Answer to Question #225215 in Calculus for Anuj

Question #225215

Question 2


(a) Given that A ~ = (x + 2y + 4z)ˆ


i + (2x - 3y- z)ˆj +(4x-y + 2z)kˆ


(i). Show that the vector field A ~ is irrotational




(ii). Find the scalar potential φ such that A~ = ∇φ, if φ(0, 0, 0) = 1


1
Expert's answer
2021-08-16T11:20:05-0400

a)

The vector field A is given as

A=(x+2y+4z)i^+(2x3yz)j^+(4xy+2z)k^A=(x+2y+4z)\hat{i}+(2x−3y−z)\hat{j}+(4x−y+2z)\hat{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=ijkδδxδδyδδz(x+2y+4z)(2x3yz)(4xy+2z)\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=[δδy(4xy+2z)δδz(2x3yz)]\nabla\times A=[ \frac{\delta}{\delta y}(4x-y+2z)-\frac{\delta}{\delta z}(2x-3y-z)]


j^[δδx(4xy+2z)δδz(x+2y+4z)]-\hat{j}[\frac{\delta}{\delta x}(4x-y+2z)-\frac{\delta}{\delta z}(x+2y+4z)]


+k^[δδx(2x3yz)δδy(x+2y+4z)]+\hat{k}[\frac{\delta}{\delta x}(2x-3y-z)-\frac{\delta}{\delta y}(x+2y+4z)]


×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

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=ϕxi^+ϕyj^+ϕzk^∇ϕ=\frac{∂ϕ}{∂x}\hat{i}+\frac{∂ϕ}{∂y}\hat{j}+\frac{∂ϕ}{∂z}\hat{k}

And the vector field is,

A=(x+2y+4z)i^+(2x3yz)j^+(4xy+2z)k^A=(x+2y+4z)\hat{i}+(2x−3y−z)\hat{j}+(4x−y+2z)\hat{k}

As A=∇ϕ

Comparing both the above equations

ϕx=x+2y+4zϕy=2x3yzϕz=4xy+2z\frac{∂ϕ}{∂x}=x+2y+4z\\\frac{∂ϕ}{∂y}=2x−3y−z\\\frac{∂ϕ}{∂z}=4x−y+2z

The value of ϕ can be found by integrating the above equations

ϕx=x+2y+4zϕ=(x+2y+4z)dxϕ=dϕϕ=(x+2y+4z)dxϕ=xdx+2ydx+4zdxϕ=x22+2yx+4zx\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

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=x22+2yx+4zx+c(y,z)ϕ=\frac{x^2}{2}+2yx+4zx+c(y,z)

On differentiating this equation with respect to y,

ϕy=y[x22+2yx+4zx+c(y,z)]ϕy=0+2x+0+cy(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)

And from the previous equation,

ϕy=2x3yz\frac{∂ϕ}{∂y}=2x−3y−z

Comparing the two equations

cy(y,z)=3yzcy(y,z)=−3y−z

Since, cy(y,z)=c(y,z)yc(y,z)=cydyc(y,z)=3ydyzdyc(y,z)=3y22zy+cSince, \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

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=x22+2yx+4zx3y22yz+c(z)ϕ=\frac{x^2}{2}+2yx+4zx−\frac{3y^2}{2}−yz+c(z)

On differentiating this equation with respect to z,

ϕz=0+0+4x0y+c(z)ϕz=4xy+c(z)\frac{∂ϕ}{∂z}=0+0+4x−0−y+c'(z)\\\frac{∂ϕ}{∂z}=4x−y+c'(z)

Comparing with ϕz=4xy+2z\frac{∂ϕ}{∂z}=4x−y+2z obtained before

c(z)=2zc(z)=2zdzc(z)=2z22+cc(z)=z2+cc'(z)=2z\\c(z)=∫2z dz\\c(z)=2\frac{z^2}{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=x22+2yx+4zx3y22yz+z2+cϕ=\frac{x^2}{2}+2yx+4zx−\frac{3y^2}2−yz+z^2+c

The final constant of integration can be found using boundary conditions,

ϕ at (0,0,0)=1ϕ \space at\space (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=x22+2yx+4zx3y22yz+z2+1ϕ=\frac{x^2}2+2yx+4zx−\frac{3y^2}{2}−yz+z^2+1


This is the required scalar potential


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