VMOU MT-04 SOLVED ASSIGNMENT | MA/M.SC. MT- 04(Differential Geometry & Tensors) | July-2024 & January-2025

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.
  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.
Section-B
(Short Answer Questions)
Note :- Answer any two questions. Each answer should be given in 200 words. Each question carries 4 marks.
  1. Prove that the torsion of the two Bertrand curves has the same sign and their product is constant.
  2. Show that the vector B i B i B^(i)B^iBi of variable magnitude suffers a parallel displacement along a curve C if and only if :
( B l B , j i B i B , j l ) d x i d s = 0 B l B , j i B i B , j l d x i d s = 0 (B^(l)B_(,j)^(i)-B^(i)B_(,j)^(l))(dx^(i))/(ds)=0\left(B^l B_{, j}^i-B^i B_{, j}^l\right) \frac{d x^i}{d s}=0(BlB,jiBiB,jl)dxids=0
  1. Prove that an entity whose inner product with an arbitrary tension is a tensor is itself a tensor.
  2. 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.
  1. (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 θ , y = u sin θ , z = 1 2 a u 2 x = u cos θ , y = u sin θ , z = 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^2x=ucosθ,y=usinθ,z=12au2
  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 θ , ϕ θ , ϕ theta,phi\theta, \phiθ,ϕ respectively with a fixed directions then:
sin θ sin ϕ d θ d ϕ = ± k τ sin θ sin ϕ d θ d ϕ = ± k τ (sin theta)/(sin phi)*(d theta)/(d phi)=+-(k)/( tau)\frac{\sin \theta}{\sin \phi} \cdot \frac{d \theta}{d \phi}= \pm \frac{k}{\tau}sinθsinϕdθdϕ=±kτ

Answer:

Question:-01(a)

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 ) y=f(x)y = f(x)y=f(x), the tangent line at a point ( x 0 , y 0 ) ( x 0 , y 0 ) (x_(0),y_(0))(x_0, y_0)(x0,y0) on the curve is given by:
y y 0 = m ( x x 0 ) , y y 0 = m ( x x 0 ) , y-y_(0)=m(x-x_(0)),y – y_0 = m(x – x_0),yy0=m(xx0),
where m m mmm is the slope of the tangent line at ( x 0 , y 0 ) ( x 0 , y 0 ) (x_(0),y_(0))(x_0, y_0)(x0,y0). The slope m m mmm is found using the derivative of f ( x ) f ( x ) f(x)f(x)f(x) at x 0 x 0 x_(0)x_0x0:
m = f ( x 0 ) . m = f ( x 0 ) . m=f^(‘)(x_(0)).m = f'(x_0).m=f(x0).


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:

  1. Radius of Curvature: The radius of the osculating circle is equal to the reciprocal of the curvature ( k k kkk) of the curve at the point:
    R = 1 k , R = 1 k , R=(1)/(k),R = \frac{1}{k},R=1k,
    where k k kkk 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 RRR 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.

Formulas:

