Osculating Circle and Osculating Sphere

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Osculating Circle and Osculating Sphere

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Osculating Circle

Let \( C:\vec{r}=\vec{r}( s )\) be a space curve with neighboring points \( P,Q,R\) on \( C\) , then osculating circle at \( P\) is defined as limiting position of circle through \( P,Q,R\) as \( Q,R\) approaches to\( P\) .
In mathematical notation
Osculating circle at \( P = \displaystyle \lim_{Q,R\to P} \)circlePQR

Note:

1. Osculating circle lies in the osculating plane
2. Osculating circle at \( P\) has in general three point contact with the curve at \( P\) .
3. Tangent of osculating circle and space curve are same at point of contact

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Osculating Sphere

Let\( C:\vec{r}=\vec{r}( s )\)be a space curve. If \( P,Q,R,S\) are four neighboring points on\( C\), then osculating sphere at\( P\)is defined as limiting position of sphere through\( P,Q,R,S\) as\( Q,R,S\) approaches to\( P\).
\( \displaystyle \lim_{Q,R,S\to P} \)= spherePQRS= Osculating sphere at \( P\)

Osculating sphere at P has four point contact with the curve at at P.

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Center of Osculating Circle

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Let \( C: \vec{r}=\vec{r}(s) \) be a space curve and \( S: (\vec{r}-c)^2=a^2 \) be a sphere
Then intersection of osculating plane and sphere at P is osculating circle at P.

Thus, equation of osculating circle at P is
\( F(s): (\vec{r}(s)-c)^2-a^2=0 \) ...(i)
Since, osculating circle has three point contact with the curve, we have
\( F'(s)=0, F''(s)=0 \)
Thus, by successive differentiation of (i) w. r. to s, we get
\( F'(s)=0 \)
or\( 2 (\vec{r}(s)-c).\vec{t} =0 \)
or\( (\vec{r}(s)-c).\vec{t} =0 \) ...(ii)
Agaun,
\( F''(s)=0 \)
or\( (\vec{r}(s)-c).(\kappa \vec{n})+(\vec{t}).\vec{t} =0 \)
or\( (\vec{r}(s)-c).(\kappa \vec{n})+1 =0 \)
or\( (\vec{r}(s)-c).(\kappa \vec{n})=-1 \)
or\( (\vec{r}(s)-c).\vec{n}= \frac{-1}{\kappa} \)
or\( (\vec{r}(s)-c).\vec{n}= - \rho \) ...(iii)
From (ii), we see that \( (\vec{r}(s)-c) \) is perpendicular to \( \vec{t} \)
Thus,\( (\vec{r}(s)-c) \) is parallel to \( \vec{n} \)
So,
\( (\vec{r}(s)-c)= \lambda \vec{n} \) ...(A)
Now,
Taking dot product on both sides of (A) by \( \vec{n} \), we get
\( (\vec{r}(s)-c). \vec{n}= \lambda \vec{n}. \vec{n} \)
or\( (\vec{r}(s)-c). \vec{n}= \lambda \) ...(iv)
Compairing (iii) and (iv), we get
\( \lambda = - \rho \)
Hence, from (A), the center of osculating circle is
\( (\vec{r}(s)-c)= \lambda \vec{n} \)
or\( (\vec{r}(s)-c)= - \rho \vec{n} \)
or \( \vec{r}(s) +\rho \vec{n}= c \)
or\( c=\vec{r} +\rho \vec{n} \)
This completes the solution.

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Note

1. The locus of center of osculating circle is
   \( \vec{r_1}=\vec{r} +\rho \vec{n} \)
   where \( \vec{r_1} \) and \( \vec{r_1} \) are corresponding points on \( C_1 \) and \( C \) respectively or it can also be written as
   \( c=\vec{r} +\rho \vec{n} \) where c is center of osculating circle
   or it can also be written as
   \( R=\vec{r} +\rho \vec{n} \) where R is center of osculating circle
2. The center of osculating circle lies on principal normal.
3. Radius of osculating circle is
   \( |c- \vec{r}|=\rho \)

In the figure below, the locus of center of osculating circle is shown

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Properties on Osculating Circle

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Let \( C: \vec{r}=\vec{r}(s) \) be a space curve and \( C_1 \) be its locus of center of osculating circle then

1. tangent to \( C_1 \) lies in normal plane of space curve \( C \)
2. in case space curve \( C \) has constant curvature , then curvature of \( C_1 \) is also constant and torsion of \( C_1 \) varies inversely as that of \( C \) .

