Teja

Subrings, Ideals, Quotient Rings & Euclidean Rings Subrings

In analogy with the concept of a subgroup of a group, we now introduce the concept of a subring. If (R, +, •) is a ring then a non-empty subset of R with the induced operations +, * as in R can be a ring. Such a ring is called a subring of the ring R.

Definition. (Subring). Let (R,+,•) be a ring and S be a non-empty subset of R. If (S, +, •) is also a ring with respect to the two operations +, • in R then (S, +, •) is a subring of R.

The binary operations in S thus defined are the induced operations in S from R.

Definition. Let (F,+, • ) be a field and (S, +, •) be a subring ofF. If (S, +, •) is a field then we say that S is a subfield of F. If (S, +, •) is an integral domain then we say that S is a subdomain of F.

Note. 1. If (S’, +, •) is a subring of the ring (R, +, •) then (S, + ) is a subgroup of the (R, +) group. Hence zero element in R is also zero element in S.

If (S, +, •) is a subfield of the field (F. +. •) then (i) (S, + ) is a subgroup of (F, + ) group and is a subgroup of (F-{0}) group.

Subrings, Ideals, Quotient Rings, And Euclidean Rings Theorems

Examples of Subrings, Ideals, Quotient Rings, And Euclidean Rings Subrings

Example1. The set of even integers is a subring of (Z, +,”•) ring or integral domain.

Example 2. (Z, +, •), (Q, +, •) are subrings of the field of real numbers (R, +, •).

Example3. Let (Q, +,•) be the ring of rational numbers. Then S = {sr2|a Zj is a non­empty subset of Q and (S, +) is a subgroup of the group (Q, +).

But for 1 / 2 ∈S we have (1 / 2). (1 / 2). = (1 / 4)  S and hence is not a binary operation in S. Thus (S, +, •) is not a subring of (Q, +,•)

Example4. Let (R, +, •) be a ring and 0 e R be the zero elements of R. Then S = {0} is a non-empty subset of R so that (S, +, •) is itself a ring. Therefore (S, +, •) is a subring of R.

{0} is called trivial subring and (R, +,•) is called improper subring of R.

Example5. For each positive integer n, the set nz = (0,± n,± 2n,± 3n,…….. } is a subring of Z. ‘

Example6. The set of Gaussian integers Z, i] = {a+b ia,b∈Z,i2 =-1} is a subring of the complex number field C.

Theorem 1. (Subring Test). Let S be a non-empty subset of a ring R. Then S is a subring of R if and only if a-b in S and a/b in S for all a/b in S.

Proof. Let S be a subring of R.

We now prove that a-b e S and ab e S V a,b e S . Since S’ is a subring of R, S is a ring with respect to the addition and multiplication operations in R

a,b ∈ S=>a,-b ∈ S => a + (-b) = a-b <=S and a,b ∈ S => ab ∈ S

Let a-b ∈ S and ab ∈ S ∀a,b ∈ S.

We now prove that S is a ring.

Since S is a nonempty subset of the commutative group (R, +) with the condition a-b ∈ S and ab ∈ S; by group theory (S, +) is a commutative subgroup of (R, +).

Since ab ∈ S∀a, b ∈ S, multiplication (•) is a binary operation in S.

Also, a,b,c ∈ S => a,b,c ∈ R => a (bc) = (ab) c

Further a, b,c ∈ S=>a,b,c ∈ R => a (b + c) = ab + ac and (b + c) a = ba + ca

(S, +, •) is a ring and hence (S, +, •) is a subring of R.

Note. Every subring contains at least zero elements of the ring.

Examples Of Ideals And Euclidean Rings In Algebra

Theorem 2. (Subfield Test). Let K be a non-empty subset of a field F. Then K is a Subfield of F if and only if a,b ∈ K => a-b ∈ K and a ∈ K,b ≠ 0 ∈ K => ab-1 ∈ K

Example 1. nZ is a subdomain of Z.

we know that (Z,+, •) where Z = the set of all integers is an integral domain.

For a fixed n∈  Z we have nZ ={ nx /x ∈Z }

0∈Z is Zero element and n0 =0=> 0 ∈nZ.

∴ nZ ≠Φ and n Z⊂ Z

Let  x,y ∈  Z. Then nx, ny ∈  nZ.

nx-ny =n(x-y) ∈ nZ ;  ( x-y ∈ Z)

Also (nx) (ny) =n(x n y) ∈ nZ;  (  x. n y ∈ Z ) n Z is a subdomain of Z.

Example 2. Z is not a subfield of Q. For 2,3 ∈ Z and 3 ≠  0 => 3-1 = (1 /3)∈ Q.

But 2.3-1 = (2/3)∈ Z

Example 3. The Unity element of a ring need not be the same as the unity element of a subring.

Consider Z6 = {0, 1, 2, 3, 4, 5} the ring with unity element 1.

For the subring s = { 0,2,4 } ; we have 0. 4 = 4. 0 =0; 2. 4 = 4. 2 = 2; 4. 4 = 4. 4 = 4

=> 4 is the unity element of S. Hence the unity of Z6 ≠ unity of S.

Applications Of Quotient Rings And Ideals In Algebra

Theorem 3. The intersection of two subrings of an R is a subring of R.

Prood: let \(S_1, S_2\) be two subrings of R. Let \(0 \in R\) be the zero element.

Since every subring contains at least zero elements of the ring, \(0 \in S_1\) and \(0 \in S_2.\)

∴ \(0 \in S_1 \cap S_2\) and hence \(S_1 \cap S_2 \neq \phi\) and \(S_1 \cap S_2\) subset R.

Let a, b \(\in S_1 \cap S_2\). Then a, b \(\in S_1\) and \(a, b \in S_2\).

a, b \(\in S_1\) and \(S_1\) is a subring of R \(\Rightarrow a-b \in S_1\) and \(a b \in S_1\)

a, b \(\in S_2\) and \(S_2\) is a subring of R

⇒ \(a-b \in S_2\) and \(a, b \in S_2\)

From (1) and (2) we have a, b\( \in S_1 \cap S_2 \Rightarrow a-b \in S_1 \cap S_2\) and \(a b \in S_1 \cap S_2\)

∴ \(S_1 \cap S_2\) is a subring of R.

