IGNOU BPHCT-133 Solved Assignment | 1st January, 2025 to 31st December, 2025 | ELECTRICITY AND MAGNETISM | BSCG CBCS

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IGNOU BPHCT-133 Solved Assignment | 1st January, 2025 to 31st December, 2025 | ELECTRICITY AND MAGNETISM | BSCG CBCS Sample Solution

BPHCT-133 Solved Assignment

PART A
  1. a) If u → u → vec(u)\overrightarrow{\mathbf{u}}u→ is a constant vector show that ∇ → × ( u → × r → ) = 2 u → ∇ → × ( u → × r → ) = 2 u → vec(grad)xx( vec(u)xx vec(r))=2 vec(u)\vec{\nabla} \times(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{r}})=\mathbf{2} \overrightarrow{\mathbf{u}}∇→×(u→×r→)=2u→.
    b) Determine the work done by a force F → = ( x + 2 y ) i ^ + ( x − y ) j ^ F → = ( x + 2 y ) i ^ + ( x − y ) j ^ vec(F)=(x+2y) hat(i)+(x-y) hat(j)\overrightarrow{\mathbf{F}}=(x+2 y) \hat{\mathbf{i}}+(x-y) \hat{\mathbf{j}}F→=(x+2y)i^+(x−y)j^ in taking a particle along the curve x ( t ) = 2 cos ⁡ t ; y ( t ) = 4 sin ⁡ t x ( t ) = 2 cos ⁡ t ; y ( t ) = 4 sin ⁡ t x(t)=2cos t;y(t)=4sin tx(t)=2 \cos t ; y(t)=4 \sin tx(t)=2cos⁡t;y(t)=4sin⁡t from t = 0 t = 0 t=0t=0t=0 to t = π 4 t = π 4 t=(pi)/(4)t=\frac{\pi}{4}t=π4.
    c) Using Stokes’ theorem, prove that curl of a conservative force field is zero everywhere.
    d) Determine the directional derivative of the scalar field ϕ = ln ⁡ ( x 2 + y 2 + z 2 ) ϕ = ln ⁡ x 2 + y 2 + z 2 phi=ln(x^(2)+y^(2)+z^(2))\phi=\ln \left(x^2+y^2+z^2\right)ϕ=ln⁡(x2+y2+z2) in the direction ( i ^ + 2 j ^ − k ^ ) ( i ^ + 2 j ^ − k ^ ) ( hat(i)+2 hat(j)- hat(k))(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})(i^+2j^−k^) at the point ( 1 , − 1 , 2 ) ( 1 , − 1 , 2 ) (1,-1,2)(1,-1,2)(1,−1,2).
  2. a) Explain with the help of diagrams what spherically and cylindrically symmetric charge distributions are. What is the electric field at a point inside a hollow metallic sphere of radius R R RRR having volume charge density ρ ρ rho\rhoρ ?
b) Determine the electrostatic force and electrostatic field on a charged particle located at A A AAA in the Figure given below due to the charged particles situated at B B BBB and C C CCC. The value of the charge on each of these particles is indicated in the Figure.
original image
Express your result both in the unit vector notation and as magnitude.
c) Two particles carrying 4C and – 2C charges are placed on a 1 m long straight wire. Determine the point on the line joining these particles where the electric potential is zero with reference to the positively charged particle.
PART B
3. a) Explain the phenomenon of polarisation of a dielectric. Show that, when a dielectric material is filled between the plates of a capacitor, the value of capacitance increases by factor of K K KKK, the dielectric constant of the material.
b) The energy of a capacitor is 4.0 μ J 4.0 μ J 4.0 muJ4.0 \mu \mathrm{~J}4.0μ J after it has been charged by a 1.5 V battery. Calculate its energy when it is charged by a 6.0 V battery.
c) A horizontal, straight wire carrying 12.0 A current from west to east is in the earth’s magnetic field B B B\mathbf{B}B. At this place, B B B\mathbf{B}B is parallel to the surface of the earth, points to the north and its magnitude is 0.04 mT . Determine the magnetic force on 1 m length of the wire. If mass of this length of wire is 50 g , calculate the value of current in the wire so that its weight is balanced by the magnetic force.
4. a) A current is flowing in an infinitely long straight wire. Using Biot-Savart law, show that the resultant magnetic field at a point along a line perpendicular to the wire is inversely proportional to the distance of the point from the wire.
b) Using Maxwell’s equations in free space, derive the wave equation for the electric and magnetic field vectors.
c) The expression of the electric field associated with an electromagnetic wave in vacuum is given by
E → = ( 800 Vm − 1 ) x ^ sin ⁡ ( 2 π × 10 8 t + k z ) E → = 800 Vm − 1 x ^ sin ⁡ 2 π × 10 8 t + k z vec(E)=(800Vm^(-1)) hat(x)sin(2pi xx10^(8)t+kz)\overrightarrow{\mathbf{E}}=\left(800 \mathrm{Vm}^{-1}\right) \hat{\mathbf{x}} \sin \left(2 \pi \times 10^8 t+k z\right)E→=(800Vm−1)x^sin⁡(2π×108t+kz)
Determine the wave number, frequency, the direction of propagation and the magnitude and direction of the magnetic field associated with the wave.

Answer:

Question:-1

a) If u → u → vec(u)\overrightarrow{\mathbf{u}}u→ is a constant vector, show that ∇ → × ( u → × r → ) = 2 u → ∇ → × ( u → × r → ) = 2 u → vec(grad)xx( vec(u)xx vec(r))=2 vec(u)\vec{\nabla} \times(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{r}})=\mathbf{2} \overrightarrow{\mathbf{u}}∇→×(u→×r→)=2u→.

Answer:

We are given that u → u → vec(u)\overrightarrow{\mathbf{u}}u→ is a constant vector, and we need to show that:
∇ → × ( u → × r → ) = 2 u → ∇ → × ( u → × r → ) = 2 u → vec(grad)xx( vec(u)xx vec(r))=2 vec(u)\vec{\nabla} \times (\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{r}}) = 2 \overrightarrow{\mathbf{u}}∇→×(u→×r→)=2u→

Step 1: Use the vector triple product identity

We start by recalling the vector identity for the curl of a cross product:
∇ → × ( A × B ) = A ( ∇ → ⋅ B ) − B ( ∇ → ⋅ A ) + ( B ⋅ ∇ → ) A − ( A ⋅ ∇ → ) B ∇ → × ( A × B ) = A ( ∇ → ⋅ B ) − B ( ∇ → ⋅ A ) + ( B ⋅ ∇ → ) A − ( A ⋅ ∇ → ) B vec(grad)xx(AxxB)=A( vec(grad)*B)-B( vec(grad)*A)+(B* vec(grad))A-(A* vec(grad))B\vec{\nabla} \times (\mathbf{A} \times \mathbf{B}) = \mathbf{A} (\vec{\nabla} \cdot \mathbf{B}) – \mathbf{B} (\vec{\nabla} \cdot \mathbf{A}) + (\mathbf{B} \cdot \vec{\nabla}) \mathbf{A} – (\mathbf{A} \cdot \vec{\nabla}) \mathbf{B}∇→×(A×B)=A(∇→⋅B)−B(∇→⋅A)+(B⋅∇→)A−(A⋅∇→)B
For our case, we have:
  • A = u → A = u → A= vec(u)\mathbf{A} = \overrightarrow{\mathbf{u}}A=u→ (which is a constant vector),
  • B = r → B = r → B= vec(r)\mathbf{B} = \overrightarrow{\mathbf{r}}B=r→ (the position vector).
Substituting these into the identity:
∇ → × ( u → × r → ) = u → ( ∇ → ⋅ r → ) − r → ( ∇ → ⋅ u → ) + ( r → ⋅ ∇ → ) u → − ( u → ⋅ ∇ → ) r → ∇ → × ( u → × r → ) = u → ( ∇ → ⋅ r → ) − r → ( ∇ → ⋅ u → ) + ( r → ⋅ ∇ → ) u → − ( u → ⋅ ∇ → ) r → vec(grad)xx( vec(u)xx vec(r))= vec(u)( vec(grad)* vec(r))- vec(r)( vec(grad)* vec(u))+( vec(r)* vec(grad)) vec(u)-( vec(u)* vec(grad)) vec(r)\vec{\nabla} \times (\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{r}}) = \overrightarrow{\mathbf{u}} (\vec{\nabla} \cdot \overrightarrow{\mathbf{r}}) – \overrightarrow{\mathbf{r}} (\vec{\nabla} \cdot \overrightarrow{\mathbf{u}}) + (\overrightarrow{\mathbf{r}} \cdot \vec{\nabla}) \overrightarrow{\mathbf{u}} – (\overrightarrow{\mathbf{u}} \cdot \vec{\nabla}) \overrightarrow{\mathbf{r}}∇→×(u→×r→)=u→(∇→⋅r→)−r→(∇→⋅u→)+(r→⋅∇→)u→−(u→⋅∇→)r→