Proof

1. The locus of center of osculating circle is denoted by \( C_1 \) and defined by
   \( \vec{r_1}=\vec{r} +\rho \vec{n} \) (i)
   where \( \vec{r_1} \) and \( \vec{r} \) are corresponding points on \( C_1 \) and \( C \) respectively Now, differentiating (i), w. r. to. s, we get
   \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho (\tau \vec{b}- \kappa \vec{t})\)
   or\( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \rho \kappa \vec{t}\)
   or\( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \vec{t}\)
   or\( \vec{t_1} \frac{ds_1}{ds}=\rho' \vec{n}+ \rho \tau \vec{b} \)
   or (ii)
   It shows that tangent to \( C_1 \) lies in normal plane of space curve \( C \).
   
2. Next, for second property, in case space curve \( C \) has constant curvature, i.e. if \( \kappa \) is constant, then
   \( \kappa \) = constant
   or\( \rho \) = constant
   or\( \rho' \) = 0
   Hence, equation (ii) becomes
   \( \vec{t_1} \frac{ds_1}{ds}=\rho' \vec{n}+ \rho \tau \vec{b} \)
   or\( \vec{t_1} \frac{ds_1}{ds}= \rho \tau \vec{b} \)
   Taking magnitude we get
   \( \frac{ds_1}{ds}= \rho \tau \) (A)
   Substituting the value of \( \frac{ds_1}{ds}= \rho \tau \), we also get
   \( \vec{t_1} \frac{ds_1}{ds}= \rho \tau \vec{b} \)
   or\( \vec{t_1} \rho \tau= \rho \tau \vec{b} \)
   or\( \vec{t_1} = \vec{b} \) (iii)
   Differentiating (iii) w. r. to. s, we get
   \( \kappa_1 \vec{n_1} \frac{ds_1}{ds}= - \tau \vec{n} \)
   Taking magnitude, we get
   \( \kappa_1 \frac{ds_1}{ds}= \tau \)
   or\( \frac{ds_1}{ds}= \frac{\tau }{\kappa_1} \) (B)
   Substituting value of \( \frac{ds_1}{ds}= \frac{\tau }{\kappa_1} \), we also get
   \( \kappa_1 \vec{n_1} \frac{ds_1}{ds}= - \tau \vec{n} \)
   or\( \kappa_1 \vec{n_1} \frac{\tau }{\kappa_1} = - \tau \vec{n} \)
   or\( \vec{n_1} = - \vec{n} \) (iv)
   Equating (A) and (B), we get
   \( \rho \tau= \frac{\tau }{\kappa_1} \)
   or\( \rho = \frac{1}{\kappa_1} \)
   or \( \kappa_1 = \kappa \) = constant
   Thus, in case space curve C has constant curvature then curvature of \( C_1 \) is also constant.
   Again
   Taking cross product between \( \vec{t_1} \) and \( \vec{n_1} \) we get
   \( \vec{t_1} x \vec{n_1}= \vec{b} x (-) \vec{n} \)
   or\( \vec{b_1} = \vec{t} \) (v)
   Now differentiating (v) w. r. to. s, we get
   \( - \tau_1 \vec{n_1} \frac{ds_1}{ds} = \kappa \vec{n} \)
   Taking magnitude, we get
   \( \tau_1 \frac{ds_1}{ds} = \kappa \)
   or \( \frac{ds_1}{ds} = \frac{\kappa}{\tau_1} \) ...(C)
   Equating (A) and (C) we get
   \( \rho \tau= \frac{\kappa}{\tau_1} \)
   or \( \tau_1= \frac{\kappa}{ \rho \tau} \)
   or\( \tau_1= \frac{\kappa^2}{ \tau} \)
   This shows that torsion of \( C_1 \) varies inversely as that of \( C \) 2nd relation established

