1. (i). Write down the equation of tangent line to a curve at a given point.
   (ii). Define the oscillating circle.
   (iii). Write down the formula to the curvature of evolute.
   (iv). State Mennier’s theorem.

2. Prove that the torsion of the two Bertrand curves has the same sign and their product is constant.
3. Show that the vector $B^i$$B^i$B^(i)B^i$B^i$ of variable magnitude suffers a parallel displacement along a curve C if and only if :

4. Prove that an entity whose inner product with an arbitrary tension is a tensor is itself a tensor.
5. If surface of sphere is a two dimensional Riemannian space. Compute the Christoffel symbols.

6. (i). State and prove Meunienis theorem.
   (ii). Find the Geodesic curvature of the curve $u=$$u=$u=u=$u=$ constant on the surface $x=u\cos \theta ,y=u\sin \theta ,z=\frac{1}{2}a{u}^2$$x=u\cos \theta ,y=u\sin \theta ,z=\frac{1}{2}a{u}^2$x=u cos theta,y=u sin theta,z=(1)/(2)au^(2)x=u \cos \theta, y=u \sin \theta, z=\frac{1}{2} a u^2$x=u\cos \theta ,y=u\sin \theta ,z=\frac{1}{2}a{u}^2$
7. (i). Find the angle between two tangential direction on the surface in the terms of direction ratio.
   (ii). If the tangent and binormal at a point of curve make angles $\theta ,\varphi$$\theta ,\varphi$theta,phi\theta, \phi$\theta ,\varphi$ respectively with a fixed directions then:

Answer:

Question:-01(a)

Write down the equation of the tangent line to a curve at a given point.

Answer:

General Formula

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Question:-01(b)

Define the oscillating circle.

Answer:

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Key Features of the Osculating Circle:

1. Radius of Curvature: The radius of the osculating circle is equal to the reciprocal of the curvature ($k$$k$kk$k$) of the curve at the point:
   $$R=\frac{1}{k},$$$$R=\frac{1}{k},$$R=(1)/(k),R = \frac{1}{k},$$R=\frac{1}{k},$$
   where $k$$k$kk$k$ is the curvature.
2. Center of the Osculating Circle: The center of the circle, called the center of curvature, lies along the normal to the curve at the given point, at a distance $R$$R$RR$R$ from the point.
3. Curvature Match: The osculating circle has the same curvature as the curve at the point of tangency.
4. Best Approximation: Among all circles passing through the point, the osculating circle best approximates the curve locally. Its radius and position ensure this close fit.

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Formulas:

• Curvature:$$k=\frac{|x^{\prime }(t)y^{″}(t)-y^{\prime }(t)x^{″}(t)|}{{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2)}^{3/2}}.$$$$k=\frac{|x^{\prime }(t)y^{″}(t)-y^{\prime }(t)x^{″}(t)|}{{\left(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2\right)}^{3/2}}.$$k=(|x^(‘)(t)y^(″)(t)-y^(‘)(t)x^(″)(t)|)/((x^(‘)(t)^(2)+y^(‘)(t)^(2))^(3//2)).k = \frac{|x'(t)y”(t) – y'(t)x”(t)|}{\left(x'(t)^2 + y'(t)^2\right)^{3/2}}.$$k=\frac{|x^{\prime }(t)y^{″}(t)-y^{\prime }(t)x^{″}(t)|}{{(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2)}^{3/2}}.$$
• Radius of Curvature:$$R=\frac{1}{k}.$$$$R=\frac{1}{k}.$$R=(1)/(k).R = \frac{1}{k}.$$R=\frac{1}{k}.$$
• Center of the Osculating Circle:$$(x_c,y_c)=(x_0-R\frac{y^{\prime }(t)}{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}},y_0+R\frac{x^{\prime }(t)}{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}}),$$$$\left(x_c,y_c\right)=\left(x_0-R\frac{y^{\prime }(t)}{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}},y_0+R\frac{x^{\prime }(t)}{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}}\right),$$(x_(c),y_(c))=(x_(0)-R(y^(‘)(t))/(sqrt(x^(‘)(t)^(2)+y^(‘)(t)^(2))),y_(0)+R(x^(‘)(t))/(sqrt(x^(‘)(t)^(2)+y^(‘)(t)^(2)))),\left(x_c, y_c\right) = \left(x_0 – R \frac{y'(t)}{\sqrt{x'(t)^2 + y'(t)^2}}, y_0 + R \frac{x'(t)}{\sqrt{x'(t)^2 + y'(t)^2}}\right),$$(x_c,y_c)=(x_0-R\frac{y^{\prime }(t)}{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}},y_0+R\frac{x^{\prime }(t)}{\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}}),$$where $(x_0,y_0)$$(x_0,y_0)$(x_(0),y_(0))(x_0, y_0)$(x_0,y_0)$ is the point of tangency and $R$$R$RR$R$ is the radius of curvature.

