Vector Calculus

Operators

1. Vector Differential Operator∇

The operator ∇ = \(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\) is defined such that

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

where Φ is a scalar point function. The symbol ∇is read as del or nabla

Note 1. If Φ is a scalar point function ∇ Φ =grad Φ \(\sum \mathbf{i} \frac{\partial \phi}{\partial x}\)

2. ∇ Φ is a vector point function

2. Scalar Differential Operator A. ∇

The operator a. ∇ \(=(\mathbf{a} \cdot \mathbf{i}) \frac{\partial}{\partial x}+(\mathbf{a} \cdot \mathbf{j}) \frac{\partial}{\partial y}+(\mathbf{a} \cdot \mathbf{k}) \frac{\partial}{\partial z}\) is defined such that

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

⇒ \(\begin{equation}
\text { and }(\mathbf{a} \cdot \bar{\nabla}) \overline{\mathbf{f}}=(\mathbf{a} \cdot \mathbf{i}) \frac{\partial \overline{\mathbf{f}}}{\partial x}+(\mathbf{a} \cdot \mathbf{j}) \frac{\partial \overline{\mathbf{f}}}{\partial y}+(\mathbf{a} \cdot \mathbf{k}) \frac{\partial \overline{\mathbf{f}}}{\partial z}
\end{equation}\)

Examples Of Vector Scalar Operator ∇ In Divergence

3. Vector Differential Operator A X ∇

The operator \(\mathbf{a} \times \nabla\)=\((\mathbf{a} \times \mathbf{i}) \frac{\partial}{\partial x}+(\mathbf{a} \times \mathbf{j}) \frac{\partial}{\partial \nu}+(\mathbf{a} \times \mathbf{k}) \frac{\partial}{\partial z}\) is defined such that

(1)\((\mathbf{a} \times \nabla) \phi\)=\((\mathbf{a} \times \mathbf{i}) \frac{\partial \phi}{\partial x}+(\mathbf{a} \times \mathbf{j}) \frac{\partial \phi}{\partial y}+(\mathbf{a} \times \mathbf{k}) \frac{\partial \phi}{\partial z}\)

(2) \((\mathbf{a} \times \nabla) \cdot \mathbf{f}\)=\((\mathbf{a} \times \mathbf{i}) \cdot \frac{\partial \mathbf{f}}{\partial x}+(\mathbf{a} \times \mathbf{j}) \cdot \frac{\partial \mathbf{f}}{\partial y}+(\mathbf{a} \times \mathbf{k}) \cdot \frac{\partial \mathbf{f}}{\partial z}\)

(3) \((\bar{a} \times \nabla) \times \bar{f}\)=\((\bar{a} \times \bar{i}) \times \frac{\partial \hat{f}}{\partial x}+(\bar{a} \times j) \times \frac{\partial \bar{f}}{\partial y}+(\bar{a} \times \bar{k}) \times \frac{\partial \bar{f}}{\partial z}\)

Differential Operators Vector Scalar Operator ∇

4. Scalar Differential Operator ∇

The operator ∇.\(=\mathbf{i} \cdot \frac{\partial}{\partial x}+\mathbf{j} \cdot \frac{\partial}{\partial y}+\mathbf{k} \cdot \frac{\partial}{\partial z}\) is defined such that  ∇.f\(=\mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{f}}{\partial y^*}+\mathbf{k} \cdot \frac{\partial \mathbf{f}}{\partial z}\)

Note. ∇.f is a scalar point function. and is given by ∇.f \(=\sum \mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}\)

5. Vector Differential Operator ∇ X

The operator ∇ x \(=\mathbf{i} \times \frac{\partial}{\partial x}+\mathbf{j} \times \frac{\partial}{\partial y}+\mathbf{k} \times \frac{\partial}{\partial z}\) is defined such that

∇ x f \(=\mathbf{i} \times \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{f}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{f}}{\partial z}\)

Note. ∇ x f is a vector function and is defined as ∇ x f \(=\sum \mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}\)

Differential Operators Divergence Of A Vector

Let f be any continuously differentiable vector point function. Then \(\mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{f}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{f}}{\partial z}\) is calle Divergence of f and is written as div f.

∴ div f=\(\mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{f}}{\partial y}+\mathbf{k} \cdot \frac{\partial \mathbf{f}}{\partial z}\)

∴ \(=\left(\mathbf{i} \cdot \frac{\partial}{\partial x}+\mathbf{j} \cdot \frac{\partial}{\partial y}+\mathbf{k} \cdot \frac{\partial}{\partial z}\right) \mathbf{f}\)

=\(\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right) . \mathbf{f}\)

∴ div = ∇ .f

Note 1. ∇is an operator and when operated on a vector function f gives the scalar function ∇ . f.

Note 2. If f is a scalar point function then ∇.f has no meaning

Divergence Of A Vector Field Explained

Differential Operators Solenoidal Vector

If div f = 0 then f is said to be a solenoidal vector point function.

Physical Interpretation c f divergence

Consider a small parallelopiped with center P = (x, y, z) and sides parallel to the coordinate axes and having magnitudes Δx, Δy, Δz within a moving fluid.

Let V= v₁i + v₂ j + v₃k be the velocity at any point.

Then we can prove that,

The total gain in volume of fluid per unit volume per unit time ∇. V = div V.

This gives the physical Interpretation of divergence

Differential Operators Level Surface L

Let f be a scalar point function and c be a real number. Q is a point in the domain S, such that Q ∈ S ⇒ f(Q) = c then s is called the level surface.

For different values c,f constitute a family of level surfaces in three-dimensional space.

Note 1. If f is a scalar point function and P is any given point, then the surface S such that Q ∈ S => f(Q) = c = f(P) defines a level surface through the point P.

Note2. If P and Q are two points on a level surface f, then f(P) = f(Q).

Theorem 1. The directional derivative of a scalar Φ at a point P (x,y, z) in the direction of a unit vector c is e. Grad Φ or e.∇Φ).

Proof: We know that

⇒ \(\frac{d \phi}{d s}=\frac{\partial \phi}{\partial x} \frac{d x}{d s}+\frac{\partial \phi}{\partial y} \frac{d y}{d s}+\frac{\partial \phi}{\partial s} \frac{d z}{d s}\)

⇒ \(=\left(\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right) \cdot\left(\mathbf{i} \frac{d x}{\partial s}+\mathbf{j} \frac{d y}{\partial s}+\mathbf{k} \frac{d z}{\partial s}\right)\) = ∇ Φ.e =(grade Φ).e

⇒ The directional derivative of Φ in the direction of e

= (∇ Φ) . e = (grad Φ) . e

Differential Operators Level Surface L

Theorem 2. ∇Φ is a vector normal to the level surface Φ(x,y,z) = c where c is a constant.

Proof: Let P (x, y, z) be a point on the surface and T be the unit tangent vector at P

Then the position vector of P is r = xi + yj + zk

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

Now \(\phi(x, y, z)=c \Rightarrow \frac{\partial \phi}{\partial s}=0\)

But \(\frac{d \phi}{\partial s}=\frac{\partial \phi}{\partial x} \frac{\partial x}{\partial s}+\frac{\partial \phi}{\partial y} \frac{\partial y}{\partial s}+\frac{\partial \phi}{\partial s} \frac{\partial z}{\partial s}\)

0= \(\left(\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right) \cdot\left(\mathbf{i} \frac{\partial x}{\partial s}+\mathbf{j} \frac{\partial y}{\partial s}+\mathbf{k} \frac{\partial z}{\partial s}\right) \Rightarrow 0=\nabla \phi \cdot \frac{\partial \bar{r}}{\partial s}\)

0 = \(\nabla \phi \cdot \frac{\partial \mathbf{r}}{\partial s}\) where \(\mathbf{T}\) is the unit tangent vector.

⇒ \(\nabla \phi\) is perpendicular to the tangent plane.

⇒ \(\nabla \phi\) is normal to the level surface \(\phi(x, y, z)=\mathrm{C}\)

Theorem 3.  If N is a unit vector normal to the level surface Φ (x, y, z) = C at point P (x, y, z) in the direction Φ increasing and p is the distance along this normal, then grad Φ =\(\nabla \phi\)=\(\frac{\partial \phi}{\partial \mathbf{N}} \mathbf{N}\)

Proof: Since grad Φ is normal to Φ (x, y, z) = C there exists a scalar p such that grad Φ = ∇ Φ= pN

The directional derivative of Φ in the direction of N is \(\frac{\partial \phi}{\partial \mathbf{N}}=\nabla \phi \cdot \mathbf{N}\)

= (pN). N = p (N . N) = p N is the unit vector.