For a curve given parametrically by ( x ( t ) , y ( t ) ) ( x ( t ) , y ( t ) ) (x(t),y(t))(x(t), y(t))(x(t),y(t)):
  • Curvature: k = | x ( t ) y ( t ) y ( t ) x ( t ) | ( x ( t ) 2 + y ( t ) 2 ) 3 / 2 . k = | x ( t ) y ( t ) y ( t ) x ( t ) | x ( t ) 2 + y ( t ) 2 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=|x(t)y(t)y(t)x(t)|(x(t)2+y(t)2)3/2.
  • Radius of Curvature: R = 1 k . R = 1 k . R=(1)/(k).R = \frac{1}{k}.R=1k.
  • Center of the Osculating Circle: ( x c , y c ) = ( x 0 R y ( t ) x ( t ) 2 + y ( t ) 2 , y 0 + R x ( t ) x ( t ) 2 + y ( t ) 2 ) , x c , y c = x 0 R y ( t ) x ( t ) 2 + y ( t ) 2 , y 0 + R x ( t ) x ( t ) 2 + y ( t ) 2 , (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),(xc,yc)=(x0Ry(t)x(t)2+y(t)2,y0+Rx(t)x(t)2+y(t)2),where ( x 0 , y 0 ) ( x 0 , y 0 ) (x_(0),y_(0))(x_0, y_0)(x0,y0) is the point of tangency and R R RRR 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 ) ) (x(t),y(t))(x(t), y(t))(x(t),y(t)), and let its curvature be k ( t ) k ( t ) k(t)k(t)k(t) (with radius of curvature 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)).
The curvature of the evolute, denoted as k evolute ( t ) k evolute ( t ) k_(“evolute”)(t)k_{\text{evolute}}(t)kevolute(t), is given by:
k evolute ( t ) = k ( t ) | R ( t ) ρ ( t ) | , k evolute ( t ) = k ( t ) | R ( t ) ρ ( t ) | , k_(“evolute”)(t)=(k(t))/(|R(t)-rho(t)|),k_{\text{evolute}}(t) = \frac{k(t)}{|R(t) – \rho(t)|},kevolute(t)=k(t)|R(t)ρ(t)|,
where:
  • 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.
  • ρ ( t ) ρ ( t ) rho(t)\rho(t)ρ(t) is the radius of the osculating circle measured along the evolute (distance to the center of curvature).

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.
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 curvature k n k n k_(n)k_nkn of a surface in a given direction is equal to the cosine-weighted average of the principal curvatures k 1 k 1 k_(1)k_1k1 and k 2 k 2 k_(2)k_2k2. Specifically:
k n = k 1 cos 2 θ + k 2 sin 2 θ , k n = k 1 cos 2 θ + k 2 sin 2 θ , k_(n)=k_(1)cos^(2)theta+k_(2)sin^(2)theta,k_n = k_1 \cos^2\theta + k_2 \sin^2\theta,kn=k1cos2θ+k2sin2θ,
where:
  • k 1 k 1 k_(1)k_1k1 and k 2 k 2 k_(2)k_2k2 are the principal curvatures of the surface at the point under consideration.
  • θ θ theta\thetaθ is the angle between the given direction and the direction of the first principal curvature ( k 1 k 1 k_(1)k_1k1).


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 α ( s ) α ( s ) alpha(s)\boldsymbol{\alpha}(s)α(s) and β ( s ) β ( s ) beta(s)\boldsymbol{\beta}(s)β(s), parameterized by their arc length s s sss, with the following properties:
  1. β ( s ) = α ( s ) + λ N ( s ) β ( s ) = α ( s ) + λ N ( s ) beta(s)=alpha(s)+lambda N(s)\boldsymbol{\beta}(s) = \boldsymbol{\alpha}(s) + \lambda \boldsymbol{N}(s)β(s)=α(s)+λN(s), where N ( s ) N ( s ) N(s)\boldsymbol{N}(s)N(s) is the principal normal of α ( s ) α ( s ) alpha(s)\boldsymbol{\alpha}(s)α(s) and λ λ lambda\lambdaλ is a constant.
  2. The principal normal directions of α ( s ) α ( s ) alpha(s)\boldsymbol{\alpha}(s)α(s) and β ( s ) β ( s ) beta(s)\boldsymbol{\beta}(s)β(s) coincide at corresponding points.

Definitions of Curvature and Torsion

For a space curve α ( s ) α ( s ) alpha(s)\boldsymbol{\alpha}(s)α(s):
  • The curvature is κ ( s ) κ ( s ) kappa(s)\kappa(s)κ(s).
  • The torsion is τ ( s ) τ ( s ) tau(s)\tau(s)τ(s).
For the Bertrand mate β ( s ) β ( s ) beta(s)\boldsymbol{\beta}(s)β(s):
  • Let its curvature and torsion be κ β ( s ) κ β ( s ) kappa _(beta)(s)\kappa_\beta(s)κβ(s) and τ β ( s ) τ β ( s ) tau _(beta)(s)\tau_\beta(s)τβ(s), respectively.

