91Ó°ÊÓ

The gradient of a function is $\mathit{v}$is defined as$\mathbf{∇}\mathit{v}$. The $∇$operator is defined as .

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

Compute the gradient of function $v$, as

$$\begin{gathered} ∇×(∇v)=∇×\left(\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k\right)\left(v\right) \\ =\frac{∂v}{∂x}i+\frac{∂v}{∂y}j+\frac{∂v}{∂z}k \end{gathered}$$

Compute curl of gradient of function $v$.

$$\begin{gathered} ∇×(∇v)=∇×\left(\frac{∂v}{∂x}i+\frac{∂v}{∂y}j+\frac{∂v}{∂z}k\right) \\ =\left|\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ \frac{∂v}{∂x}& \frac{∂v}{∂y}& \frac{∂v}{∂z}\end{array}\right| \\ =i\left(\frac{{∂}^{2}v}{∂y∂z}-\frac{{∂}^{2}v}{∂z∂y}\right)-j\left(\frac{{∂}^{2}v}{∂z∂x}-\frac{{∂}^{2}v}{∂x∂z}\right)+ki\left(\frac{{∂}^{2}v}{∂x∂y}-\frac{{∂}^{2}v}{∂y∂x}\right) \end{gathered}$$

Compute curl of gradient of function 0.

The function is given as $f(x,y,z)=x^2{y}^3{z}^4$is computed as follows:

Compute the gradient of function $f(x,y,z)=x^2{y}^3{z}^4,$as

$$\begin{gathered} ∇f(x,y,z)=\left(\frac{∂}{∂x}i+\frac{∂}{∂y}j+\frac{∂}{∂z}k\right)\left(x^2{y}^3{z}^4\right) \\ =\frac{∂\left(x^2{y}^3{z}^4\right)}{∂x}i+\frac{∂\left(x^2{y}^3{z}^4\right)}{∂y}j+\frac{∂\left(x^2{y}^3{z}^4\right)}{∂z}k \\ =2x{y}^3{z}^{4}i+3x^2{y}^2{z}^4+j4x^2{y}^3{z}^3 \end{gathered}$$

Compute curl of gradient of function $f(x,y,z)$

role="math" localid="1657353951851" $$\begin{gathered} ∇×(∇f)=∇×\left(\frac{∂f}{∂x}i+\frac{∂f}{∂y}j+\frac{∂f}{∂z}k\right) \\ =\left|\begin{array}{ccc}i& j& k\\ \frac{∂}{∂x}& \frac{∂}{∂y}& \frac{∂}{∂z}\\ 2x{y}^3{z}^4& 3x^2{y}^2{z}^4& 4x^2{y}^3{z}^3\end{array}\right| \\ =i(12x^2{y}^2{z}^3-12x^2{y}^2{z}^3)-j(8x{y}^3{z}^3-8x{y}^3{z}^3)+k(6x{y}^2{z}^4-6x{y}^2{z}^4) \\ =0 \end{gathered}$$

Thus, curl of gradient is always 0.