Question

find the divergence of each of the following vector fields at all points where they are defined.
(a) \\( \text{div} \left\langle \frac{x}{(x^2 + y^2 + z^2)^{1.5}}, \frac{y}{(x^2 + y^2 + z^2)^{1.5}}, \frac{z}{(x^2 + y^2 + z^2)^{1.5}} \
ight\
angle = \\)
(b) \\( \text{div} \left\langle 5e^{xy}, 4\cos(xy), 5e^{\ln(x^2 + y^2 + 2)} \
ight\
angle = \\)
(c) \\( \text{div} \left\langle 4x^2 - \sin(xz), 4, - (\sin(xz)) \
ight\
angle = \\)

Explanation:

Step1: Recall divergence formula

For vector field $\langle P,Q,R
angle$, $\text{div} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$

Part (a)

Let $r^2 = x^2+y^2+z^2$, so $P=\frac{x}{r^3}$, $Q=\frac{y}{r^3}$, $R=\frac{z}{r^3}$

Step2: Compute $\frac{\partial P}{\partial x}$

$\frac{\partial P}{\partial x} = \frac{r^3 - x \cdot 3r^2 \cdot \frac{x}{r}}{r^6} = \frac{r^2 - 3x^2}{r^5}$

Step3: Compute $\frac{\partial Q}{\partial y}$ and $\frac{\partial R}{\partial z}$

Similarly: $\frac{\partial Q}{\partial y} = \frac{r^2 - 3y^2}{r^5}$, $\frac{\partial R}{\partial z} = \frac{r^2 - 3z^2}{r^5}$

Step4: Sum partial derivatives

$\text{div} = \frac{(r^2-3x^2)+(r^2-3y^2)+(r^2-3z^2)}{r^5} = \frac{3r^2 - 3(x^2+y^2+z^2)}{r^5} = 0$

Part (b)

$P=5e^{xy}$, $Q=4\cos(xy)$, $R=5(x^2+y^2+2)$ (since $e^{\ln A}=A$)

Step5: Compute partial derivatives

$\frac{\partial P}{\partial x}=5y e^{xy}$, $\frac{\partial Q}{\partial y}=-4x\sin(xy)$, $\frac{\partial R}{\partial z}=0$

Step6: Sum partial derivatives

$\text{div}=5y e^{xy} - 4x\sin(xy)$

Part (c)

$P=4x^2 - \sin(xz)$, $Q=4$, $R=-\sin(xz)$

Step7: Compute partial derivatives

$\frac{\partial P}{\partial x}=8x - z\cos(xz)$, $\frac{\partial Q}{\partial y}=0$, $\frac{\partial R}{\partial z}=-x\cos(xz)$

Step8: Sum partial derivatives

$\text{div}=8x - z\cos(xz) - x\cos(xz)$

Answer:

(a) $0$
(b) $5y e^{xy} - 4x\sin(xy)$
(c) $8x - \cos(xz)(x + z)$