Answer to Question #184483 in Calculus for Njabulo

Question

11. (Section 7.9 and Chapter 8) Consider the 3–dimensional vector field F defined by F (x, y, z) = ￾ 12x 2 y 2 + 2z 2 + 1, 8x 3 y − 3z, 4xz − 3y − 3
 (a) Write down the Jacobian matrix JF (x, y, z). (2) (b) Determine the divergence div F(x, y, z). (2) (c) Determine curl F(x, y, z). (2) (d) Give reasons why F has a potential function. (Refer to the relevant definitions and theorems in the study guide.) (2) (e) Find a potential function of F, using the method of Example 7.9.1. Note, however that that example concerns a 2-dimensional vector field, so you will have to adapt the method to be suitable for a 3-dimensional vector field. Pay special attention to the notation that you use for derivatives of functions of more than one variable.

Answer 1

Given field,

$F(x,y,z)=(12x^2y^2+2z^2+1,8x^3y-3z,4xz-3y-3)$

or $\vec{F}=(12x^2y^2+2z^2+1)\hat{i}+(8x^3y-3z)\hat{j}+(4xz-3y-3)\hat{k}$

(1) Jacobian Matrix $J(F(x,y,z)=\begin{bmatrix} \dfrac{df_x}{dx} & \dfrac{df_x}{dy}& \dfrac{df_x}{dz} \\\\ \dfrac{df_y}{dx}& \dfrac{df_y}{dy}& \dfrac{df_y}{dz}\\\\ \dfrac{df_z}{dx} & \dfrac{df_z}{dy}& \dfrac{df_z}{dz} \end{bmatrix}$

=$\begin{bmatrix} 24xy^2& 24x^2y& 4z \\\\ 24xy^2 & 8x^3& -3\\\\ 4z& -3&0 \end{bmatrix}$

(2) $div F(x, y, z)$

$= (\dfrac{d}{dx}\hat{i}+\dfrac{d}{dy}\hat{j}+\dfrac{d}{dz}\hat{k}).((12x^2y^2+2z^2+1)\hat{i}+(8x^3y-3z)\hat{j}+(4xz-3y-3)\hat{k})$

$=24xy^2+8x^3+4x$

(3) $curl F(x,y,z)=\begin{bmatrix} \hat{i} & \hat{j} &\hat{k}\\\\ \dfrac{d}{dx} &\dfrac{d}{dy}&\dfrac{d}{dz} \\\\ 12x^2y^2+2z^2+1& 8x^3y-3z& 4xz-3y-3 \end{bmatrix}$

$=\hat{i}(-3-(-3))-\hat{j}(4z-4z)+\hat{k}(24x^2y-24x^2y)\\ =0$

(4) F is a potential function because, Whenever There exist a field, There exist a potential associated

with it. We can calculate the potential function F by the integraing the given vector field.

(5) As we know-

Potential function is given by-

$F_v=-\int F.dr$

$=\int ((12x^2y^2+2z^2+1)\hat{i}+(8x^3y-3z)\hat{j}+(4xz-3y-3)\hat{k})(dx\hat{i}+dy\hat{j}+dz\hat{z})$

$=(4x^3y+2z^2x+x+4x^3y^2-3yz+2x^2z-3xy-3x)$

$=8x^3y^2+2x^2z+2xz^2-3xy-3yz-2x$