Hence grad \(\phi=p \mathbf{N}=\frac{\partial \phi}{\partial \mathbf{N}} \mathbf{N}\)

Note. The magnitude of grad Φ=|grad Φ| =p\(=\frac{\partial \phi}{\partial \mathbf{N}}\)

Theorems On Directional Derivative Of A Scalar Function

Theorem 4. Grad Φ is a vector in the direction in which the maximum value of\(\frac{\partial \phi}{\partial s}\)occurs.

Proof: The directional derivative of Φ at a point P in the direction of the unit vector \(\overline{\mathbf{e}}\) is \(\frac{\partial \phi}{\partial s}\) =e.(grad) Φ  \(=\vec{e} \cdot\left(\frac{\partial \phi}{\partial \mathbf{N}} \mathbf{N}\right)\)=\(\frac{\partial \phi}{\partial \mathbf{N}}(\bar{e} \cdot \mathbf{N})\)

Where N is the unit normal vector at P to the surface

∴ \(\frac{\partial \phi}{\partial s}\)=\(\frac{\partial \phi}{\partial \mathbf{N}} \cos (\bar{e}, \mathbf{N})\)

This will be maximum when cos (\(\overline{\mathbf{e}}\) , N) = 1

⇒ Angle between e and N is zero

⇒  e is along the normal N

Therefore the directional derivative is maximum along the normal to the surface.

∴ The greatest value of the directional derivative of Φ =|grad Φ |

Note. Since ∇Φ\(=\frac{\partial \phi}{\partial \mathbf{N}} \mathbf{N}\)

∇ Φ in magnitude and direction represents the maximum rate of increase of Φ

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.

Differential Operators Exercise5(a)

1. Prove that

1. \(\nabla\left(\frac{1}{r}\right)\)\(=-\frac{\bar{r}}{r^3}\)
2. grad r= \(\frac{\mathbf{r}}{|\mathbf{r}|}\)r²=x²+y²+z

Solution:

⇒ \(|\mathbf{r}|=r=\sqrt{x^2+y^2+z^2} \Rightarrow r^2=x^2+y^2+z^2\)

∴ 2 r \(\frac{d \mathbf{r}}{d x}=2 x \Rightarrow \frac{d \mathbf{r}}{d t}=\frac{x}{r}\)

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

⇒ \(\nabla\left(\frac{1}{r}\right)=\sum \mathbf{i} \frac{\partial}{\partial x}\left(\frac{1}{r}\right)=-\frac{1}{r^2} \sum \mathbf{i} \frac{d \mathbf{r}}{d x}=-\frac{1}{r^2} \sum \frac{x \mathbf{i}}{r}=-\frac{(x \mathbf{i}+y \mathbf{j}+z \mathbf{k})}{r^3}=-\frac{\mathbf{r}}{r^3}\)

2.  If=\(\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right) \mathbf{i}+\left(z \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial z}\right) \mathbf{j}+\left(x \frac{\partial f}{\partial y}-y \frac{\partial f}{\partial x}\right) \mathbf{k}\)

prove that

1. F.r=0 
2. F. grad f=0

Solution:

1. \(\mathbf{F} . \mathbf{r}=\left[\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right) \mathbf{i}+\left(z \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial z}\right) \mathbf{j}+\left(x \frac{\partial f}{\partial y}-y \frac{\partial f}{\partial x}\right) \mathbf{k}\right]\)

= \(x\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right)+y\left(z \frac{\partial f}{\partial x}-x \frac{\partial f}{\partial z}\right)+z\left(x \frac{\partial f}{\partial y}-y \frac{\partial f}{\partial x}\right)\)

∴ \(\mathbf{F} \cdot \mathbf{r}=0\)

2. \(\mathbf{F}=\sum \mathbf{i}\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right) \text{grad} f=\nabla f=\sum \mathbf{i} \frac{\partial f}{\partial x}\)

⇒ \(\mathbf{F} \cdot \text{grad} f=\sum \frac{\partial f}{\partial x}\left(y \frac{\partial f}{\partial z}-z \frac{\partial f}{\partial y}\right)=0\)

Examples Of Differential Operator Solutions For Exercise 5(A)

3. Find grad f at the point (1,1,-2) where

1. f= x³+y³+3xyz
   
2. f= x²y+y²x+z²

Solution:

1. \(\text{grad} f=\nabla f=\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\)

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

At (1,1,-2), \(\text{grad} f=-3 \bar{i}-3 \bar{j}+3 \bar{k}\)

2. grad f = \(\nabla f=\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\)

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

At (1,1,-2), \(\text{grad} f=3 \bar{i}+3 \bar{j}-4 \bar{k}\)

4. Find the directional derivative of

1. Φ = xy+yz+zx at A in the direction of \(\overrightarrow{\mathrm{AB}}\), where A= (1,2,-1) ; B=(-1,2,3)
2. Φ=xyz at (1,1,1) in the direction of the vector i+j+k

Solution:

⇒ \(\overrightarrow{\mathbf{A B}}=\mathbf{i}(-1-1)+\mathbf{j}(2-2)+\mathbf{k}(3+1)=-2 \mathbf{i}+4 \mathbf{k}\)

Unit vector along \(\mathbf{A B}, \mathbf{e}=\frac{-2}{2 \sqrt{5}}(\mathbf{i}-2 \mathbf{k})\)

∴ Directional derivative along \(\mathbf{A B}=\mathbf{e}. \nabla \phi\)

= \(-\frac{1}{\sqrt{5}}(\mathbf{i}-2 \mathbf{k}) \cdot[\mathbf{i}(y+z)+\mathbf{j}(z+x)+\mathbf{k}(x+y)]=\sqrt{5}\) at \(\mathbf{A}(1,2,-1)\).

5. Find the angle between the surfaces x²+y²+z²=9 and x²+y²-z=3 at (2,-1,2) [Hint: The angle between the surfaces is the angle between the normals]

Solution:

Given

x²+y²+z²=9 and x²+y²-z=3 at (2,-1,2)

Let f = \(x^2+y^2+z^2-9 ; \quad \phi=x^2+y^2-z-3\) then \(\nabla f=2 x \mathbf{i}+2 \boldsymbol{j}+2 z \mathbf{k} ; \quad \nabla \phi=2 x \mathbf{i}+2 \boldsymbol{y} \mathbf{j}-\mathbf{k}\)

At (2,-1,2), \(\nabla f=4 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}\)

⇒\(\nabla \phi=4 \mathbf{i}-2 \mathbf{j}-\mathbf{k}\)

Angle between the surfaces = Angle between the normals

= \(\text{Cos}^{-1} \frac{\nabla f . \nabla \phi}{|\nabla f||\nabla \phi|}=\text{Cos}^{-1} \frac{8}{3 \sqrt{21}}\)

Exercise 5(A) Calculus Differential Operator Explanation

6. Find the angle between the surfaces xy²z=3x+z² and 3x²-y²+2z=1 at(1,-2,1)

Solution:

Given

y²z=3x+z² and 3x²-y²+2z=1 at(1,-2,1)

f = \(x y^2 z-3 x-z^2 \text { and } \phi=3 x^2-y^2+2 z-1\)

⇒ \(\nabla f=\mathbf{i}\left[\left(y^2 z\right)-3\right]+\mathbf{j}(2 x y z)+\mathbf{k}\left(x y^2-2 z\right)\)

and \(\nabla \phi=\mathbf{i}(6 x)-\mathbf{j}(2 y)+2 \mathbf{k}\)

at (1,-2,1), \(\nabla f=\mathbf{i}-4 \mathbf{j}+2 \mathbf{k} \text { and } \nabla \phi=6 \mathbf{i}+4 \mathbf{j}+2 \mathbf{k}\)

∴ Angle between the surfaces = \(\cos ^{-1} \frac{\nabla f \cdot \nabla \phi}{|\nabla f||\nabla \phi|}\)

= \(\cos ^{-1}\left(\frac{6-16+4}{\sqrt{21} \sqrt{56}}\right)=\cos ^{-1}\left(\frac{-3}{7 \sqrt{6}}\right)\)

Differential Operators Exercise 5(b)

1. prove that

1. div r=3 
2. curl r=0 
3. div (r×a)=0
4. curl(r×a)=-2a
5. grad(r.a) =a

Solution:

1. \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

div r = \(\frac{\partial}{\partial x}(x)+\frac{\partial}{\partial y}(y)+\frac{\partial}{\partial z}(z)=1+1+1=3\)

2. curl \(\bar{r}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x & y & z\end{array}\right|=0\)

3. div \((\mathbf{r} \times \mathbf{a})=\sum \mathbf{i} \cdot \frac{\partial}{\partial x}(\mathbf{r} \times \mathbf{a})\)