Differential Operators Solved Problems

Example.1. If f and g are two scaler point functions, prove that

1. div ( f∇g) =f ∇²g+∇f .∇g
2. div (f ∇g) =div (g∇f)=f ∇²-g ∇²f

Solution:

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

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

∴ \(\nabla:(f \nabla g)=\frac{\partial}{\partial x}\left(f \frac{\partial g}{\partial x}\right)+\frac{\partial}{\partial y}\left(f \frac{\partial g}{\partial y}\right)+\frac{\partial}{\partial z}\left(f \frac{\partial g}{\partial z}\right)\)

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

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

∴ \(\text{div}(f \nabla g)=f \nabla^2 g+\nabla f . \nabla g\)

Differential Operators Vector Identities Scalar Potential

Another Method: Let ∇g =A.   By Theorem 1

⇒ \(\nabla \cdot(f \mathbf{A})=(\nabla f) \cdot \mathbf{A}+f(\nabla \cdot \mathbf{A})\)

= \((\nabla f) \cdot(\nabla g)+f[\nabla \cdot(\nabla g)\)

= \((\nabla f) \cdot(\nabla g)+f\left(\nabla^2 g\right)\)

We have proved that \(\text{div}(f \nabla g)=f \nabla^2 g+\nabla f . \nabla g\)…..(1)

Interchanging f and \(g \text{div}(g \nabla f)=g \nabla^2 f+\nabla g. \nabla f\)….(2)

(1) – (2) : \(\text{div}(f \nabla g) \text{div}(g \nabla f)=f \nabla^2 g-g \nabla^2 f\)

Example.2. Prove that div{(r×a) ×b)=-2(a.b) where a and b are constant vectors.

Solution: div{(r×a)× b}=div [(r.b)a-a(a.b)r=div(r.b)a-div(a.b)r By theorem 1

= \((\mathbf{r} \cdot \mathbf{b}) \text{div} \mathbf{a}+\mathbf{a} \cdot \text{grad}(\mathbf{r} \cdot \mathbf{b})-\{(\mathbf{a} \cdot \mathbf{b}) \text{div} \mathbf{r}+\mathbf{r} \cdot \text{grad}(\mathbf{a} \cdot \mathbf{b})\}\)

= \(0+\mathbf{a} \cdot \text{grad}(\mathbf{r} \cdot \mathbf{b})-(\mathbf{a} \cdot \mathbf{b}) 3+0=+\mathbf{a} \cdot \Sigma \mathbf{i} \frac{\partial}{\partial x}(\mathbf{r} \cdot \mathbf{b})-3(\mathbf{a} \cdot \mathbf{b})\)

= \(+\mathbf{a} \cdot \Sigma \mathbf{i}\left(\frac{\partial r}{\partial x} \cdot \mathbf{b}\right)-3(\mathbf{a} \cdot \mathbf{b})=+\mathbf{a} \Sigma \mathbf{i}(\mathbf{i} \cdot \mathbf{b})-3(\mathbf{a} \cdot \mathbf{b})\)

= \(\mathbf{a} \cdot \mathbf{b}-3(\mathbf{a} \cdot \mathbf{b})=-2(\mathbf{a} \cdot \mathbf{b})\)

Example.3. Prove that curl[(r×a)×b]=b×a where a and b are constant vectors.

Solution:

curl\([(\mathbf{r} \times \mathbf{a}) \times \mathbf{b}]=\text{curl}[(\mathbf{r} . \mathbf{b}) \mathbf{a}-(\mathbf{a} . \mathbf{b}) \mathbf{r}]\)

= \(\text{curl}(\mathbf{r} \cdot \mathbf{b}) \mathbf{a}-\text{curl}(\mathbf{a} \cdot \mathbf{b}) \mathbf{r}=(\mathbf{r} . \mathbf{b}) \text{curl} \mathbf{a}+\text{grad}(\mathbf{r} \cdot \mathbf{b}) \times \mathbf{a}-(\bar{a} \cdot \bar{b}) \text{curl} \bar{r}\)

= \(0+\nabla(r . \bar{b}) \times \bar{a}-0 \quad=\mathbf{b} \times \mathbf{a}\) (because \(\nabla(\bar{r} \cdot \bar{b})=\bar{b}\))

Example.4. Prove that curl (f grad Φ) =(grad f) × (grad Φ)

Solution:  curl (f grad Φ)= ∇×(f ∇ Φ)=f curl (grad Φ) + (grad f)  × (grad Φ) [by Theorem 2]

= grad f× grad Φ

∴ curl ∇ Φ=0

Examples Of Vector Identities In Irrotational Fields

Example.5. Prove that div(∇Φ ×∇f) =0

Solution:  div(A×B) = B . curl A-A.  curl B

∴ div (∇Φ ×∇f)=∇f.curl (∇Φ)−∇Φ. curl (∇f) = 0

(since Curl grad Φ = 0 = Curl grad f)

Example.6. Prove that \(\nabla \cdot\left(\frac{\mathbf{r}}{r^3}\right)\)=0

Solution:

We know that \(\nabla \cdot(\phi \mathbf{a})=\nabla \phi \cdot \mathbf{a}+\phi(\nabla \cdot \mathbf{a})\)

Take \(\phi=\frac{1}{r^3}\). Then, \(\nabla \cdot\left(\frac{1}{r^3} \mathbf{r}\right)=\nabla\left(\frac{1}{r^3}\right) \cdot \mathbf{r}+\frac{1}{r^3}(\nabla \cdot \mathbf{r})=\left(-3 r^{-5}\right)(\mathbf{r} \cdot \mathbf{r})+\frac{1}{r^3}(3)\)

Using \(\nabla\left(r^n\right)=n r^{n-2} \mathbf{r} ; \nabla \cdot(\bar{r})=3\)

= \(\frac{-3}{r^5}\left(r^2\right)+\frac{3}{r^3}=\frac{-3}{r^3}+\frac{3}{r^3}=0\)

Example.7. If Φ(x,y,z) is a solution of the Laplace equation, Then ∇Φ is both solenoidal and irrotational. 

Solution:

Given

Φ(x,y,z) is a solution of the Laplace equation

∇² (Φ) =0 ⇒∇ .(∇Φ) = 0.

∴ ∇Φ is solenoidal.

We know that ∇×(∇Φ) = 0 is always true. Thus ∇Φ is always irrational.

Example.8. Show that the vector (x²−yz)i + (y²− zx)j +(z² − xy)k is irrotational and find its scalar potential.

Solution:

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

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

∴ \(\mathbf{f}\) is irrotational. Then there exists \(\phi\) such that \(\mathbf{f}=\nabla \phi\).

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

Comparing components, we get

⇒ \(\frac{\partial \phi}{\partial x}=x^2-y z \Rightarrow \phi=\int\left(x^2-y z\right) d x=\frac{x^3}{3}-x y z+f_1(y, z)\)……(1)

⇒ \(\frac{\partial \phi}{\partial y}=y^2-z x \Rightarrow \phi=\frac{y^3}{3}-x y z+f_2(z, x)\)….(2)

⇒ \(\frac{\partial \phi}{\partial z}=z^2-x y \Rightarrow \phi=\frac{z^3}{3}-x y z+f_3(x, y)\)……(3)

From (1), (2), (3), \(\phi=\frac{1}{3}\left(x^3+y^3+z^3\right)-x y z+\) constant, which is the required scalar potential.

Applications Of Scalar Potential In Vector Calculus Problems

Example.9. Find constants a,b,c, so that the vector A= (x+2y+az)i+(bx-3y-z)j+(4x+cy+2z)k= is irrotational. Also, find Φ  such that A=∇Φ.

