Differential Operators Solved Problems Gradient Divergence Curl

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Differential Operators Solved Problems

Example. 1. Prove that \( \nabla \mathrm{f}(r)\)=\(\mathbf{f}^{\prime}(r) \frac{\overline{\mathbf{r}}}{r} \) Hence prove that

  1. \(\nabla r=\frac{\overline{\mathbf{r}}}{r}\)
  2. \(\text { (2) } \nabla\left(\frac{1}{r}\right)=\frac{-\overline{\mathbf{r}}}{r^3}\)
  3. \(\text { (3) } \nabla\left(r^3\right)=3 r \overline{\mathrm{r}}\)
  4. \(\text { (4) } \overline{\mathrm{a}} \nabla\left(\frac{1}{r}\right)=\frac{-\overline{\mathbf{a}} \cdot \overline{\mathbf{r}}}{r^3}\)

Solution. \(\nabla f(r)=\sum \frac{i \delta}{\partial x}\{\vec{f}(r)\}=\sum i f^{\prime}(r) \frac{d r}{d x}=\sum i f^{\prime}(r) \frac{x}{r}=\frac{f^{\prime}(r)}{r} \sum x i\)

(1)  f(r) = r ⇒ \(\frac{f^{\prime}(r)}{r}\)=\(\frac{1}{r}\)

∴ ∇r\(=\frac{\mathbf{r}}{r}\)

(2) Taking f(r)=\(\frac{1}{r}\), We get f'(r)\(=\frac{-1}{r^2}\)

⇒\(\frac{f^{\prime}(r)}{r}\)=\(\frac{-1}{r^3}\)

∴ \(\nabla\left(\frac{1}{r}\right)\)=\(\frac{-\bar{r}}{r^3}\)

(3) f(r) =r3 ⇒ f'(r)=3r2⇒ \(\frac{f^{\prime}(r)}{r}\) = 3r ⇒∇(r3)\(=3 r \bar{r}\)

(4) Clear from (3) and (4)

Differential Operators Gradient Divergence Curl Examples

Example. 2. If a – x + y + z, b = x² + y² + z², c = xy + yz + zx; prove that [grad a, grad b, grad c] 

Solution:

Given

a- x + y + z, b = x² + y² + z², c = xy + yz + zx

a=x+y+z

∴ \(\frac{\partial a}{\partial x}=1, \frac{\partial a}{\partial y}=1, \frac{\partial a}{\partial z}=1\)

∴ \(\text{grad} a=\nabla a=\mathbf{i} \frac{\partial a}{\partial x}+\mathbf{j} \frac{\partial a}{\partial y}+\mathbf{k} \frac{\partial a}{\partial z}=\mathbf{i}+\mathbf{j}+\mathbf{k}\)

Given, \(b=x^2+y^2+z^2\)

∴ \(\frac{\partial b}{\partial x}=2 x, \frac{\partial b}{\partial y}=2 y, \frac{\partial b}{\partial z}=2 z\)

∴ \(\text{grad} b=\nabla b=\mathbf{i} \frac{\partial b}{\partial x}+\mathbf{j} \frac{\partial b}{\partial y}+\mathbf{k} \frac{\partial b}{\partial z}=2 x \mathbf{i}+2 y \mathbf{j}+2 z \mathbf{k}\)

Again, \(c=x y+y z+z x \quad \frac{\partial c}{\partial x}=y+z, \frac{\partial c}{\partial y}=z+x, \frac{\partial c}{\partial z}=x+y\)

⇒ \(\text{grad} c=\nabla c=\mathbf{i} \frac{\partial c}{\partial x}+\mathbf{j} \frac{\partial c}{\partial y}+\mathbf{k} \frac{\partial c}{\partial z}=(y+z) \mathbf{i}+(z+x) \mathbf{j}+(x+y) \mathbf{k}\)

∴ \((\text{grad} a) \cdot(\text{grad} b) \times(\text{grad} c)=\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 x & 2 y & 2 z \\
y+z & z+x & x+y
\end{array}\right|\)

On simplification \([\text{grad} a, \text{grad} b, \text{grad} c]=0\)

Example. 3. Prove that ∇ r = r/r, if r = xi + yj + zk and r =|r|

Solution: If r =xi + yj + zk then r = |r| =  \(\sqrt{\left(x^2+y^2+z^2\right)} \Rightarrow r^2=x^2+y^2+z^2 \)

⇒ \(2 r \frac{\partial r}{\partial x}=2 x, 2 r \frac{\partial r}{\partial y}=2 y, 2 r \frac{\partial r}{\partial z}=2 z\)

⇒ \(\nabla r=\mathbf{i} \frac{\partial r}{\partial x}+\mathbf{j} \frac{\partial r}{\partial y}+\mathbf{k} \frac{\partial r}{\partial z}=\frac{x \mathbf{i}}{r}+\frac{y \mathbf{j}}{r}+\frac{z \mathbf{k}}{r}=\frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{r}=\frac{\mathbf{r}}{r} \)

Gradient Divergence Curl Detailed Examples

Example. 4. Prove that \(\nabla\left(r^n\right)=n r^{n-2} \mathbf{r}\)

Solution: r= xi+yj+zk  ;  \(r=|\mathbf{r}|=\sqrt{x^2+y^2+z^2}\)

D. W.r. to \(x \text { partially } 2 r \frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r} \text {, similarly, } \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r}\)

⇒ \(\nabla\left(r^n\right)=\sum \mathbf{i} \frac{\partial}{\partial x}\left(r^n\right)=\sum \mathbf{i} n r^{n-1} \frac{\partial r}{\partial x}=\sum \mathbf{i} n r^{n-1} \frac{x}{r}=n r^{n-2} \sum x \mathbf{i}=n r^{n-2} \mathbf{r}\)

Differential Operators Note : from the above result, we can have

  1. \(\nabla\left(\frac{1}{r}\right)=-\frac{\mathbf{r}}{r^3} \text { taking } n=-1\)
  2. \(\nabla(r)=\frac{\mathbf{r}}{r} \text { taking } n=1\)

Example. 5. Prove that \(\nabla(\log |r|)=\frac{\mathbf{r}}{r^2} \)

Solution:  We have r= xi+yj+zk    ;  \(r=|\mathbf{r}|=\sqrt{x^2+y^2+z^2}\)

⇒\(\frac{\partial r}{\partial x}=\frac{x}{r}, \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r}\)

⇒ \(\nabla(\log |r|)=\sum i \frac{\partial}{\partial x} \log |r|=\sum \mathbf{i} \frac{1}{r} \frac{\partial r}{\partial x}=\sum \mathbf{i} \frac{1}{r} \frac{x}{r}=\frac{1}{r^2} \sum x \mathbf{i}=\frac{\mathbf{r}}{r^2} \)

 

 

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