⇒ \(\sum \mathbf{i} \cdot\left(\frac{\partial \mathbf{r}}{\partial x} \times \mathbf{a}\right)=\sum \mathbf{i} \cdot(\mathbf{i} \times \mathbf{a})=\sum[\mathbf{i} \mathbf{a}]=0\)

4. Let \(\mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k} ; \quad \mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)

curl \((\mathbf{r} \times \mathbf{a})=\nabla \times(\mathbf{r} \times \mathbf{a})=\sum \mathbf{i} \times \frac{\partial}{\partial x}(\mathbf{r} \times \mathbf{a})\)

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

= \(\sum\left[\left(a_1\right) \mathbf{i}-\mathbf{a}\right]=\left(a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}\right)-3 \mathbf{a}=\bar{a}-3 \bar{a}=-2 \mathbf{a}\)

Differential Operators Exercise Step-By-Step Solutions

2. If =x²yi-2xzj+2yzk,find

1. div f
2. curl f

Solution:

1. div f = \(\frac{\partial}{\partial x}\left(x^2 y\right)+\frac{\partial}{\partial y}(-2 x z)+\frac{\partial}{\partial z}(2 y z)=2 x y+2 y\)

2. curl \((curl \mathrm{f})=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 y & -2 x z & 2 y z\end{array}\right|=2 \mathrm{i}(z+x)-\mathrm{k}\left(2 z+x^2\right)\)

curl(curl f) = \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 z+2 x & 0 & -2 z-x^2
\end{array}\right|=2 \mathrm{j}(1+x)\)

3.

1. Show that 3y⁴z²i+4x³z²j-3x²y²k is solenoidal.
2. if f= (x+3y)i+y-2z)j+(x+pz)k is a solenoidal find p?

Solution:

If f is a solenoidal then div f=0

∴ \(\frac{\partial}{\partial x}(x+3 y)+\frac{\partial}{\partial y}(y-2 z)+\frac{\partial}{\partial z}(x+p z)=0\)

1+1+p=0

⇒ p=-2

4. Prove that f= (sin y+z)i+ (x cosy-z)j+9x-y0k is irrotational.

Solution: If f is irrotational then curl f=0

5. If Φ =2x³y²z find div (grad Φ).

Solution:

Given

If Φ =2x³y²z

Grad \(\phi=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}=6 x^2 y^2 z^4 \mathbf{i}+4 x^3 y z^4 \mathbf{j}+8 x^3 y^2 z^3 \mathbf{k}\)

∴ div\((\text{grad} \phi)=\frac{\partial}{\partial x}\left(6 x^2 y^2 z^4\right)+\frac{\partial}{\partial y}\left(4 x^3 y z^4\right)+\frac{\partial}{\partial z}\left(8 x^3 y^2 z^3\right)\)

= \(12 x y^2 z^4+4 x^3 z^4+24 x^3 y^2 z^2\)

6. Find curl f if e ^x+y+z (i+j+k).

Solution:

f = \(e^{x+y+z}(\mathbf{i}+\mathbf{j}+\mathbf{k})\)

curl \(\mathbf{f}=\nabla \times \mathbf{f}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
e^{x+y+z} & e^{x+y+z} & e^{x+y+z}
\end{array}\right|\)

= \(\mathrm{i}\left(e^{x+y+z}-e^{x+y+z}\right)-e t c=0\)

Examples Of Differential Operator Problems

7. If f=(x+y+1) i+j-(x+y)k prove that f. curl f=0

Solution:

Given

f=(x+y+1) i+j-(x+y)k

curl \(\mathbf{f}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x+y+1 & -1 & -(x+y)
\end{array}\right|\)

= \(\mathbf{i}(0-1)-\mathbf{j}(-1-0)+\mathbf{k}(0-1)=-\mathbf{i}+\mathbf{j}-\mathbf{k}\)

∴ \(f . \text{curl} \mathbf{f}=-(x+y+1)+1+x+y=0\)

8. If A=2yzi-x²yj+xz²k and Φ=2x²yz³ find  (A×∇)Φ.

Solution:

Given

A=2yzi-x²yj+xz²k and Φ=2x²yz³

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

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

= \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
.2 y z & -x^2 y & x z^2 \\
4 x y z^3 & 2 x^2 z^3 & 6 x^2 y z^2
\end{array}\right|\)

Simplify to get the answer.

9. If Φ=x²-y² show that ∇²Φ=0

Solution:

Given

Φ=x²-y²

⇒ \(\nabla^2 \phi=\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}+\frac{\partial^2 \phi}{\partial z^2}=\frac{\partial}{\partial x}(2 x)+\frac{\partial}{\partial y}(-2 y)+0=2-2=0\)

10. Show that ∇²(x/r³)=0

Solution:

⇒ \(\nabla^2\left(\frac{x}{r^3}\right)=\sum \frac{\partial^2}{\partial x^2}\left(\frac{x}{r^3}\right)=\sum \frac{\partial}{\partial x}\left[\frac{\partial}{\partial x}\left(\frac{x}{r^3}\right)\right]\)

Now \(\frac{\partial}{\partial x}\left[\frac{\partial}{\partial x}\left(\frac{x}{r^3}\right)\right]=\frac{\partial}{\partial x}\left[\frac{1}{r^3}-\frac{3 x}{r^4} \frac{\partial r}{\partial x}\right]\)

= \(\frac{\partial}{\partial x}\left[\frac{1}{r^3}-\frac{3 x}{r^4}\left(\frac{x}{r}\right)\right] \cdot\)

(because \(r^2=x^2+y^2+z^2\) gives \(\frac{\partial r}{\partial x}=\frac{x}{r}\))

= \(\frac{\partial}{\partial x}\left[\frac{1}{r^3}-\frac{3 x^2}{r^5}\right]=-\frac{3}{r^4} \frac{\partial r}{\partial x}-\frac{6 x}{r^5}+\frac{15 x^2}{r^6} \cdot \frac{\partial r}{\partial x}\)

= \(-\frac{3}{r^4}\left(\frac{x}{r}\right)-\frac{6 x}{r^5}+\frac{15 x^2}{r^6}\left(\frac{x}{r}\right)=-\frac{9 x}{r^5}+\frac{15 x^3}{r^7}\)

Again \(\frac{\partial^2}{\partial y^2}\left(\frac{x}{r^3}\right)=\frac{\partial}{\partial y}\left[\frac{\partial}{\partial y}\left(\frac{x}{r^3}\right)\right]=\frac{\partial}{\partial y}\left[-\frac{3 x}{r^4} \cdot \frac{\partial r}{\partial y}\right]\)

= \(\frac{\partial}{\partial y}\left[-\frac{3 x}{r^4}\left(\frac{y}{r}\right)\right]\)

= \(\frac{\partial}{\partial y}\left[-\frac{3 x y}{r^5}\right]=-\frac{3 x}{r^5}+\frac{15 x y}{r^6} \cdot \frac{\partial r}{\partial y}=-\frac{3 x}{r^5}+\frac{15 x y^2}{r^7}\)

Similarly \(\frac{\partial^2}{\partial z^2}\left(\frac{x}{r^3}\right)=-\frac{3 x}{r^5}+\frac{15 x z^2}{r^7}\)

∴ \(\nabla^2\left(\frac{x}{r^3}\right)=\left(\frac{\partial^2}{\partial x}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)\left(\frac{x}{r^3}\right)\)

= \(-\frac{9 x}{r^5}+\frac{15 x^3}{r^7}-\frac{3 x}{r^5}+\frac{15 x y^2}{r^7}-\frac{3 x}{r^5}+\frac{15 x z^2}{r^7}\)

= \(-\frac{15 x}{r^5}+\frac{15 x}{r^7}\left(x^2+y^2+z^2\right)=-\frac{15 x}{r^5}+\frac{15 x}{r^7}\left(r^2\right)=0\)

Understanding Exercise 5(B) In Differential Operators

11. Show that ∇²(log r)=1/r².

Solution:

⇒ \(\nabla^2(\log r)=\sum \frac{\partial^2}{\partial x^2}(\log r)=\sum \frac{\partial}{\partial x}\left(\frac{1}{r} \frac{\partial r}{\partial x}\right)=\sum \frac{\partial}{\partial x}\left(\frac{1}{r} \cdot \frac{x}{r}\right)\)

= \(\sum \frac{\partial}{\partial x}\left(\frac{x}{r^2}\right)=\sum\left[\frac{1}{r^2}-\frac{2 x}{r^3}\left(\frac{x}{r}\right)\right]\)

= \(\sum\left(\frac{1}{r^2}-\frac{2 x^2}{r^4}\right)\)