Solution:

Given vector is \(\mathbf{A}=(x+2 y+a z) \mathbf{i}+(b x-3 y-z) \mathbf{j}+(4 x+c y+2 z) \mathbf{k}\)

vector \(\mathbf{A}\) is irrotational

⇒ \(\text{curl} \mathbf{A}=\mathbf{0}\)

⇒ \(\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\partial / \partial x & \partial / \partial y & \partial / \partial z \\
x+2 y+a z & b x-3 y-z & 4 x+c y+2 z
\end{array}\right|\)= 0

⇒ \(\mathbf{i}(c+1)+\mathbf{j}(a-4)+\mathbf{k}(b-2)=0\)

⇒ \(\mathbf{i}(c+1)+\mathbf{j}(a-4)+\mathbf{k}(b-2)=0 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k}\)

Comparing both sides, c+1=0, a-4=0, b-2=0

∴ c=-1, a=4, b=2

Now \(\mathbf{A}=(x+2 y+4 z) \mathbf{i}+(2 x-3 y-z) \mathbf{j}+(4 x-y-2 z) \mathbf{k}\), on substituting the values of a, b, c

We have \(\mathbf{A}=\nabla \phi\)

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

Comparing both sides, we have

⇒ \(\frac{\partial \phi}{\partial x}=x+2 y+4 z \Rightarrow \phi=\frac{x^2}{2}+2 x y+4 z x+f_1(y, z)\)

⇒ \(\frac{\partial \phi}{\partial y}=2 x-3 y-z \Rightarrow \phi=2 x y-\frac{3 y^2}{2}-y z+f_2(z, x)\)

⇒ \(\frac{\partial \phi}{\partial z}=4 x-y+2 z \Rightarrow \phi=4 x z-y z+z^2+f_3(x, y)\)

Hence \(\phi=\frac{x^2}{2}-\frac{3 y^2}{2}+z^2+2 x y+4 z x-y z+c\) or \(\phi=\frac{x^2}{2}-\frac{3 y^2}{2}+z^2+2 x y-y z+4 z x+c\)

Differential Operators Solved Problems

Example. 1. If f = xy²i + 2x²yz j- 3yz²k find

1. div f,
2. curl f at the point (1, — 1,1)

Solution:

Given

f = xy²i + 2x²yz j- 3yz²k

1. \(\text{div} \mathbf{f}=\nabla \cdot \mathbf{f}=\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}=\frac{\partial}{\partial x}\left(x y^2\right)+\frac{\partial}{\partial y}\left(2 x^2 y z\right)+\frac{\partial}{\partial z}\left(-3 y z^2\right)\)

= \(y^2+2 x^2 z-6 y z \quad \text { at }(1,-1,1) \text{div} \mathbf{f}=1+2+6=9\)

2. \(\text{curl} \mathbf{f}=\nabla \times \mathbf{f}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
x y^2 & 2 x^2 y z & -3 y z^2
\end{array}\right|\)

= \(\mathbf{i}\left[\frac{\partial}{\partial y}\left(-3 y z^2\right)-\frac{\partial}{\partial z}\left(2 x^2 y z\right)\right]-\mathbf{j}\left[\frac{\partial}{\partial x}\left(-3 y z^2\right)-\frac{\partial}{\partial z}\left(x y^2\right)\right]\)

+ \(\mathbf{k}\left[\frac{\partial}{\partial x}\left(2 x^2 y z\right)-\frac{\partial}{\partial y}\left(x y^2\right)\right]-3 z^2-2 x^2 y) \mathbf{i}+(4 x y z-2 x y) \mathbf{k}\)

at (1,-1,1), \(\text{curl} \mathbf{f}=-\mathbf{i}-2 \mathbf{k}\)

Examples Of Laplacian And Curl In Vector Calculus

Example. 2. Find div f and curl f where f = grad(x³ + y³ + z³ – 3xyz)

Solution:

Given

f = grad(x³ + y³ + z³ – 3xyz)

f = \(\text{grad}\left(x^3+y^3+z^3-3 x y z\right)=\sum \mathbf{i} \frac{\partial}{\partial x}\left(x^3+y^3+z^3-3 x y z\right)\)

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

div \(\mathbf{f}=\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}=6 x+6 y+6 z=6(x+y+z)\)

curl \(\mathbf{f}=\nabla \times \mathbf{f}=\left|\begin{array}{ccc}
\mathbf{1} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
3 x^2-3 y z & 3 y^2-3 z x & 3 z^2-3 x y
\end{array}\right|\)

= \(\mathbf{i}(-3 x+3 x)-\mathbf{j}(-3 y+3 y)+\mathbf{k}(-3 z+3 z)=\mathbf{0}\)

Example.3. Prove that grade(r.a)=a where a is a constant vector

Solution: Let a= a₁i + a₂j + a₃k . Then i.a = a₁; j.a=a₂ ;k .a = a₃

Grad \((\mathbf{r} \cdot \mathbf{a})=\nabla(\overline{\mathbf{r}} \cdot \overline{\mathbf{a}})=\Sigma \mathbf{i} \frac{\partial}{\partial x}(\mathbf{r} \cdot \mathbf{a})\)

= \(\Sigma \mathbf{i}\left(\frac{\partial \mathbf{r}}{\partial x} \cdot \mathbf{a}\right)=\Sigma \mathbf{i}(\mathbf{i} \cdot \mathbf{a})=\mathbf{i}(\mathbf{i} \cdot \mathbf{a})+\mathbf{j}(\mathbf{j} \cdot \mathbf{a})+\mathbf{k}(\mathbf{k} \cdot \mathbf{a})\)

Grad \((\mathbf{r} \cdot \mathbf{a})=a_1 \mathbf{i}+a_2 \mathbf{j}+a_3 \mathbf{k}=\mathbf{a}\)

Example.4. If a is a constant vector, prove that curl \(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\)=\(-\frac{\mathbf{a}}{r^3}+\frac{3 \mathbf{r}}{r^5}(\mathbf{a} \cdot \mathbf{r})\)

Solution:

Given

A is a constant vector

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

⇒ \(\frac{\partial \mathbf{r}}{\partial x}=\mathbf{i}, \frac{\partial \mathbf{r}}{\partial y}=\mathbf{j}, \frac{\partial \mathbf{r}}{\partial z}=\mathbf{k} \left|\mathbf{r}^2\right|=r^2=x^2+y^2+z^2\)

⇒ \(\frac{\partial r}{\partial x}=\frac{x}{r}, \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r} \text{curl} \frac{\mathbf{a} \times \mathbf{r}}{r^3}=\Sigma \mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)\)

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

= \(\mathbf{a} \times\left[\frac{1}{r^3} \mathbf{i}-\frac{3 x}{r^5} \mathbf{r}\right]\)