Step 2: Simplify the terms

Now, let’s simplify each term in the expression:
  1. Term 1: u → ( ∇ → ⋅ r → ) u → ( ∇ → ⋅ r → ) vec(u)( vec(grad)* vec(r))\overrightarrow{\mathbf{u}} (\vec{\nabla} \cdot \overrightarrow{\mathbf{r}})u→(∇→⋅r→)
    The divergence of the position vector r → r → vec(r)\overrightarrow{\mathbf{r}}r→ is:
    ∇ → ⋅ r → = 3 ∇ → ⋅ r → = 3 vec(grad)* vec(r)=3\vec{\nabla} \cdot \overrightarrow{\mathbf{r}} = 3∇→⋅r→=3
    Therefore, the first term is:
    u → ( ∇ → ⋅ r → ) = 3 u → u → ( ∇ → ⋅ r → ) = 3 u → vec(u)( vec(grad)* vec(r))=3 vec(u)\overrightarrow{\mathbf{u}} (\vec{\nabla} \cdot \overrightarrow{\mathbf{r}}) = 3 \overrightarrow{\mathbf{u}}u→(∇→⋅r→)=3u→
  2. Term 2: r → ( ∇ → ⋅ u → ) r → ( ∇ → ⋅ u → ) vec(r)( vec(grad)* vec(u))\overrightarrow{\mathbf{r}} (\vec{\nabla} \cdot \overrightarrow{\mathbf{u}})r→(∇→⋅u→)
    Since u → u → vec(u)\overrightarrow{\mathbf{u}}u→ is a constant vector, its divergence is zero:
    ∇ → ⋅ u → = 0 ∇ → ⋅ u → = 0 vec(grad)* vec(u)=0\vec{\nabla} \cdot \overrightarrow{\mathbf{u}} = 0∇→⋅u→=0
    Thus, the second term is zero:
    r → ( ∇ → ⋅ u → ) = 0 r → ( ∇ → ⋅ u → ) = 0 vec(r)( vec(grad)* vec(u))=0\overrightarrow{\mathbf{r}} (\vec{\nabla} \cdot \overrightarrow{\mathbf{u}}) = 0r→(∇→⋅u→)=0
  3. Term 3: ( r → ⋅ ∇ → ) u → ( r → ⋅ ∇ → ) u → ( vec(r)* vec(grad)) vec(u)(\overrightarrow{\mathbf{r}} \cdot \vec{\nabla}) \overrightarrow{\mathbf{u}}(r→⋅∇→)u→
    Since u → u → vec(u)\overrightarrow{\mathbf{u}}u→ is a constant vector, the derivative of u → u → vec(u)\overrightarrow{\mathbf{u}}u→ with respect to any coordinate is zero:
    ( r → ⋅ ∇ → ) u → = 0 ( r → ⋅ ∇ → ) u → = 0 ( vec(r)* vec(grad)) vec(u)=0(\overrightarrow{\mathbf{r}} \cdot \vec{\nabla}) \overrightarrow{\mathbf{u}} = 0(r→⋅∇→)u→=0
  4. Term 4: ( u → ⋅ ∇ → ) r → ( u → ⋅ ∇ → ) r → ( vec(u)* vec(grad)) vec(r)(\overrightarrow{\mathbf{u}} \cdot \vec{\nabla}) \overrightarrow{\mathbf{r}}(u→⋅∇→)r→
    The derivative of the position vector r → r → vec(r)\overrightarrow{\mathbf{r}}r→ with respect to a coordinate is the identity matrix (i.e., ∇ → r → = I ∇ → r → = I vec(grad) vec(r)=I\vec{\nabla} \overrightarrow{\mathbf{r}} = \mathbf{I}∇→r→=I):
    ( u → ⋅ ∇ → ) r → = u → ( u → ⋅ ∇ → ) r → = u → ( vec(u)* vec(grad)) vec(r)= vec(u)(\overrightarrow{\mathbf{u}} \cdot \vec{\nabla}) \overrightarrow{\mathbf{r}} = \overrightarrow{\mathbf{u}}(u→⋅∇→)r→=u→

Step 3: Combine the terms

Now, combining all the simplified terms:
∇ → × ( u → × r → ) = 3 u → − 0 + 0 − u → ∇ → × ( u → × r → ) = 3 u → − 0 + 0 − u → vec(grad)xx( vec(u)xx vec(r))=3 vec(u)-0+0- vec(u)\vec{\nabla} \times (\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{r}}) = 3 \overrightarrow{\mathbf{u}} – 0 + 0 – \overrightarrow{\mathbf{u}}∇→×(u→×r→)=3u→−0+0−u→
∇ → × ( u → × r → ) = 2 u → ∇ → × ( u → × r → ) = 2 u → vec(grad)xx( vec(u)xx vec(r))=2 vec(u)\vec{\nabla} \times (\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{r}}) = 2 \overrightarrow{\mathbf{u}}∇→×(u→×r→)=2u→
Thus, we have shown that:
∇ → × ( u → × r → ) = 2 u → ∇ → × ( u → × r → ) = 2 u → vec(grad)xx( vec(u)xx vec(r))=2 vec(u)\vec{\nabla} \times (\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{r}}) = 2 \overrightarrow{\mathbf{u}}∇→×(u→×r→)=2u→

b) Determine the work done by a force F → = ( x + 2 y ) i ^ + ( x − y ) j ^ F → = ( x + 2 y ) i ^ + ( x − y ) j ^ vec(F)=(x+2y) hat(i)+(x-y) hat(j)\overrightarrow{\mathbf{F}}=(x+2 y) \hat{\mathbf{i}}+(x-y) \hat{\mathbf{j}}F→=(x+2y)i^+(x−y)j^ in taking a particle along the curve x ( t ) = 2 cos ⁡ t ; y ( t ) = 4 sin ⁡ t x ( t ) = 2 cos ⁡ t ; y ( t ) = 4 sin ⁡ t x(t)=2cos t;y(t)=4sin tx(t)=2 \cos t ; y(t)=4 \sin tx(t)=2cos⁡t;y(t)=4sin⁡t from t = 0 t = 0 t=0t=0t=0 to t = π 4 t = π 4 t=(pi)/(4)t=\frac{\pi}{4}t=π4.

Answer:

To determine the work done by the force F → = ( x + 2 y ) i ^ + ( x − y ) j ^ F → = ( x + 2 y ) i ^ + ( x − y ) j ^ vec(F)=(x+2y) hat(i)+(x-y) hat(j)\overrightarrow{\mathbf{F}} = (x + 2y) \hat{\mathbf{i}} + (x – y) \hat{\mathbf{j}}F→=(x+2y)i^+(x−y)j^ along the curve parameterized by x ( t ) = 2 cos ⁡ t x ( t ) = 2 cos ⁡ t x(t)=2cos tx(t) = 2 \cos tx(t)=2cos⁡t and y ( t ) = 4 sin ⁡ t y ( t ) = 4 sin ⁡ t y(t)=4sin ty(t) = 4 \sin ty(t)=4sin⁡t from t = 0 t = 0 t=0t = 0t=0 to t = π 4 t = π 4 t=(pi)/(4)t = \frac{\pi}{4}t=π4, we need to compute the line integral of the force field along the given path. The work done is given by:
W = ∫ C F → ⋅ d r → , W = ∫ C   F → ⋅ d r → , W=int _(C) vec(F)*d vec(r),W = \int_C \overrightarrow{\mathbf{F}} \cdot d\overrightarrow{\mathbf{r}},W=∫CF→⋅dr→,
where r → ( t ) = x ( t ) i ^ + y ( t ) j ^ r → ( t ) = x ( t ) i ^ + y ( t ) j ^ vec(r)(t)=x(t) hat(i)+y(t) hat(j)\overrightarrow{\mathbf{r}}(t) = x(t) \hat{\mathbf{i}} + y(t) \hat{\mathbf{j}}r→(t)=x(t)i^+y(t)j^ is the position vector, and d r → d r → d vec(r)d\overrightarrow{\mathbf{r}}dr→ is the differential displacement along the curve.

