a)
The vector field A is given as
A = ( x + 2 y + 4 z ) i ^ + ( 2 x − 3 y − z ) j ^ + ( 4 x − y + 2 z ) k ^ A=(x+2y+4z)\hat{i}+(2x−3y−z)\hat{j}+(4x−y+2z)\hat{k} A = ( x + 2 y + 4 z ) i ^ + ( 2 x − 3 y − z ) j ^ + ( 4 x − y + 2 z ) k ^
A vector field is said to be irrotational if the curl of this vector field is zero everywhere.
Thus, to be irrotational
∇×A=0
∇ × A = ∣ i j k δ δ x δ δ y δ δ z ( x + 2 y + 4 z ) ( 2 x − 3 y − z ) ( 4 x − y + 2 z ) ∣ \nabla\times A=\begin{vmatrix}
i & j& k \\
\frac{\delta}{\delta x} &
\frac{\delta}{\delta y} & \frac{\delta}{\delta z}\\(x+2y+4z) & (2x-3y-z) & (4x-y+2z)
\end{vmatrix} ∇ × A = ∣ ∣ i δ x δ ( x + 2 y + 4 z ) j δy δ ( 2 x − 3 y − z ) k δz δ ( 4 x − y + 2 z ) ∣ ∣
∇ × A = [ δ δ y ( 4 x − y + 2 z ) − δ δ z ( 2 x − 3 y − z ) ] \nabla\times A=[
\frac{\delta}{\delta y}(4x-y+2z)-\frac{\delta}{\delta z}(2x-3y-z)] ∇ × A = [ δy δ ( 4 x − y + 2 z ) − δz δ ( 2 x − 3 y − z )]
− j ^ [ δ δ x ( 4 x − y + 2 z ) − δ δ z ( x + 2 y + 4 z ) ] -\hat{j}[\frac{\delta}{\delta x}(4x-y+2z)-\frac{\delta}{\delta z}(x+2y+4z)] − j ^ [ δ x δ ( 4 x − y + 2 z ) − δz δ ( x + 2 y + 4 z )]
+ k ^ [ δ δ x ( 2 x − 3 y − z ) − δ δ y ( x + 2 y + 4 z ) ] +\hat{k}[\frac{\delta}{\delta x}(2x-3y-z)-\frac{\delta}{\delta y}(x+2y+4z)] + k ^ [ δ x δ ( 2 x − 3 y − z ) − δy δ ( x + 2 y + 4 z )]
∇ × A = i ^ [ ( − 1 ) − ( − 1 ) ] − j ^ [ ( 4 ) − ( 4 ) ] + k ^ [ ( 2 ) − ( 2 ) ] ∇ × A = i ^ ( 0 ) − j ^ ( 0 ) + k ^ ( 0 ) ∇ × A = 0 \nabla\times A=\hat{i}[(-1)-(-1)]-\hat{j}[(4)-(4)]+\hat{k}[(2)-(2)]\\\nabla\times A=\hat{i}(0)-\hat{j}(0)+\hat{k}(0)\\\nabla\times A=0 ∇ × A = i ^ [( − 1 ) − ( − 1 )] − j ^ [( 4 ) − ( 4 )] + k ^ [( 2 ) − ( 2 )] ∇ × A = i ^ ( 0 ) − j ^ ( 0 ) + k ^ ( 0 ) ∇ × A = 0
Thus, the given vector field is irrotational
b)
The vector field A is given in terms of a scalar potential as
A=∇ϕ
The gradient of the scalar potential ϕ is
∇ ϕ = ∂ ϕ ∂ x i ^ + ∂ ϕ ∂ y j ^ + ∂ ϕ ∂ z k ^ ∇ϕ=\frac{∂ϕ}{∂x}\hat{i}+\frac{∂ϕ}{∂y}\hat{j}+\frac{∂ϕ}{∂z}\hat{k} ∇ ϕ = ∂ x ∂ ϕ i ^ + ∂ y ∂ ϕ j ^ + ∂ z ∂ ϕ k ^
And the vector field is,
A = ( x + 2 y + 4 z ) i ^ + ( 2 x − 3 y − z ) j ^ + ( 4 x − y + 2 z ) k ^ A=(x+2y+4z)\hat{i}+(2x−3y−z)\hat{j}+(4x−y+2z)\hat{k} A = ( x + 2 y + 4 z ) i ^ + ( 2 x − 3 y − z ) j ^ + ( 4 x − y + 2 z ) k ^
As A=∇ϕ
Comparing both the above equations
∂ ϕ ∂ x = x + 2 y + 4 z ∂ ϕ ∂ y = 2 x − 3 y − z ∂ ϕ ∂ z = 4 x − y + 2 z \frac{∂ϕ}{∂x}=x+2y+4z\\\frac{∂ϕ}{∂y}=2x−3y−z\\\frac{∂ϕ}{∂z}=4x−y+2z ∂ x ∂ ϕ = x + 2 y + 4 z ∂ y ∂ ϕ = 2 x − 3 y − z ∂ z ∂ ϕ = 4 x − y + 2 z
The value of ϕ can be found by integrating the above equations
∂ ϕ ∂ x = x + 2 y + 4 z ∂ ϕ = ( x + 2 y + 4 z ) d x ϕ = ∫ d ϕ ϕ = ∫ ( x + 2 y + 4 z ) d x ϕ = ∫ x d x + ∫ 2 y d x + ∫ 4 z d x ϕ = x 2 2 + 2 y x + 4 z x \frac{∂ϕ}{∂x}=x+2y+4z\\∂ϕ=(x+2y+4z)dx\\ϕ=∫dϕ\\ϕ=∫(x+2y+4z)dx\\ϕ=∫x dx+∫2y dx+∫4z dx\\ϕ=\frac{x^2}{2}+2yx+4zx ∂ x ∂ ϕ = x + 2 y + 4 z ∂ ϕ = ( x + 2 y + 4 z ) d x ϕ = ∫ d ϕ ϕ = ∫ ( x + 2 y + 4 z ) d x ϕ = ∫ x d x + ∫ 2 y d x + ∫ 4 z d x ϕ = 2 x 2 + 2 y x + 4 z x