= \(\frac{3}{r^2}-\frac{2}{r^4}\left(x^2+y^2+z^2\right)=\frac{3}{r^2}-\frac{2}{r^4}\left(r^2\right)=\frac{1}{r^2}\)

Differential Operators Solved Problems

Example. 1. Find the directional derivative of f = xy + yz + zx in the direction of the vector i + 2j + 2k at the point (1, 2, 0)

Solution:

Given f = xy + yz + zx

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

= (y + z) i + (z + x) j + (x + y) k

If e is the unit vector in the direction of the vector i + 2 j + 2k, then

⇒ \(\mathbf{e}=-\frac{\mathbf{i}+2 \mathbf{j}+2 \mathbf{k}}{\sqrt{\left(1+2^2+2^2\right)}}=\frac{1}{3}(\mathbf{i}+2 \mathbf{j}+2 \mathbf{k}) \)

The directional derivative = e.∇f

= \(\frac{1}{3}(i+2 j+2 k) \cdot[(y+z) \mathbf{i}+(z+x) \mathbf{j}+(x+y) \mathbf{k}]=\frac{1}{3}(4 x+3 y+3 z)=\frac{10}{3} \text { at }(1,2,0)\)

Gradient And Level Surface Solved Examples

Example. 2. Find the directional derivative of the function xy²+ yz² +zx² along the tangent to the curve x =t¹  y = t²,z = t³  at the point (1, 1, 1).

Solution:

Given

xy²+ yz² +zx²

Here f = xy²+ yz² + zx²

⇒ \(\begin{equation} \nabla f=\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}=\left(y^2+2 x z\right) \mathbf{i}+\left(z^2+2 x y\right) \mathbf{j}+\left(x^2+2 y z\right) \mathbf{k} \end{equation}\)

= 3 (i + j + k) at (1,1,1)

Let r be the position vector of the curve x = t¹

y = t² z = t³ , then r = t¹i + t²j + t³k

⇒ \(\begin{equation} \frac{d \mathbf{r}}{d t}=\mathbf{i}+2 t \mathbf{j}+3 t^2 \mathbf{k} \end{equation}\) = (i+2j + 3k) at (1,1,1)

dr/dt is the vector along the tangent to the curve.

The unit vector along the tangent is T

⇒ \( \begin{equation} \mathbf{T}=\frac{\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}}{\sqrt{(1+4+9)}}=\frac{1}{\sqrt{14}}(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}) \end{equation}\)

The directional derivative along the tangent

⇒ \(\begin{equation} =\mathbf{T} \cdot \nabla f=\frac{1}{\sqrt{14}}(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}) \cdot 3(\mathbf{i}+\mathbf{j}+\mathbf{k})=\frac{3}{\sqrt{14}}(1+2+3)=\frac{18}{\sqrt{14}} \end{equation}\)

Example. 3. Find the directional derivative of the function f = x² – y² + 2z² at the pt P(l, 2, 3) in the direction of the line PQ where Q = (5, 0,4) 2

Solution:

Given

f = x² – y² + 2z² at the pt P(l, 2, 3) in the direction of the line PQ where Q = (5, 0,4) 2

The position vectors of P and Q with respect to the origin O are

OP =i + 2j + 3k and OQ = 5i + 4k    PQ = OQ (or)  OP = 4i- 2j + k

If e is the unit vector in the direction of PQ then e = \( \begin{equation}\frac{4 \mathbf{i}-2 \mathbf{j}+\mathbf{k}}{\sqrt{4^2+2^2+1^2}}=\frac{1}{\sqrt{21}}(4 \mathbf{i}-2 \mathbf{j}+\mathbf{k})\end{equation}\)

Again grad f = \(\begin{equation} \mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}=2 x \mathbf{i}-2 y \mathbf{j}+4 z \mathbf{k} \end{equation} \)

The directional derivative = e.∇f

⇒ \(\begin{equation} \frac{1}{\sqrt{21}}(4 \mathbf{i}-2 \mathbf{j}+\mathbf{k}) \cdot(2 x \mathbf{i}-2 y \mathbf{j}+4 z \mathbf{k})=\frac{1}{\sqrt{21}}(8 x+4 y+4 z) \end{equation}\)

⇒ \(\begin{equation} =\frac{1}{\sqrt{21}}(8+8+12)=\frac{28}{\sqrt{21}} \text { at }(1,2,3) \end{equation}\)

Applications Of Level Surfaces In Calculus Problems

Example. 4. Find the greatest value of the directional derivative of the function f = x²y z³ at (2,1,- 1)

Solution: grad f  =2xyz³i + x²z³f + 3x²yz²k =- 4i- 4j + 12k at (2,1,-1)

Greatest value of directional derivative of f =|∇f|=\(\sqrt{16+16+144}\) \(=4 \sqrt{11}\)

Example. 5. Find the directional, derivative of Φ = x²yz + 4xz² in the direction of vector 2i−j− 2k at (1,-2,-1)

Solution: Let Φ = x²yz + 4xz² be the given function

Differentiating partially, we get \(\begin{equation} \frac{\partial \phi}{\partial x}=2 x y z+4 z^2, \frac{\partial \phi}{\partial y}=x^2 z, \frac{\partial \phi}{\partial z}=x^2 y+8 x z \end{equation}\)

grad Φ= ∇Φ\(=\frac{\partial \phi}{\partial x} \mathbf{i}+\frac{\partial \phi}{\partial y} \mathbf{j}+\frac{\partial \phi}{\partial z} \mathbf{k}\)  (2xyz+4z²)i+(x²z)j+x²y+8xz)k

Given vector is 2i-j-k Let e is the unit vector its direction.

⇒ \(\begin{equation}\mathbf{e}=\frac{2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}}{\sqrt{4+1+4}}=\frac{2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}}{3}\end{equation}\)

The direction derivative of Φ  in the direction of 2i- j- 2k

⇒ \(\begin{equation} =\nabla \phi \cdot \mathbf{e}=2\left(2 x y z+4 z^2\right)-\left(x^2 z\right)-2\left(x^2 y+8 x z\right) \end{equation}\)

Its value at (1,-2,-1) = 8 + 1 + 20 = 29

Differential Operators Example 6

Example. 6. Find the angle of intersection at (4,- 3,2) of spheres x²+ y² + z² = 29 and x² + y² + z² + 4x- 6y- 8z- 47 = 0.

Solution:

Given

(4,- 3,2) of spheres x²+ y² + z² = 29 and x² + y² + z² + 4x- 6y- 8z- 47 = 0

Let f= x² + y² + z²– 29

g = x² + y² +z² + 4x- 6y- 8z- 47

Then grad f = 2xi + 2yj + 2zk and grad g = (2x + 4) i + (2y- 6) j + (2z- 8) k

The angle between two surfaces at a point is the angle between the normal to the surfaces at that point.

Let n₁= grad f at (4,- 3, 2) = 8i- + 4k

and n₂ =grad g at (4, − 3, 2) = 12i − 12j − 4k

The vectors n₁ and n₂ are along the normals to the two surfaces at (4,- 3, 2). If 0 is the angle between these vectors, then.

n₁.n₂ =|n₁| |n₂| cos0 ⇒  96 + 72−16 \(=\sqrt{(64+36+16)} \sqrt{(144+144+16)} \cos \theta\)

∴ cos θ\(=\frac{152}{\sqrt{(116)(304)}}\) cos^-1

⇒ θ=cos^-1 \(\sqrt{(19 / 29)}\)

Differential Operators In Level Surface Examples

Example: 7 Show that (1)(a.∇)Φ = a.∇Φ  (2) (a.∇) r=a

Solution:  (1) Let a= a₁i+a₂j+a₃k, Then

a. ∇=(a₁i+a₂j+a₃k ).\(\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right)\)

⇒ \(=a_1 \frac{\partial}{\partial x}+a_2 \frac{\partial}{\partial y}+a_3 \frac{\partial}{\partial z}\)

⇒ \(=\sum a_1 \frac{\partial}{\partial x}\)

∴ (a.∇) Φ \(=\left(\Sigma a_1 \frac{\partial}{\partial x}\right)(\phi)\)\(=\Sigma a_1 \frac{\partial \phi}{\partial x}\)

Again a.∇ =(a₁i+a₂j+a₃k). \(\left(\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right)\)\(=a_1 \frac{\partial \phi}{\partial x}+a_2 \frac{\partial \phi}{\partial y}+a_3 \frac{\partial \phi}{\partial z}\)

Hence (a.∇ ) Φ = a.∇Φ

(2) r=xi+yj+zk

∴\(\frac{\partial \mathbf{r}}{\partial x}=\mathbf{i}, \frac{\partial \mathbf{r}}{\partial y}=\mathbf{j}, \frac{\partial \mathbf{r}}{\partial z}=\mathbf{k}\)

∴ (a.∇) r \(=\left(\Sigma a_1 \frac{\partial}{\partial x}\right) \mathbf{r}\)

∴ \(=\Sigma a_1 \frac{\partial \mathbf{r}}{\partial x}\)= a₁i+a₂j+a₃k =a .