Step 1: Relationship Between the Curvatures

The Bertrand mate is given by:
β ( s ) = α ( s ) + λ N ( s ) . β ( s ) = α ( s ) + λ N ( s ) . beta(s)=alpha(s)+lambda N(s).\boldsymbol{\beta}(s) = \boldsymbol{\alpha}(s) + \lambda \boldsymbol{N}(s).β(s)=α(s)+λN(s).
Differentiating with respect to s s sss:
β ( s ) = α ( s ) + λ N ( s ) . β ( s ) = α ( s ) + λ N ( s ) . beta^(‘)(s)=alpha^(‘)(s)+lambdaN^(‘)(s).\boldsymbol{\beta}'(s) = \boldsymbol{\alpha}'(s) + \lambda \boldsymbol{N}'(s).β(s)=α(s)+λN(s).
Since α ( s ) = T ( s ) α ( s ) = T ( s ) alpha^(‘)(s)=T(s)\boldsymbol{\alpha}'(s) = \boldsymbol{T}(s)α(s)=T(s) (the unit tangent vector), and N ( s ) N ( s ) N^(‘)(s)\boldsymbol{N}'(s)N(s) can be expressed as:
N ( s ) = κ ( s ) T ( s ) + τ ( s ) B ( s ) , N ( s ) = κ ( s ) T ( s ) + τ ( s ) B ( s ) , N^(‘)(s)=-kappa(s)T(s)+tau(s)B(s),\boldsymbol{N}'(s) = -\kappa(s) \boldsymbol{T}(s) + \tau(s) \boldsymbol{B}(s),N(s)=κ(s)T(s)+τ(s)B(s),
where B ( s ) B ( s ) B(s)\boldsymbol{B}(s)B(s) is the binormal vector, we have:
β ( s ) = T ( s ) + λ [ κ ( s ) T ( s ) + τ ( s ) B ( s ) ] . β ( s ) = T ( s ) + λ κ ( s ) T ( s ) + τ ( s ) B ( s ) . beta^(‘)(s)=T(s)+lambda[-kappa(s)T(s)+tau(s)B(s)].\boldsymbol{\beta}'(s) = \boldsymbol{T}(s) + \lambda \left[-\kappa(s) \boldsymbol{T}(s) + \tau(s) \boldsymbol{B}(s)\right].β(s)=T(s)+λ[κ(s)T(s)+τ(s)B(s)].
Simplify:
β ( s ) = ( 1 λ κ ( s ) ) T ( s ) + λ τ ( s ) B ( s ) . β ( s ) = ( 1 λ κ ( s ) ) T ( s ) + λ τ ( s ) B ( s ) . beta^(‘)(s)=(1-lambda kappa(s))T(s)+lambda tau(s)B(s).\boldsymbol{\beta}'(s) = (1 – \lambda \kappa(s)) \boldsymbol{T}(s) + \lambda \tau(s) \boldsymbol{B}(s).β(s)=(1λκ(s))T(s)+λτ(s)B(s).
The unit tangent vector for β ( s ) β ( s ) beta(s)\boldsymbol{\beta}(s)β(s) is:
T β ( s ) = β ( s ) β ( s ) . T β ( s ) = β ( s ) β ( s ) . T_( beta)(s)=(beta^(‘)(s))/(||beta^(‘)(s)||).\boldsymbol{T}_\beta(s) = \frac{\boldsymbol{\beta}'(s)}{\|\boldsymbol{\beta}'(s)\|}.Tβ(s)=β(s)β(s).
The magnitude of β ( s ) β ( s ) beta^(‘)(s)\boldsymbol{\beta}'(s)β(s) is:
β ( s ) = ( 1 λ κ ( s ) ) 2 + ( λ τ ( s ) ) 2 . β ( s ) = ( 1 λ κ ( s ) ) 2 + ( λ τ ( s ) ) 2 . ||beta^(‘)(s)||=sqrt((1-lambda kappa(s))^(2)+(lambda tau(s))^(2)).\|\boldsymbol{\beta}'(s)\| = \sqrt{(1 – \lambda \kappa(s))^2 + (\lambda \tau(s))^2}.β(s)=(1λκ(s))2+(λτ(s))2.
The curvature of β ( s ) β ( s ) beta(s)\boldsymbol{\beta}(s)β(s) is related to κ ( s ) κ ( s ) kappa(s)\kappa(s)κ(s) as:
κ β ( s ) = κ ( s ) | 1 λ κ ( s ) | . κ β ( s ) = κ ( s ) | 1 λ κ ( s ) | . kappa _(beta)(s)=(kappa(s))/(|1-lambda kappa(s)|).\kappa_\beta(s) = \frac{\kappa(s)}{|1 – \lambda \kappa(s)|}.κβ(s)=κ(s)|1λκ(s)|.

Step 2: Relationship Between the Torsions

The torsion of β ( s ) β ( s ) beta(s)\boldsymbol{\beta}(s)β(s), denoted as τ β ( s ) τ β ( s ) tau _(beta)(s)\tau_\beta(s)τβ(s), is related to τ ( s ) τ ( s ) tau(s)\tau(s)τ(s). Specifically:
τ β ( s ) = τ ( s ) ( 1 λ κ ( s ) ) 2 . τ β ( s ) = τ ( s ) ( 1 λ κ ( s ) ) 2 . tau _(beta)(s)=(tau(s))/((1-lambda kappa(s))^(2)).\tau_\beta(s) = \frac{\tau(s)}{(1 – \lambda \kappa(s))^2}.τβ(s)=τ(s)(1λκ(s))2.

Step 3: Product of the Torsions

The product of the torsions τ ( s ) τ ( s ) tau(s)\tau(s)τ(s) and τ β ( s ) τ β ( s ) tau _(beta)(s)\tau_\beta(s)τβ(s) is:
τ ( s ) τ β ( s ) = τ ( s ) τ ( s ) ( 1 λ κ ( s ) ) 2 . τ ( s ) τ β ( s ) = τ ( s ) τ ( s ) ( 1 λ κ ( s ) ) 2 . tau(s)*tau _(beta)(s)=tau(s)*(tau(s))/((1-lambda kappa(s))^(2)).\tau(s) \cdot \tau_\beta(s) = \tau(s) \cdot \frac{\tau(s)}{(1 – \lambda \kappa(s))^2}.τ(s)τβ(s)=τ(s)τ(s)(1λκ(s))2.
Simplify:
τ ( s ) τ β ( s ) = τ ( s ) 2 ( 1 λ κ ( s ) ) 2 . τ ( s ) τ β ( s ) = τ ( s ) 2 ( 1 λ κ ( 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}.τ(s)τβ(s)=τ(s)2(1λκ(s))2.
This shows that the product is proportional to τ ( s ) 2 τ ( s ) 2 tau(s)^(2)\tau(s)^2τ(s)2, which remains constant if τ ( s ) τ ( s ) tau(s)\tau(s)τ(s) does not change.

Step 4: Signs of the Torsions

The torsions τ ( s ) τ ( s ) tau(s)\tau(s)τ(s) and τ β ( s ) τ β ( s ) tau _(beta)(s)\tau_\beta(s)τβ(s) depend on ( 1 λ κ ( s ) ) 2 ( 1 λ κ ( s ) ) 2 (1-lambda kappa(s))^(2)(1 – \lambda \kappa(s))^2(1λκ(s))2, which is always positive. Thus, the signs of τ ( s ) τ ( s ) tau(s)\tau(s)τ(s) and τ β ( s ) τ β ( s ) tau _(beta)(s)\tau_\beta(s)τβ(s) must be the same.

Conclusion

  1. The torsions of the two Bertrand curves have the same sign.
  2. The product of their torsions is constant: τ ( s ) τ β ( s ) = τ ( s ) 2 ( 1 λ κ ( s ) ) 2 . τ ( s ) τ β ( s ) = τ ( s ) 2 ( 1 λ κ ( 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}.τ(s)τβ(s)=τ(s)2(1λκ(s))2.

Question:-03

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

( B l B , j i B i B , j l ) d x i d s = 0 B l B , j i B i B , j l d x i d s = 0 (B^(l)B_(,j)^(i)-B^(i)B_(,j)^(l))(dx^(i))/(ds)=0\left(B^l B_{, j}^i – B^i B_{, j}^l\right) \frac{d x^i}{d s} = 0(BlB,jiBiB,jl)dxids=0

Answer:

To show that the vector B i B i B^(i)B^iBi of variable magnitude suffers a parallel displacement along a curve C C CCC if and only if:
( B l B , j i B i B , j l ) d x j d s = 0 , B l B , j i B i B , j l d x j d s = 0 , (B^(l)B_(,j)^(i)-B^(i)B_(,j)^(l))(dx^(j))/(ds)=0,\left(B^l B_{, j}^i – B^i B_{, j}^l\right) \frac{dx^j}{ds} = 0,(BlB,jiBiB,jl)dxjds=0,
we proceed step by step.

Step 1: Definition of Parallel Displacement

A vector B i B i B^(i)B^iBi undergoes parallel displacement along a curve C C CCC if its covariant derivative along the tangent to the curve is zero. This condition is written as:
D B i D s = 0 , D B i D s = 0 , (DB^(i))/(Ds)=0,\frac{D B^i}{Ds} = 0,DBiDs=0,
where s s sss is the arc length parameter along the curve, and D D s D D s (D)/(Ds)\frac{D}{Ds}DDs is the covariant derivative along the curve.
Using the covariant derivative definition:
D B i D s = d B i d s + Γ j k i B j d x k d s , D B i D s = d B i d s + Γ j k i B j d x k d s , (DB^(i))/(Ds)=(dB^(i))/(ds)+Gamma_(jk)^(i)B^(j)(dx^(k))/(ds),\frac{D B^i}{Ds} = \frac{d B^i}{ds} + \Gamma^i_{jk} B^j \frac{dx^k}{ds},DBiDs=dBids+ΓjkiBjdxkds,
where Γ j k i Γ j k i Gamma_(jk)^(i)\Gamma^i_{jk}Γjki are the Christoffel symbols, d x k d s d x k d s (dx^(k))/(ds)\frac{dx^k}{ds}dxkds is the tangent vector to the curve, and d B i d s d B i d s (dB^(i))/(ds)\frac{dB^i}{ds}dBids is the ordinary derivative of B i B i B^(i)B^iBi along the curve.
Thus, the parallel displacement condition becomes:
(1) d B i d s + Γ j k i B j d x k d s = 0. (1) d B i d s + Γ j k i B j d x k d s = 0. {:(1)(dB^(i))/(ds)+Gamma_(jk)^(i)B^(j)(dx^(k))/(ds)=0.:}\frac{d B^i}{ds} + \Gamma^i_{jk} B^j \frac{dx^k}{ds} = 0. \tag{1}(1)dBids+ΓjkiBjdxkds=0.