= \(\frac{\mathbf{a} \times \mathbf{i}}{r^3}-\frac{3 x(\mathbf{a} \times \mathbf{r})}{r^5}\)

∴ \(\mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=\mathbf{i} \times\left[\frac{\mathbf{a} \times \mathbf{i}}{r^3}-\frac{3 x}{r^5}(\mathbf{a} \times \mathbf{r})\right]=\frac{\mathbf{i} \times(\mathbf{a} \times \mathbf{i})}{r^3}-\frac{3 x}{r^5} \mathbf{i} \times(\mathbf{a} \times \mathbf{r})\)

= \(\frac{(\mathbf{i} . \mathbf{i}) \mathbf{a}-(\mathbf{i} \cdot \mathbf{a}) \mathbf{i}}{r^3}-\frac{3 x}{r^5}[(\mathbf{i} . \mathbf{r}) \mathbf{a}-(\mathbf{i} . \mathbf{a}) \mathbf{r}]\)

let \(\mathbf{a}=a_1 \mathbf{i}+a_2 \mathbf{j}+a \mathbf{k}\)

⇒ \(\mathbf{i} \cdot \mathbf{a}=a_1 ; \mathbf{j} \cdot \mathbf{a}=a_2 ; \mathbf{k} \cdot \mathbf{a}=a_3\)

∴ \(i \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=\frac{\left(\mathbf{a}-a_1 \mathbf{i}\right)}{r^3}-\frac{3 x}{r^5}\left(x \mathbf{a}-a_1 \mathbf{r}\right)\)

∴ \(\sum \mathbf{i} \times \frac{\partial}{\partial x}\left(\frac{\mathbf{a} \times \mathbf{r}}{r^3}\right)=\sum\left(\frac{\mathbf{a}-a_1 \mathbf{i}}{r^3}\right)-\frac{3}{r^5} \sum\left(x^2 \mathbf{a}-a_1 x \mathbf{r}\right)\)

= \(\frac{3 \mathbf{a}-\mathbf{a}}{r^3}-\frac{3 \mathbf{a}}{r^5}\left(x^2+y^2+z^2\right)+\frac{3 \mathbf{r}}{r^5}\left(a_1 x+a_2 y+a_3 z\right)\)

= \(\frac{2 \mathbf{a}}{r^3}-\frac{3 \mathbf{a} r^2}{r^5}+\frac{3 \mathbf{r}}{r^3}(\mathbf{r} \cdot \mathbf{a})=-\frac{\mathbf{a}}{r^3}+\frac{3 \mathbf{r}}{r^5}(\mathbf{r} \cdot \mathbf{a})\)

Theorems Involving Laplacian Operator And Curl With Examples

Example.5. Prove thatt ∇ × f(r)r=0

Solution:

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

= \(\sum \mathbf{i}\left\{\frac{\partial}{\partial y} f(r) z-\frac{\partial}{\partial z} f(r) y\right\}=\sum \mathbf{i}\left\{f^{\prime}(r) \frac{\partial r}{\partial y} z-f^{\prime}(r) \frac{\partial r}{\partial z} y\right\}\)

= \(\sum \mathbf{i} \frac{f^{\prime}(r)}{r}\{y z-z y\}=0\)

Example.6. Prove that (f × ∇).r =0

Solution:

∴ \((\mathbf{f} \times \nabla) \cdot \mathbf{r}=\left\{\mathbf{f} \times \Sigma \mathbf{i} \frac{\partial}{\partial x}\right\} \cdot \mathbf{r}=\mathbf{f} \times\left\{\Sigma \mathbf{i} \cdot \frac{\partial \mathbf{r}}{\partial x}\right\}=\left\{\Sigma(\mathbf{f} \times \mathbf{i}) \cdot \frac{\partial \mathbf{r}}{\partial x}\right\}=\Sigma(\mathbf{f} \times \mathbf{i}) \cdot \mathbf{i}=0\)

Example.7. Prove that (f ×∇)×r=-2f

Solution:

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

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

= \(\Sigma(\mathbf{f} \times \mathbf{i}) \times \mathbf{i}=\Sigma\{(\mathbf{f} . \mathbf{i}) \mathbf{i}-\mathbf{f}\}=(\mathbf{f} . \mathbf{i}) \mathbf{i}+(\mathbf{f} . \mathbf{j}) \mathbf{j}+(\mathbf{f} . \mathbf{k}) \mathbf{k}-3 \mathbf{f}\)

= \(\mathbf{f}-3 \mathbf{f}=-2 \mathbf{f}\)

Example.8. Show that \(\nabla^2\left(\frac{1}{r}\right)\)=0

Solution:

r = \(|\mathbf{r}|=\sqrt{x^2+\dot{y}^2+z^2} . \quad 2 r \frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}, \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r} \text {. }\)

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

= \(\sum \frac{r^3(-1)+x\left(3 r^2\right) \frac{\partial r}{\partial x}}{r^6}=\sum \frac{-r^3+x\left(3 r^2\right) \frac{x}{r}}{r^6}=\sum \frac{-r^3+3 x^2 r}{r^6}\)

= \(\sum\left(\frac{-1}{r^3}+\frac{3 x^2}{r^5}\right)=\frac{-3}{r^3}+\frac{3 r^2}{r^5}=\frac{-3}{r^3}+\frac{3}{r^3}=0\)

Note: \(\frac{1}{r}\) satisfies the Laplace equation \(\nabla^2 f=0\) Thus \(\frac{1}{r}\) is a harmonic function.

Laplacian And Curl Operator Detailed Examples With Solutions

Example.9. Find div F and Curl F where F= x²zi- 2y³z³j + xy²zk at (1,—1,1)

Solution:

Given \(\mathbf{F}=x^2 z \mathbf{i}-2 y^3 z^3 \mathbf{j}+x y^2 z \mathbf{k}\)

div \(\mathbf{F}=\frac{\partial}{\partial x}\left(x^2 z\right)+\frac{\partial}{\partial y}\left(-2 y^3 z^3\right)+\frac{\partial}{\partial z}\left(x y^2 z\right)=2 x z-6 y^2 z^3+x y^2\)

div \(\mathbf{F}\) at (1,-1,1) = \(2(1)(1)-6(-1)^2(1)^3+1(-1)^2=2-6+1=-3\)

Curl \(\mathbf{F}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x^2 z & -2 y^3 z^3 & x y^2 z\end{array}\right|=\mathbf{i}\left(2 x y z+6 y^3 z^2\right)-\mathbf{j}\left(y^2 z-x^2\right)+\mathbf{k}(0-0)\)

∴ Curl \(\mathbf{F}\) at (1,-1,1) = \(\mathbf{i}[2(-1)(1))]-\mathbf{j}(1-1)=-8 \mathbf{i}\)

Example.10. Show that \(\nabla^2[f(r)]\)=\(f^{\prime \prime}(r)+\frac{2}{r} f^{\prime}(r)=\frac{d^2 f}{d r^2}+\frac{2}{r} \frac{d f}{d r}\)

