Question

) For a scalar field ( , , ) ,
n n n
φ x y z = x + y + z show that (∇φ) r = nφ
r r
. , where n is a non-zero
real constant.

Answer 1

Answer on Question #62036 - Math - Vector Calculus

Question

For a scalar field

$$\varphi = x ^ {n} + y ^ {n} + z ^ {n},$$

where $n$ is a non-zero real constant,

show that

$$(\nabla \varphi , \vec {r}) = n \varphi$$

Solution

If $\vec{r} = x\vec{i} + y\vec{j} + z\vec{k}$ and $\varphi = x^n + y^n + z^n$, then

$$\begin{array}{l} \nabla \varphi = \frac {\partial \varphi}{\partial x} \vec{i} + \frac {\partial \varphi}{\partial y} \vec{j} + \frac {\partial \varphi}{\partial z} \vec{k} = \\ = \frac {\partial (x ^ {n} + y ^ {n} + z ^ {n})}{\partial x} \vec{i} + \frac {\partial (x ^ {n} + y ^ {n} + z ^ {n})}{\partial y} \vec{j} + \frac {\partial (x ^ {n} + y ^ {n} + z ^ {n})}{\partial z} \vec{k} = \\ = n x ^ {n - 1} \vec{i} + n y ^ {n - 1} \vec{j} + n z ^ {n - 1} \vec{k}; \end{array}$$ $$\begin{array}{l} (\nabla \varphi , \vec{r}) = \left(n x ^ {n - 1} \vec{i} + n y ^ {n - 1} \vec{j} + n z ^ {n - 1} \vec{k}, \quad x \vec{i} + y \vec{j} + z \vec{k}\right) = \\ = n x ^ {n - 1} x + n y ^ {n - 1} y + n z ^ {n - 1} z = n (x ^ {n} + y ^ {n} + z ^ {n}) = n \varphi. \end{array}$$

www.AssignmentExpert.com