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    ;  \(r=|\mathbf{r}|=\sqrt{x^2+y^2+z^2}\)

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

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

Derivative Of A Vector Function Vector With Constant Magnitude

Theorem:1  The necessary and sufficient condition that f(t) is a vector of constant magnitude is f · df/dt= 0.

Proof : (1) The condition is necessary

Let f(t) be a vector of constant magnitude. Then f(t). f(t) = |f(t)|² = const Differentiating w.r. to t we get

⇒ \(f \cdot \frac{d f}{d t}+\frac{d f}{d t} \cdot f=0 \Rightarrow f \cdot \frac{d f}{d t}=0\)

(2) Condition is sufficient \(\text { Let } f \cdot \frac{d f}{d t}=0 \text { then } \frac{1}{2} \frac{d}{d t}(f . f)=0 \Rightarrow \text { f.f }=\text { const. }\)

Continuous Magnitude Vector Function Properties

Derivative of a Vector Function Vectors With Direction Constant

Theorem: The necessary and sufficient condition for f(t) to have constant direction is \(\mathrm{f} \times \frac{\mathrm{df}}{\mathrm{dt}}=0 \text { Proof : Let } \mathrm{f}(t)=f(t) \mathbf{F}(t) \)

Where f(t) denotes the magnitude of f(t) and F(t), is a vector function with unit magnitude for every value of t :

Now, \(\mathbf{f}=f \mathbf{F} \quad \Rightarrow \quad \frac{d \mathbf{f}}{d t}=f \frac{d \mathbf{F}}{d t}+\frac{d f}{d t} \mathbf{F} \)

∴ \(\mathbf{f} \times \frac{d \mathbf{F}}{d t}=f \mathbf{F} \times\left(f \frac{d \mathbf{f t}}{d t}+\frac{d f}{d t} \mathbf{F}\right)=f^2 \mathbf{F} \times \frac{d \mathbf{F}}{d t}\)

∵ F × F = 0

(1) The condition is necessary

Let the direction of f be constant. Then F = const ⇒ \(\frac{d F}{d t}=0 \Rightarrow f \times \frac{d f}{d t}=0\)

(2) The condition is sufficient

Let \(\mathrm{f} \times \frac{d \mathrm{f}}{d t}=0 \text { i.e. } f^2 \mathrm{~F} \times \frac{d \mathrm{~F}}{d t}=0 \Rightarrow \mathbf{F} \times \frac{d \mathrm{~F}}{d t}=0 \quad[because f \neq 0]\)

Also F being of constant magnitude \(\mathrm{F} \cdot \frac{d \mathrm{~F}}{d t}=0\)

Hence from (1) and (2) we have \(\mathrm{F} \cdot \frac{d \mathrm{~F}}{d t}=0\) ⇒ F = constant i.e. the direction of f remains the same.

Derivative of a Vector Function Composite Vector Function

Theorem: Let s be a scalar function defined over the domain S and differentiable at t ∈ S. If f is a vector function differentiable at s(t) in the range of functions then the composite function f(s) is differentiable at t and \(f[s(t)]^{\prime}=\mathrm{f}^{\prime}[s(t)] s^{\prime}(t) \text { or } \frac{d f}{d t}=\frac{d f}{d x} \frac{d s}{d t}\)

Proof: Since f(s) and s(t) are differentiable, we have \(\frac{d \Delta}{d t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\delta s}{\delta t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\Delta(t+\delta t)-s(t)}{\delta t} \text { and } \frac{d f}{d s}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\delta \mathrm{f}}{\delta s}=\mathrm{L}_{\delta t \rightarrow 0} \frac{\mathrm{r}(t+\delta t)-\mathrm{f}(t)}{\delta t}\)

Now \({Lt}_{\delta t \rightarrow 0} \frac{\mathrm{f}[s(t+\delta t)]-\mathrm{f}[s(t)]}{\delta t}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathrm{f}[s(t+\delta t)]-\mathrm{f}[a(t)]}{\lambda(t+\delta t)-s(t)} \cdot \frac{\Delta(t+\delta t)-\Delta(t)}{\delta t} .\)

= \(\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathrm{f}(s+\delta t)-\mathrm{f}(s)}{\delta t}\)

⇒ \(\mathrm{Lt}_{\delta t \rightarrow 0} \frac{s(t+\delta t)-s(t)}{d t}=\frac{d r}{d s} \cdot \frac{d a}{d t}\)

Thus f[s(t)] is differentiable at t and \(\frac{d f}{d t}=\frac{d f}{d s}-\frac{d s}{d t}\)

Understanding Continuous Magnitude In Vector Functions

Derivative of a Vector Function Velocity And Acceleration Of A Particle

If r is the position vector of a point P in space and if r(t) = x(t)i+ y(t)j +z(t)k is a function of time t, then \(\frac{d \mathbf{r}}{d t}\) will represent the velocity of the point P at any time and it is directed by v.

v = \(\frac{d \mathbf{r}}{d t}\) = velocity of P at any time

Similarly,\( \frac{d v}{d t}=\frac{d^2 r}{d t^2}\) represents the acceleration of the particle at any time and is denoted by a

a = \( \frac{d v}{d t}=\frac{d^2 r}{d t^2}\) = accelaration of P at any time.

Differential Operators Theorems

Theorem1. If the vector f is expressible in terms of its  Cartesian components  f₁,f₂,f₃ as f= f₁i+f₂j+f₃k, Then ∇.f\(=\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial v}+\frac{\partial f_3}{\partial z}\)

Proof:

Given f=f₁i+f₂j+f₃k

∴ div f =∇ .f\(=\mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \cdot \frac{\partial \mathbf{f}}{\partial v}+\mathbf{k} \cdot \frac{\partial \mathbf{f}}{\partial \mathbf{z}}\)

⇒ \(=\mathbf{i} \cdot\left(\mathbf{i} \frac{\partial f_1}{\partial x}+\mathbf{j} \frac{\partial f_2}{\partial x}+\mathbf{k} \frac{\partial f_3}{\partial x}\right)+\mathbf{j} \cdot\left(\mathbf{i} \frac{\partial f_1}{\partial y}+\mathbf{j} \frac{\partial f_2}{\partial y}+\mathbf{k} \frac{\partial f_3}{\partial y}\right)\)\(+\mathbf{k} \cdot\left(\mathbf{i} \frac{\partial f_1}{\partial z}+\mathbf{j} \frac{\partial f_2}{\partial z}+\mathbf{k} \cdot \frac{\partial f_3}{\partial z}\right)\)

=\(\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}\)

Theorem 2. Prove that \(\text{div}(\mathbf{f} \pm \mathbf{g})=\text{div} . \mathbf{f} \pm \text{div} . \mathbf{g}\)

Proof:

⇒ \(\text{div}(\mathbf{f} \pm \mathbf{g})=\Sigma \mathbf{i} \cdot \frac{\partial}{\partial x}(\mathbf{f} \pm \mathbf{g})=\Sigma \mathbf{i} \cdot\left(\frac{\partial \mathbf{f}}{\partial x} \pm \frac{\partial \mathbf{g}}{\partial x}\right)\)

= \(\Sigma \mathbf{i} \cdot \frac{\partial \mathbf{f}}{\partial x} \pm \Sigma \mathbf{i} \cdot \frac{\partial \mathbf{g}}{\partial x}=\text{div} \mathbf{f} \pm \text{div} \mathbf{g}\)

Differential Operator’s Theorems And Proofs

Differential Operators The Laplacian Operator∇²

⇒ \(\nabla . \nabla \phi=\Sigma i \cdot \frac{\partial}{\partial x}\left(i \frac{\partial \phi}{\partial x}+j \frac{\partial \phi}{\partial y}+k \frac{\partial \phi}{\partial z}\right)=\sum \frac{\partial^2 \phi}{\partial x^2}\)

= \(\frac{\partial^2 \phi}{\partial x^2}+\frac{\partial^2 \phi}{\partial y^2}+\frac{\partial^2 \phi}{\partial z^2}=\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right) \phi=\nabla^2 \phi\)

Thus the operator \(\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\) is called the Laplacian operator.

Note: \(\nabla^2 \phi=0\) is called Laplacian equation.