Step 1: Parameterize the Force Field

Substitute the parametric equations x ( t ) = 2 cos ⁡ t x ( t ) = 2 cos ⁡ t x(t)=2cos tx(t) = 2 \cos tx(t)=2cos⁡t and y ( t ) = 4 sin ⁡ t y ( t ) = 4 sin ⁡ t y(t)=4sin ty(t) = 4 \sin ty(t)=4sin⁡t into the force field components:
  • F x = x + 2 y = 2 cos ⁡ t + 2 ⋅ 4 sin ⁡ t = 2 cos ⁡ t + 8 sin ⁡ t F x = x + 2 y = 2 cos ⁡ t + 2 ⋅ 4 sin ⁡ t = 2 cos ⁡ t + 8 sin ⁡ t F_(x)=x+2y=2cos t+2*4sin t=2cos t+8sin tF_x = x + 2y = 2 \cos t + 2 \cdot 4 \sin t = 2 \cos t + 8 \sin tFx=x+2y=2cos⁡t+2⋅4sin⁡t=2cos⁡t+8sin⁡t,
  • F y = x − y = 2 cos ⁡ t − 4 sin ⁡ t F y = x − y = 2 cos ⁡ t − 4 sin ⁡ t F_(y)=x-y=2cos t-4sin tF_y = x – y = 2 \cos t – 4 \sin tFy=x−y=2cos⁡t−4sin⁡t.
Thus, the force field along the curve is:
F → ( t ) = ( 2 cos ⁡ t + 8 sin ⁡ t ) i ^ + ( 2 cos ⁡ t − 4 sin ⁡ t ) j ^ . F → ( t ) = ( 2 cos ⁡ t + 8 sin ⁡ t ) i ^ + ( 2 cos ⁡ t − 4 sin ⁡ t ) j ^ . vec(F)(t)=(2cos t+8sin t) hat(i)+(2cos t-4sin t) hat(j).\overrightarrow{\mathbf{F}}(t) = (2 \cos t + 8 \sin t) \hat{\mathbf{i}} + (2 \cos t – 4 \sin t) \hat{\mathbf{j}}.F→(t)=(2cos⁡t+8sin⁡t)i^+(2cos⁡t−4sin⁡t)j^.

Step 2: Compute the Differential Displacement

The position vector is:
r → ( t ) = x ( t ) i ^ + y ( t ) j ^ = 2 cos ⁡ t i ^ + 4 sin ⁡ t j ^ . r → ( t ) = x ( t ) i ^ + y ( t ) j ^ = 2 cos ⁡ t i ^ + 4 sin ⁡ t j ^ . vec(r)(t)=x(t) hat(i)+y(t) hat(j)=2cos t hat(i)+4sin t hat(j).\overrightarrow{\mathbf{r}}(t) = x(t) \hat{\mathbf{i}} + y(t) \hat{\mathbf{j}} = 2 \cos t \hat{\mathbf{i}} + 4 \sin t \hat{\mathbf{j}}.r→(t)=x(t)i^+y(t)j^=2cos⁡ti^+4sin⁡tj^.
Differentiate with respect to t t ttt to find the velocity vector (tangent to the curve):
d r → d t = d d t ( 2 cos ⁡ t ) i ^ + d d t ( 4 sin ⁡ t ) j ^ = − 2 sin ⁡ t i ^ + 4 cos ⁡ t j ^ . d r → d t = d d t ( 2 cos ⁡ t ) i ^ + d d t ( 4 sin ⁡ t ) j ^ = − 2 sin ⁡ t i ^ + 4 cos ⁡ t j ^ . (d vec(r))/(dt)=(d)/(dt)(2cos t) hat(i)+(d)/(dt)(4sin t) hat(j)=-2sin t hat(i)+4cos t hat(j).\frac{d\overrightarrow{\mathbf{r}}}{dt} = \frac{d}{dt} (2 \cos t) \hat{\mathbf{i}} + \frac{d}{dt} (4 \sin t) \hat{\mathbf{j}} = -2 \sin t \hat{\mathbf{i}} + 4 \cos t \hat{\mathbf{j}}.dr→dt=ddt(2cos⁡t)i^+ddt(4sin⁡t)j^=−2sin⁡ti^+4cos⁡tj^.
The differential displacement is:
d r → = d r → d t d t = ( − 2 sin ⁡ t i ^ + 4 cos ⁡ t j ^ ) d t . d r → = d r → d t d t = ( − 2 sin ⁡ t i ^ + 4 cos ⁡ t j ^ ) d t . d vec(r)=(d vec(r))/(dt)dt=(-2sin t hat(i)+4cos t hat(j))dt.d\overrightarrow{\mathbf{r}} = \frac{d\overrightarrow{\mathbf{r}}}{dt} dt = (-2 \sin t \hat{\mathbf{i}} + 4 \cos t \hat{\mathbf{j}}) dt.dr→=dr→dtdt=(−2sin⁡ti^+4cos⁡tj^)dt.

Step 3: Compute the Dot Product

The integrand of the line integral is the dot product F → ⋅ d r → F → ⋅ d r → vec(F)*d vec(r)\overrightarrow{\mathbf{F}} \cdot d\overrightarrow{\mathbf{r}}F→⋅dr→:
F → ⋅ d r → = ( 2 cos ⁡ t + 8 sin ⁡ t ) ( − 2 sin ⁡ t ) + ( 2 cos ⁡ t − 4 sin ⁡ t ) ( 4 cos ⁡ t ) . F → ⋅ d r → = ( 2 cos ⁡ t + 8 sin ⁡ t ) ( − 2 sin ⁡ t ) + ( 2 cos ⁡ t − 4 sin ⁡ t ) ( 4 cos ⁡ t ) . vec(F)*d vec(r)=(2cos t+8sin t)(-2sin t)+(2cos t-4sin t)(4cos t).\overrightarrow{\mathbf{F}} \cdot d\overrightarrow{\mathbf{r}} = (2 \cos t + 8 \sin t) (-2 \sin t) + (2 \cos t – 4 \sin t) (4 \cos t).F→⋅dr→=(2cos⁡t+8sin⁡t)(−2sin⁡t)+(2cos⁡t−4sin⁡t)(4cos⁡t).
Calculate each term:
  • First component: ( 2 cos ⁡ t + 8 sin ⁡ t ) ( − 2 sin ⁡ t ) = − 4 cos ⁡ t sin ⁡ t − 16 sin 2 ⁡ t ( 2 cos ⁡ t + 8 sin ⁡ t ) ( − 2 sin ⁡ t ) = − 4 cos ⁡ t sin ⁡ t − 16 sin 2 ⁡ t (2cos t+8sin t)(-2sin t)=-4cos t sin t-16sin^(2)t(2 \cos t + 8 \sin t)(-2 \sin t) = -4 \cos t \sin t – 16 \sin^2 t(2cos⁡t+8sin⁡t)(−2sin⁡t)=−4cos⁡tsin⁡t−16sin2⁡t,
  • Second component: ( 2 cos ⁡ t − 4 sin ⁡ t ) ( 4 cos ⁡ t ) = 8 cos 2 ⁡ t − 16 sin ⁡ t cos ⁡ t ( 2 cos ⁡ t − 4 sin ⁡ t ) ( 4 cos ⁡ t ) = 8 cos 2 ⁡ t − 16 sin ⁡ t cos ⁡ t (2cos t-4sin t)(4cos t)=8cos^(2)t-16 sin t cos t(2 \cos t – 4 \sin t)(4 \cos t) = 8 \cos^2 t – 16 \sin t \cos t(2cos⁡t−4sin⁡t)(4cos⁡t)=8cos2⁡t−16sin⁡tcos⁡t.
Combine:
F → ⋅ d r → = ( − 4 cos ⁡ t sin ⁡ t − 16 sin 2 ⁡ t ) + ( 8 cos 2 ⁡ t − 16 sin ⁡ t cos ⁡ t ) . F → ⋅ d r → = ( − 4 cos ⁡ t sin ⁡ t − 16 sin 2 ⁡ t ) + ( 8 cos 2 ⁡ t − 16 sin ⁡ t cos ⁡ t ) . vec(F)*d vec(r)=(-4cos t sin t-16sin^(2)t)+(8cos^(2)t-16 sin t cos t).\overrightarrow{\mathbf{F}} \cdot d\overrightarrow{\mathbf{r}} = (-4 \cos t \sin t – 16 \sin^2 t) + (8 \cos^2 t – 16 \sin t \cos t).F→⋅dr→=(−4cos⁡tsin⁡t−16sin2⁡t)+(8cos2⁡t−16sin⁡tcos⁡t).
Simplify:
= 8 cos 2 ⁡ t − 16 sin 2 ⁡ t − 4 cos ⁡ t sin ⁡ t − 16 sin ⁡ t cos ⁡ t = 8 cos 2 ⁡ t − 16 sin 2 ⁡ t − 20 sin ⁡ t cos ⁡ t . = 8 cos 2 ⁡ t − 16 sin 2 ⁡ t − 4 cos ⁡ t sin ⁡ t − 16 sin ⁡ t cos ⁡ t = 8 cos 2 ⁡ t − 16 sin 2 ⁡ t − 20 sin ⁡ t cos ⁡ t . =8cos^(2)t-16sin^(2)t-4cos t sin t-16 sin t cos t=8cos^(2)t-16sin^(2)t-20 sin t cos t.= 8 \cos^2 t – 16 \sin^2 t – 4 \cos t \sin t – 16 \sin t \cos t = 8 \cos^2 t – 16 \sin^2 t – 20 \sin t \cos t.=8cos2⁡t−16sin2⁡t−4cos⁡tsin⁡t−16sin⁡tcos⁡t=8cos2⁡t−16sin2⁡t−20sin⁡tcos⁡t.