Solution:

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

⇒ \(r^2=x^2+ y^2+z^2 ; 2 r \frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}\)

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

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

= \(\Sigma \frac{r \frac{\partial}{\partial x}\left(f^{\prime}(r) x\right)-x f^{\prime}(r) \frac{\partial r}{\partial x}}{r^2}=\Sigma \frac{r\left(f^{\prime \prime}(r) \frac{\partial r}{\partial x} x+f^{\prime}(r)\right)-x f^{\prime}(r) \frac{x}{r}}{r^2}\)

= \(\Sigma \frac{r\left(f^{\prime \prime}(r) \frac{x}{r} x+f^{\prime}(r)\right)-\frac{x^2 f^{\prime}(r)}{r}}{r^2}=\Sigma \frac{f^{\prime \prime}(r) x^2+r f^{\prime}(r)-\frac{x^2 f^{\prime}(r)}{r}}{r^2}\)

= \(\frac{f^{\prime \prime}(r)}{r^2} \Sigma x^2+\frac{f^{\prime}(r)}{r} \Sigma 1-\frac{f^{\prime}(r)}{r^3} \Sigma x^2=\frac{f^{\prime \prime}(r)}{r^2} r^2+\frac{3 f^{\prime}(r)}{r}-\frac{f^{\prime}(r)}{r^3} r^2\)

= \(f^{\prime \prime \prime}(r)+2 \frac{f^{\prime}(r)}{r}\)

Example.11.(1) Find div F and curl F where F= rⁿ r̄.
(2) Show that rⁿ r is irrotational. Find for which value of n it is solenoidal.

Solution:

We have \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) and \(r^2=x^2+y^2+z^2\) where \(r=|\mathbf{r}|\)

D.V. r to x partially, \(2 r \frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}\).

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

div \(\mathbf{F}=\text{div}\left(r^n \mathbf{r}\right)=\sum i \cdot \frac{\partial}{\partial x}\left(r^n \mathbf{r}\right)\)

= \(\sum i \cdot\left[r^n \frac{\partial \mathbf{r}}{\partial x}+n r^{n-1} \frac{\partial r}{\partial x} \mathbf{r}\right]=\sum i \cdot\left[r^n(i)+n \cdot r^{n-1} \frac{x}{r} \mathbf{r}\right]\)

= \(\sum\left[r^n+n r^{n-2} x^2\right]\)

(because \(i \cdot \mathbf{i}=1\) and \( i \cdot \mathbf{r}=x)\)

= \(3 r^n+n r^n=(n+3) r^n\)

If \(\mathbf{F}\) is solenoidal, we have div f{F}=0

(n+3) \(r^n=0 \Rightarrow n=-3\)

Put n=-3, we get \(\mathbf{F}=r^n \mathbf{r}=r^{-3} \mathbf{r}=\frac{\mathbf{r}}{r^3}\) is solenoidal.

2. Curl \(\left(r^n \mathbf{r}\right)=\sum \mathbf{i} \times \frac{\partial}{\partial x}\left(r^2 \mathbf{r}\right)=\sum \mathbf{i} \times\left(\frac{r^n \partial \mathbf{r}}{\partial x}+\mathbf{r} n r^{n-1} \frac{\partial r}{\partial x}\right)\)

= \(r^n \sum \mathbf{i} \times \mathbf{i}+n r^{n-1} \sum \mathbf{i} \times \mathbf{r}\left(\frac{x}{r}\right)=r^n(\mathbf{0})+n r^{n-2} \sum \mathbf{i} \times(x \mathbf{i}+y \mathbf{j}+z \mathbf{k}) x\)

= \(n r^{n-2} \sum x y \mathbf{k}-x z \mathbf{i}=\mathbf{0}+\mathbf{0}=\mathbf{0}\)

∴\( r^n \mathbf{r}\) is irrotational.

Examples Of Laplacian And Curl In Vector Calculus

Example.12. Find l,m,n so that the vector F=(2x+3y+lz)i+(mx+2y+3z)j=(2x+xy+3z)k is irrotational

Solution:

We have Curl \(\mathbf{F}=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
2 x+3 y+l z & m x+2 y+3 z & 2 x+n y+3 z
\end{array}\right|\)

= \(\mathbf{i}[n-3]-\mathbf{j}[2-l]+\mathbf{k}[m-3]\)

Since f{F} is irrotational curl \(\mathbf{F}=\mathbf{O} \Rightarrow n=3, l=2, m=3\)

Example.13. If F= xy²i +2x²yzj-34z²k find div F at (1,—1,1)

Solution:

Given \(\mathbf{F}=x y^2 \mathbf{i}+2 x^2 y z \mathbf{j}-3 y z \mathbf{k}=\mathrm{F}_1 \mathbf{i}+\mathrm{F}_2 \mathbf{j}+\mathrm{F}_3 \mathbf{k}\)

div \(\mathbf{F}=\frac{\partial \mathrm{F}_1}{\partial x}+\frac{\partial \mathrm{F}_2}{\partial y}+\frac{\partial \mathrm{F}_3}{\partial z}=\left(y^2+2 x^2 z-6 y z\right)\)

(div F) at (1,-1,1) = \((-1)^2+2(1)^2(1)-6(-1)(1)=1+2+6=9\)

Example.14. Prove that F = y³z²i-3x²z⁵j-15x⁵y⁴k is solenoidal 

Solution:

Let \(\mathbf{F}=\mathrm{F}_1 \mathbf{i}+\mathrm{F}_2 \mathbf{j}+\mathrm{F}_3 \mathbf{k}\)

where \(\mathrm{F}_1=y^3 z^2, \mathrm{~F}_2=-3 x^2 z^2, \mathrm{~F}_3=-15 x^5 y^4\)

div \(\mathbf{F}=\frac{\partial \mathrm{F}_1}{\partial x}+\frac{\partial \mathrm{F}_2}{\partial y}+\frac{\partial \mathrm{F}_3}{\partial z}=0\)

∴ f{F} is a solenoidal vector.

Example.15. Show that ∇²rⁿ=n(n+1)r^n-2

Solution:

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

⇒ \(r^2=x^2+y^2+z^2\)

D. W. r. to x partially

2r \(\frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r} \text {. Similarly } \frac{\partial r}{\partial y}=\frac{y}{r}, \frac{\partial r}{\partial z}=\frac{z}{r}\)

We have \(\nabla^2 r^n=+\sum \frac{\partial^2}{\partial x^2}\left(r^n\right)\)

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

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

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

= \(n\left[\sum\left\{(n-2) r^{n-4} x^2+r^{n-2}\right\}\right]=n\left[(n-2) r^{n-4} r^2+3 r^{n-2}\right]\)

= \(n\left[(n-2+3) r^{n-2}\right]=n(n+1) r^{n-2}\)

Hence the result.