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Question:-01(c)

Write down the formula to the curvature of evolute.

Answer:

Formula for the Curvature of the Evolute:

• $k(t)$$k(t)$k(t)k(t)$k(t)$ is the curvature of the original curve.
• $R(t)=1/k(t)$$R(t)=1/k(t)$R(t)=1//k(t)R(t) = 1/k(t)$R(t)=1/k(t)$ is the radius of curvature of the original curve.
• $\rho (t)$$\rho (t)$rho(t)\rho(t)$\rho (t)$ is the radius of the osculating circle measured along the evolute (distance to the center of curvature).

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Explanation

1. Evolute: The evolute of a curve is the locus of the centers of curvature of the original curve.
2. Geometry: The curvature of the evolute depends on how the curvature of the original curve changes relative to the distance from the curve to its center of curvature.

Question:-01(d)

State Mennier’s theorem.

Answer:

Statement of Meunier’s Theorem:

• $k_1$$k_1$k_(1)k_1$k_1$ and $k_2$$k_2$k_(2)k_2$k_2$ are the principal curvatures of the surface at the point under consideration.
• $\theta$$\theta$theta\theta$\theta$ is the angle between the given direction and the direction of the first principal curvature ($k_1$$k_1$k_(1)k_1$k_1$).

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Question:-02

Prove that the torsion of the two Bertrand curves has the same sign and their product is constant.

Answer:

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Bertrand Curves

1. $\mathit{\beta }(s)=\mathit{\alpha }(s)+\lambda \mathit{N}(s)$$\mathit{\beta }(s)=\mathit{\alpha }(s)+\lambda \mathit{N}(s)$beta(s)=alpha(s)+lambda N(s)\boldsymbol{\beta}(s) = \boldsymbol{\alpha}(s) + \lambda \boldsymbol{N}(s)$\mathit{\beta }(s)=\mathit{\alpha }(s)+\lambda \mathit{N}(s)$, where $\mathit{N}(s)$$\mathit{N}(s)$N(s)\boldsymbol{N}(s)$\mathit{N}(s)$ is the principal normal of $\mathit{\alpha }(s)$$\mathit{\alpha }(s)$alpha(s)\boldsymbol{\alpha}(s)$\mathit{\alpha }(s)$ and $\lambda$$\lambda$lambda\lambda$\lambda$ is a constant.
2. The principal normal directions of $\mathit{\alpha }(s)$$\mathit{\alpha }(s)$alpha(s)\boldsymbol{\alpha}(s)$\mathit{\alpha }(s)$ and $\mathit{\beta }(s)$$\mathit{\beta }(s)$beta(s)\boldsymbol{\beta}(s)$\mathit{\beta }(s)$ coincide at corresponding points.

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Definitions of Curvature and Torsion

• The curvature is $\kappa (s)$$\kappa (s)$kappa(s)\kappa(s)$\kappa (s)$.
• The torsion is $\tau (s)$$\tau (s)$tau(s)\tau(s)$\tau (s)$.

• Let its curvature and torsion be ${\kappa }_{\beta }(s)$${\kappa }_{\beta }(s)$kappa _(beta)(s)\kappa_\beta(s)${\kappa }_{\beta }(s)$ and ${\tau }_{\beta }(s)$${\tau }_{\beta }(s)$tau _(beta)(s)\tau_\beta(s)${\tau }_{\beta }(s)$, respectively.

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Step 1: Relationship Between the Curvatures

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Step 2: Relationship Between the Torsions

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Step 3: Product of the Torsions

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Step 4: Signs of the Torsions

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Conclusion

1. The torsions of the two Bertrand curves have the same sign.
2. The product of their torsions is constant:$$\tau (s)\cdot {\tau }_{\beta }(s)=\frac{\tau (s{)}^2}{(1-\lambda \kappa (s){)}^2}.$$$$\tau (s)\cdot {\tau }_{\beta }(s)=\frac{\tau (s{)}^2}{(1-\lambda \kappa (s){)}^2}.$$tau(s)*tau _(beta)(s)=(tau(s)^(2))/((1-lambda kappa(s))^(2)).\tau(s) \cdot \tau_\beta(s) = \frac{\tau(s)^2}{(1 – \lambda \kappa(s))^2}.$$\tau (s)\cdot {\tau }_{\beta }(s)=\frac{\tau (s{)}^2}{(1-\lambda \kappa (s){)}^2}.$$