Differential Operators Curl Of A Vector

Definition: Let \(\mathbf{f}\) be any continuously differentiable vector point function. Then the vector function defined by \(\mathbf{i} \times \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{f}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{f}}{\partial z}\) is called the Curl of \(\mathbf{f}\) and is denoted by \(\nabla \times \mathbf{f}\)

∴ Curl \(\mathbf{f}=\mathbf{i} \times \frac{\partial \mathbf{f}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{f}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{f}}{\partial z}=\left(\mathbf{i} \times \frac{\partial}{\partial x}+\mathbf{j} \times \frac{\partial}{\partial y}+\mathbf{k} \times \frac{\partial}{\partial z}\right) \mathbf{f}\)

= \(\left(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\right) \times \mathbf{f}=\nabla \times \mathbf{f}\)

Theorem: If f is a differentiable vector point function given by f=f₁i+f₂j+f₃k then, curl f 

Proof:

Curl \(\left.\mathbf{f}=\nabla \times \mathbf{f}=\sum \mathbf{i} \times \frac{\partial \mathbf{t}}{\partial x}=\sum \mathbf{i} \times \frac{\partial}{\partial x} f_1 \mathbf{i}+f_2 \mathbf{j}+f_3 \mathbf{k}\right)\)

= \(\sum\left(\frac{\partial f_2}{\partial x} \mathbf{k}-\frac{\partial f_3 \mathbf{j}}{\partial x}\right)=\left(\frac{\partial f_2}{\partial x} \mathbf{k}-\frac{\partial f_3}{\partial x} \mathbf{j}\right)+\left(\frac{\partial f_3}{\partial y} \mathbf{i}-\frac{\partial f_1}{\partial y} \mathbf{k}\right)+\left(\frac{\partial f_1}{\partial z} \mathbf{j}-\frac{\partial f_2}{\partial z} \mathbf{i}\right)\)

= \(\mathbf{i}\left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z}\right)+\mathbf{j}\left(\frac{\partial f_1}{\partial z}-\frac{\partial f_3}{\partial x}\right)+\mathbf{k}\left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right)\)

Note 1. The expression for Curl f can be well remembered if we treat \(\nabla\) as an operative vector quantity that is Curl \(\mathbf{f}=\left|\begin{array}{lll}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ f_1 & f_2 & f_3\end{array}\right|\)

2. If f is a constant vector then \(\text{Curl} \mathbf{f}=\mathbf{0}\)

Laplacian Operator ∇² Vector Identities

Differential Operators Irrotational Vector

A vector point function f is said to be irrotational if. Curl f = 0 i.e. ∇x f = 0

Physical Interpretation of curl If w is the angular velocity of a rigid body rotating about a fixed axis and v is the velocity of any point P(x,y,z) on the body, then w =1/2 curl v . Thus the angular velocity of rotation at any point is equal to half the curl of the velocity vector. This justifies the use of the word “curl of a vector”.’

Theorem Prove that curl (A ± B) = curl A ± curl B

Proof: 

Curl \((\mathrm{A} \pm \mathrm{B})=\nabla \times(\mathrm{A} \pm \mathrm{B})=\sum \mathbf{i} \times \frac{\partial}{\partial x}(\mathrm{~A} \pm \mathrm{B})=\sum \mathbf{i} \times\left(\frac{\partial \mathrm{A}}{\partial x} \pm \frac{\partial \mathrm{B}}{\partial x}\right)\)

= \(\sum \mathbf{i} \times \frac{\partial \mathrm{A}}{\partial x} \pm \sum \mathbf{i} \times \frac{\partial \mathrm{B}}{\partial x}=\nabla \times \mathrm{A} \pm \nabla \times \mathrm{B}=\text{curl} \mathrm{A} \pm \text{curl} \mathrm{B}\)

Theorem: The differential operators’ grad, div, and curl are invariant with respect to any Cartesian coordinate frame.

Proof: Let P (x, y, z) be a point in the coordinate frame OXYZ and i, j, k be the right-handed unit vectors along the axes. Let (x’, y’,z’ ) be the coordinates of P.w.r. to the new coordinate frame O’ X’ Y’ Z’.

If \(\mathbf{i}^{\prime}, \mathbf{j}^{\prime}, \mathbf{k}^{\prime}\) are the right handed unit vectors along \(\overrightarrow{\mathrm{OX}^{\prime}}, \overrightarrow{\mathrm{OY}^{\prime}}, \overrightarrow{\mathrm{OZ}^{\prime}}\), then \(\overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{OO}^{\prime}+} \overrightarrow{\mathrm{O}^{\prime} \mathrm{P}^{\prime}}\)

⇒ \(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}=\overrightarrow{\mathrm{OO}^{\prime}}+x^{\prime} \mathbf{i}^{\prime}+y^{\prime} \mathbf{j}^{\prime}+z^{\prime} \mathbf{k}^{\prime}\)

As \(x^{\prime}, y^{\prime}, z^{\prime}\) are the functions of x y z differentiating partially with respect to x, y, z we get \(\mathbf{i}=\mathbf{i}^{\prime} \frac{\partial x^{\prime}}{\partial x}+\mathbf{j}^{\prime} \frac{\partial y^{\prime}}{\partial x}+\mathbf{k}^{\prime} \frac{\partial z^{\prime}}{\partial x}\)…..(1)

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

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

Now taking the dot product of (1), (2), (3) with \(\mathbf{i}^{\prime}, \mathbf{j}^{\prime}, \mathbf{k}^{\prime}\) we get \(\mathbf{i}. \mathbf{i}^{\prime}=\frac{\partial x^{\prime}}{\partial x}, \mathbf{i} . \mathbf{j}^{\prime} \frac{\partial y^{\prime}}{\partial x}, \mathbf{i} \cdot \mathbf{k}^{\prime} \frac{\partial z^{\prime}}{\partial x}\)

⇒ \(\mathbf{j} \cdot \mathbf{i}^{\prime}=\frac{\partial x^{\prime}}{\partial y}, \mathbf{j} \cdot \mathbf{j} \frac{\partial y^{\prime}}{\partial y}, \mathbf{j} \cdot \mathbf{k} \frac{\partial z^{\prime}}{\partial y}\)

⇒ \(\mathbf{k} \cdot \mathbf{i}^{\prime} \frac{\partial x^{\prime}}{\partial z}, \mathbf{k} \cdot \mathbf{j}^{\prime} \frac{\partial y^{\prime}}{\partial z}, \mathbf{k} \cdot \mathbf{k} \frac{\partial z^{\prime}}{\partial z}\)

Now by the principle of partial differentiation

1. \(\frac{\partial}{\partial y}=\frac{\partial}{\partial x^{\prime}} \frac{\partial x^{\prime}}{\partial x}+\frac{\partial}{\partial y^{\prime}} \frac{\partial y^{\prime}}{\partial x}+\frac{\partial}{\partial z^{\prime}} \frac{\partial z^{\prime}}{\partial x}=(i.i)\frac{\partial}{\partial x^{\prime}}+ (i. \left.\mathbf{j}^{\prime}\right) \frac{\partial}{\partial y^{\prime}}+(i. \left.\mathbf{k}^{\prime}\right) \frac{\partial}{\partial z^{\prime}} \quad\)……. from (4)

2. \(\frac{\partial}{\partial y}=\frac{\partial}{\partial x^{\prime}} \frac{\partial x^{\prime}}{\partial z}+\frac{\partial}{\partial y^{\prime}} \frac{\partial y^{\prime}}{\partial z}+\frac{\partial}{\partial z^{\prime}} \frac{\partial z^{\prime}}{\partial z}=\left(\mathbf{k} \cdot \mathbf{i}^{\prime}\right) \frac{\partial}{\partial x^2}+\left(\mathbf{j} \cdot \mathbf{j}^{\prime}\right) \frac{\partial}{\partial y^1}+\left(\mathbf{j} \cdot \mathbf{k}^{\prime}\right) \frac{\partial}{\partial z} \quad \cdots\) from (5)

3. \(\frac{\partial}{\partial z}=\frac{\partial}{\partial x^{\prime}} \frac{\partial x^{\prime}}{\partial z}+\frac{\partial}{\partial y^{\prime}} \frac{\partial y^{\prime}}{\partial z}+\frac{\partial}{\partial z^{\prime}} \frac{\partial z^{\prime}}{\partial z}=\left(\mathbf{k} \cdot \mathbf{i}^{\prime}\right) \frac{\partial}{\partial x^{\prime}}+(\mathbf{k} \cdot \mathbf{j}) \frac{\partial}{\partial y^{\prime}}+\left(\mathbf{k} \cdot \mathbf{k}^{\prime}\right) \frac{\partial}{\partial z^{\prime}} \ldots\) from (6)