Note : When n=-1, \(\nabla^2\left(r^n\right)=(-1)(-1+1) r^{-1-2}=0\) .

∴ \(\nabla^2\left(\frac{1}{r}\right)=0\)

Derivative of a Vector Function Partial Differentiation

So far the reader has studied the differentiation of vector functions in one variable. However, a vector may be a function of more than one scalar variable.

Let f be the vector function of scalar variables p, q, t over a domain S, then we write f = f(p, q, t).

Treating t as the variable and p, q as constants; f can be considered as the vector function of the scalar variable t over a domain S.

⇒ \(\text { If Lt } \mathrm{L}_{\delta t \rightarrow 0} \frac{\mathrm{I}(p, q, t+\delta t)-\mathrm{f}(p, q, t)}{\delta \mathrm{q}}\)

exists then f is said to have partial derivative w.r.to ’ t ’ and is denoted by \(\frac{\partial \mathrm{O}^{\delta_2}}{\partial \mathrm{I}}\)

Similarly, treating q as the variable and taking p, and t as constants

⇒ \(\frac{\partial \mathrm{f}}{\partial q}=\mathrm{Lt}_{\delta t \rightarrow 0} \frac{\mathrm{f}(p, q+\delta q, t)-\mathrm{f}(p, q, t)}{\delta q} \text { etc. }\)

Let A, B, φ be the functions of more than one scalar variable. Then the following can be verified.

1. \(\frac{\partial}{\partial t}(\phi \mathbf{A})=\frac{\partial \phi}{\partial t} \mathbf{A}+\phi \frac{\partial \mathbf{A}}{\partial t}\)

2. If λ is a constant then \(\frac{\partial}{\partial t}(\lambda \mathbf{A})=\lambda \frac{\partial \mathbf{A}}{\partial t}\)

3. If c is a constant vector then \(\frac{\partial}{\pi}(\phi \mathbf{c})=\frac{\partial \phi}{\partial t} \mathbf{c}\)

4. \(\frac{\partial}{\partial t}(\mathbf{A} \pm \mathbf{B})=\frac{\partial \mathbf{A}}{\partial t} \pm \frac{\partial \mathrm{B}}{\partial t} 5 \cdot \frac{\partial}{\partial t}(\mathbf{A} \cdot \mathbf{B})=\frac{\partial \mathrm{A}}{\partial t} \cdot \mathbf{B}+\mathbf{A} \cdot \frac{\partial \mathrm{B}}{\partial t}\)

5. \(\frac{\partial}{\partial t}(\mathbf{A} \times \mathbf{B})=\frac{\partial \mathbf{A}}{\partial t} \times \mathbf{B}+\mathbf{A} \times \frac{\partial \mathrm{B}}{\partial t}\)

Let f = f₁ i + f₂ j + f₃ k where f₁, f₂, f₃ are differentiable scalar functions of more than one variable t. Then

⇒ \(\frac{\mu_i}{\partial t}=\mathrm{i} \frac{\partial_1}{\partial t}+\mathrm{j} \frac{\partial_2}{\partial t}+\mathrm{k} \frac{\partial_2}{\partial t}\)

Higher partial derivatives are derivatives defined as in the Calculus of real variables. Thus, for instant

⇒ \(\frac{\partial^2 \mathrm{r}}{\partial t^2}=\frac{\partial}{\partial t}\left(\frac{\partial \mathrm{f}}{\partial t}\right), \frac{\partial^2 \mathrm{r}}{\partial t \partial_P}=\frac{\partial}{\partial t}\left(\frac{\partial \mathrm{r}}{\partial \mathrm{p}}\right) \text { etc. }\)

Derivative Of A Vector Function With Scalar Multiplication

Derivative of a Vector Function Solved Problems

Example 1. If \(\mathbf{f}=\left(2 x^2 y-x^4\right) \mathbf{i}+\left(e^{x y}-y \sin x\right) \mathbf{j}+\left(x^2 \cos y\right) \mathbf{k} \text {. find } \frac{\partial^2 f}{\partial x^2} \text { and } \frac{\partial^2 f}{\partial x \partial_y}\)

Solution. Given f = (2x²y − x⁴) i + (e^xy − y sin x) j + (x² cos y) k

⇒ \(\frac{\partial \mathrm{f}}{\partial x}\)= (4xy − 4x³)i + (y e^xy − y cos x) j + (2x cos y)k

⇒ \(\frac{\partial^2 \mathrm{f}}{\partial x^2}\)= (4y − 12x²)i + (y² e^xy + y sin x) j + (2cos y)k

⇒ \(\frac{\partial^2 \mathrm{f}}{x \partial y y}=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)=4 x \mathbf{i}+\left(e^{x y}+y^2 e^{x y}-\cos x\right) \mathbf{j}-2 x \sin , y \mathbf{k} \)

Example. 2. If A = 2x² i − 3yzj + xz²k φ = 2z − x³y then find

⇒ \(\mathbf{A} \cdot\left[\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right] \text { and } \mathbf{A} \times\left[\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right] \text { at }(1,-1,1)\)

Solution:

Given

A = 2x² i − 3yzj + xz²k φ = 2z − x³y

At(1,-1,1), \(\mathbf{A}=2 \mathbf{i}+3 \mathbf{j}+\mathbf{k} \text { and } \frac{\partial \phi}{\partial x}=+3, \frac{\partial \phi}{\partial y}=-1, \frac{\partial \phi}{\partial z}=2\)

⇒ \(\mathbf{A} \cdot\left[\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right]_{(1,-1,1)}=(2 \mathbf{i}+3 \mathbf{j}+\mathbf{k}) \cdot(+3 \mathbf{i}-\mathbf{j}+2 \mathbf{k})=+6-3+2=5\)

and \(\mathbf{A} \times\left[\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial y}+\mathbf{k} \frac{\partial \phi}{\partial z}\right]=\left|\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 1 \\
3 & -1 & 2
\end{array}\right|=7 \mathbf{i}-\mathbf{j}-11 \mathbf{k}\)

Differential Operators Scalar Point Function

Let S be a domain in space. If to each point P∈S there corresponds a scalar Φ (P) then Φ is called a scalar point function over the domain S.

Example Let Φ (P) be the density at any point P of a material body occupying copying a region R. Then Φ is a scalar point function defined for the region R.

Example Let f(p) be the temperature at any point of a body occupying a certain region S.
Then f is a scalar point function defined for the region S.

Differential Operators Vector Point Function

Let S be a domain in space, If to each point P∈S there corresponds a vector f (p), then f is called a vector point function over S.

Example. Let f (p) denote the velocity of a particle at a point P in a region R then f is a vector point function defined in the region R.

Note 1. If OXYZ be a frame of reference in space then a point P= (x, y, z). Then we can write Φ (P) = Φ (x, y, z) so that a scalar point function appears as a scalar function of three variables.