Step 4: Set Up and Evaluate the Integral

The work done is:
W = ∫ t = 0 t = π / 4 ( 8 cos 2 ⁡ t − 16 sin 2 ⁡ t − 20 sin ⁡ t cos ⁡ t ) d t . W = ∫ t = 0 t = π / 4   ( 8 cos 2 ⁡ t − 16 sin 2 ⁡ t − 20 sin ⁡ t cos ⁡ t ) d t . W=int_(t=0)^(t=pi//4)(8cos^(2)t-16sin^(2)t-20 sin t cos t)dt.W = \int_{t=0}^{t=\pi/4} (8 \cos^2 t – 16 \sin^2 t – 20 \sin t \cos t) dt.W=∫t=0t=π/4(8cos2⁡t−16sin2⁡t−20sin⁡tcos⁡t)dt.
Split the integral:
W = 8 ∫ 0 π / 4 cos 2 ⁡ t d t − 16 ∫ 0 π / 4 sin 2 ⁡ t d t − 20 ∫ 0 π / 4 sin ⁡ t cos ⁡ t d t . W = 8 ∫ 0 π / 4   cos 2 ⁡ t d t − 16 ∫ 0 π / 4   sin 2 ⁡ t d t − 20 ∫ 0 π / 4   sin ⁡ t cos ⁡ t d t . W=8int_(0)^(pi//4)cos^(2)tdt-16int_(0)^(pi//4)sin^(2)tdt-20int_(0)^(pi//4)sin t cos tdt.W = 8 \int_0^{\pi/4} \cos^2 t \, dt – 16 \int_0^{\pi/4} \sin^2 t \, dt – 20 \int_0^{\pi/4} \sin t \cos t \, dt.W=8∫0π/4cos2⁡tdt−16∫0π/4sin2⁡tdt−20∫0π/4sin⁡tcos⁡tdt.

Integral 1: ∫ 0 π / 4 cos 2 ⁡ t d t ∫ 0 π / 4   cos 2 ⁡ t d t int_(0)^(pi//4)cos^(2)tdt\int_0^{\pi/4} \cos^2 t \, dt∫0π/4cos2⁡tdt

Use the identity cos 2 ⁡ t = 1 + cos ⁡ 2 t 2 cos 2 ⁡ t = 1 + cos ⁡ 2 t 2 cos^(2)t=(1+cos 2t)/(2)\cos^2 t = \frac{1 + \cos 2t}{2}cos2⁡t=1+cos⁡2t2:
∫ cos 2 ⁡ t d t = ∫ 1 + cos ⁡ 2 t 2 d t = 1 2 t + 1 4 sin ⁡ 2 t . ∫ cos 2 ⁡ t d t = ∫ 1 + cos ⁡ 2 t 2 d t = 1 2 t + 1 4 sin ⁡ 2 t . intcos^(2)tdt=int(1+cos 2t)/(2)dt=(1)/(2)t+(1)/(4)sin 2t.\int \cos^2 t \, dt = \int \frac{1 + \cos 2t}{2} \, dt = \frac{1}{2} t + \frac{1}{4} \sin 2t.∫cos2⁡tdt=∫1+cos⁡2t2dt=12t+14sin⁡2t.
Evaluate from 0 to π / 4 π / 4 pi//4\pi/4π/4:
[ 1 2 t + 1 4 sin ⁡ 2 t ] 0 π / 4 = ( 1 2 ⋅ π 4 + 1 4 sin ⁡ π 2 ) − 0 = π 8 + 1 4 ⋅ 1 = π 8 + 1 4 . 1 2 t + 1 4 sin ⁡ 2 t 0 π / 4 = 1 2 ⋅ π 4 + 1 4 sin ⁡ π 2 − 0 = π 8 + 1 4 ⋅ 1 = π 8 + 1 4 . [(1)/(2)t+(1)/(4)sin 2t]_(0)^(pi//4)=((1)/(2)*(pi)/(4)+(1)/(4)sin((pi)/(2)))-0=(pi)/(8)+(1)/(4)*1=(pi)/(8)+(1)/(4).\left[ \frac{1}{2} t + \frac{1}{4} \sin 2t \right]_0^{\pi/4} = \left( \frac{1}{2} \cdot \frac{\pi}{4} + \frac{1}{4} \sin \frac{\pi}{2} \right) – 0 = \frac{\pi}{8} + \frac{1}{4} \cdot 1 = \frac{\pi}{8} + \frac{1}{4}.[12t+14sin⁡2t]0π/4=(12⋅π4+14sin⁡π2)−0=π8+14⋅1=π8+14.

Integral 2: ∫ 0 π / 4 sin 2 ⁡ t d t ∫ 0 π / 4   sin 2 ⁡ t d t int_(0)^(pi//4)sin^(2)tdt\int_0^{\pi/4} \sin^2 t \, dt∫0π/4sin2⁡tdt

Use the identity sin 2 ⁡ t = 1 − cos ⁡ 2 t 2 sin 2 ⁡ t = 1 − cos ⁡ 2 t 2 sin^(2)t=(1-cos 2t)/(2)\sin^2 t = \frac{1 – \cos 2t}{2}sin2⁡t=1−cos⁡2t2:
∫ sin 2 ⁡ t d t = ∫ 1 − cos ⁡ 2 t 2 d t = 1 2 t − 1 4 sin ⁡ 2 t . ∫ sin 2 ⁡ t d t = ∫ 1 − cos ⁡ 2 t 2 d t = 1 2 t − 1 4 sin ⁡ 2 t . intsin^(2)tdt=int(1-cos 2t)/(2)dt=(1)/(2)t-(1)/(4)sin 2t.\int \sin^2 t \, dt = \int \frac{1 – \cos 2t}{2} \, dt = \frac{1}{2} t – \frac{1}{4} \sin 2t.∫sin2⁡tdt=∫1−cos⁡2t2dt=12t−14sin⁡2t.
Evaluate:
[ 1 2 t − 1 4 sin ⁡ 2 t ] 0 π / 4 = ( 1 2 ⋅ π 4 − 1 4 sin ⁡ π 2 ) − 0 = π 8 − 1 4 ⋅ 1 = π 8 − 1 4 . 1 2 t − 1 4 sin ⁡ 2 t 0 π / 4 = 1 2 ⋅ π 4 − 1 4 sin ⁡ π 2 − 0 = π 8 − 1 4 ⋅ 1 = π 8 − 1 4 . [(1)/(2)t-(1)/(4)sin 2t]_(0)^(pi//4)=((1)/(2)*(pi)/(4)-(1)/(4)sin((pi)/(2)))-0=(pi)/(8)-(1)/(4)*1=(pi)/(8)-(1)/(4).\left[ \frac{1}{2} t – \frac{1}{4} \sin 2t \right]_0^{\pi/4} = \left( \frac{1}{2} \cdot \frac{\pi}{4} – \frac{1}{4} \sin \frac{\pi}{2} \right) – 0 = \frac{\pi}{8} – \frac{1}{4} \cdot 1 = \frac{\pi}{8} – \frac{1}{4}.[12t−14sin⁡2t]0π/4=(12⋅π4−14sin⁡π2)−0=π8−14⋅1=π8−14.

