Question

Consider the 3–dimensional vector field F defined by F (x, y,z) = 12x²y² +2z² +1, 8x³y −3z, 4xz −3y −3.

(a) Write down the Jacobian matrix J_F (x, y,z).

(b) Determine the divergence div F (x, y,z). (c) Determine curl F (x, y,z). (d) Give reasons why F has a potential function.

(e) Find a potential function of F.

Answer 1

Ans:-

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

or $\overrightarrow F=$ $(12x ^2 y^ 2+2z^ 2+1)\hat i+(8x^ 3 y−3z)\hat j+(4xz−3y−3)\hat k$

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

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

$(ii)$ $divF(x,y,z)$

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

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

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

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

$(iv)$

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 integrating the given vector field.

$(v)$ As we know -

Potential function is given by-

$F _v =−∫F.dr$

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

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

$=8x ^3y^ 2+2x^ 2z+2xz^ 2−3xy−3yz−2x$