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Center of Osculating Sphere

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Let \( C: \vec{r}=\vec{r}(s) \) be a space curve and \( S: (\vec{r}-c)^2=a^2 \) be a sphere
Then intersection of osculating sphere at P with the curve is osculating sphere at P.
Thus, equation of osculating sphere at P is
\( F(s): (\vec{r}(s)-c)^2-a^2=0 \) (i)
Since, osculating sphere has four point contact with the curve, we have
\( F'(s)=0, F''(s)=0, F'''(s)=0 \)
Thus, by successive differentiation of (i) w. r. to s, we get
\( F'(s)=0 \)
or\( 2 (\vec{r}(s)-c).\vec{t} =0 \)
or\( (\vec{r}(s)-c).\vec{t} =0 \) (ii)
Again,
\( F''(s)=0 \)
or\( (\vec{r}(s)-c).(\kappa \vec{n})+(\vec{t}).\vec{t} =0 \)
or\( (\vec{r}(s)-c).(\kappa \vec{n})+1 =0 \)
or\( (\vec{r}(s)-c).(\kappa \vec{n})=-1 \)
or\( (\vec{r}(s)-c).\vec{n}= \frac{-1}{\kappa} \)
or\( (\vec{r}(s)-c).\vec{n}= - \rho \) (iii)
Again
\( F'''(s)=0 \)
or \( (\vec{r}(s)-c).(\tau \vec{b}- \kappa \vec{t} ) + \vec{t}.\vec{n}= - \rho' \)
or\( (\vec{r}(s)-c).(\tau \vec{b}- \kappa \vec{t} ) = - \rho' \)
or\( \tau (\vec{r}(s)-c).\vec{b} - \kappa (\vec{r}(s)-c). \vec{t} = - \rho' \)
From (i), \( (\vec{r}(s)-c).\vec{t} =0 \) भएकोले
or\( \tau (\vec{r}(s)-c).\vec{b} = - \rho' \)
or\( (\vec{r}(s)-c).\vec{b} = - \rho' \frac{1}{\tau } \)
or\( (\vec{r}(s)-c).\vec{b} = - \sigma \rho' \) (iv)
From (ii), we see that \( (\vec{r}(s)-c) \) is perpendicular to \( \vec{t} \)
Thus,\( (\vec{r}(s)-c) \) lies in normal plane So,
\( (\vec{r}(s)-c)= \lambda \vec{n}+ \mu \vec{b} \) (A)
Now,
Taking dot product on both sides of (A) by \( \vec{n} \), we get
\( (\vec{r}(s)-c). \vec{n}= (\lambda \vec{n}+ \mu \vec{b}).\vec{n} \)
or\( (\vec{r}(s)-c). \vec{n}= \lambda \) (v)
Compairing (iii) and (v), we get
\( \lambda = - \rho \)
Again,
Taking dot product on both sides of (A) by \( \vec{b} \), we get
\( (\vec{r}(s)-c). \vec{b}= (\lambda \vec{n}+ \mu \vec{b}).\vec{b} \)
or\( (\vec{r}(s)-c). \vec{b}= \mu \) (vi)
Compairing (iv) and (vi), we get
\( \mu = - \sigma \rho' \)
Hence, from (A), the center of osculating sphere is
\( (\vec{r}(s)-c)= \lambda \vec{n}+ \mu \vec{b} \)
or\( \vec{r}(s)-c=- \rho \vec{n} - \sigma \rho' \vec{b} \)
or\( \vec{r}(s)+ \rho \vec{n} + \sigma \rho' \vec{b} = c \)
or\( c= \vec{r} + \rho \vec{n} + \sigma \rho' \vec{b} \)
This completes the solution.

Note

1. The locus of center of osculating sphere is
   \( \vec{r_1}= \vec{r} + \rho \vec{n} + \sigma \rho' \vec{b} \)
   where \( \vec{r_1} \) and \( \vec{r} \) are corresponding points on \( C_1 \) and \( C \) respectively or it can also be written as
   \( c= \vec{r} + \rho \vec{n} + \sigma \rho' \vec{b} \) where c is center of osculating sphere or it can also be written as
   \( R= \vec{r} + \rho \vec{n} + \sigma \rho' \vec{b} \) where R is center of osculating sphere
2. The center of osculating spheree is
   \( |c- \vec{r}|=|\rho \vec{n} + \sigma \rho' \vec{b}|= \sqrt{\rho^2+ (\sigma \rho')^2} \)
   

In the figure below, locus of center of osculating sphere is shown.

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Properties on Osculating Sphere

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Let \( C: \vec{r}=\vec{r}(s) \) be a space curve and \( C_1 \) be its locus of center of osculating sphere then

1. tangent, binormal and principal normal to \( C_1 \) are respectively parallel to binormal, tangent and principal normal to the space curve \( C \).
2. product of curvatures at corresponding points of the curves \( C \) and \( C_1 \) is equal to the product of torsions at these points.