Question:-03

Show that the vector $B^i$$B^i$B^(i)B^i$B^i$ of variable magnitude suffers a parallel displacement along a curve $C$$C$CC$C$ if and only if:

Answer:

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Step 1: Definition of Parallel Displacement

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Step 2: Expand the Condition

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Step 3: Rewrite the Derivatives

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Step 4: Symmetry of the Christoffel Symbols

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Step 5: Final Interpretation

Question:-04

Prove that an entity whose inner product with an arbitrary tension is a tensor is itself a tensor.

Answer:

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Step 1: Definitions

Tensor Transformation Law:

Inner Product:

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Step 2: Problem Statement

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Step 3: Scalar Transformation

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Step 4: Conclusion

Question:-05

If the surface of a sphere is a two-dimensional Riemannian space, compute the Christoffel symbols.

Answer:

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Step 1: Parametrize the Sphere

• $\theta \in [0,\pi ]$$\theta \in [0,\pi ]$theta in[0,pi]\theta \in [0, \pi]$\theta \in [0,\pi ]$: polar angle (latitude).
• $\varphi \in [0,2\pi )$$\varphi \in [0,2\pi )$phi in[0,2pi)\phi \in [0, 2\pi)$\varphi \in [0,2\pi )$: azimuthal angle (longitude).

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Step 2: Metric of the Sphere

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Step 3: Christoffel Symbols

Inverse Metric

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Non-Zero Components of the Christoffel Symbols

