17. Divergence, Curl and Potentials
d. Differential Identities
2. Second Order Differential Operators and Identities
b. The Curl of a Gradient (\(\vec\nabla\times\vec\nabla f\))
Let's start with an example and an exercise.
Compute the curl of the gradient of the function \(f=x^2y^3z^4\)
The gradient is \[ \vec\nabla f=\left\langle 2xy^3z^4,3x^2y^2z^4,4x^2y^3z^3\right\rangle \] So the curl of the gradient is \[\begin{aligned} \vec\nabla\times\vec\nabla f &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\[2pt] \partial_x & \partial_y & \partial_z \\[2pt] 2xy^3z^4 & 3x^2y^2z^4 & 4x^2y^3z^3 \end{vmatrix} \\[2pt] &=\hat\imath(12x^2y^2z^3-12x^2y^2z^3) -\hat\jmath(8xy^3z^3-8xy^3z^3) +\hat k(6xy^2z^4-6xy^2z^4) \\ &=\langle 0,0,0\rangle \end{aligned}\]
Compute the curl of the gradient of the function \(g=x\sin(y)-z\cos(x)\).
\(\vec\nabla\times\vec\nabla g=\langle 0,0,0\rangle\)
The gradient is \[ \vec\nabla g=\left\langle \sin(y)+z\sin(x),x\cos(y),-\cos(x)\right\rangle \] So the curl of the gradient is \[\begin{aligned} \vec\nabla\times\vec\nabla g &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\[2pt] \partial_x & \partial_y & \partial_z \\ \sin(y)+z\sin(x) & x\cos(y) & -\cos(x) \end{vmatrix} \\[2pt] &=\hat\imath(0-0) -\hat\jmath(\sin(x)-\sin(x)) +\hat k(\cos(y)-\cos(y)) \\ &=\langle 0,0,0\rangle \end{aligned}\]
It is no coincidence that in both problems the answer was \(\vec 0=\langle 0,0,0\rangle\). As long as the assumptions of Clairaut's Theorem are satisfied (so that mixed partial derivatives are equal) the curl of a gradient is always \(\vec 0\).
If all second partial derivatives of \(f\) are continuous functions, then \[ \vec\nabla\times\vec\nabla f=\vec 0 \]
The general gradient is \[ \vec\nabla f =\left\langle \partial_x f,\partial_y f,\partial_z f\right\rangle \] So the curl of the gradient is \[\begin{aligned} \vec\nabla\times\vec\nabla f &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\[2pt] \partial_x & \partial_y & \partial_z \\[2pt] \partial_x f & \partial_y f & \partial_z f \end{vmatrix} \\[2pt] &=\hat\imath (\partial_y\partial_z f-\partial_z\partial_y f) -\hat\jmath (\partial_x\partial_z f-\partial_z\partial_x f) \\ &\qquad+\hat k (\partial_x\partial_y f-\partial_y\partial_x f) \\ &=\langle 0,0,0\rangle \end{aligned}\] provided the mixed partial derivatives are equal.
In this class, we will always be dealing with continuous functions. So you can always assume the curl of a gradient is zero.
This identity is used to help determine when a vector field has a scalar potential as discussed at the end of this chapter. This identity is also used to solve the equations of electrostatics.