Differential Operators Exercise 5(c)

Differential Operators Exercise 5(c)

1. If A=2xz²i-yzj+3xz³ k and Φ =x²yz find

1. ∇× (∇Φ)
   
2. ∇×(∇×A)
3. ∇×(∇ΦA) at (1,1,1,)

Solution:

Given

A=2xz²i-yzj+3xz³ k and Φ =x²y

1. \(\nabla \phi=2 x y z \mathbf{i}+x^2 z \mathbf{j}+x^2 y \mathbf{k}\)

⇒ \(\nabla \times(\nabla \phi)\) = \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x y z & x^2 z & x^2 y
\end{array}\right|\)

= \(\mathbf{i}\left(x^2-x^2\right)-\mathbf{j}(2 x y-2 x y)+\mathbf{k}(2 x z-2 x z)=0\)

2. \(\nabla. \mathbf{A}=2 z^2-z+9 x z^2\)

⇒ \(\nabla \times(\nabla \times \mathbf{A})=\nabla(\nabla \cdot \mathbf{A})-\nabla^2 \mathbf{A}\)

= \(\sum \mathbf{i} \frac{\partial}{\partial x}\left(2 z^2-z+9 x z^2\right)-\sum \frac{\partial^2 \mathbf{A}}{\partial x^2}\)

= \(\mathbf{i}\left(9 z^2\right)+\mathbf{k}(4 z-1+18 x z)-(4 x \mathbf{i}+18 x z \mathbf{k})\)

= \(\mathbf{i}\left(9 z^2-4 x\right)+\mathbf{k}(4 z-1)\)

At (1,1,1)  \(\nabla \times(\nabla \times \mathbf{A})=5 \mathbf{i}+3 \mathbf{k}\)

Examples Of Solutions For Exercise 5(C) In Calculus

2. if A= 3xyz²i+2xy³j-x²yzk and Φ =3x²-yz  find

1. ∇. (ΦA)
   
2. grad. (∇Φ)at(1,-1,1)

Solution:

Given

A= 3xyz²i+2xy³j-x²yzk and Φ =3x²-yz

1. \(\nabla \cdot(\phi \mathbf{A})=(\nabla \phi) \cdot \mathbf{A}+(\nabla \cdot \mathbf{A}) \phi\)

Now \(\nabla \phi=6 x \mathbf{i}-z \mathbf{j}-y \mathbf{k}\)

⇒ \(\nabla \cdot \mathbf{A}=3 y z^2+6 x y^2-x^2 y\) At (1,-1,1)

∴ \((\nabla \phi) \cdot \mathbf{A}+(\nabla \cdot \mathbf{A}) \phi\)

= \((6 \mathbf{i}-\mathbf{j}+\mathbf{k}) \cdot(-3 \mathbf{i}-2 \mathbf{j}+\mathbf{k})+(-3+6+1)(3+1)\)

= (-18+2+1)+16=1 .

2. \(\text{div}(\nabla \phi)=\nabla \cdot(\nabla \phi)=6 x-z-y\)=6(1)-1+1 at (1,-1,1)]

3. If A=(3x²y-z)i+(xz³+y⁴)j-2x³z²k find grad div A at (2,-1,0).

Solution:

Given

A=(3x²y-z)i+(xz³+y⁴)j-2x³z²k

⇒ \(\text{div} \mathbf{A}=\nabla \cdot \mathbf{A}=\frac{\partial}{\partial x}\left(3 x^2 y-z\right)+\frac{\partial}{\partial y}\left(x z^3+y^4\right)+\frac{\partial}{\partial z}\left(-2 x^3 z^2\right)\)

= \(6 x y+4 y^3-4 x^3 z\)

grad\((\text{div} \mathbf{A})=\nabla(\nabla \cdot \mathbf{A})=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)\left(6 x y+4 y^3-4 x^3 z\right)\)

= \(\mathbf{i}\left(6 y-12 x^2 z\right)+\mathbf{j}\left(6 x+12 y^2\right)+\mathbf{k}\left(-4 x^3\right)\)