1. ${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }$${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }$Gamma_(phi phi)^(theta)\Gamma^\theta_{\phi\phi}${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }$:
   Using the definition of Christoffel symbols:
   $${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\frac{1}{2}{g}^{\theta \theta }(\frac{\partial g_{\varphi \varphi }}{\partial \theta }-\frac{\partial g_{\theta \varphi }}{\partial \varphi }-\frac{\partial g_{\varphi \theta }}{\partial \varphi }).$$$${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\frac{1}{2}{g}^{\theta \theta }\left(\frac{\partial g_{\varphi \varphi }}{\partial \theta }-\frac{\partial g_{\theta \varphi }}{\partial \varphi }-\frac{\partial g_{\varphi \theta }}{\partial \varphi }\right).$$Gamma_(phi phi)^(theta)=(1)/(2)g^(theta theta)((delg_(phi phi))/(del theta)-(delg_(theta phi))/(del phi)-(delg_(phi theta))/(del phi)).\Gamma^\theta_{\phi\phi} = \frac{1}{2} g^{\theta\theta} \left( \frac{\partial g_{\phi\phi}}{\partial \theta} – \frac{\partial g_{\theta\phi}}{\partial \phi} – \frac{\partial g_{\phi\theta}}{\partial \phi} \right).$${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\frac{1}{2}{g}^{\theta \theta }(\frac{\partial g_{\varphi \varphi }}{\partial \theta }-\frac{\partial g_{\theta \varphi }}{\partial \varphi }-\frac{\partial g_{\varphi \theta }}{\partial \varphi }).$$
   Since $g_{\theta \varphi }=g_{\varphi \theta }=0$$g_{\theta \varphi }=g_{\varphi \theta }=0$g_(theta phi)=g_(phi theta)=0g_{\theta\phi} = g_{\phi\theta} = 0$g_{\theta \varphi }=g_{\varphi \theta }=0$, this reduces to:
   $${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\frac{1}{2}\frac{1}{R^2}\frac{\partial }{\partial \theta }(R^2{\sin }^2\theta ).$$$${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\frac{1}{2}\frac{1}{R^2}\frac{\partial }{\partial \theta }\left(R^2{\sin }^2\theta \right).$$Gamma_(phi phi)^(theta)=(1)/(2)(1)/(R^(2))(del)/(del theta)(R^(2)sin^(2)theta).\Gamma^\theta_{\phi\phi} = \frac{1}{2} \frac{1}{R^2} \frac{\partial}{\partial \theta} \left( R^2 \sin^2\theta \right).$${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\frac{1}{2}\frac{1}{R^2}\frac{\partial }{\partial \theta }(R^2{\sin }^2\theta ).$$
   Compute the derivative:
   $$\frac{\partial }{\partial \theta }(R^2{\sin }^2\theta )=R^2\cdot 2\sin \theta \cos \theta .$$$$\frac{\partial }{\partial \theta }\left(R^2{\sin }^2\theta \right)=R^2\cdot 2\sin \theta \cos \theta .$$(del)/(del theta)(R^(2)sin^(2)theta)=R^(2)*2sin theta cos theta.\frac{\partial}{\partial \theta} \left( R^2 \sin^2\theta \right) = R^2 \cdot 2 \sin\theta \cos\theta.$$\frac{\partial }{\partial \theta }(R^2{\sin }^2\theta )=R^2\cdot 2\sin \theta \cos \theta .$$
   Thus:
   $${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\frac{1}{2R^2}\cdot R^2\cdot 2\sin \theta \cos \theta =\sin \theta \cos \theta .$$$${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\frac{1}{2R^2}\cdot R^2\cdot 2\sin \theta \cos \theta =\sin \theta \cos \theta .$$Gamma_(phi phi)^(theta)=(1)/(2R^(2))*R^(2)*2sin theta cos theta=sin theta cos theta.\Gamma^\theta_{\phi\phi} = \frac{1}{2R^2} \cdot R^2 \cdot 2 \sin\theta \cos\theta = \sin\theta \cos\theta.$${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\frac{1}{2R^2}\cdot R^2\cdot 2\sin \theta \cos \theta =\sin \theta \cos \theta .$$
2. ${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }$${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }$Gamma_(phi theta)^(phi)\Gamma^\phi_{\phi\theta}${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }$ and ${\mathrm{\Gamma }}_{\theta \varphi }^{\varphi }$${\mathrm{\Gamma }}_{\theta \varphi }^{\varphi }$Gamma_(theta phi)^(phi)\Gamma^\phi_{\theta\phi}${\mathrm{\Gamma }}_{\theta \varphi }^{\varphi }$:
   These are symmetric (${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }={\mathrm{\Gamma }}_{\theta \varphi }^{\varphi }$${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }={\mathrm{\Gamma }}_{\theta \varphi }^{\varphi }$Gamma_(phi theta)^(phi)=Gamma_(theta phi)^(phi)\Gamma^\phi_{\phi\theta} = \Gamma^\phi_{\theta\phi}${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }={\mathrm{\Gamma }}_{\theta \varphi }^{\varphi }$), so we compute one:
   $${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }=\frac{1}{2}{g}^{\varphi \varphi }(\frac{\partial g_{\varphi \varphi }}{\partial \theta }-\frac{\partial g_{\varphi \theta }}{\partial \varphi }+\frac{\partial g_{\theta \varphi }}{\partial \varphi }).$$$${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }=\frac{1}{2}{g}^{\varphi \varphi }\left(\frac{\partial g_{\varphi \varphi }}{\partial \theta }-\frac{\partial g_{\varphi \theta }}{\partial \varphi }+\frac{\partial g_{\theta \varphi }}{\partial \varphi }\right).$$Gamma_(phi theta)^(phi)=(1)/(2)g^(phi phi)((delg_(phi phi))/(del theta)-(delg_(phi theta))/(del phi)+(delg_(theta phi))/(del phi)).\Gamma^\phi_{\phi\theta} = \frac{1}{2} g^{\phi\phi} \left( \frac{\partial g_{\phi\phi}}{\partial \theta} – \frac{\partial g_{\phi\theta}}{\partial \phi} + \frac{\partial g_{\theta\phi}}{\partial \phi} \right).$${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }=\frac{1}{2}{g}^{\varphi \varphi }(\frac{\partial g_{\varphi \varphi }}{\partial \theta }-\frac{\partial g_{\varphi \theta }}{\partial \varphi }+\frac{\partial g_{\theta \varphi }}{\partial \varphi }).$$
   Again, since $g_{\varphi \theta }=g_{\theta \varphi }=0$$g_{\varphi \theta }=g_{\theta \varphi }=0$g_(phi theta)=g_(theta phi)=0g_{\phi\theta} = g_{\theta\phi} = 0$g_{\varphi \theta }=g_{\theta \varphi }=0$, this reduces to:
   $${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }=\frac{1}{2}\frac{1}{R^2{\sin }^2\theta }\frac{\partial }{\partial \theta }(R^2{\sin }^2\theta ).$$$${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }=\frac{1}{2}\frac{1}{R^2{\sin }^2\theta }\frac{\partial }{\partial \theta }\left(R^2{\sin }^2\theta \right).$$Gamma_(phi theta)^(phi)=(1)/(2)(1)/(R^(2)sin^(2)theta)(del)/(del theta)(R^(2)sin^(2)theta).\Gamma^\phi_{\phi\theta} = \frac{1}{2} \frac{1}{R^2 \sin^2\theta} \frac{\partial}{\partial \theta} \left( R^2 \sin^2\theta \right).$${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }=\frac{1}{2}\frac{1}{R^2{\sin }^2\theta }\frac{\partial }{\partial \theta }(R^2{\sin }^2\theta ).$$
   Using the same derivative as above:
   $${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }=\frac{1}{2R^2{\sin }^2\theta }\cdot R^2\cdot 2\sin \theta \cos \theta =\frac{\cos \theta }{\sin \theta }.$$$${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }=\frac{1}{2R^2{\sin }^2\theta }\cdot R^2\cdot 2\sin \theta \cos \theta =\frac{\cos \theta }{\sin \theta }.$$Gamma_(phi theta)^(phi)=(1)/(2R^(2)sin^(2)theta)*R^(2)*2sin theta cos theta=(cos theta)/(sin theta).\Gamma^\phi_{\phi\theta} = \frac{1}{2R^2 \sin^2\theta} \cdot R^2 \cdot 2 \sin\theta \cos\theta = \frac{\cos\theta}{\sin\theta}.$${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }=\frac{1}{2R^2{\sin }^2\theta }\cdot R^2\cdot 2\sin \theta \cos \theta =\frac{\cos \theta }{\sin \theta }.$$
3. ${\mathrm{\Gamma }}_{\theta \theta }^{\varphi }$${\mathrm{\Gamma }}_{\theta \theta }^{\varphi }$Gamma_(theta theta)^(phi)\Gamma^\phi_{\theta\theta}${\mathrm{\Gamma }}_{\theta \theta }^{\varphi }$:
   This component vanishes because $g_{\varphi \varphi }$$g_{\varphi \varphi }$g_(phi phi)g_{\phi\phi}$g_{\varphi \varphi }$ does not depend on $\varphi$$\varphi$phi\phi$\varphi$.
4. ${\mathrm{\Gamma }}_{\theta \theta }^{\theta }$${\mathrm{\Gamma }}_{\theta \theta }^{\theta }$Gamma_(theta theta)^(theta)\Gamma^\theta_{\theta\theta}${\mathrm{\Gamma }}_{\theta \theta }^{\theta }$:
   This component vanishes because $g_{\theta \theta }$$g_{\theta \theta }$g_(theta theta)g_{\theta\theta}$g_{\theta \theta }$ does not depend on $\theta$$\theta$theta\theta$\theta$ or $\varphi$$\varphi$phi\phi$\varphi$.

