grad of scalar field question Question
If $P(r)$ is a point $r = \rho \cos \phi \, \boldsymbol i + \rho \sin \phi \boldsymbol \, \boldsymbol j + z \, \boldsymbol k$ and a scalar field $V = {\rho}^2 \, z \sin(2 \phi)$ exists in space, using cylindrical polar coordinates ($\rho,\phi,z$) determine $\operatorname{grad} V$ at the point at which $\rho = 1, \, \phi = \dfrac{\pi}{4}, \, z = 2$
My Answer
Use the standard formula for grad in cylindrical polar coordinates:
$\operatorname{grad} \, V = \dfrac{\partial V}{\partial \rho} \boldsymbol I + \dfrac{1}{\rho} \dfrac{\partial V}{\partial \phi} \boldsymbol J + \dfrac{\partial V}{\partial z} \boldsymbol K$
we get:
$\operatorname{grad} \, V = 2 \, \rho \, z \sin \phi \, \boldsymbol I + \rho \, z \cos \phi \, \boldsymbol J + \rho^2 \sin \phi \, \boldsymbol K$
and substituting in the values of $\rho = 1, \, \phi = \dfrac{\pi}{2} \, \text{ and }\, z = 2$ gives:
$\operatorname{grad} \, V = \dfrac{4}{\sqrt{2}} \boldsymbol I \, + \dfrac{2}{\sqrt{2}} \boldsymbol J \, + \, \dfrac{1}{\sqrt{2}} \boldsymbol K$
Book Answer
$4 \boldsymbol I \, + \, \boldsymbol K$
I've not used the point P(r) in any way. How have I misunderstood the question?
Thanks,
Mitch.

$\nabla V = \left( \frac{\partial V}{\partial \rho},\frac{1}{\rho}\frac{\partial V}{\partial \phi}, \frac{\partial V}{\partial z}\right) = \operatorname{grad} V = \left(2 \rho z \sin 2 \phi, 2 \rho z \cos 2 \phi, \rho ^2 \sin 2 \phi  \right)$.
Substituting $\rho = 1, \phi = \frac{\pi}{4}, z=2$, we get:
$\nabla V\left(1,\frac{\pi}{4},2\right) = (4,0,1)$.
I think the mention of $P(\mathbf{r})$ is just to make it clear that we're in cylindrical coordinates - it's just the Cartesian coordinates $(x,y,z)$ expressed in terms of $\rho,\phi,z$ i.e. $x = \rho \cos \phi, \ y = \rho \sin \phi, \ z=z$.
