Question

Consider the 3-dimensional vector field F defined by F(x,y,z)=(2xyz,x²z+2yz²,x²y+2y²z+e^z).

1.write down the Jacobian matrix jf(x,y,z).

2.determine divF (x,y,z).

3.determine curl F (x,y,z).

4.does F have a potential function? Give reasons for your answer, referring to the relevant definitions and theorems in the study guide.

5.find a potential function of F .

Answer 1

1.

$$F(x,y,z)=(2xyz,x^2z+2yz^2,x^2y+2y^2z+e^z)$$

$$\dfrac{\partial F_x}{\partial x}=2yz, \dfrac{\partial F_y}{\partial x}=2xz, \dfrac{\partial F_z}{\partial x}=2xy$$

$$\dfrac{\partial F_x}{\partial y}=2xz, \dfrac{\partial F_y}{\partial y}=2z^2, \dfrac{\partial F_z}{\partial y}=x^2+4yz$$

$$\dfrac{\partial F_x}{\partial z}=2xy, \dfrac{\partial F_y}{\partial z}=x^2+4yz, \dfrac{\partial F_z}{\partial z}=2y^2+e^z$$

Jacobian Matrix

$$J(F(x,y,z))=\begin{vmatrix} 2yz & 2xz & 2xy \\ 2xz & 2z^2 & x^2+4yz \\ 2xy & x^2+4yz & 2y^2+e^z \\ \end{vmatrix}$$

$$=2yz\begin{vmatrix} 2z^2 & x^2+4yz \\ x^2+4yz & 2y^2+e^z \end{vmatrix}$$

$$-2xz\begin{vmatrix} 2xz & x^2+4yz \\ 2xy & 2y^2+e^z \end{vmatrix}$$

$$+2xy\begin{vmatrix} 2xz & 2z^2 \\ 2xy & x^2+4yz \end{vmatrix}$$

$$=8y^3z^3+4yz^3e^z-2x^4yz-16x^2y^2z^2-32y^3z^3$$

$$-8x^2y^2z^2-4x^2z^2e^z+4x^4yz+16x^2y^2z^2$$

$$+4x^4yz+16x^2y^2z^2-8x^2y^2z^2$$

$$=6x^4yz-24y^3z^3-4x^2z^2e^z+4yz^3e^z$$

2.

$$\nabla=\langle\dfrac{\partial }{\partial x},\dfrac{\partial }{\partial y},\dfrac{\partial }{\partial z}\rangle$$

$$divF=\nabla F=2yz+2z^2+2y^2+e^z$$

3.

$$curl F(x,y,z)=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \dfrac{\partial }{\partial x} & \dfrac{\partial }{\partial y} & \dfrac{\partial }{\partial z} \\ 2xyz & x^2z+2yz^2 & x^2y+2y^2z+e^z \end{vmatrix}$$

$$=\vec{i}(x^2+4yz-x^2-4yz)$$

$$-\vec{j}(2xy-2xy)$$

$$+\vec{k}(2xz-2xz)=\vec{0}$$

4. Let $F(x,y,z)$  be a vector field in space on a simply connected domain. If  $curl F(x,y,z)=0,$ then $F$ is conservative.

There exists a function $f$ such that $F=\nabla f.$In this situation $f$ is called

a potential function for $F.$

5.

$$f_x=2xyz=>f=x^2yz+g(y,z)$$

$$f_y=x^2z+g_y=x^2z+2yz^2$$

$$g_y=2yz^2$$

$$g=y^2z^2+h(z)$$

$$f=x^2yz+y^2z^2+h(z)$$

$$f_z=x^2y+2y^2z+h'(z)=x^2y+2y^2z+e^z$$

$$h'(z)=e^z=>h(z)=e^z+C$$

A potential function of F is

$$f=x^2yz+y^2z^2+e^z+C$$