Proof
The locus of center of osculating sphere is denoted by \( C_1 \) and defined by
\( \vec{r_1}=\vec{r} +\rho \vec{n}+ \sigma \rho' \vec{b} \) (i)
where \( \vec{r_1} \) and \( \vec{r} \) are corresponding points on \( C_1 \) and \( C \) respectively Now, differentiating (i), w. r. to. s, we get
\( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho (\tau \vec{b}- \kappa \vec{t})+ (\sigma \rho')' \vec{b}+(\sigma \rho') (- \tau \vec{n}) \)
or \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \rho \kappa \vec{t} + (\sigma \rho')' \vec{b}-\sigma \rho' \tau \vec{n} \)
or \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \vec{t} + (\sigma \rho')' \vec{b}- \rho' \vec{n} \)
or \( \vec{t_1} \frac{ds_1}{ds}= \rho \tau \vec{b}+ \frac{d}{ds}(\sigma \rho') \vec{b} \)
or \( \vec{t_1} \frac{ds_1}{ds}= [\rho \tau + \frac{d}{ds}(\sigma \rho')] \vec{b} \) (ii)
or or or or Taking magnitude we get
\( \frac{ds_1}{ds}= [\rho \tau + \frac{d}{ds}(\sigma \rho')] \) (A)
Substituting the value of \( \frac{ds_1}{ds}= [\rho \tau + \frac{d}{ds}(\sigma \rho')] \), we also get
\( \vec{t_1} \frac{ds_1}{ds}= [\rho \tau + \frac{d}{ds}(\sigma \rho')] \vec{b} \)
or or \( \vec{t_1} [\rho \tau + \frac{d}{ds}(\sigma \rho')] = [\rho \tau + \frac{d}{ds}(\sigma \rho')] \vec{b} \)
or \( \vec{t_1} = \vec{b} \) (iii)
Differentiating (iii) w. r. to. s, we get
\( \kappa_1 \vec{n_1} \frac{ds_1}{ds}= - \tau \vec{n} \)
Taking magnitude, we get
\( \kappa_1 \frac{ds_1}{ds}= \tau \)
or \( \frac{ds_1}{ds}= \frac{\tau }{\kappa_1} \) (B)
Substituting value of \( \frac{ds_1}{ds}= \frac{\tau }{\kappa_1} \), we also get
\( \kappa_1 \vec{n_1} \frac{ds_1}{ds}= - \tau \vec{n} \)
or \( \kappa_1 \vec{n_1} \frac{\tau }{\kappa_1} = - \tau \vec{n} \)
or \( \vec{n_1} = - \vec{n} \) (iv)
Again
Taking cross product between \( \vec{t_1} \) and \( \vec{n_1} \) we get
\( \vec{t_1} \times \vec{n_1}= \vec{b} \times (-) \vec{n} \)
or or \( \vec{b_1} = \vec{t} \) (v)
Now differentiating (v) w. r. to. s, we get
\( - \tau_1 \vec{n_1} \frac{ds_1}{ds} = \kappa \vec{n} \)
Taking magnitude, we get
\( \tau_1 \frac{ds_1}{ds} = \kappa \)
or \( \frac{ds_1}{ds} = \frac{\kappa}{\tau_1} \) (C)
Here,

• shows that tangent to \( C_1 \) is parallel to binormalto \( C \).
• shows that principal normal to \( C_1 \) is parallel to principal normal to \( C \).
• shows that binormal to \( C_1 \) is parallel tangent to \( C \).

Thus first relation established.
Next,
Equating (B) and (C) we get
\( \frac{\tau }{\kappa_1}= \frac{\kappa}{\tau_1} \)
or \( \tau \tau_1= \kappa \kappa_1 \)
This shows that product of curvatures at corresponding points of the curves \( C \) and \( C_1 \) is equal to the product of torsions at these points.
Thus 2nd relation established