Integral 3: ∫ 0 π / 4 sin ⁡ t cos ⁡ t d t ∫ 0 π / 4   sin ⁡ t cos ⁡ t d t int_(0)^(pi//4)sin t cos tdt\int_0^{\pi/4} \sin t \cos t \, dt∫0π/4sin⁡tcos⁡tdt

Use substitution: let u = sin ⁡ t u = sin ⁡ t u=sin tu = \sin tu=sin⁡t, so d u = cos ⁡ t d t d u = cos ⁡ t d t du=cos tdtdu = \cos t \, dtdu=cos⁡tdt:
  • At t = 0 t = 0 t=0t = 0t=0, u = 0 u = 0 u=0u = 0u=0,
  • At t = π / 4 t = π / 4 t=pi//4t = \pi/4t=π/4, u = sin ⁡ π 4 = 2 2 u = sin ⁡ π 4 = 2 2 u=sin((pi)/(4))=(sqrt2)/(2)u = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}u=sin⁡π4=22.
∫ 0 π / 4 sin ⁡ t cos ⁡ t d t = ∫ 0 2 / 2 u d u = [ u 2 2 ] 0 2 / 2 = 1 2 ( 2 2 ) 2 = 1 2 ⋅ 2 4 = 1 4 . ∫ 0 π / 4   sin ⁡ t cos ⁡ t d t = ∫ 0 2 / 2   u d u = u 2 2 0 2 / 2 = 1 2 2 2 2 = 1 2 ⋅ 2 4 = 1 4 . int_(0)^(pi//4)sin t cos tdt=int_(0)^(sqrt2//2)udu=[(u^(2))/(2)]_(0)^(sqrt2//2)=(1)/(2)((sqrt2)/(2))^(2)=(1)/(2)*(2)/(4)=(1)/(4).\int_0^{\pi/4} \sin t \cos t \, dt = \int_0^{\sqrt{2}/2} u \, du = \left[ \frac{u^2}{2} \right]_0^{\sqrt{2}/2} = \frac{1}{2} \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{1}{2} \cdot \frac{2}{4} = \frac{1}{4}.∫0π/4sin⁡tcos⁡tdt=∫02/2udu=[u22]02/2=12(22)2=12⋅24=14.

Combine:

W = 8 ( π 8 + 1 4 ) − 16 ( π 8 − 1 4 ) − 20 ⋅ 1 4 . W = 8 π 8 + 1 4 − 16 π 8 − 1 4 − 20 ⋅ 1 4 . W=8((pi)/(8)+(1)/(4))-16((pi)/(8)-(1)/(4))-20*(1)/(4).W = 8 \left( \frac{\pi}{8} + \frac{1}{4} \right) – 16 \left( \frac{\pi}{8} – \frac{1}{4} \right) – 20 \cdot \frac{1}{4}.W=8(π8+14)−16(π8−14)−20⋅14.
  • First term: 8 ( π 8 + 1 4 ) = π + 2 8 π 8 + 1 4 = π + 2 8((pi)/(8)+(1)/(4))=pi+28 \left( \frac{\pi}{8} + \frac{1}{4} \right) = \pi + 28(π8+14)=π+2,
  • Second term: − 16 ( π 8 − 1 4 ) = − 16 ( π 8 − 2 8 ) = − 16 ⋅ π − 2 8 = − 2 π + 4 − 16 π 8 − 1 4 = − 16 π 8 − 2 8 = − 16 ⋅ π − 2 8 = − 2 π + 4 -16((pi)/(8)-(1)/(4))=-16((pi)/(8)-(2)/(8))=-16*(pi-2)/(8)=-2pi+4-16 \left( \frac{\pi}{8} – \frac{1}{4} \right) = -16 \left( \frac{\pi}{8} – \frac{2}{8} \right) = -16 \cdot \frac{\pi – 2}{8} = -2\pi + 4−16(π8−14)=−16(π8−28)=−16⋅π−28=−2π+4,
  • Third term: − 20 ⋅ 1 4 = − 5 − 20 ⋅ 1 4 = − 5 -20*(1)/(4)=-5-20 \cdot \frac{1}{4} = -5−20⋅14=−5.
Total:
W = ( π + 2 ) + ( − 2 π + 4 ) − 5 = π − 2 π + 2 + 4 − 5 = − π + 1. W = ( π + 2 ) + ( − 2 π + 4 ) − 5 = π − 2 π + 2 + 4 − 5 = − π + 1. W=(pi+2)+(-2pi+4)-5=pi-2pi+2+4-5=-pi+1.W = (\pi + 2) + (-2\pi + 4) – 5 = \pi – 2\pi + 2 + 4 – 5 = -\pi + 1.W=(π+2)+(−2π+4)−5=π−2π+2+4−5=−π+1.

Final Answer

The work done is:
W = 1 − π . W = 1 − π . W=1-pi.W = 1 – \pi.W=1−π.

c) Using Stokes’ theorem, prove that curl of a conservative force field is zero everywhere.

Answer:

To prove that the curl of a conservative force field is zero everywhere using Stokes’ theorem, let’s proceed step by step with clarity and rigor. A conservative force field is one that can be expressed as the gradient of a scalar potential function, and we’ll use this property along with Stokes’ theorem to establish the result.

Step 1: Define a Conservative Force Field

A force field F → F → vec(F)\overrightarrow{\mathbf{F}}F→ is conservative if there exists a scalar potential function V V VVV such that:
F → = − ∇ V , F → = − ∇ V , vec(F)=-grad V,\overrightarrow{\mathbf{F}} = -\nabla V,F→=−∇V,
where ∇ V = ∂ V ∂ x i ^ + ∂ V ∂ y j ^ + ∂ V ∂ z k ^ ∇ V = ∂ V ∂ x i ^ + ∂ V ∂ y j ^ + ∂ V ∂ z k ^ grad V=(del V)/(del x) hat(i)+(del V)/(del y) hat(j)+(del V)/(del z) hat(k)\nabla V = \frac{\partial V}{\partial x} \hat{\mathbf{i}} + \frac{\partial V}{\partial y} \hat{\mathbf{j}} + \frac{\partial V}{\partial z} \hat{\mathbf{k}}∇V=∂V∂xi^+∂V∂yj^+∂V∂zk^ is the gradient of V V VVV. The negative sign is a convention in physics (e.g., gravitational or electric fields), but for mathematical generality, we can also consider F → = ∇ V F → = ∇ V vec(F)=grad V\overrightarrow{\mathbf{F}} = \nabla VF→=∇V without loss of generality. The key property of a conservative field is that its line integral around any closed path is zero:
∮ C F → ⋅ d r → = 0 , ∮ C F → ⋅ d r → = 0 , oint_(C) vec(F)*d vec(r)=0,\oint_C \overrightarrow{\mathbf{F}} \cdot d\overrightarrow{\mathbf{r}} = 0,∮CF→⋅dr→=0,
for any closed curve C C CCC.

Step 2: Recall Stokes’ Theorem

Stokes’ theorem relates the line integral of a vector field around a closed curve C C CCC to the surface integral of the curl of that field over a surface S S SSS bounded by C C CCC:
∮ C F → ⋅ d r → = ∬ S ( ∇ × F → ) ⋅ d S → , ∮ C F → ⋅ d r → = ∬ S   ( ∇ × F → ) ⋅ d S → , oint_(C) vec(F)*d vec(r)=∬_(S)(grad xx vec(F))*d vec(S),\oint_C \overrightarrow{\mathbf{F}} \cdot d\overrightarrow{\mathbf{r}} = \iint_S (\nabla \times \overrightarrow{\mathbf{F}}) \cdot d\overrightarrow{\mathbf{S}},∮CF→⋅dr→=∬S(∇×F→)⋅dS→,
where:
  • C C CCC is the closed boundary of the surface S S SSS,
  • ∇ × F → ∇ × F → grad xx vec(F)\nabla \times \overrightarrow{\mathbf{F}}∇×F→ is the curl of the vector field F → F → vec(F)\overrightarrow{\mathbf{F}}F→,
  • d S → d S → d vec(S)d\overrightarrow{\mathbf{S}}dS→ is the differential area vector on the surface S S SSS.

