Question

Find, $\mathbf{\nabla }\mathbf{r}$, where$\mathbf{r}\mathbf{=}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}$ , using (6.7) and also using (6.3). Show that your results are the same by using (4.11) and (4.12).

Short answer

The solution to this problem is $\nabla r=e_r$.

Step-by-step solution

Given Information

The given information is$r=\sqrt{x^2+y^2}$

Definition of Scalar field.

In mathematics and physics, scalar field or scalar-valued function referred to a scalar value to everypointin aspace– possiblyphysical space. The scalar may either be a (dimensionless)mathematical numberor aphysical quantity.

Find the solution.

Use the following equation:

$$\begin{gathered} \nabla f=e_r\frac{\partial f}{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial f}{\partial \theta }+e_z\frac{\partial f}{\partial z} \\ \nabla r=\frac{ar}{ar}{e}_r \\ =e_r \end{gathered}$$

Use the following equation:

$$\begin{gathered} \nabla \varphi =grad\varphi \\ =i\frac{\partial \varphi }{\partial x}+j\frac{\partial \varphi }{\partial y}+k\frac{\partial \varphi }{\partial z}. \end{gathered}$$

Find the value of r.

$$\begin{gathered} r=\sqrt{x^2+y^2}.\nabla r \\ =\frac{1}{\sqrt{x^2+y^2}}\left(xi+yj\right) \\ =\cos \theta i+\sin \theta \\ =e_r \end{gathered}$$

Hence, by using both the equation (1) and (2), the answer is same $\nabla r=e_r$.