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Theorems

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1. Show that principal normal to a curve is normal to the locus of center of curvature at points where curvature is stationary.
   or
   Show that principal normal to a curve is normal to the locus of center of osculating circle at points where curvature is stationary.
   or
   Show that principal normal to a curve is normal to the locus of center of circle of curvature at points where curvature is stationary.
   center of curvature =center of osculating circle=center of circle of curvature Proof
   The locus of center of osculating circle is denoted by \( C_1 \) and defined by
   \( \vec{r_1}=\vec{r} +\rho \vec{n} \) (i)
   where \( \vec{r_1} \) and \( \vec{r} \) are corresponding points on \( C_1 \) and \( C \) respectively
   Now, differentiating (i), w. r. to. s, we get
   \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho (\tau \vec{b}- \kappa \vec{t})\)
   or\( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \rho \kappa \vec{t}\)
   or\( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \vec{t}\)
   or\( \vec{t_1} \frac{ds_1}{ds}=\rho' \vec{n}+ \rho \tau \vec{b} \) (ii)
   At the points where curvature is stationary, we have
   \( \kappa \) = constant
   or\( \rho \) = constant
   or\( \rho' \) = 0
   Thus, from (ii), we get
   \( \vec{t_1} \frac{ds_1}{ds}=\rho' \vec{n}+ \rho \tau \vec{b} \)
   or\( \vec{t_1} \frac{ds_1}{ds}= \rho \tau \vec{b} \)
   Operating dot product on both side by \( \vec{n} \), we get
   \( \vec{t_1} \frac{ds_1}{ds}= \rho \tau \vec{b} \)
   or\( \vec{n}.\vec{t_1} \frac{ds_1}{ds}= \rho \tau \vec{n}. \vec{b} \)
   or\( \vec{n}.\vec{t_1} \frac{ds_1}{ds}= 0 \)
   It shows that, principal normal to a curve is normal to the locus of center of osculating circle at points where curvature is stationary.

2. Show that radius of osculating circle of a circular helix is equal to radius of osculating sphere.

   Proof
   The radius a of osculating circle is
   radius a of osculating circle = \( \rho \) (i)
   The radius a of osculating sphere is
   \( \sqrt{\rho^2 + (\sigma \rho')^2} \) (ii)
   Since, the curvature κ of circular helix is constant, we have
   \( \rho \) = constant
   or \( \rho ' = 0 \)
   Then, radius of osculating sphere is
   radius a of osculating sphere = \( \sqrt{\rho^2 + (\sigma \rho')^2} \)
   or radius a of osculating sphere = \( \sqrt{\rho^2 + (\sigma \times 0)^2} \)
   or radius a of osculating sphere = \( \sqrt{\rho^2} \)
   or radius a of osculating sphere = \( \rho \)
   Thus, radius of osculating sphere of a circular helix is equal to radius of osculating circle.

3. Show that tangent to the locus of center of osculating sphere is parallel to the binormal of the curve at corresponding points.

   Proof
   Let \( C: \vec{r}=\vec{r}(s) \) be a space curve and \( C_1 \) be its locus of center of osculating sphere then \( C_1 \) and defined by
   \( \vec{r_1}=\vec{r} +\rho \vec{n}+ \sigma \rho' \vec{b} \) (i)
   where \( \vec{r_1} \) and \( \vec{r} \) are corresponding points on \( C_1 \) and \( C \) respectively
   Now, differentiating (i), w. r. to. s, we get
   \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho (\tau \vec{b}- \kappa \vec{t})+ (\sigma \rho')' \vec{b}+(\sigma \rho') (- \tau \vec{n}) \)
   or \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \rho \kappa \vec{t} + (\sigma \rho')' \vec{b}-\sigma \rho' \tau \vec{n} \)
   or \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \vec{t} + (\sigma \rho')' \vec{b}- \rho' \vec{n} \)
   or \( \vec{t_1} \frac{ds_1}{ds}= \rho \tau \vec{b}+ \frac{d}{ds}(\sigma \rho') \vec{b} \)
   or \( \vec{t_1} \frac{ds_1}{ds}= [\rho \tau + \frac{d}{ds}(\sigma \rho')] \vec{b} \) (ii)
   This shows that, tangent to the locus of center of osculating sphere is parallel to the binormal of the curve at corresponding points.

4. Show that for a curve of constant curvature center of osculating sphere coincides with center of osculating circle.

   Proof
   The locus of center of osculating sphere is
   \( \vec{r_1}=\vec{r} +\rho \vec{n}+ \sigma \rho' \vec{b} \) (i)
   If curvature κ is constant, then
   \( \rho \) = constant
   or \( \rho' =0 \)
   Thus locus of center of osculating sphere becomes
   \( \vec{r_1}=\vec{r} +\rho \vec{n}+ \sigma \rho' \vec{b} \)
   or \( \vec{r_1}=\vec{r} +\rho \vec{n}+ \sigma \times 0 \times \vec{b} \)
   or \( \vec{r_1}=\vec{r} +\rho \vec{n} \)
   This is also the locus of center of osculating circle
   Hence, center of spherical curvature of a curve of constant curvature coincides with center of circle of curvature.