Step 3: Apply Stokes’ Theorem to a Conservative Field

Since F → F → vec(F)\overrightarrow{\mathbf{F}}F→ is conservative, we know:
∮ C F → ⋅ d r → = 0 , ∮ C F → ⋅ d r → = 0 , oint_(C) vec(F)*d vec(r)=0,\oint_C \overrightarrow{\mathbf{F}} \cdot d\overrightarrow{\mathbf{r}} = 0,∮CF→⋅dr→=0,
for any closed curve C C CCC. By Stokes’ theorem, this implies:
∮ C F → ⋅ d r → = ∬ S ( ∇ × F → ) ⋅ d S → = 0. ∮ C F → ⋅ d r → = ∬ S   ( ∇ × F → ) ⋅ d S → = 0. oint_(C) vec(F)*d vec(r)=∬_(S)(grad xx vec(F))*d vec(S)=0.\oint_C \overrightarrow{\mathbf{F}} \cdot d\overrightarrow{\mathbf{r}} = \iint_S (\nabla \times \overrightarrow{\mathbf{F}}) \cdot d\overrightarrow{\mathbf{S}} = 0.∮CF→⋅dr→=∬S(∇×F→)⋅dS→=0.
Thus, for any surface S S SSS bounded by any closed curve C C CCC, the surface integral of the curl is zero:
∬ S ( ∇ × F → ) ⋅ d S → = 0. ∬ S   ( ∇ × F → ) ⋅ d S → = 0. ∬_(S)(grad xx vec(F))*d vec(S)=0.\iint_S (\nabla \times \overrightarrow{\mathbf{F}}) \cdot d\overrightarrow{\mathbf{S}} = 0.∬S(∇×F→)⋅dS→=0.

Step 4: Argue that the Curl is Zero Everywhere

To conclude that ∇ × F → = 0 ∇ × F → = 0 grad xx vec(F)=0\nabla \times \overrightarrow{\mathbf{F}} = 0∇×F→=0 everywhere, consider the implications of the surface integral being zero for any surface S S SSS. Suppose, for contradiction, that ∇ × F → ≠ 0 ∇ × F → ≠ 0 grad xx vec(F)!=0\nabla \times \overrightarrow{\mathbf{F}} \neq 0∇×F→≠0 at some point P P PPP in the domain. Since the curl is a continuous vector field (assuming F → F → vec(F)\overrightarrow{\mathbf{F}}F→ is sufficiently smooth, which is typical for physical force fields), if ∇ × F → ≠ 0 ∇ × F → ≠ 0 grad xx vec(F)!=0\nabla \times \overrightarrow{\mathbf{F}} \neq 0∇×F→≠0 at P P PPP, there exists a small neighborhood around P P PPP where ∇ × F → ∇ × F → grad xx vec(F)\nabla \times \overrightarrow{\mathbf{F}}∇×F→ is non-zero and has a consistent direction (due to continuity).
Now, choose a small surface S S SSS passing through P P PPP with boundary C C CCC, such that the normal vector d S → d S → d vec(S)d\overrightarrow{\mathbf{S}}dS→ aligns with ∇ × F → ∇ × F → grad xx vec(F)\nabla \times \overrightarrow{\mathbf{F}}∇×F→ in this neighborhood. For example, if ∇ × F → ∇ × F → grad xx vec(F)\nabla \times \overrightarrow{\mathbf{F}}∇×F→ has a positive z z zzz-component at P P PPP, orient S S SSS in the x y x y xyxyxy-plane around P P PPP with d S → d S → d vec(S)d\overrightarrow{\mathbf{S}}dS→ pointing along k ^ k ^ hat(k)\hat{\mathbf{k}}k^. The dot product ( ∇ × F → ) ⋅ d S → ( ∇ × F → ) ⋅ d S → (grad xx vec(F))*d vec(S)(\nabla \times \overrightarrow{\mathbf{F}}) \cdot d\overrightarrow{\mathbf{S}}(∇×F→)⋅dS→ would then be positive over this small surface, and since ∇ × F → ∇ × F → grad xx vec(F)\nabla \times \overrightarrow{\mathbf{F}}∇×F→ is non-zero in a neighborhood, the integral:
∬ S ( ∇ × F → ) ⋅ d S → , ∬ S   ( ∇ × F → ) ⋅ d S → , ∬_(S)(grad xx vec(F))*d vec(S),\iint_S (\nabla \times \overrightarrow{\mathbf{F}}) \cdot d\overrightarrow{\mathbf{S}},∬S(∇×F→)⋅dS→,
would be positive (or negative, depending on orientation), not zero. This contradicts the result from Stokes’ theorem that the integral is zero for a conservative field.
Since this contradiction arises for any point P P PPP where we assume ∇ × F → ≠ 0 ∇ × F → ≠ 0 grad xx vec(F)!=0\nabla \times \overrightarrow{\mathbf{F}} \neq 0∇×F→≠0, it follows that ∇ × F → ∇ × F → grad xx vec(F)\nabla \times \overrightarrow{\mathbf{F}}∇×F→ cannot be non-zero anywhere. Thus, we must have:
∇ × F → = 0 , ∇ × F → = 0 , grad xx vec(F)=0,\nabla \times \overrightarrow{\mathbf{F}} = 0,∇×F→=0,
everywhere in the domain.

Step 5: Verify with the Definition of Curl

To reinforce this, compute the curl explicitly. Let F → = − ∇ V = − ( ∂ V ∂ x i ^ + ∂ V ∂ y j ^ + ∂ V ∂ z k ^ ) F → = − ∇ V = − ∂ V ∂ x i ^ + ∂ V ∂ y j ^ + ∂ V ∂ z k ^ vec(F)=-grad V=-((del V)/(del x)( hat(i))+(del V)/(del y)( hat(j))+(del V)/(del z)( hat(k)))\overrightarrow{\mathbf{F}} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{\mathbf{i}} + \frac{\partial V}{\partial y} \hat{\mathbf{j}} + \frac{\partial V}{\partial z} \hat{\mathbf{k}} \right)F→=−∇V=−(∂V∂xi^+∂V∂yj^+∂V∂zk^). The curl is:
∇ × F → = | i ^ j ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z − ∂ V ∂ x − ∂ V ∂ y − ∂ V ∂ z | . ∇ × F → = i ^ j ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z − ∂ V ∂ x − ∂ V ∂ y − ∂ V ∂ z . grad xx vec(F)=|[ hat(i), hat(j), hat(k)],[(del)/(del x),(del)/(del y),(del)/(del z)],[-(del V)/(del x),-(del V)/(del y),-(del V)/(del z)]|.\nabla \times \overrightarrow{\mathbf{F}} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ -\frac{\partial V}{\partial x} & -\frac{\partial V}{\partial y} & -\frac{\partial V}{\partial z} \end{vmatrix}.∇×F→=|i^j^k^∂∂x∂∂y∂∂z−∂V∂x−∂V∂y−∂V∂z|.
Expanding this:
  • x x xxx-component: ∂ ∂ y ( − ∂ V ∂ z ) − ∂ ∂ z ( − ∂ V ∂ y ) = − ∂ 2 V ∂ y ∂ z + ∂ 2 V ∂ z ∂ y = 0 ∂ ∂ y − ∂ V ∂ z − ∂ ∂ z − ∂ V ∂ y = − ∂ 2 V ∂ y ∂ z + ∂ 2 V ∂ z ∂ y = 0 (del)/(del y)(-(del V)/(del z))-(del)/(del z)(-(del V)/(del y))=-(del^(2)V)/(del y del z)+(del^(2)V)/(del z del y)=0\frac{\partial}{\partial y} \left( -\frac{\partial V}{\partial z} \right) – \frac{\partial}{\partial z} \left( -\frac{\partial V}{\partial y} \right) = -\frac{\partial^2 V}{\partial y \partial z} + \frac{\partial^2 V}{\partial z \partial y} = 0∂∂y(−∂V∂z)−∂∂z(−∂V∂y)=−∂2V∂y∂z+∂2V∂z∂y=0,
  • y y yyy-component: ∂ ∂ z ( − ∂ V ∂ x ) − ∂ ∂ x ( − ∂ V ∂ z ) = − ∂ 2 V ∂ z ∂ x + ∂ 2 V ∂ x ∂ z = 0 ∂ ∂ z − ∂ V ∂ x − ∂ ∂ x − ∂ V ∂ z = − ∂ 2 V ∂ z ∂ x + ∂ 2 V ∂ x ∂ z = 0 (del)/(del z)(-(del V)/(del x))-(del)/(del x)(-(del V)/(del z))=-(del^(2)V)/(del z del x)+(del^(2)V)/(del x del z)=0\frac{\partial}{\partial z} \left( -\frac{\partial V}{\partial x} \right) – \frac{\partial}{\partial x} \left( -\frac{\partial V}{\partial z} \right) = -\frac{\partial^2 V}{\partial z \partial x} + \frac{\partial^2 V}{\partial x \partial z} = 0∂∂z(−∂V∂x)−∂∂x(−∂V∂z)=−∂2V∂z∂x+∂2V∂x∂z=0,
  • z z zzz-component: ∂ ∂ x ( − ∂ V ∂ y ) − ∂ ∂ y ( − ∂ V ∂ x ) = − ∂ 2 V ∂ x ∂ y + ∂ 2 V ∂ y ∂ x = 0 ∂ ∂ x − ∂ V ∂ y − ∂ ∂ y − ∂ V ∂ x = − ∂ 2 V ∂ x ∂ y + ∂ 2 V ∂ y ∂ x = 0 (del)/(del x)(-(del V)/(del y))-(del)/(del y)(-(del V)/(del x))=-(del^(2)V)/(del x del y)+(del^(2)V)/(del y del x)=0\frac{\partial}{\partial x} \left( -\frac{\partial V}{\partial y} \right) – \frac{\partial}{\partial y} \left( -\frac{\partial V}{\partial x} \right) = -\frac{\partial^2 V}{\partial x \partial y} + \frac{\partial^2 V}{\partial y \partial x} = 0∂∂x(−∂V∂y)−∂∂y(−∂V∂x)=−∂2V∂x∂y+∂2V∂y∂x=0,
assuming V V VVV has continuous second partial derivatives (i.e., ∂ 2 V ∂ x ∂ y = ∂ 2 V ∂ y ∂ x ∂ 2 V ∂ x ∂ y = ∂ 2 V ∂ y ∂ x (del^(2)V)/(del x del y)=(del^(2)V)/(del y del x)\frac{\partial^2 V}{\partial x \partial y} = \frac{\partial^2 V}{\partial y \partial x}∂2V∂x∂y=∂2V∂y∂x, etc.). This confirms ∇ × F → = 0 ∇ × F → = 0 grad xx vec(F)=0\nabla \times \overrightarrow{\mathbf{F}} = 0∇×F→=0, consistent with our conclusion from Stokes’ theorem.

