Section-A
(Very Short Answer Type Questions)
Note:- Answer all questions. As per the nature of the question you delimit your answer in one word, one sentence or maximum up to 30 words. Each question carries 1 (one) mark.
(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.
Section-B
(Short Answer Questions)
Note :- Answer any two questions. Each answer should be given in 200 words. Each question carries 4 marks.
Prove that the torsion of the two Bertrand curves has the same sign and their product is constant.
Show that the vector B^(i)B^i of variable magnitude suffers a parallel displacement along a curve C if and only if :
Prove that an entity whose inner product with an arbitrary tension is a tensor is itself a tensor.
If surface of sphere is a two dimensional Riemannian space. Compute the Christoffel symbols.
Section-C
(Long Answer Questions)
Note :- Answer any one question. Each answer should be given in 800 words. Each question carries 08 marks.
(i). State and prove Meunienis theorem.
(ii). Find the Geodesic curvature of the curve u=u= constant on the surface 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
(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,phi\theta, \phi respectively with a fixed directions then:
Write down the equation of the tangent line to a curve at a given point.
Answer:
The equation of the tangent line to a curve at a given point can be determined as follows:
General Formula
If a curve is described by y=f(x)y = f(x), the tangent line at a point (x_(0),y_(0))(x_0, y_0) on the curve is given by:
y-y_(0)=m(x-x_(0)),y – y_0 = m(x – x_0),
where mm is the slope of the tangent line at (x_(0),y_(0))(x_0, y_0). The slope mm is found using the derivative of f(x)f(x) at x_(0)x_0:
m=f^(‘)(x_(0)).m = f'(x_0).
Question:-01(b)
Define the oscillating circle.
Answer:
The oscillating circle (or osculating circle) at a point on a curve is the circle that best approximates the curve near that point. It shares the same tangent, curvature, and radius of curvature as the curve at the given point. The term "osculating" comes from the Latin word osculare, meaning "to kiss," as the circle "kisses" the curve at that point.
Key Features of the Osculating Circle:
Radius of Curvature: The radius of the osculating circle is equal to the reciprocal of the curvature (kk) of the curve at the point:
R=(1)/(k),R = \frac{1}{k},
where kk is the curvature.
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 RR from the point.
Curvature Match: The osculating circle has the same curvature as the curve at the point of tangency.
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.
Formulas:
For a curve given parametrically by (x(t),y(t))(x(t), y(t)):
Center of the Osculating Circle:(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),where (x_(0),y_(0))(x_0, y_0) is the point of tangency and RR is the radius of curvature.
Question:-01(c)
Write down the formula to the curvature of evolute.
Answer:
The curvature of the evolute of a given curve is related to the curvature of the original curve as follows:
Formula for the Curvature of the Evolute:
Let the original curve be parameterized by (x(t),y(t))(x(t), y(t)), and let its curvature be k(t)k(t) (with radius of curvature R(t)=1//k(t)R(t) = 1/k(t)).
The curvature of the evolute, denoted as k_(“evolute”)(t)k_{\text{evolute}}(t), is given by:
R(t)=1//k(t)R(t) = 1/k(t) is the radius of curvature of the original curve.
rho(t)\rho(t) is the radius of the osculating circle measured along the evolute (distance to the center of curvature).
Explanation
Evolute: The evolute of a curve is the locus of the centers of curvature of the original curve.
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.
This formula accounts for the relationship between the geometry of the evolute and the original curve.
Question:-01(d)
State Mennier’s theorem.
Answer:
Meunier’s Theorem is a fundamental result in the geometry of surfaces in differential geometry. It relates the normal curvature of a surface to the principal curvatures at a given point. The theorem states:
Statement of Meunier’s Theorem:
The normal curvaturek_(n)k_n of a surface in a given direction is equal to the cosine-weighted average of the principal curvaturesk_(1)k_1 and k_(2)k_2. Specifically:
k_(1)k_1 and k_(2)k_2 are the principal curvatures of the surface at the point under consideration.
theta\theta is the angle between the given direction and the direction of the first principal curvature (k_(1)k_1).
Question:-02
Prove that the torsion of the two Bertrand curves has the same sign and their product is constant.
Answer:
To prove that the torsion of two Bertrand curves has the same sign and their product is constant, we proceed as follows:
Bertrand Curves
Two curves are said to be Bertrand curves if each point on one curve corresponds to a point on the other such that the principal normal vectors of the curves at corresponding points are the same.
Let the two Bertrand curves be alpha(s)\boldsymbol{\alpha}(s) and beta(s)\boldsymbol{\beta}(s), parameterized by their arc length ss, with the following properties:
beta(s)=alpha(s)+lambda N(s)\boldsymbol{\beta}(s) = \boldsymbol{\alpha}(s) + \lambda \boldsymbol{N}(s), where N(s)\boldsymbol{N}(s) is the principal normal of alpha(s)\boldsymbol{\alpha}(s) and lambda\lambda is a constant.
The principal normal directions of alpha(s)\boldsymbol{\alpha}(s) and beta(s)\boldsymbol{\beta}(s) coincide at corresponding points.
Definitions of Curvature and Torsion
For a space curve alpha(s)\boldsymbol{\alpha}(s):
The curvature is kappa(s)\kappa(s).
The torsion is tau(s)\tau(s).
For the Bertrand mate beta(s)\boldsymbol{\beta}(s):
Let its curvature and torsion be kappa _(beta)(s)\kappa_\beta(s) and tau _(beta)(s)\tau_\beta(s), respectively.
This shows that the product is proportional to tau(s)^(2)\tau(s)^2, which remains constant if tau(s)\tau(s) does not change.
Step 4: Signs of the Torsions
The torsions tau(s)\tau(s) and tau _(beta)(s)\tau_\beta(s) depend on (1-lambda kappa(s))^(2)(1 – \lambda \kappa(s))^2, which is always positive. Thus, the signs of tau(s)\tau(s) and tau _(beta)(s)\tau_\beta(s) must be the same.
Conclusion
The torsions of the two Bertrand curves have the same sign.
The product of their torsions is constant: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}.
Question:-03
Show that the vector B^(i)B^i of variable magnitude suffers a parallel displacement along a curve CC if and only if:
A vector B^(i)B^i undergoes parallel displacement along a curve CC if its covariant derivative along the tangent to the curve is zero. This condition is written as:
(DB^(i))/(Ds)=0,\frac{D B^i}{Ds} = 0,
where ss is the arc length parameter along the curve, and (D)/(Ds)\frac{D}{Ds} is the covariant derivative along the curve.
where Gamma_(jk)^(i)\Gamma^i_{jk} are the Christoffel symbols, (dx^(k))/(ds)\frac{dx^k}{ds} is the tangent vector to the curve, and (dB^(i))/(ds)\frac{dB^i}{ds} is the ordinary derivative of B^(i)B^i along the curve.
Thus, the parallel displacement condition becomes: