Vector Calculus

Subrings, Ideals, Quotient Rings & Euclidean Rings Subrings Solved Problems

Example. 1. Prove that S1 = {0, 3}, S2 ={0,2, 4}  are subrings of Z6 = {0,1, 2,3,4 5} with respect to addition and multiplication of residue classes.

Solution. Since (Z6,+,•) is a ring, from the property R4  of the ring we have -0 = 6,-2 = 4,-3 = 3, -4 = 2. ={ 0, 3.} is a non-empty subset of Z6

From the above tables a,b ∈ S1 =>a-b∈ S1 and a.’b ∈ S1

By the theorem (1), S1 is a subring of Z6.

S2 = {0, 2, 4} is a non-empty subset of Z6

From the above tables ; a,b ∈ S2 =>a-b ∈ S2 and a.b ∈ S2

S2 is a subring of Z6

We can see that S1 ∩S2 = { 6 } is the trivial subring. But \(S_1 \cup S_2=\{\overline{0}, \overline{2},\overline{3}, \overline{4}\} \text { is not a }\)

subring of Z6, because 2,3\( S_1 \cup S_2 \Rightarrow \overline{2}+\overline{3}=\overline{5} \& S_1 \cup S_2 \)

Subrings And Ideals In Ring Of Endomorphisms Of Abelian Groups

Example. 2. Show that the set of matrices \(\left(\begin{array}{ll}
a & b \\
d & c
\end{array}\right)\) is a subring of the ring of 2×2 matrices whose elements are integers

Solution. Let R = \(\left\{\left(\begin{array}{ll}
a & b \\
d & c
\end{array}\right) \mid a, b, c, d \in Z\right\}\) be the ring of 2×2 matrices and

S = \(\left\{\left(\begin{array}{ll}
a & b \\
0 & c
\end{array}\right) \mid 0, a, b, c \in Z\right\}\) Then 5 # <|) and S cR .

Let A, B eS so that A= \(\left(\begin{array}{ll}
a_1 & b_1 \\
0 & c_1
\end{array}\right), B=\left(\begin{array}{cc}
a_2 & b_2 \\
0 & c_2
\end{array}\right)\) where 0, ah by, c{, a2, b2i c2 e Z.

∴ \(A-B=\left(\begin{array}{cc}
a_1-a_2 & b_1-b_2 \\
0 & c_1-c_2
\end{array}\right) \text { and } A B=\left(\begin{array}{cc}
a_1 a_2 & a_1 b_2+b_1 c_2 \\
0 & c_1 c_2
\end{array}\right) \text {. }\)

Sincea \(1-a_2, b_1-b_2, c_1-c_2, a_1 a_2, a_1 b_2+b_1 c_2, c_1 c_2 \in Z\)

Note: If R is a commutative ring then S is an ideal of R.

Understanding Euclidean Rings And Endomorphism Rings

Example. 3. Let R be a ring and a ∈ R be a fixed element. Then prove that S =(x  ∈ R | ax = 0} [is a subring of R.

Solution. If 0 ∈R is the zero element of r and a ∈ R, we have a0 = 0 => 0 ∈ S

S ≠ Φ and S⊂ R

Let x,y∈ S . Then x, y e R and ax = 0, ay= 0.

Now a(x-y) = ax-ay = 0- 0 = x-y ∈ S.

Also a (xy) = (ax)y = 0y = 0 => xy ∈ S. Hence S is a subring of R.

Notation. Let R be a ring and a∈ R be a fixed element. The intersection of the family of subrings containing ‘a’ is a subring of R. This subring is denoted by Ra and is called the subring of R generated by ‘a’.

Example. 4. If R is a ring and C (R) = (x € R \ xa — ax ∀ a∈ R] then prove that C (R) is a subring of R.

Solution.  For 0 ∈ R, the zero element of the ring, we have 0a = a0 ∀ a ∈R.

By the definition of C (R), 0 ∈C (R). C(R)≠ Φ and C (R)⊂R

Let x,y ∈ C(R)

Then x,y ∈ R and xa = ax, ya = a y V a ∈R

∀a∈R, a(x-y)-ax-ay =xa-ya =(x-y) a

Also, ∀ a ∈R,a(xy) = (ax)y = (xa)y = x (ay) = x (ya) = (xy) a …. (1)

Subrings, Ideals, Quotient Rings & Euclidean Rings Principal Ideal

If a commutative ring with unity from Theorem (3) Art! 2.2 we observed that for a given a ∈ R the set {ra | r ∈ #} is an ideal in R that contains the element ‘a’.

Definition. Let R be a commutative ring with unity and a ∈ R The ideal {ra |r ∈ R) of all multiples of ‘a’ is called the principal ideal generated by ‘a’ and is denoted by (a) or (a).

An ideal Uofthe ring R is a principal ideal =>U = (ci) = {ra ) r ∈ It} for some a∈R

example. 1 The null ideal or trivial ideal {0}’ofaring# is the principal ideal generated by the zero element of R. That is null ideal = (0).

example. 2. The unit ideal or improper ideal R of a ring R is the principal ideal generated by the unity element T of the ring R. That is R =(l).

example.3. Z is a commutative ring with a unity element. The principal ideal generated by 2 ∈ Z = (2) = [2n| n ∈ Z} = the set of all even integers.

example. 4. A field has only null ideal = (0) and unit ideal =F =(l) which are principal ideals.

Definition. (Principal ideal ring). A commutative ring R with unity is a principal ‘ ideal ring if every ideal in R is a principal ideal

D is a principal ideal domain => every ideal U in D is in the form U (a) for some a ∈D.

Principal Ideal Theorem In Euclidean Rings Explained

Theorem. 1. A field is a principal ideal ring

Proof. A field F has only two ideals, namely, U = (0) and U = F = {l).

But U = (0) and U = (l) are principal ideals. the field F is a principal ideal ring.

Theorem. 2. ring of integers the principal ideal ring, (or) Every ideal Z is of Z is a principal ideal.

Proof. Let U be the ideal of Z and U = {0}. Then U is generated by the zero element.

.’. U = (0) is a principal ideal. Let U be an ideal of Z and U≠ (0).

there exists a ∈U so that a ≠  0,

Since u ⊂ Z, one of a, -a must be a positive integer.

∴ the set of positive integers U+ in U is non-empty. by the well-ordering principle, U+ has the least member, say, b.

We now prove that U = {b) = the principal ideal generated by ‘b’.

Let x∈U. Since x,b are integers and b≠0 there exist q,r ∈ Z

such that x = bq+r; 0 <,r<b (Division algorithm).

b ∈ U,q ∈ Z and u is an ideal => bq ∈ U. x∈U,bq ∈U => x-bq =r ∈U .

Now r ∈U 0 <r < b and b is the least member in U+ =>r = 0.

x-bq =r=>x-bq = 0 => x = bq .

Hence x∈U=>x=bq for q∈Z =>U = (b).

∴  every ideal U of Z is a principal ideal. Hence Z  is a principal ideal ring.

Examples Of Principal Ideal Theorem In Subrings And Quotient Rings

Note. 1. Principal ideal rings that are also integral domains, such as rings of integers Z  are called principal ideal domains

2. If Z is the ring of integers then the principal ideal generated by a∈ Z is ] {aq \q∈Z} = (a).

Subrings, Ideals, Quotient Rings & Euclidean Rings Ideas

The concept of an ideal of a ring is analogous to that of a normal subgroup of a group. Some of the subrings which we call ideals play a very important role as the normal subgroups
in group theory.

Definition. (Ideal). Let. (R, +,• ) be a ring. A non-empty subset U of R is called a two-sided ideal or ideal if (1) a,b ∈ U=>a-b∈U and (2) a ∈ U and r ∈ R=> ar,r a∈ U.

(Ideal). A subring U of a ring R is called a (two-sided) ideal of R if, for every r ∈ R and every a ∈ U, both ra and ar are in U.

Note

1. A subring u of the ring R satisfying rU⊂ U and Ur⊂U for all r ∈ R is an ideal.
2. The (2) condition of ideal is stronger than the (2) condition of a subring.
3. The condition (1), namely, a,b∈ U => a -b∈U is called module property.
4. If U is an ideal of the ring R,+, • ) then (U, +) is a normal subgroup of the commutative group (R, +). Hence zero element in R is zero element in U.

Definition. A non-empty subset U of a ring R is called a right ideal if.

(1) a,b∈U=>a-b ∈U and(2)a ∈U,r ∈R=>ar ∈U . A non-empty subset U of a ring R is called a left ideal if (1) a,b ∈U a-b ell and a ∈U,r∈R=>ra ∈U .

Note.

1. An ideal is both a left and a right ideal.
2. For commutative rings left ideals coincide with right ideals.

Theorem 1. If U is an ideal of a ring R with unity element and 1∈ U then U  =R.

Proof: by the definition ideal  U⊂R.

Also x ∈R=>x.1 eR =>x.1∈U for x∈R, 1 ∈u (Def. of ideal) =>x ∈U .

R⊂U and hence U ⊂ R

Theorems On Ideals In Subrings And Quotient Rings

Theorem 2. A field has no proper non-trivial ideals, (or) The ideals of a field f are only {0} and F itself

Proof. Let U be an ideal of F so that U ≠ {0}

We now prove that U = F

By the definition of ideal, U⊂ F …… (1)

Let a∈ U and a≠ 0.

For a (≠ 0) ∈F there exists a -1  ∈  F so that aa -1 = 1

x ∈F => x 1 ∈ F=>x . 1 ∈U for xeF, 1∈U

=> x∈U .-. F ∈U … (2)

From (1) and (2): U = F Hence ideals of F are either {o} orF.

Theorem 3. If R is a commutative ring and a ∈ R then Ra = {ra\r| ∈ R \ is an ideal of R.

Proof. For 0 ∈ R, 0a = 0∈Ra. Ra≠ Φand R⊂R.

Let x,y∈Ra. Then x =r 1 a, y = r2a where r1,r2 ∈ R  :. x,y ∈ Ra=>x-y ∈ R

x. y = r1a-r2a=(r1-r2)a= ra where r =r1~r2 ∈ R.

Let x ∈ Ra and r ∈R.

x.r = (r1a)r ( :.x = r1a Where r∈R) =r1(ar) = r1(ra) (By R5 and R is commutative )

=(r1r) a-r’a where r’ = r1r ∈ R .

Since R is commutative, x.r=r.x.

∴ x∈Ra, r ∈R=>xr = rx ∈Ra  ……. (2)

Hence from (1) and (2): Ra is an ideal of R.

Note. 1. If R is a commutative ring and a e R then aR = { ar \ r e R } is an ideal of R.

2. If R is a r in and a ∈ R then Ra is a left ideal and aR is a right ideal.

Theorem 4. A commutative ring R with unity element is a field if R have no proper ideals.

Proof. Since the Ring R has no proper nontrivial ideals, the ideals of R are {o} and R only.

To prove that R is a field we have to show that every a (≠ 0) ∈ R has a multiplicative inverse. We know that aR = {ar|r ∈ R} is an ideal of R.

Since a ≠ 0 .aR ≠ {0} and hence aR =R (By hypothesis)

1 ∈ R => 1 ∈ a R =>1 = ab for some b≠ 0 ∈ R . Since R is conimutatiye,1= ab= ba .

a ≠ 0 ∈ R has a multiplicative inverse b∈ R . Hence R is a field.

Note. If R is a ring with a unity element and R has no proper nontrivial ideals then R is a division ring.

Examples Of Ideals In Quotient Rings And Euclidean Domains

Theorem. 5. The intersection of two ideals of a ring R is an ideal of R.

Proof. Let U1, and U2 be two ideals of the ring R.

If 0 ∈ R is the zero element, then 0∈ U1 and 0 ∈ U2 .