Final Answer

Using Stokes’ theorem, we’ve shown that for a conservative force field F → F → vec(F)\overrightarrow{\mathbf{F}}F→, where ∮ C F → ⋅ d r → = 0 ∮ C F → ⋅ d r → = 0 oint_(C) vec(F)*d vec(r)=0\oint_C \overrightarrow{\mathbf{F}} \cdot d\overrightarrow{\mathbf{r}} = 0∮CF→⋅dr→=0, the surface integral ∬ S ( ∇ × F → ) ⋅ d S → = 0 ∬ S   ( ∇ × F → ) ⋅ d S → = 0 ∬_(S)(grad xx vec(F))*d vec(S)=0\iint_S (\nabla \times \overrightarrow{\mathbf{F}}) \cdot d\overrightarrow{\mathbf{S}} = 0∬S(∇×F→)⋅dS→=0 for any surface S S SSS. Since this holds for all surfaces, ∇ × F → ∇ × F → grad xx vec(F)\nabla \times \overrightarrow{\mathbf{F}}∇×F→ must be zero everywhere. Thus:
∇ × F → = 0 , ∇ × F → = 0 , grad xx vec(F)=0,\nabla \times \overrightarrow{\mathbf{F}} = 0,∇×F→=0,
everywhere in the domain, proving that the curl of a conservative force field is zero.

d) Determine the directional derivative of the scalar field ϕ = ln ⁡ ( x 2 + y 2 + z 2 ) ϕ = ln ⁡ x 2 + y 2 + z 2 phi=ln(x^(2)+y^(2)+z^(2))\phi=\ln \left(x^2+y^2+z^2\right)ϕ=ln⁡(x2+y2+z2) in the direction ( i ^ + 2 j ^ − k ^ ) ( i ^ + 2 j ^ − k ^ ) ( hat(i)+2 hat(j)- hat(k))(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})(i^+2j^−k^) at the point ( 1 , − 1 , 2 ) ( 1 , − 1 , 2 ) (1,-1,2)(1,-1,2)(1,−1,2).

Answer:

To determine the directional derivative of the scalar field ϕ = ln ⁡ ( x 2 + y 2 + z 2 ) ϕ = ln ⁡ ( x 2 + y 2 + z 2 ) phi=ln(x^(2)+y^(2)+z^(2))\phi = \ln(x^2 + y^2 + z^2)ϕ=ln⁡(x2+y2+z2) at the point ( 1 , − 1 , 2 ) ( 1 , − 1 , 2 ) (1,-1,2)(1, -1, 2)(1,−1,2) in the direction of the vector i ^ + 2 j ^ − k ^ i ^ + 2 j ^ − k ^ hat(i)+2 hat(j)- hat(k)\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} – \hat{\mathbf{k}}i^+2j^−k^, follow these steps:

1. Compute the Gradient of ϕ ϕ phi\phiϕ

The gradient of a scalar field ϕ ϕ phi\phiϕ, denoted by ∇ ϕ ∇ ϕ grad phi\nabla \phi∇ϕ, is the vector of partial derivatives of ϕ ϕ phi\phiϕ with respect to x x xxx, y y yyy, and z z zzz:
∇ ϕ = ( ∂ ϕ ∂ x , ∂ ϕ ∂ y , ∂ ϕ ∂ z ) ∇ ϕ = ∂ ϕ ∂ x , ∂ ϕ ∂ y , ∂ ϕ ∂ z grad phi=((del phi)/(del x),(del phi)/(del y),(del phi)/(del z))\nabla \phi = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right)∇ϕ=(∂ϕ∂x,∂ϕ∂y,∂ϕ∂z)
The scalar field is ϕ ( x , y , z ) = ln ⁡ ( x 2 + y 2 + z 2 ) ϕ ( x , y , z ) = ln ⁡ ( x 2 + y 2 + z 2 ) phi(x,y,z)=ln(x^(2)+y^(2)+z^(2))\phi(x, y, z) = \ln(x^2 + y^2 + z^2)ϕ(x,y,z)=ln⁡(x2+y2+z2). To find the partial derivatives, apply the chain rule.
∂ ϕ ∂ x = 1 x 2 + y 2 + z 2 ⋅ 2 x = 2 x x 2 + y 2 + z 2 ∂ ϕ ∂ x = 1 x 2 + y 2 + z 2 ⋅ 2 x = 2 x x 2 + y 2 + z 2 (del phi)/(del x)=(1)/(x^(2)+y^(2)+z^(2))*2x=(2x)/(x^(2)+y^(2)+z^(2))\frac{\partial \phi}{\partial x} = \frac{1}{x^2 + y^2 + z^2} \cdot 2x = \frac{2x}{x^2 + y^2 + z^2}∂ϕ∂x=1x2+y2+z2⋅2x=2xx2+y2+z2
∂ ϕ ∂ y = 1 x 2 + y 2 + z 2 ⋅ 2 y = 2 y x 2 + y 2 + z 2 ∂ ϕ ∂ y = 1 x 2 + y 2 + z 2 ⋅ 2 y = 2 y x 2 + y 2 + z 2 (del phi)/(del y)=(1)/(x^(2)+y^(2)+z^(2))*2y=(2y)/(x^(2)+y^(2)+z^(2))\frac{\partial \phi}{\partial y} = \frac{1}{x^2 + y^2 + z^2} \cdot 2y = \frac{2y}{x^2 + y^2 + z^2}∂ϕ∂y=1x2+y2+z2⋅2y=2yx2+y2+z2
∂ ϕ ∂ z = 1 x 2 + y 2 + z 2 ⋅ 2 z = 2 z x 2 + y 2 + z 2 ∂ ϕ ∂ z = 1 x 2 + y 2 + z 2 ⋅ 2 z = 2 z x 2 + y 2 + z 2 (del phi)/(del z)=(1)/(x^(2)+y^(2)+z^(2))*2z=(2z)/(x^(2)+y^(2)+z^(2))\frac{\partial \phi}{\partial z} = \frac{1}{x^2 + y^2 + z^2} \cdot 2z = \frac{2z}{x^2 + y^2 + z^2}∂ϕ∂z=1x2+y2+z2⋅2z=2zx2+y2+z2
So, the gradient of ϕ ϕ phi\phiϕ is:
∇ ϕ = ( 2 x x 2 + y 2 + z 2 , 2 y x 2 + y 2 + z 2 , 2 z x 2 + y 2 + z 2 ) ∇ ϕ = 2 x x 2 + y 2 + z 2 , 2 y x 2 + y 2 + z 2 , 2 z x 2 + y 2 + z 2 grad phi=((2x)/(x^(2)+y^(2)+z^(2)),(2y)/(x^(2)+y^(2)+z^(2)),(2z)/(x^(2)+y^(2)+z^(2)))\nabla \phi = \left( \frac{2x}{x^2 + y^2 + z^2}, \frac{2y}{x^2 + y^2 + z^2}, \frac{2z}{x^2 + y^2 + z^2} \right)∇ϕ=(2xx2+y2+z2,2yx2+y2+z2,2zx2+y2+z2)

