kappa \)
Substituting value of \(\frac{ds_1}{ds}= (c-s) \kappa \) , we get
\( \vec{t_1}= \vec{n} \) (A)
Now, equation of rectifying developable of the curve \( C\) is
\( ( R-\vec{r} ).\vec{n}=0 \)
(ii)
And, equation of polar developable of \(C_1\) is
\( ( R-\vec{r}_1).\vec{t}_1=0 \)
or\( ( R-\vec{r}-( c-s )\vec{t} ).\vec{t}_1=0\)
or\( ( R-\vec{r}-( c-s )\vec{t} ).\vec{n}=0 \)
or\( ( R-\vec{r} ).\vec{n}=0 \) (iii)
Here, we showed that (ii) and (iii) are identical.
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