:. 0∈U1∩U2 and hence U1∩ U2 ≠φ

Let a,b ∈ U1 ∩ U2 and r ∈ R. Then a b∈U and a,b ∈U2.

a,b ∈ U1,r ∈R and U1 is an ideal =>a-b ∈U1 and ar, ra ∈ U1……………………….1

a,b ∈U2,r ∈R and U2 is an ideal=> a-b ∈U2 and ar, ra ∈U2………………………….2

From (1) and (2): a-b ∈ U1 ∩ U2 and ar, ra ∈ U1∩U2

Hence U1∩ U2 is an ideal of R.

Remark: The union of two ideals of a ring R need not be an ideal of R

For the ring Z of integers, A = { 2n | n∈ Z) and B = {3n\n ∈ Z } are two ideals.

But, for 2,3 ∈ A∪B, 3-2 = 1 ≠ A∪B.   A∪B is not an ideal of Z

Understanding The Properties Of Ideals In Algebraic Structures

Theorem 6. If U₁, and U₂ are two ideals of a ring R then U₁∪U₂is ideal of R if and only if U₁⊂ U₂ or U₂⊂U₁

Proof. Let U₁ ∪ U₂ be an ideal of R. We now prove that U₁⊂ U₂ or U₂⊂U₁

If possible, suppose that U₁⊄ U₂ or U₂⊄U₁

Since U₁⊄ U₂ there exists an element a ∈ U₁ and a ∈ U₂.

Since U₂⊄U₁ there exists an element b ∈ U₂ and ∈ U₁.

a ∈ U₁ and b∈ U₂  => a, b ∈ U₁⊂ U₂

a,b ∈ U₁⊂ U₂ and U₁⊂ U₂ is an ideal =>a-b ∈ U₁ ∪ U₂

=> a-b ∈ U₁  or a-b ∈ U₂

But a-b ∈ U₁=>a- {a-b) = b ∈U₁ …………………..1

a-b ∈ U₂ => b+(a-b)=a ∈ U₂    ………………………2

Both (1) and (2) contradict a ∉U₂ and b∉U₁

Our supposition is wrong.

Hence U₁⊂ U₂ or U₂⊂U₁

Conversely, let U₁⊂ U₂ or U₂⊂U₁

Then U₁ ∪ U₂  = U2 or U₁and is an ideal.

Note. If S₁, and S₂ are two subrings of a ring prove that S₁ u S₂ is also a subring iff either S_{1 ⊆} S₂ or S₂ ⊆S₁

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

Derivative of a Vector Function Solved Problems

Example. 1 . If r = a cos ti + a sin tj + at tan θk find \( \left|\frac{d r}{d t} \times \frac{d^2}{d t^2}\right| \text { and } \mid \frac{d r}{d t} \frac{d^2 t}{d t^2} \frac{d^2}{d t^2}\)

Solution.

Given

r = a cos ti + a sin tj + at tan θk ∴ dtr = −a sin ti + a cos tj + a tan θk

⇒ \(\frac{d^2 r}{d t^2}=-a \cos t \mathbf{i}-a \sin t \mathbf{j}+0 . \quad \cdots \frac{d^3 r}{d t^2}=a \sin t \mathbf{i}-a \cos t \mathbf{j}\)

∴ \(\frac{d r}{d t} \times \frac{d^2 r}{d t^2}\)

= \(\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \\-a \sin t & a \cos t & a \tan \theta \\-a \cos t & -a \sin t & 0
\end{array}\right|=a^2 \sin t \tan \theta \mathrm{i}-a^2 \cos t \tan \theta \mathbf{j}+a^2 \mathbf{k}\)

∴ \(\left|\frac{\mathbf{d r}}{d t} \times \frac{d^2 r}{d t^2}\right|\)

= \(a^2 \sqrt{\left(\sin ^2 t \tan ^2 \theta+\cos ^2 t \tan ^2 \theta+1\right)}=a^2 \sqrt{\left(\tan ^2 \theta+1\right)}\)

= \(d^2 \sec \theta \text { And }\left[\frac{d t}{d t} \frac{d^2 t}{d t^2} \frac{d^2 c}{d t^2}\right]= \)

⇒\(\left(\frac{d r}{d t} \times \frac{d^2 r}{d t^2}\right) \cdot \frac{d^2 r}{d t^3}\)

= (a² sin t tan θi − a² cos t tan θj + a²k) · (a sin ti− a cos tk) = a³ sin² t tan θ+ a³ cos² t tan θ = a³ tan θ

Derivative Of A Vector Function With Scalar Variable

Example. 2. If A = t²i−tj+(2t+1)k and B = (2t−3)i+j−tk find

(1) (A × B)0

(2) (| A + B|)0 · atit = 1

Solution.