Also, we know that \(\left.\begin{array}{l}
\mathbf{i}^{\prime}=\left(\mathbf{i}^{\prime} \cdot \mathbf{i}\right) \mathbf{i}+\left(\mathbf{i}^{\prime} \cdot \mathbf{j}\right) \mathbf{j}^{\prime}+\left(\mathbf{i}^{\prime} \cdot \mathbf{k}\right) \mathbf{k} \\
\mathbf{j}^{\prime}=\left(\mathbf{j}^{\prime} \cdot \mathbf{i}\right) \mathbf{i}+(\mathbf{j} \cdot \mathbf{j}) \mathbf{j}+(\mathbf{j} \cdot \mathbf{k}) \mathbf{k} \\
\mathbf{k}^{\prime}=\left(\mathbf{k}^{\prime} \cdot \mathbf{i}\right) \mathbf{i}+\left(\mathbf{k}^{\prime} \cdot \mathbf{j}\right) \mathbf{j}+\left(\mathbf{k}^{\prime} \cdot \mathbf{k}\right) \mathbf{k}
\end{array}\right\} \cdots\)….(A)

1. To Prove grad is invariant

Multiplying (1), (2) by \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) and adding we get \(\mathbf{i} \frac{\partial}{\partial x}+\mathbf{j} \frac{\partial}{\partial y}+\mathbf{k} \frac{\partial}{\partial z}\)

= \(\left[\mathbf{i}\left(\mathbf{i} \cdot \mathbf{i}^{\prime}\right)+\mathbf{j}\left(\mathbf{j} \cdot \mathbf{i}^{\prime}\right)+\right. \left.\mathbf{k}\left(\mathbf{k} \cdot \mathbf{i}^{\prime}\right)\right] \frac{\partial}{\partial x^{\prime}}+\left[\mathbf{i}\left(\mathbf{i} \cdot \mathbf{j}^{\prime}\right)+\mathbf{j}\left(\mathbf{j} \cdot \mathbf{j}^{\prime}\right)+\mathbf{k}\left(\mathbf{k} \cdot \mathbf{j}^{\prime}\right)\right] \frac{\partial}{\partial y^{\prime}}\)

+ \(\left[\mathbf{i}\left(\mathbf{i} \cdot \mathbf{k}^{\prime}\right)+\mathbf{j}\left(\mathbf{j} \cdot \mathbf{k}^{\prime}\right)+\mathbf{k}\left(\mathbf{k} \cdot \mathbf{k}^{\prime}\right)\right] \frac{\partial}{\partial z^{\prime}}=\mathbf{i}^{\prime} \frac{\partial}{\partial x^{\prime}}+\mathbf{j}^{\prime} \frac{\partial}{\partial y^{\prime}}+\mathbf{k}^{\prime} \frac{\partial}{\partial z^{\prime}}\)

Therefore grad is invariant.

2. To prove div is invariant

Taking dot product of (4), (5), (6) with \(\mathbf{i}, \mathbf{j}, \mathbf{k}\)

Taking the cross product of (4), (5), (6) with \(\mathbf{i}, \mathbf{j}, \mathbf{k}\)

⇒ \(\mathbf{i} \times \frac{\partial}{\partial x}=\left(\mathbf{i} . \mathbf{i}^{\prime}\right) \mathbf{i} \times \frac{\partial}{\partial x^{\prime}}+\left(\mathbf{i} \cdot \mathbf{j}^{\prime}\right) \mathbf{i} \times \frac{\partial}{\partial y^{\prime}}+\left(\mathbf{i} \cdot \mathbf{k}^{\prime}\right) \mathbf{i} \times \frac{\partial}{\partial z^{\prime}}\)

⇒ \(\mathbf{j} \times \frac{\partial}{\partial y}=\left(\mathbf{j} \cdot \mathbf{i}^{\prime}\right) \mathbf{j} \times \frac{\partial}{\partial x^{\prime}}+\left(\mathbf{j} \cdot \mathbf{j}^{\prime}\right) \mathbf{j} \times \frac{\partial}{\partial y^{\prime}}+\left(\mathbf{j} \cdot \mathbf{k}^{\prime}\right) \mathbf{k} \times \frac{\partial}{\partial z^{\prime}}\)

⇒ \(\mathbf{k} \times \frac{\partial}{\partial z}=\left(\mathbf{k} \cdot \mathbf{i}^{\prime}\right) \mathbf{k} \times \frac{\partial}{\partial x^{\prime}}+\left(\mathbf{k} \cdot \mathbf{j}^{\prime}\right) \mathbf{k} \times \frac{\partial}{\partial y^{\prime}}+\left(\mathbf{k} \cdot \mathbf{k}^{\prime}\right) \mathbf{k} \times \frac{\partial}{\partial z^{\prime}}\)

Adding vertically and applying (A) \(\mathbf{i} \times \frac{\partial}{\partial x}+\mathbf{j} \times \frac{\partial}{\partial y}+\mathbf{k} \times \frac{\partial}{\partial z}=\mathbf{i}^{\prime} \times \frac{\partial}{\partial x^{\prime}}+\mathbf{j}^{\prime} \times \frac{\partial}{\partial y^{\prime}}+\mathbf{k}^{\prime} \times \frac{\partial}{\partial z^{\prime}} \text {. }\)

Thus curl is invariant.

Vector Calculus Identities Involving Laplacian Operator

Differential Operators Vector Identities

Theorem1. If A is a differential vector function and Φ is a differentiable scalar function, then prove that div ( Φ A)=(grade Φ).A+ Φdiv A or ∇. ( Φ A) + Φ (∇.A)

Proof: div\((\phi \mathbf{A})=\nabla \cdot \phi(\mathbf{A})=\Sigma \mathbf{i} \cdot \frac{\partial}{\partial x}(\phi \mathbf{A})=\sum \mathbf{i} \cdot\left(\frac{\partial \phi}{\partial x} \mathbf{A}+\phi \frac{\partial \mathbf{A}}{\partial x}\right)\)

= \(\sum\left(\mathbf{i} \frac{\partial \phi}{\partial x}\right) \cdot \mathbf{A}+\sum\left(\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}\right) \phi\)

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

Theorem 2. Prove that curl (Φ A)=(gradeΦ) × A + Φ curl A or ∇× (Φ A)= (∇Φ) × A +Φ(∇ ×A)

Proof: curl\((\phi \mathbf{A})=\nabla \times(\phi \mathbf{A})=\Sigma \mathbf{i} \times \frac{\partial}{\partial x}(\phi \mathbf{A})\)

= \(\sum \mathbf{i} \times\left(\frac{\partial \phi}{\partial x} \mathbf{A}+\phi \frac{\partial \mathbf{A}}{\partial x}\right)=\sum\left(\mathbf{i} \frac{\partial \phi}{\partial x}\right) \times \mathbf{A}+\sum\left(\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}\right) \phi=\nabla \phi \times \mathbf{A}+(\nabla \times \mathbf{A}) \phi\)

Theorem 3. Prove that grad(A.B) = (B .∇) A+(A.∇) B+B × curl A+A × curl B

Proof: Now \(\mathbf{A} \times \text{curl} \mathbf{B}=\mathbf{A} \times(\nabla \times \mathbf{B})\)

= \(\mathbf{A} \times \sum \mathbf{i} \times \frac{\partial \mathbf{B}}{\partial x}=\sum \mathbf{A} \times\left(\mathbf{i} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)

= \(\sum\left\{\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{i}-(\mathbf{A} \cdot \mathbf{i}) \frac{\partial \mathbf{B}}{\partial x}\right\}=\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)-\left(\mathbf{A} \cdot \sum \mathbf{i} \frac{\partial}{\partial x}\right) \mathbf{B}\)

∴ \(\mathbf{A} \times \text{curl} \mathbf{B}=\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)-(\mathbf{A} \cdot \nabla) \mathbf{B}\)

⇒ \(\mathbf{A} \times \text{curl} \mathbf{B}+(\mathbf{A} \cdot \nabla) \mathbf{B}=\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)\)

Similarly \(\mathbf{B} \times \text{curl} \mathbf{A}=\sum \mathbf{i}\left(\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial x}\right)-(\mathbf{B} \cdot \nabla) \mathbf{A}\)

⇒ \(\mathbf{B} \times \text{curl} \mathbf{A}+(\mathbf{B} \cdot \nabla) \mathbf{A}=\sum \mathbf{i}\left(\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial x}\right)\)