5. Show that tangent to the locus of center of osculating sphere passes through the center of osculating circle.

   Proof The locus of center of osculating sphere is defined by
   \( \vec{r_1}=\vec{r} +\rho \vec{n}+ \sigma \rho' \vec{b} \) (i)
   Differentiating (i) w. r. to. s, we get
   \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho (\tau \vec{b}- \kappa \vec{t})+ (\sigma \rho')' \vec{b}+(\sigma \rho') (- \tau \vec{n}) \)
   or \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \rho \kappa \vec{t} + (\sigma \rho')' \vec{b}-\sigma \rho' \tau \vec{n} \)
   or \( \vec{t_1} \frac{ds_1}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \vec{t} + (\sigma \rho')' \vec{b}- \rho' \vec{n} \)
   or
   or \( \vec{t_1} \frac{ds_1}{ds}= \rho \tau \vec{b}+ \frac{d}{ds}(\sigma \rho') \vec{b} \)
   or \( \vec{t_1} \frac{ds_1}{ds}= [\rho \tau + \frac{d}{ds}(\sigma \rho')] \vec{b} \)
   or \( \vec{t_1} = [\rho \tau + \frac{d}{ds}(\sigma \rho')] \frac{ds}{ds_1} \vec{b} \) (ii)
   The equation of tangent to \( C_1 \) is
   \( R= \vec{r_1} + \lambda \vec{t_1} \) (iii)
   Now, substituting (i) and (ii) in (iii) we get
   \( R= \vec{r_1} + \lambda \vec{t_1} \)
   or \( R= [\vec{r} +\rho \vec{n}+ \sigma \rho' \vec{b}] + \lambda [\rho \tau + \frac{d}{ds}(\sigma \rho')] \frac{ds}{ds_1} \vec{b} \)
   or \( R= \vec{r} +\rho \vec{n}+ \sigma \rho' \vec{b} + \lambda [\rho \tau + \frac{d}{ds}(\sigma \rho')] \frac{ds}{ds_1} \vec{b} \)
   or
   If we take,
   \( \lambda = \frac{-\sigma \rho'}{\rho \tau + \frac{d}{ds}(\sigma \rho')} \frac{ds_1}{ds} \) (iv)
   Then,
   \( R= \vec{r} +\rho \vec{n}\)
   Since, position vector of center of osculating circle is
   \( R= \vec{r} +\rho \vec{n}\),
   The tangent to the locus of center of osculating sphere passes through the center of osculating circle.

6. If a curve lies on a sphere, show that \( \rho \) and \( \sigma \) are connected by the relation \( \rho \tau + \frac{d}{ds}(\sigma \rho')=0 \)

   or

   Show that necessary and sufficient condition for a curve to lie on a surface of sphere is \( \rho \tau + \frac{d}{ds}(\sigma \rho')=0 \)

   Proof
   The locus of center of osculating sphere is defined by
   \( c=\vec{r} +\rho \vec{n}+ \sigma \rho' \vec{b} \) (i)
   Differentiating (i) w. r. to. s, we get
   \( \frac{dc}{ds}=\vec{t} +\rho' \vec{n}+ \rho (\tau \vec{b}- \kappa \vec{t})+ (\sigma \rho')' \vec{b}+(\sigma \rho') (- \tau \vec{n}) \)
   or \( \frac{dc}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \rho \kappa \vec{t} + (\sigma \rho')' \vec{b}-\sigma \rho' \tau \vec{n} \)
   or \( \frac{dc}{ds}=\vec{t} +\rho' \vec{n}+ \rho \tau \vec{b}- \vec{t} + (\sigma \rho')' \vec{b}- \rho' \vec{n} \)
   or \( \frac{dc}{ds}= \rho \tau \vec{b}+ \frac{d}{ds}(\sigma \rho') \vec{b} \)
   or \( \frac{dc}{ds}= [\rho \tau + \frac{d}{ds}(\sigma \rho')] \vec{b} \)
   Necessary condition
   Assume that the curve lie on the surface of sphere, then
   the position of the center c is constant
   Thus,
   \( \frac{dc}{ds}=0 \)
   or \( [\rho \tau + \frac{d}{ds}(\sigma \rho')] \vec{b}=0 \)
   or \( [\rho \tau + \frac{d}{ds}(\sigma \rho')]=0 \)
   Sufficient condition
   Assume that
   \( [\rho \tau + \frac{d}{ds}(\sigma \rho')]=0 \)
   Then
   \( \frac{dc}{ds}=0 \)
   It shows that, center of osculating sphere is constant, thus center is independent of point on the curve, Hence, curve lies on the surface of sphere.

Relation of locus of Osculating Circle and Osculating Sphere

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