Since this was a partial differential with respect to x, the constant c thus
should be a function of y and/or z
Thus,
ϕ = x 2 2 + 2 y x + 4 z x + c ( y , z ) ϕ=\frac{x^2}{2}+2yx+4zx+c(y,z) ϕ = 2 x 2 + 2 y x + 4 z x + c ( y , z )
On differentiating this equation with respect to y,
∂ ϕ ∂ y = ∂ ∂ y [ x 2 2 + 2 y x + 4 z x + c ( y , z ) ] ∂ ϕ ∂ y = 0 + 2 x + 0 + c y ( y , z ) \frac{∂ϕ}{∂y}=\frac{∂}{∂}y[\frac{x^2}{2}+2yx+4zx+c(y,z)]\\\frac{∂ϕ}{∂y}=0+2x+0+cy(y,z) ∂ y ∂ ϕ = ∂ ∂ y [ 2 x 2 + 2 y x + 4 z x + c ( y , z )] ∂ y ∂ ϕ = 0 + 2 x + 0 + cy ( y , z )
And from the previous equation,
∂ ϕ ∂ y = 2 x − 3 y − z \frac{∂ϕ}{∂y}=2x−3y−z ∂ y ∂ ϕ = 2 x − 3 y − z
Comparing the two equations
c y ( y , z ) = − 3 y − z cy(y,z)=−3y−z cy ( y , z ) = − 3 y − z
S i n c e , c y ( y , z ) = ∂ c ( y , z ) ∂ y c ( y , z ) = ∫ c y d y c ( y , z ) = ∫ − 3 y d y − ∫ z d y c ( y , z ) = − 3 y 2 2 − z y + c Since, \space\\cy(y,z)=\frac{∂c(y,z)}{∂y}\\c(y,z)=∫cyd_y\\c(y,z)=∫−3y dy−∫z dy\\c(y,z)=−3\frac{y^2}{2}−zy+c S in ce , cy ( y , z ) = ∂ y ∂ c ( y , z ) c ( y , z ) = ∫ cy d y c ( y , z ) = ∫ − 3 y d y − ∫ z d y c ( y , z ) = − 3 2 y 2 − zy + c
Here, c is another constant of integration, which is now a function of z
Substituting this constant value in the equation for the scalar potential,
ϕ = x 2 2 + 2 y x + 4 z x − 3 y 2 2 − y z + c ( z ) ϕ=\frac{x^2}{2}+2yx+4zx−\frac{3y^2}{2}−yz+c(z) ϕ = 2 x 2 + 2 y x + 4 z x − 2 3 y 2 − yz + c ( z )
On differentiating this equation with respect to z,
∂ ϕ ∂ z = 0 + 0 + 4 x − 0 − y + c ′ ( z ) ∂ ϕ ∂ z = 4 x − y + c ′ ( z ) \frac{∂ϕ}{∂z}=0+0+4x−0−y+c'(z)\\\frac{∂ϕ}{∂z}=4x−y+c'(z) ∂ z ∂ ϕ = 0 + 0 + 4 x − 0 − y + c ′ ( z ) ∂ z ∂ ϕ = 4 x − y + c ′ ( z )
Comparing with ∂ ϕ ∂ z = 4 x − y + 2 z \frac{∂ϕ}{∂z}=4x−y+2z ∂ z ∂ ϕ = 4 x − y + 2 z obtained before
c ′ ( z ) = 2 z c ( z ) = ∫ 2 z d z c ( z ) = 2 z 2 2 + c c ( z ) = z 2 + c c'(z)=2z\\c(z)=∫2z dz\\c(z)=2\frac{z^2}{2}+c\\c(z)=z^2+c c ′ ( z ) = 2 z c ( z ) = ∫ 2 z d z c ( z ) = 2 2 z 2 + c c ( z ) = z 2 + c
c here is the final constant of integration
∂ϕ/∂z=0+0+4x-0-y+c'(z),∂ϕ/∂z=4x-y+c'(z)Comparing with ∂ϕ/∂z=4x-y+2z obtained before,c'(z)=2z, c(z)=∫2z dz, c(z)=2z2/2+c, c(z)=z2+c is the final constant of integration.
Thus, the scalar potential becomes
ϕ = x 2 2 + 2 y x + 4 z x − 3 y 2 2 − y z + z 2 + c ϕ=\frac{x^2}{2}+2yx+4zx−\frac{3y^2}2−yz+z^2+c ϕ = 2 x 2 + 2 y x + 4 z x − 2 3 y 2 − yz + z 2 + c
The final constant of integration can be found using boundary conditions,
ϕ a t ( 0 , 0 , 0 ) = 1 ϕ \space at\space (0,0,0)=1 ϕ a t ( 0 , 0 , 0 ) = 1
Thus, substitute x=0, y=0 and z=0 in the scalar potential equation and we
get
c=1
Thus
ϕ = x 2 2 + 2 y x + 4 z x − 3 y 2 2 − y z + z 2 + 1 ϕ=\frac{x^2}2+2yx+4zx−\frac{3y^2}{2}−yz+z^2+1 ϕ = 2 x 2 + 2 y x + 4 z x − 2 3 y 2 − yz + z 2 + 1
This is the required scalar potential
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