4.3 The Operator \(\nabla \)
If a scalar function \(\Phi (x,y,z)\) has continuous first partial derivatives, its gradient is \[\grad \Phi = \frac {\partial \Phi }{\partial x}\,\mathbf i + \frac {\partial \Phi }{\partial y}\,\mathbf j + \frac {\partial \Phi }{\partial z}\,\mathbf k = \left (\mathbf i\frac {\partial }{\partial x} + \mathbf j \frac {\partial }{\partial y} + \mathbf k \frac {\partial }{\partial z}\right ) \Phi .\] The bracket is a vector differential operator, called del or nabla: \[\nabla = \mathbf i \frac {\partial }{\partial x} + \mathbf j\frac {\partial } {\partial y} + \mathbf k \frac {\partial }{\partial z} .\] It has the form of a vector but is an operator, and means something only when it acts on a function.
Properties
Let \(f\) and \(g\) be differentiable functions and \(\alpha \) a constant. Then
- 1.
- \(\nabla (\alpha f) = \alpha \nabla f\);
- 2.
- \(\nabla (f + g) = \nabla f + \nabla g\);
- 3.
- \(\nabla (fg) = f\nabla g + g \nabla f\);
- 4.
- \(\nabla \left (\frac {f}{g}\right ) = \frac {g\nabla f - f \nabla g}{g^2}\) where \(g\neq 0\).
Each follows by applying the corresponding one-variable rule to every component. For example, for property 4,
Proof. The \(\mathbf i\)-component of \(\nabla (fg^{-1})\) is, by the product and chain rules, \[\frac {\partial }{\partial x}(fg^{-1}) = \frac {\partial f}{\partial x}g^{-1} - fg^{-2}\frac {\partial g}{\partial x} = \frac {1}{g}\frac {\partial f}{\partial x} - \frac {f}{g^2}\frac {\partial g}{\partial x} ,\] and the \(\mathbf j\)- and \(\mathbf k\)-components are the same with \(y\) and \(z\) in place of \(x\). Collecting, \[\nabla \left (\frac {f}{g}\right ) = \frac {1}{g}\nabla f - \frac {f}{g^2}\nabla g = \frac {g \nabla f - f \nabla g }{g^2} .\] □
Example 4.3.1. Prove that \(\nabla r^{n} = n r^{n-2}\mathbf r\), where \(\mathbf r = x\,\mathbf i + y\,\mathbf j + z\,\mathbf k\) and \(r = |\mathbf r| = \sqrt {x^2 + y^2 + z^2}\).
Solution. By the chain rule, \(\frac {\partial }{\partial x}r^n = n r^{n-1} \frac {\partial r}{\partial x}\), and \[\frac {\partial r}{\partial x} = \frac {\partial }{\partial x} (x^2 + y^2 + z^2)^{1/2} = \frac {1}{2}\cdot 2x\cdot (x^2 + y^2 + z^2)^{-1/2} = \frac {x}{r} .\] Similarly \(\frac {\partial r}{\partial y} = \frac {y}{r}\) and \(\frac {\partial r}{\partial z} = \frac {z}{r}\). So \[\frac {\partial r^n}{\partial x} = nr^{n-1} \frac {x}{r} = nr^{n-2}x ,\qquad \frac {\partial r^n}{\partial y} = nr^{n-2}y ,\qquad \frac {\partial r^n}{\partial z}= nr^{n-2}z ,\] and \[\nabla r^n = nr^{n-2}x \,\mathbf i + nr^{n-2} y \,\mathbf j + nr^{n-2} z\,\mathbf k = nr^{n-2} (x\,\mathbf i + y \,\mathbf j + z\,\mathbf k) = nr^{n-2}\mathbf r .\] In particular \(\nabla r = \frac {\mathbf r}{r}\), the unit vector pointing away from the origin.
Example 4.3.2. Find \(\nabla \left (xe^{yz}\right )\) and \(\nabla \left (\frac {x}{y^2 + z^2}\right )\) at the point \((1,1,0)\).
Solution. \(\nabla (xe^{yz}) = \langle e^{yz},\ xze^{yz},\ xye^{yz}\rangle \), which is \(\langle 1, 0, 1\rangle \) at \((1,1,0)\). For the quotient, with \(f = x\) and \(g = y^2 + z^2\), property 4 gives \[\nabla \left (\frac fg\right ) = \frac {g\langle 1,0,0\rangle - x\langle 0, 2y, 2z\rangle }{g^2} = \frac {\langle 1, 0, 0\rangle - \langle 0,2,0\rangle }{1} = \langle 1,-2,0\rangle \] at \((1,1,0)\), where \(g = 1\).
Questions on this section
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