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Summary of Non-Zero Christoffel Symbols

1. ${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\sin \theta \cos \theta$${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\sin \theta \cos \theta$Gamma_(phi phi)^(theta)=sin theta cos theta\Gamma^\theta_{\phi\phi} = \sin\theta \cos\theta${\mathrm{\Gamma }}_{\varphi \varphi }^{\theta }=\sin \theta \cos \theta$,
2. ${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }={\mathrm{\Gamma }}_{\theta \varphi }^{\varphi }=\frac{\cos \theta }{\sin \theta }$${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }={\mathrm{\Gamma }}_{\theta \varphi }^{\varphi }=\frac{\cos \theta }{\sin \theta }$Gamma_(phi theta)^(phi)=Gamma_(theta phi)^(phi)=(cos theta)/(sin theta)\Gamma^\phi_{\phi\theta} = \Gamma^\phi_{\theta\phi} = \frac{\cos\theta}{\sin\theta}${\mathrm{\Gamma }}_{\varphi \theta }^{\varphi }={\mathrm{\Gamma }}_{\theta \varphi }^{\varphi }=\frac{\cos \theta }{\sin \theta }$.

Question:-06

(a) State and prove Meunienis theorem.

Answer:

Statement of Meunier’s Theorem:

• $k_n$$k_n$k_(n)k_n$k_n$ is the normal curvature of the surface in a given direction.
• $k_1$$k_1$k_(1)k_1$k_1$ and $k_2$$k_2$k_(2)k_2$k_2$ are the principal curvatures of the surface at the point.
• $\theta$$\theta$theta\theta$\theta$ is the angle between the given direction and the direction of the first principal curvature ($k_1$$k_1$k_(1)k_1$k_1$).