At (2,-1,0) \(\text{grad}(\text{div} \mathbf{A})=-6 \mathbf{i}+24 \mathbf{j}-32 \mathbf{k}\)

Exercise 5(C) Differential Operators Worked Example

4. If f and Φ are differential scalar point functions, show that f∇ × ∇Φ is solenoidal.

Solution:

Given

f and Φ are differential scalar point functions

⇒ \(\text{div}(\mathbf{A} \times \mathbf{B})=\mathbf{B} \cdot \text{curl} \mathbf{A}-\mathbf{A} \cdot \text{curl} \mathbf{B}\)

⇒ \(\text{div}(\nabla f \times \nabla \varphi)=\nabla \varphi \cdot \text{curl} \nabla f-\nabla f \text{curl} \nabla \varphi\)

= \(\nabla \phi \cdot(\text{curl} \text{grad} f)-\nabla f \cdot(\text{curl} \text{grad} \phi)=0\)

5. If f=x²yz,g= xy-3z² find div(grad f × grad g).

Solution:

Given

f=x²yz,g= xy-3z²

⇒ \(\text{grad} f=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)\left(x^2 y z\right)=2 x y z \mathbf{i}+x^2 z \mathbf{j}+x^2 y \mathbf{k}\)

⇒ \(\text{grad} g=\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)\left(x y-3 z^2\right)=y \mathbf{i}+x \mathbf{j}-6 z \mathbf{k}\)

∴ \(\text{div}(\text{grad} f \times \text{grad} g)=\nabla .(\text{grad} f \times \text{grad} g)\)

= \((\text{grad} g) \cdot[\nabla \times \text{grad} f]-(\text{grad} f) \cdot[\nabla \times \text{grad} g]\)

∴ \(\nabla \times \text{grad} f=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x y z & x^2 z & x^2 y
\end{array}\right|\)

= \(\mathbf{i}\left(x^2-x^2\right)-\mathbf{j}(2 x y-2 x y)+\mathbf{k}(2 x z-2 x z)\)

curl(grad f)=0

Similarly \(\text{curl}(\text{grad} g)=0\)

∴ \(\text{div}[(\text{grad} f) \times \text{grad} g]=0\)

6. If a is a constant vector, prove that

1. ∇(a.f)=(a.∇)f+a × curl f
   
2. ∇(a×f)=-a. curl f
   
3. ∇×(a×f)=a div f-(a.∇)f
   
4. div \(\begin{equation}
   \frac{\bar{r}}{r}
   \end{equation}\)=\(\begin{equation}=\frac{2}{r}\end{equation}\)

Solution:

1. \(\bar{a} \times(\nabla \times \mathbf{F})=\mathbf{a} \times\left(\sum \mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x}\right)=\sum\left[\mathbf{a} \times\left(\mathbf{i} \times \frac{\partial \mathbf{F}}{\partial x}\right)\right]\)

= \(\sum\left[\left(\mathbf{a} \cdot \frac{\partial \mathbf{F}}{\partial x}\right) \mathbf{i}-(\mathbf{a} \cdot \mathbf{i}) \frac{\partial \mathbf{F}}{\partial x}\right]=\sum \mathbf{i} \frac{\partial}{\partial x}(\mathbf{a} \cdot \mathbf{F})-\sum\left(\mathbf{a} \cdot \mathbf{i} \frac{\partial}{\partial x}\right) \mathbf{F}\)

⇒ \(\mathbf{a} \times(\nabla \times \mathbf{F})=\nabla(\mathbf{a} \cdot \mathbf{F})-(\mathbf{a} \cdot \nabla) \mathbf{F}\)

∴ \(\nabla(\mathbf{a} \cdot \mathbf{F})=(\mathbf{a} \cdot \nabla) \mathbf{F}+\mathbf{a} \times(\nabla \times \mathbf{F})\)

2. \(\nabla \cdot(\mathbf{a} \times \mathbf{f})=\sum \mathbf{i} \cdot\left[\frac{\partial}{\partial x}(\mathbf{a} \times \mathbf{f})\right]=\sum \mathbf{i} \cdot\left[\mathbf{a} \times \frac{\partial \mathbf{f}}{\partial x}\right]=\sum \mathbf{i} \cdot\left[-\frac{\partial \mathbf{f}}{\partial x} \times \mathbf{a}\right]\)