Adding (1) and (2) we have

∴ \(\mathbf{A} \times \text{curl} \mathbf{B}+(\mathbf{A} \cdot \nabla) \mathbf{B}+\mathbf{B} \times \text{curl} \mathbf{A}+(\mathbf{B} \cdot \nabla) \mathbf{A}\)

= \(\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}\right)+\sum \mathbf{i}\left(\mathbf{B} \cdot \frac{\partial \mathbf{A}}{\partial x}\right)=\sum \mathbf{i}\left(\mathbf{A} \cdot \frac{\partial \mathbf{B}}{\partial x}+\frac{\partial \mathbf{A}}{\partial x} \cdot \mathbf{B}\right)\)

= \(\sum \mathbf{i} \frac{\partial}{\partial x}(\mathbf{A} \cdot \mathbf{B})=\nabla(\mathbf{A} \cdot \mathbf{B})=\text{grad}(\mathbf{A} \cdot \mathbf{B})\)

Theorems For Laplacian Operator In Scalar And Vector Fields

Theorem 4. Prove that div (A×B)= B.curlA-A, curl B ∇ . (A×B) =B.(∇×A) – A . (∇×B)

Proof: \(\text{div}(\mathbf{A} \times \mathbf{B})=\nabla \cdot(\mathbf{A} \times \mathbf{B})\)

= \(\sum \mathbf{i} \cdot \frac{\partial}{\partial x}(\mathbf{A} \times \mathbf{B})=\sum \mathbf{i} \cdot\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)

= \(\sum \mathbf{i} .\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}\right)+\sum \mathbf{i} .\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)

= \(\sum\left(\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}\right) \cdot \mathbf{B}-\left(\sum \mathbf{i} \times \frac{\partial \mathbf{B}}{\partial x}\right) \cdot \mathbf{A}\)

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

Note. If \(\overline{\mathrm{A}}\) and \(\overline{\mathrm{B}}\) are irrotational then \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}\) is solenoidal.

⇒ \(\overline{\mathrm{A}}, \overline{\mathrm{B}}\) are irrotational.

⇒ \(\nabla \times \overline{\mathrm{A}}=\overline{0}, \nabla \times \overline{\mathrm{B}}=\overline{0}\) substituting in the theorem, \(\nabla(\overline{\mathrm{A}} \times \overline{\mathrm{B}})=\overline{\mathrm{B}} \cdot(\nabla \times \overline{\mathrm{A}})-\overline{\mathrm{A}} \cdot(\nabla \cdot \overline{\mathrm{B}})=0\)

∴ \(\overline{\mathrm{A}} \times \overline{\mathrm{B}}\) is solenoidal.

Theorem 5. Prove that curl(A×B) = Adiv B-B div A +(B.∇)A-(A.∇)B

Proof: curl\((\mathbf{A} \times \mathbf{B})=\nabla \times(\mathbf{A} \times \mathbf{B})\)

= \(\sum \mathbf{i} \times \frac{\partial}{\partial x}(\mathbf{A} \times \mathbf{B})=\sum \mathbf{i} \times\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)

= \(\sum \mathbf{i} \times\left(\frac{\partial \mathbf{A}}{\partial x} \times \mathbf{B}\right)+\sum \mathbf{i} \times\left(\mathbf{A} \times \frac{\partial \mathbf{B}}{\partial x}\right)\)

= \(\sum\left\{(\mathbf{i} \cdot \mathbf{B}) \frac{\partial \mathbf{A}}{\partial x}-\left(\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}\right) \mathbf{B}\right\}+\sum\left\{\left(\mathbf{i} \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{A}-(\mathbf{i} \cdot \mathbf{A}) \frac{\partial \mathbf{B}}{\partial x}\right\}\)

= \(\sum(\mathbf{B} \cdot \mathbf{i}) \frac{\partial \mathbf{A}}{\partial x}-\sum\left(\mathbf{i} \cdot \frac{\partial \mathbf{A}}{\partial x}\right) \mathbf{B}+\sum\left(\mathbf{i} \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{A}-\sum(\mathbf{A} \cdot \mathbf{i}) \frac{\partial \mathbf{B}}{\partial x}\)

= \(\left(\mathbf{B} \cdot \sum i \frac{\partial}{\partial x}\right) \mathbf{A}-\left(\sum i \cdot \frac{\partial \mathbf{A}}{\partial x}\right) \mathbf{B}+\left(\sum i \cdot \frac{\partial \mathbf{B}}{\partial x}\right) \mathbf{A}-\left(\mathbf{A} \cdot \sum i \frac{\partial}{\partial x}\right) \mathbf{B}\)

= \((\mathbf{B} \cdot \nabla) \mathbf{A}-(\nabla \cdot \mathbf{A}) \mathbf{B}+(\nabla \cdot \mathbf{B}) \mathbf{A}-(\mathbf{A} \cdot \nabla) \mathbf{B}\)

= \(\mathbf{A} \text{div} \mathbf{B}-\mathbf{B} \text{div} \mathbf{A}+(\mathbf{B} \cdot \nabla) \mathbf{A}-(\mathbf{A} \cdot \nabla) \mathbf{B}\)

Theorem 6. Prove that curl grad Φ=0

Proof: \(\text{grad} \phi=\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\)

curl\((\text{grad} \phi)=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z}
\end{array}\right|\)

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

Note : Since Curl \((\text{grad} \phi)=0\) grad \(\phi\) is always irrotational.

Understanding Theorems Involving ∇² In Differential Operators

Theorem 7. Prove that div. curl f=0

Proof:

Let \(\mathbf{f}=f_1 \mathbf{i}+f_2 \mathbf{j}+f_3 \mathbf{k}\)

∴ \(\text{curl}=\nabla \times \mathbf{f}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
f_1 & f_2 & f_3
\end{array}\right|\)

= \(\left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z}\right) \mathbf{i}-\left(\frac{\partial f_3}{\partial x}-\frac{\partial f_1}{\partial z}\right) \mathbf{j}+\left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right) \mathbf{k}.\)

∴ \(\text{div}(\text{curl} \mathbf{f})=\nabla \cdot(\nabla \times \mathbf{f})\)

= \(\frac{\partial}{\partial x}\left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z}\right)-\frac{\partial}{\partial y}\left(\frac{\partial f_3}{\partial x}-\frac{\partial f_1}{\partial z}\right)+\frac{\partial}{\partial z}\left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right)\)

= \(\frac{\partial^2 f_3}{\partial x \partial y}-\frac{\partial^2 f_2}{\partial x \partial z}-\frac{\partial^2 f_3}{\partial y \partial x}+\frac{\partial^2 f_1}{\partial y \partial z}+\frac{\partial^2 f_2}{\partial x \partial z}-\frac{\partial^2 f_1}{\partial z \partial y}=0\)

Note. Since div. (curl \(\mathbf{f})=0\), curl \(\mathbf{f}\) is always solenoidal.

Theorem 8. Prove that ∇× (∇×A)=∇(∇.A) − ∇² A

Proof: \(\nabla \times(\nabla \times \mathbf{A})=\sum \mathbf{i} \times \frac{\partial}{\partial x}(\nabla \times \mathbf{A})\)

Now, \(\mathbf{i} \times \frac{\partial}{\partial x}(\nabla \times \mathbf{A})=\mathbf{i} \times \frac{\partial}{\partial x}\left(\mathbf{i} \times \frac{\partial \mathbf{A}}{\partial x}+\mathbf{j} \times \frac{\partial \mathbf{A}}{\partial y}+\mathbf{k} \times \frac{\partial \mathbf{A}}{\partial z}\right)\)

= \(\mathbf{i} \times\left(\mathbf{i} \times \frac{\partial^2 \mathbf{A}}{\partial x^2}+\mathbf{j} \times \frac{\partial^2 \mathbf{A}}{\partial x \partial y}+\mathbf{k} \times \frac{\partial^2 \mathbf{A}}{\partial x \partial z}\right)\)

= \(\mathbf{i} \times\left(\mathbf{i} \times \frac{\partial^2 \mathbf{A}}{\partial x^2}\right)+\mathbf{i} \times\left(\mathbf{j} \times \frac{\partial^2 \mathbf{A}}{\partial x \partial y}\right)+\mathbf{i} \times\left(\mathbf{k} \times \frac{\partial^2 \mathbf{A}}{\partial x \partial z}\right)\)

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

Differential Operators Scalar Potential Of An Irritational Vector

Definition: a vector f is said to be irrotational if curl f=0 If f is irrotational, there will always exist a scalar function φ(x,y,z) such that f= grade Φ.This Φ is called the Scalar potential of f.
it is easy to prove that, if f= grad Φ, then curl f=0. Hence ∇× f=0  ⇔ there exists a scalar function Φ such that f=∇Φ.