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Proof of Meunier’s Theorem:

Step 1: Curvature of a Surface

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Step 2: Principal Curvatures and Directions

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Step 3: General Direction Vector

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Step 4: Apply the Shape Operator

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Step 5: Compute the Normal Curvature

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Conclusion:

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Geometric Interpretation:

• The normal curvature $k_n$$k_n$k_(n)k_n$k_n$ in a given direction is a weighted average of the principal curvatures $k_1$$k_1$k_(1)k_1$k_1$ and $k_2$$k_2$k_(2)k_2$k_2$.
• The weights are ${\cos }^2\theta$${\cos }^2\theta$cos^(2)theta\cos^2\theta${\cos }^2\theta$ and ${\sin }^2\theta$${\sin }^2\theta$sin^(2)theta\sin^2\theta${\sin }^2\theta$, which depend on how closely the direction aligns with the principal directions.

Question:-06(b)

Find the Geodesic curvature of the curve $u=\text{constant}$$u=\text{constant}$u=”constant”u = \text{constant}$u=\text{constant}$ on the surface $x=u\cos \theta$$x=u\cos \theta$x=u cos thetax = u \cos \theta$x=u\cos \theta$, $y=u\sin \theta$$y=u\sin \theta$y=u sin thetay = u \sin \theta$y=u\sin \theta$, $z=\frac{1}{2}a{u}^2$$z=\frac{1}{2}a{u}^2$z=(1)/(2)au^(2)z = \frac{1}{2} a u^2$z=\frac{1}{2}a{u}^2$

Answer:

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Step 1: Understand the Surface and Curve

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Step 2: Compute the First Fundamental Form

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Step 3: Unit Tangent Vector to the Curve

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Step 4: Compute the Geodesic Curvature

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Final Answer

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Question:-07

(a) Find the angle between two tangential directions on the surface in terms of direction ratios.

Answer:

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Solution By Steps

Step 1: Understand the Problem

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Step 2: Compute the Dot Product of Two Tangential Directions

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Step 3: Compute the Magnitudes of $\mathbf{a}$$\mathbf{a}$a\mathbf{a}$\mathbf{a}$ and $\mathbf{b}$$\mathbf{b}$b\mathbf{b}$\mathbf{b}$

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Step 4: Express the Cosine of the Angle

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Final Answer

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Question:-07(b)

If the tangent and binormal at a point of curve make angles $\theta ,\varphi$$\theta ,\varphi$theta,phi\theta, \phi$\theta ,\varphi$ respectively with fixed directions, then:

Answer:

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Step 1: Geometric Setup

• The tangent vector $\mathbf{T}$$\mathbf{T}$T\mathbf{T}$\mathbf{T}$,
• The normal vector $\mathbf{N}$$\mathbf{N}$N\mathbf{N}$\mathbf{N}$,
• The binormal vector $\mathbf{B}$$\mathbf{B}$B\mathbf{B}$\mathbf{B}$.

• $k$$k$kk$k$ is the curvature, which measures the rate of change of $\mathbf{T}$$\mathbf{T}$T\mathbf{T}$\mathbf{T}$ with respect to $s$$s$ss$s$,
• $\tau$$\tau$tau\tau$\tau$ is the torsion, which measures the rate of change of $\mathbf{B}$$\mathbf{B}$B\mathbf{B}$\mathbf{B}$ with respect to $s$$s$ss$s$.

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Step 2: Angles with Fixed Directions

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Step 3: Differentiating $\cos \theta$$\cos \theta$cos theta\cos\theta$\cos \theta$ and $\cos \varphi$$\cos \varphi$cos phi\cos\phi$\cos \varphi$

For $\cos \theta =\mathbf{T}\cdot \mathbf{a}$$\cos \theta =\mathbf{T}\cdot \mathbf{a}$cos theta=T*a\cos\theta = \mathbf{T} \cdot \mathbf{a}$\cos \theta =\mathbf{T}\cdot \mathbf{a}$:

For $\cos \varphi =\mathbf{B}\cdot \mathbf{a}$$\cos \varphi =\mathbf{B}\cdot \mathbf{a}$cos phi=B*a\cos\phi = \mathbf{B} \cdot \mathbf{a}$\cos \varphi =\mathbf{B}\cdot \mathbf{a}$:

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Step 4: Relationship Between $\theta$$\theta$theta\theta$\theta$ and $\varphi$$\varphi$phi\phi$\varphi$

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Final Answer: