Answer to Question #225215 in Calculus for Anuj

Question

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

Answer 1

a)

The vector field A is given as

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

A vector field is said to be irrotational if the curl of this vector field is zero everywhere.

Thus, to be irrotational

∇×A=0

$\nabla\times A=\begin{vmatrix} i & j& k \\ \frac{\delta}{\delta x} & \frac{\delta}{\delta y} & \frac{\delta}{\delta z}\\(x+2y+4z) & (2x-3y-z) & (4x-y+2z) \end{vmatrix}$

$\nabla\times A=[ \frac{\delta}{\delta y}(4x-y+2z)-\frac{\delta}{\delta z}(2x-3y-z)]$

$-\hat{j}[\frac{\delta}{\delta x}(4x-y+2z)-\frac{\delta}{\delta z}(x+2y+4z)]$

$+\hat{k}[\frac{\delta}{\delta x}(2x-3y-z)-\frac{\delta}{\delta y}(x+2y+4z)]$

$\nabla\times A=\hat{i}[(-1)-(-1)]-\hat{j}[(4)-(4)]+\hat{k}[(2)-(2)]\\\nabla\times A=\hat{i}(0)-\hat{j}(0)+\hat{k}(0)\\\nabla\times A=0$

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

$∇ϕ=\frac{∂ϕ}{∂x}\hat{i}+\frac{∂ϕ}{∂y}\hat{j}+\frac{∂ϕ}{∂z}\hat{k}$

And the vector field is,

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

As A=∇ϕ

Comparing both the above equations

$\frac{∂ϕ}{∂x}=x+2y+4z\\\frac{∂ϕ}{∂y}=2x−3y−z\\\frac{∂ϕ}{∂z}=4x−y+2z$

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

$\frac{∂ϕ}{∂x}=x+2y+4z\\∂ϕ=(x+2y+4z)dx\\ϕ=∫dϕ\\ϕ=∫(x+2y+4z)dx\\ϕ=∫x dx+∫2y dx+∫4z dx\\ϕ=\frac{x^2}{2}+2yx+4zx$

Since this was a partial differential with respect to x, the constant c thus

should be a function of y and/or z

Thus,

$ϕ=\frac{x^2}{2}+2yx+4zx+c(y,z)$

On differentiating this equation with respect to y,

$\frac{∂ϕ}{∂y}=\frac{∂}{∂}y[\frac{x^2}{2}+2yx+4zx+c(y,z)]\\\frac{∂ϕ}{∂y}=0+2x+0+cy(y,z)$

And from the previous equation,

$\frac{∂ϕ}{∂y}=2x−3y−z$

Comparing the two equations

$cy(y,z)=−3y−z$

$Since, \space\\cy(y,z)=\frac{∂c(y,z)}{∂y}\\c(y,z)=∫cyd_y\\c(y,z)=∫−3y dy−∫z dy\\c(y,z)=−3\frac{y^2}{2}−zy+c$

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,

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

On differentiating this equation with respect to z,

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

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

$c'(z)=2z\\c(z)=∫2z dz\\c(z)=2\frac{z^2}{2}+c\\c(z)=z^2+c$

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

$ϕ=\frac{x^2}{2}+2yx+4zx−\frac{3y^2}2−yz+z^2+c$

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

$ϕ \space at\space (0,0,0)=1$

Thus, substitute x=0, y=0 and z=0 in the scalar potential equation and we

 get

c=1

Thus

$ϕ=\frac{x^2}2+2yx+4zx−\frac{3y^2}{2}−yz+z^2+1$

This is the required scalar potential