= – \(\sum \mathbf{i} \cdot\left(\frac{\partial \mathbf{f}}{\partial x} \times \mathbf{a}\right)=-\sum\left(\mathbf{i} \times \frac{\partial \mathbf{f}}{\partial x}\right) \cdot \mathbf{a}=-(\nabla \times \mathbf{f}) \cdot \mathbf{a}=-\mathbf{a} \cdot(\text{curl} \mathbf{f})\)

3. \(\nabla \times(\mathbf{a} \times \mathbf{f})=\sum \mathbf{i} \times\left[\frac{\partial}{\partial x}(\mathbf{a} \times \mathbf{f})\right]=\sum \mathbf{i} \times\left(\mathbf{a} \times \frac{\partial \mathbf{f}}{\partial x}\right)\)

= \(\sum\left[\left(\mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}\right) \mathbf{a}-(\mathbf{i} \cdot \mathbf{a}) \frac{\partial \mathbf{f}}{\partial x}\right]=\mathbf{a} \sum\left(\mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}\right)-\sum(\mathbf{a} \cdot \mathbf{i}) \frac{\partial \mathbf{f}}{\partial x}\)

= \(\mathbf{a}(\text{div} \mathbf{f})-\mathbf{a} \cdot\left(\sum \mathbf{i} \frac{\partial}{\partial x}\right) \mathbf{f}=\mathbf{a} \text{div} \mathbf{f}-(\mathbf{a} \cdot \nabla) \mathbf{f}\)

3. \( \nabla \times(\mathbf{a} \times \mathbf{f})=\sum \mathbf{i} \times\left[\frac{\partial}{\partial x}(\mathbf{a} \times \mathbf{f})\right]=\sum \mathbf{i} \times\left(\mathbf{a} \times \frac{\partial \mathbf{f}}{\partial x}\right)\)

= \(\sum\left[\left(\mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}\right) \mathbf{a}-(\mathbf{i} \cdot \mathbf{a}) \frac{\partial \mathbf{f}}{\partial x}\right]=\mathbf{a} \sum\left(\mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}\right)-\sum(\mathbf{a} \cdot \mathbf{i}) \frac{\partial \mathbf{f}}{\partial x}\)

= \(\mathbf{a}(\text{div} \mathbf{f})-\mathbf{a} \cdot\left(\sum \mathbf{i} \frac{\partial}{\partial x}\right) \mathbf{f}=\mathbf{a} \text{div} \mathbf{f}-(\mathbf{a} \cdot \nabla) \mathbf{f}\)

Differential Operator Application In Exercise 5(C)

7. prove that div (A×r)=r.curlA.

Solution:

div (A×r) = \(\mathbf{r} \cdot \text{curl} \mathbf{A}-\mathbf{A} \cdot \text{curl} \mathbf{r}\)

= \(\mathbf{r} \cdot \text{curl} \mathbf{A}-\mathbf{A} \cdot(\mathbf{O})\)

= \(\mathbf{r} \cdot \text{curl} \mathbf{A}\)

8. Prove that curl (Φ grad Φ)=0

Solution:

⇒ \(\text{curl}(\phi \text{grad} \phi)=\nabla \times(\phi \nabla \phi)=(\nabla \phi) \times(\nabla \phi)+\phi[\nabla \times(\nabla \phi)]=0+0=0\)

9. If A and B are irrotational, prove that A×B is solenoidal.

Solution:

Given A and B are irrotational

⇒ \(\nabla \times \mathbf{A}=0\) and \(\nabla \times \mathbf{B}=0\)

Now \(\nabla \cdot(\mathbf{A} \times \mathbf{B})=\mathbf{B} \cdot(\nabla \times \mathbf{A})-\mathbf{A} \cdot(\nabla \times \mathbf{B})=0\)

⇒ \(\mathbf{A} \times \mathbf{B}\) is solenoidal.