2. Evaluate the Gradient at the Point ( 1 , − 1 , 2 ) ( 1 , − 1 , 2 ) (1,-1,2)(1, -1, 2)(1,−1,2)

Substitute x = 1 x = 1 x=1x = 1x=1, y = − 1 y = − 1 y=-1y = -1y=−1, and z = 2 z = 2 z=2z = 2z=2 into the gradient:
∇ ϕ ( 1 , − 1 , 2 ) = ( 2 ( 1 ) 1 2 + ( − 1 ) 2 + 2 2 , 2 ( − 1 ) 1 2 + ( − 1 ) 2 + 2 2 , 2 ( 2 ) 1 2 + ( − 1 ) 2 + 2 2 ) ∇ ϕ ( 1 , − 1 , 2 ) = 2 ( 1 ) 1 2 + ( − 1 ) 2 + 2 2 , 2 ( − 1 ) 1 2 + ( − 1 ) 2 + 2 2 , 2 ( 2 ) 1 2 + ( − 1 ) 2 + 2 2 grad phi(1,-1,2)=((2(1))/(1^(2)+(-1)^(2)+2^(2)),(2(-1))/(1^(2)+(-1)^(2)+2^(2)),(2(2))/(1^(2)+(-1)^(2)+2^(2)))\nabla \phi(1, -1, 2) = \left( \frac{2(1)}{1^2 + (-1)^2 + 2^2}, \frac{2(-1)}{1^2 + (-1)^2 + 2^2}, \frac{2(2)}{1^2 + (-1)^2 + 2^2} \right)∇ϕ(1,−1,2)=(2(1)12+(−1)2+22,2(−1)12+(−1)2+22,2(2)12+(−1)2+22)
First, compute the denominator:
x 2 + y 2 + z 2 = 1 2 + ( − 1 ) 2 + 2 2 = 1 + 1 + 4 = 6 x 2 + y 2 + z 2 = 1 2 + ( − 1 ) 2 + 2 2 = 1 + 1 + 4 = 6 x^(2)+y^(2)+z^(2)=1^(2)+(-1)^(2)+2^(2)=1+1+4=6x^2 + y^2 + z^2 = 1^2 + (-1)^2 + 2^2 = 1 + 1 + 4 = 6x2+y2+z2=12+(−1)2+22=1+1+4=6
Now, compute the gradient components:
∇ ϕ ( 1 , − 1 , 2 ) = ( 2 6 , − 2 6 , 4 6 ) = ( 1 3 , − 1 3 , 2 3 ) ∇ ϕ ( 1 , − 1 , 2 ) = 2 6 , − 2 6 , 4 6 = 1 3 , − 1 3 , 2 3 grad phi(1,-1,2)=((2)/(6),(-2)/(6),(4)/(6))=((1)/(3),(-1)/(3),(2)/(3))\nabla \phi(1, -1, 2) = \left( \frac{2}{6}, \frac{-2}{6}, \frac{4}{6} \right) = \left( \frac{1}{3}, \frac{-1}{3}, \frac{2}{3} \right)∇ϕ(1,−1,2)=(26,−26,46)=(13,−13,23)

3. Normalize the Direction Vector

The given direction vector is v = i ^ + 2 j ^ − k ^ = ( 1 , 2 , − 1 ) v = i ^ + 2 j ^ − k ^ = ( 1 , 2 , − 1 ) v= hat(i)+2 hat(j)- hat(k)=(1,2,-1)\mathbf{v} = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} – \hat{\mathbf{k}} = (1, 2, -1)v=i^+2j^−k^=(1,2,−1). To normalize this vector, compute its magnitude:
| v | = 1 2 + 2 2 + ( − 1 ) 2 = 1 + 4 + 1 = 6 | v | = 1 2 + 2 2 + ( − 1 ) 2 = 1 + 4 + 1 = 6 |v|=sqrt(1^(2)+2^(2)+(-1)^(2))=sqrt(1+4+1)=sqrt6|\mathbf{v}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}|v|=12+22+(−1)2=1+4+1=6
Now, normalize the vector:
v ^ = 1 | v | v = 1 6 ( 1 , 2 , − 1 ) v ^ = 1 | v | v = 1 6 ( 1 , 2 , − 1 ) hat(v)=(1)/(|v|)v=(1)/(sqrt6)(1,2,-1)\hat{\mathbf{v}} = \frac{1}{|\mathbf{v}|} \mathbf{v} = \frac{1}{\sqrt{6}} (1, 2, -1)v^=1|v|v=16(1,2,−1)

4. Compute the Directional Derivative

The directional derivative of ϕ ϕ phi\phiϕ in the direction of the unit vector v ^ v ^ hat(v)\hat{\mathbf{v}}v^ is given by the dot product of the gradient of ϕ ϕ phi\phiϕ and the unit vector v ^ v ^ hat(v)\hat{\mathbf{v}}v^:
D v ^ ϕ = ∇ ϕ ⋅ v ^ D v ^ ϕ = ∇ ϕ ⋅ v ^ D_( hat(v))phi=grad phi* hat(v)D_{\hat{\mathbf{v}}} \phi = \nabla \phi \cdot \hat{\mathbf{v}}Dv^ϕ=∇ϕ⋅v^
Substitute the values of ∇ ϕ ( 1 , − 1 , 2 ) ∇ ϕ ( 1 , − 1 , 2 ) grad phi(1,-1,2)\nabla \phi(1, -1, 2)∇ϕ(1,−1,2) and v ^ v ^ hat(v)\hat{\mathbf{v}}v^:
D v ^ ϕ = ( 1 3 , − 1 3 , 2 3 ) ⋅ 1 6 ( 1 , 2 , − 1 ) D v ^ ϕ = 1 3 , − 1 3 , 2 3 ⋅ 1 6 ( 1 , 2 , − 1 ) D_( hat(v))phi=((1)/(3),(-1)/(3),(2)/(3))*(1)/(sqrt6)(1,2,-1)D_{\hat{\mathbf{v}}} \phi = \left( \frac{1}{3}, \frac{-1}{3}, \frac{2}{3} \right) \cdot \frac{1}{\sqrt{6}} (1, 2, -1)Dv^ϕ=(13,−13,23)⋅16(1,2,−1)
Compute the dot product:
D v ^ ϕ = 1 6 ( 1 3 × 1 + − 1 3 × 2 + 2 3 × ( − 1 ) ) D v ^ ϕ = 1 6 1 3 × 1 + − 1 3 × 2 + 2 3 × ( − 1 ) D_( hat(v))phi=(1)/(sqrt6)((1)/(3)xx1+(-1)/(3)xx2+(2)/(3)xx(-1))D_{\hat{\mathbf{v}}} \phi = \frac{1}{\sqrt{6}} \left( \frac{1}{3} \times 1 + \frac{-1}{3} \times 2 + \frac{2}{3} \times (-1) \right)Dv^ϕ=16(13×1+−13×2+23×(−1))
D v ^ ϕ = 1 6 ( 1 3 − 2 3 − 2 3 ) D v ^ ϕ = 1 6 1 3 − 2 3 − 2 3 D_( hat(v))phi=(1)/(sqrt6)((1)/(3)-(2)/(3)-(2)/(3))D_{\hat{\mathbf{v}}} \phi = \frac{1}{\sqrt{6}} \left( \frac{1}{3} – \frac{2}{3} – \frac{2}{3} \right)Dv^ϕ=16(13−23−23)
D v ^ ϕ = 1 6 ( − 3 3 ) = − 1 6 D v ^ ϕ = 1 6 − 3 3 = − 1 6 D_( hat(v))phi=(1)/(sqrt6)(-(3)/(3))=(-1)/(sqrt6)D_{\hat{\mathbf{v}}} \phi = \frac{1}{\sqrt{6}} \left( -\frac{3}{3} \right) = \frac{-1}{\sqrt{6}}Dv^ϕ=16(−33)=−16
Thus, the directional derivative of ϕ ϕ phi\phiϕ at the point ( 1 , − 1 , 2 ) ( 1 , − 1 , 2 ) (1,-1,2)(1, -1, 2)(1,−1,2) in the direction of the vector ( i ^ + 2 j ^ − k ^ ) ( i ^ + 2 j ^ − k ^ ) ( hat(i)+2 hat(j)- hat(k))(\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} – \hat{\mathbf{k}})(i^+2j^−k^) is:
D v ^ ϕ = − 1 6 D v ^ ϕ = − 1 6 D_( hat(v))phi=(-1)/(sqrt6)D_{\hat{\mathbf{v}}} \phi = \frac{-1}{\sqrt{6}}Dv^ϕ=−16
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