Step 2: Expand the Condition

Multiply equation (1) by B l B l B^(l)B^lBl to introduce a symmetric product:
B l ( d B i d s + Γ j k i B j d x k d s ) = 0. B l d B i d s + Γ j k i B j d x k d s = 0. B^(l)((dB^(i))/(ds)+Gamma_(jk)^(i)B^(j)(dx^(k))/(ds))=0.B^l \left(\frac{d B^i}{ds} + \Gamma^i_{jk} B^j \frac{dx^k}{ds}\right) = 0.Bl(dBids+ΓjkiBjdxkds)=0.
Similarly, consider the equation for another index l l lll instead of i i iii:
B i ( d B l d s + Γ j k l B j d x k d s ) = 0. B i d B l d s + Γ j k l B j d x k d s = 0. B^(i)((dB^(l))/(ds)+Gamma_(jk)^(l)B^(j)(dx^(k))/(ds))=0.B^i \left(\frac{d B^l}{ds} + \Gamma^l_{jk} B^j \frac{dx^k}{ds}\right) = 0.Bi(dBlds+ΓjklBjdxkds)=0.
Subtract these two equations:
B l d B i d s B i d B l d s + B l Γ j k i B j d x k d s B i Γ j k l B j d x k d s = 0. B l d B i d s B i d B l d s + B l Γ j k i B j d x k d s B i Γ j k l B j d x k d s = 0. B^(l)(dB^(i))/(ds)-B^(i)(dB^(l))/(ds)+B^(l)Gamma_(jk)^(i)B^(j)(dx^(k))/(ds)-B^(i)Gamma_(jk)^(l)B^(j)(dx^(k))/(ds)=0.B^l \frac{d B^i}{ds} – B^i \frac{d B^l}{ds} + B^l \Gamma^i_{jk} B^j \frac{dx^k}{ds} – B^i \Gamma^l_{jk} B^j \frac{dx^k}{ds} = 0.BldBidsBidBlds+BlΓjkiBjdxkdsBiΓjklBjdxkds=0.

Step 3: Rewrite the Derivatives

Using the chain rule, the derivative of B i B i B^(i)B^iBi along the curve is:
d B i d s = B i x j d x j d s = B , j i d x j d s , d B i d s = B i x j d x j d s = B , j i d x j d s , (dB^(i))/(ds)=(delB^(i))/(delx^(j))(dx^(j))/(ds)=B_(,j)^(i)(dx^(j))/(ds),\frac{d B^i}{ds} = \frac{\partial B^i}{\partial x^j} \frac{dx^j}{ds} = B^i_{,j} \frac{dx^j}{ds},dBids=Bixjdxjds=B,jidxjds,
where B , j i B , j i B_(,j)^(i)B^i_{,j}B,ji is the partial derivative of B i B i B^(i)B^iBi with respect to x j x j x^(j)x^jxj.
Substitute this into the previous equation:
B l B , j i d x j d s B i B , j l d x j d s + B l Γ j k i B j d x k d s B i Γ j k l B j d x k d s = 0. B l B , j i d x j d s B i B , j l d x j d s + B l Γ j k i B j d x k d s B i Γ j k l B j d x k d s = 0. B^(l)B_(,j)^(i)(dx^(j))/(ds)-B^(i)B_(,j)^(l)(dx^(j))/(ds)+B^(l)Gamma_(jk)^(i)B^(j)(dx^(k))/(ds)-B^(i)Gamma_(jk)^(l)B^(j)(dx^(k))/(ds)=0.B^l B^i_{,j} \frac{dx^j}{ds} – B^i B^l_{,j} \frac{dx^j}{ds} + B^l \Gamma^i_{jk} B^j \frac{dx^k}{ds} – B^i \Gamma^l_{jk} B^j \frac{dx^k}{ds} = 0.BlB,jidxjdsBiB,jldxjds+BlΓjkiBjdxkdsBiΓjklBjdxkds=0.

Step 4: Symmetry of the Christoffel Symbols

The Christoffel symbols satisfy the symmetry property Γ j k i = Γ k j i Γ j k i = Γ k j i Gamma_(jk)^(i)=Gamma_(kj)^(i)\Gamma^i_{jk} = \Gamma^i_{kj}Γjki=Γkji. Use this to group terms:
( B l B , j i B i B , j l ) d x j d s + ( B l Γ j k i B i Γ j k l ) B j d x k d s = 0. B l B , j i B i B , j l d x j d s + B l Γ j k i B i Γ j k l B j d x k d s = 0. (B^(l)B_(,j)^(i)-B^(i)B_(,j)^(l))(dx^(j))/(ds)+(B^(l)Gamma_(jk)^(i)-B^(i)Gamma_(jk)^(l))B^(j)(dx^(k))/(ds)=0.\left(B^l B^i_{,j} – B^i B^l_{,j}\right) \frac{dx^j}{ds} + \left(B^l \Gamma^i_{jk} – B^i \Gamma^l_{jk}\right) B^j \frac{dx^k}{ds} = 0.