Similarly, f (P)= f (x, y, z) so that a vector point function f can be considered as a vector function of three variables x, y,z.

2. If r denotes the position vector ofP.w.r. to the origin o then Φ (P) and f(P) may be written as Φ (r) and f (r).

Differential Operators Scalar Point Function

Example 1. Let O be the origin in space. For each point P in the space, there corresponds to a unique real number equal to OP. The function so defined on the space is called a distance function. It is denoted by r.

For P = (x,y, z) , r (P) = OP \(=\sqrt{\left(x^2+y^2+z^2\right)}\)

Example 2. Let O be the origin in space S . For each PeS there corresponds a unique vector OP. The function so defined on the space S is called the position vector function. It is denoted by r.

For P =(x,y,z) r(P)= OP= xi+yj +zk

Differential Operators Delta Neighbourhood

Let P be a point in space and δ> O . The set of all the points Q such that PQ < δ is called δ- nbd of P. If P is deleted from δ- nbd of P then it is called deleted δ- nbd of P.

Differential Operators Limit

(1) Let Φ be a scalar point function defined on a deleted-nbd of P and I ∈ R . If for each ∈> o , there exists δ> O , such that

⇒ \(O<Q P<\delta \Rightarrow|\phi(Q)-l|<\varepsilon\) then we say \(\underset{Q \rightarrow P}{L t} \phi=l\)

(2) Let f be a vector point function defined on a deleted-nbd of P and I be a vector. If for each ϵ > 0 there exists δ > 0 such that O < QP < δ =>| f (Q)- 1 | < ϵ then we say  Lt f = I

Q→P

Differential Operators Vector Point Function

Differential Operators Continutinty

(1) Let Φ be defined on a nbd of. p If each ϵ > 0 there exists δ > 0 such that

⇒ \(\mathrm{O}<\mathrm{QP}<\delta \Rightarrow|\phi(\mathrm{Q})-\phi(\mathrm{P})|<\varepsilon\) then we say that \(\phi\) is continuous at \(\mathbf{P}\).

(2) Let f be defined on a nbd of P. If for each ϵ < 0, there exists δ> 0 such that

⇒ \(\mathrm{O}<\mathrm{QP}<\delta \Rightarrow|\mathbf{f}(\mathrm{Q})-\mathbf{f}(\mathrm{P})|<\varepsilon\) then we say that \(\mathbf{f}\) is continuous at \(\mathbf{P}.\)

Directional Derivative At A Point

(1) Let P be a point in space and L be a ray through P in the direction of unit vector e. Let a scalar point function Φ be defined in a nbd D of P. Let Q ≠ P and Q∈L ⋂D.

⇒ \(\text { If } \underset{Q \rightarrow P}{\mathrm{Lt}} \frac{\phi(\mathrm{Q})-\phi(\mathrm{P})}{\mathrm{QP}}\) exists then we say that the limit is the directional derivative of Φ at P in the direction e. It is denoted by

⇒ \(\frac{\partial \phi}{\partial e} or \frac{\partial \phi}{\partial \mathrm{s}}\) where s = \(\mathrm{PQ}\)

(2) Let P be a point in space and L be a ray through P in the direction of the unit Vector e. Let a vector point function f be defined in a nbd D of P. Let Q ≠P and Q∈L ⋂D

⇒ \(\text { If } \underset{Q \rightarrow P}{\mathrm{Lt}} \frac{\mathbf{f}(\mathrm{Q})-\mathbf{f}(\mathrm{P})}{\mathrm{QP}}\)exists then we say that the limit is the directional derivative of f

At P in the direction of e. It is denoted by \(\frac{\partial \boldsymbol{f}}{\partial \bar{e}}\) or \(\frac{\partial \mathbf{f}}{\partial s}\) where s=\(\mathrm{PQ}\).

Differential Operators Scalar Vector Point Function Directional Derivative At A Point Note

Note 1. If e = i = unit vector along OX then

(1) \(\frac{\partial \phi}{\partial \mathrm{e}}\)=\(\frac{\partial \phi}{\partial \mathrm{i}} or \frac{\partial \phi}{\partial x}\)

(2) \(\frac{\partial f}{\partial e}\)=\(\frac{\partial f}{\partial i} or \frac{\partial f}{\partial x}\)

If e = j = unit vector along OY, then

(1) \(\frac{\partial \phi}{\partial \mathrm{e}}\)=\(\frac{\partial \phi}{\partial \mathrm{j}} or \frac{\partial \phi}{\partial y}\)

(2) \(\frac{\partial \mathbf{f}}{\partial \mathrm{e}}\)=\(\frac{\partial \mathbf{f}}{\partial \mathrm{j}} or \frac{\partial \mathrm{f}}{\partial y}\)

If e = k = unit vector along OZ, then

(1) \(\frac{\partial \phi}{\partial \mathrm{e}}\)=\(\frac{\partial \phi}{\partial \mathbf{k}} or \frac{\partial \phi}{\partial z}\)

(2) \(\frac{\partial \mathbf{f}}{\partial \mathrm{e}}\)=\(\frac{\partial \mathbf{f}-}{\partial \mathbf{k}} or \frac{\partial \mathbf{f}}{\partial z}\)

Properties Of Scalar And Vector Point Functions

Note 2. If φ land ψ are scalar point functions and f and g are vector functions having directional derivatives at P in the direction of unit vector e then the following results hold

(1)\(\frac{\partial}{\partial s}(\phi \pm \Psi)=\frac{\partial \phi}{\partial s} \pm \frac{\partial \Psi}{\partial s}\)

(2)\(\frac{\partial}{\partial s}(\phi \Psi)=\phi \frac{\partial \Psi}{\partial s}+\Psi \frac{\partial \phi}{\partial s}\)

(3)\(\frac{\partial}{\partial s}(\mathbf{f} \pm \mathbf{g})=\frac{\partial \mathbf{f}}{\partial s} \pm \frac{\partial \mathbf{g}}{\partial s}\)

(4)\(\frac{\partial}{\partial s} \text { (f.g) }=\frac{\partial \mathbf{f}}{\partial s} \cdot \mathbf{g}+\mathbf{f} \cdot \frac{\partial \mathbf{g}}{\partial s}\)

(5)\(\frac{\partial}{\partial s}(\mathbf{f} \times \mathbf{g})=\frac{\partial \mathbf{f}}{\partial s} \times \mathbf{g}+\mathbf{f} \times \frac{\partial \mathrm{g}}{\partial s}\)

(6)\(\frac{\partial}{\partial s}(\phi \mathbf{f})=\phi \frac{\partial \mathbf{f}}{\partial s}+\frac{\partial \phi}{\partial s} \mathbf{f} \)

Note 3. If f = f₁i + f₂j + f₃k and f₁,f₂,f₃ having directional derivatives at P. in the direction of e, then \( \frac{\partial \mathbf{f}}{\partial s}=\mathbf{i} \frac{\partial \mathbf{f}_1}{\partial s}+\mathbf{j} \frac{\partial \mathbf{f}_2}{\partial s}+\mathbf{k} \frac{\partial \mathbf{f}_3}{\partial s}\)

Differential Operators Scalar Vector Point Function Theorem

Theorem 1: If r is the position vector function and e is a unit vector then \(\frac{\partial \mathbf{r}}{\partial e}=\mathbf{e}\)

Proof: Let P be a point in the domain of r. Let Q ∈P and Q∈L where L is the ray through P in the direction of e.

⇒ \(\frac{\partial \mathbf{r}}{\partial \mathbf{e}}=\text{Lt}_{Q \rightarrow P} \frac{\mathbf{r}(\mathrm{Q})-\mathbf{r}(\mathrm{P})}{\mathrm{QP}}=\text{Lt}_{Q \rightarrow P} \frac{\mathbf{O Q}-\mathbf{O P}}{\mathrm{QP}}=\text{Lt}_{Q \rightarrow P} \frac{\mathbf{P Q}}{\mathbf{P Q}}\)

= \(\underset{Q \rightarrow P}{\mathrm{Lt}} \frac{(\mathrm{PQ}) \mathbf{e}}{\mathrm{QP}}=\text{Lt}_{Q \rightarrow P} \mathbf{e}=\mathbf{e}\)

Gradient Of Scalar And Vector Point Functions Examples

Note 1. \(\frac{\partial \mathbf{r}}{\partial x}\)=\(\frac{\partial \mathbf{r}}{\partial \mathbf{i}}\)= i  \(\frac{\partial \mathbf{r}}{\partial y}\)=\(\frac{\partial \mathbf{r}}{\partial \mathbf{j}}\)=j  and    \(\frac{\partial \mathbf{r}}{\partial z}\)=\(\frac{\partial \mathbf{r}}{\partial \mathbf{k}}\) =k

Note 2. If r=xi+yj+zk      then  \(\frac{\partial \mathbf{r}}{\partial \mathbf{e}}\)=\(\frac{\partial \mathbf{r}}{\partial s}\)=\(\frac{\partial x}{\partial s} \mathbf{i}+\frac{\partial y}{\partial s} \mathbf{j}+\frac{\partial z}{\partial s} \mathbf{k}\)  = e

Example. If \(r=|\mathbf{r}| \text { prove that } \frac{\partial r}{\partial \mathbf{e}}\)=\(\frac{\mathbf{r} . \mathbf{e}}{r}\)

Solution. We know that   r²=r²   i.e.   r.r = r²

∴ \(\frac{\partial}{\partial \mathbf{e}}(\mathbf{r}, \mathbf{r})\)=\(\frac{\partial}{\partial \mathbf{e}}\left(r^2\right)\)

⇒ 2r. \(\frac{\partial \mathbf{r}}{d \mathbf{e}}\)=\(2 r \frac{\partial r}{d \mathbf{e}}\) r. e = \(r \frac{\partial r}{d \mathrm{e}}\)

⇒ \(\frac{\partial r}{d \mathbf{e}}\)=\(\frac{\mathbf{r} \cdot \mathbf{e}}{r}\)

Differential Operators Gradient Of A Scalar Point Function

Let Φ be a scalar point function having the directional derivatives

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

Directions of i, j, and k respectively. The vector function \(\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial z}+\mathbf{k} \frac{\partial \phi}{\partial z}\)  is called the gradient of Φ .

It is written as grad Φ or ∇ Φ

∴ grad Φ = ∇ Φ = \(\mathbf{i} \frac{\partial \phi}{\partial x}+\mathbf{j} \frac{\partial \phi}{\partial z}+\mathbf{k} \frac{\partial \phi}{\partial z}\)

Note. Now r = xi + y j + zk => dr = (dx) i + (dy) j + (dz) k

If Φ is a scalar point function, then\( d \phi=\frac{\partial \phi}{\partial x} d x+\frac{\partial \phi}{\partial y} d y+\frac{\partial \phi}{\partial z} d z\)

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

Differential Operators Gradient Of A Scalar Point Function

Theorem 1:  If i and g are two scalar point functions then (1) grad(f ±g) = grad f ± grad g , grad(fg) = (grad f) g ± f (grad g)

Proof:

(1) grad (f± g)=∇(f± g) \(=\sum \mathbf{i} \frac{\partial}{\partial x}(f \pm g)\)

⇒ \(=\sum \mathbf{i}\left(\frac{\partial f}{\partial x} \pm \frac{\partial y}{\partial x}\right)\)=\(\sum \mathbf{i} \frac{\partial f}{\partial x} \pm \sum \mathbf{i} \frac{\partial y}{\partial x}\) ∇f ±∇g= grad f ± grad g

(2) grad (fg) = ∇(fg)\(=\sum \mathbf{i} \frac{\partial}{\partial x}(f g)\)

= \(\sum \mathbf{i}\left(f \frac{\partial g}{\partial x}+g \frac{\partial f}{\partial x}\right)\)

⇒ \(=\sum \mathbf{i} \frac{\partial g}{\partial x} f+\sum \mathbf{i} \frac{\partial f}{\partial x} g\)

⇒ \(=f \sum \mathbf{i} \frac{\partial g}{\partial x}+g \sum \mathbf{i} \frac{\partial f}{\partial x}\) = f∇g±g∇f

= f (grad g) + g (grad f)

grad(f ±g) = grad f ± grad g , grad(fg) = (grad f) g ± f (grad g)

Note. If f is a scalar point function and c is a constant, then∇(cf) \(=\sum \mathbf{i} \frac{\partial}{\partial x}(c f)\)

⇒ \(=\sum \mathbf{i}\left(c \frac{\partial f}{\partial x}\right)\)=\(c \sum \mathbf{i} \frac{\partial f}{\partial x}\)=\(c \nabla f\)

Theorems Involving Two Scalar Point Functions

Theorem 2: The necessary and sufficient condition for a scalar point function f to be constant is that ∇f=0

Proof: Let f(x, y, z) be a constant function.

∴ \(\frac{\partial f}{\partial x}=0, \frac{\partial f}{\partial y}=0, \frac{\partial f}{\partial z}=0 \)

∴ grad f=∇f\(=\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\) =0

Thus the condition is necessary. Conversely, Let grad f = 0.

∴ \(\nabla f\)=\(\mathbf{i} \frac{\partial f}{\partial x}+\mathbf{j} \frac{\partial f}{\partial y}+\mathbf{k} \frac{\partial f}{\partial z}\) = 0

⇒\(\frac{\partial f}{\partial x}=0, \frac{\partial f}{\partial y}=0, \frac{\partial f}{\partial z}=0\)

⇒ f is independent of x,y and z   ⇒ f is a constant

Hence the condition is sufficient

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 = \(\cos ^{-1} \frac{\nabla f \cdot \nabla \phi}{|\nabla f||\nabla \phi|}\)

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