Given A = t²i − tj + (2t + 1)k

⇒ \(\mathbf{A}^{\prime}=\frac{d \mathbf{A}}{d t}=2 t \mathbf{i}-\mathbf{j}+2 \mathbf{k} \text { and } \mathbf{B}=(2 t-3) \mathbf{i}+\mathbf{j}-t \mathbf{k}\)

⇒ \(\mathbf{B}^{\prime}=\frac{d \mathbf{B}}{d t}=2 \mathbf{i}-\mathbf{k} \text { at } t=1, \mathbf{A}=\mathbf{i}-\mathbf{j}+3 \mathbf{k}, \mathbf{B}=-\mathbf{i}+\mathbf{j}-\mathbf{k}\)

⇒ \(\frac{d \mathrm{~A}}{d t}=2 \mathrm{t}-\mathrm{j}+2 \mathrm{k}, \frac{d \mathrm{~B}}{d t}=2 \mathrm{i}-\mathrm{k}\)

∴ A × B’ = \(\left|\begin{array}{ccc} \mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & -1 & 3 \\2 & 0 & -1\end{array}\right|\)

= \(\mathbf{i}+7 \mathbf{j}+2 \mathbf{k} \quad \mathbf{A}^{\prime} \times \mathbf{B}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\2 & -1 & 2 \\-1 & 1 & -1\end{array}\right|= \)

−i + k

∴ (A × B)’ = (i + 7j + 2k) + (−i + k) = 7j + 3k (5)

A + B = t² i − tj + (2t + 1)k + (2t − 3)i + j − tk

A + B = (t² + 2t − 3) i + (1 − t)j + (t + 1)k

|A + B| = \(\sqrt{\left[\left(t^2+2 t-3\right)^2+(1-t)^2+(t+1)^2\right]}=\sqrt{\left(t^4+4 t^3-12 t+11\right)}\)

∴ \((|\mathbf{A}+\mathbf{B}|)^{\prime}=\frac{4 t^3+12 t-12}{2 \sqrt{\left(t^4+4 t^3-12 t+11\right)}} \quad ∴ \text { At } t=1,(|\mathbf{A}+\mathbf{B}|)^{\prime}=1\)

Example. 3. If r is a vector function, of the scalar t, whose modulus is r and a, b are constant vectors, differentiate the following with respect to t. \(\begin{array}{ll}\text { (1) } \frac{r}{r^2}+\frac{r b}{a \cdot r} & \text { (2) } r^3 r+a \times r\end{array}\)

Solution.

(1) \(\frac{d}{d t}\left[\frac{r}{r^2}+\frac{r \mathrm{~b}}{\mathrm{a}-\mathrm{r}}\right]=\frac{d}{d t}\left(\frac{r}{r^2}\right)+\mathrm{b} \frac{d}{d t}\left(\frac{r}{\mathrm{a} \cdot \mathrm{r}}\right)\)

⇒ \(\left.=\frac{r^2 \frac{d r}{d t}-\mathbf{r}(2 r) \frac{d r}{d t}}{r^4}+\frac{\mathbf{b}\left[(\mathbf{a} \cdot \mathbf{r}) \frac{d r}{d t}-r\left(\mathbf{a} \cdot \frac{d r}{d t}\right)\right]}{\left(\mathbf{a} \cdot \mathbf{r}^2\right.}\right)\)

⇒ \(=\frac{r^2 \mathbf{r}^{\prime}-2 r \mathbf{r}^{\prime}}{r^4}+\mathbf{b}\left[\frac{1}{(\mathbf{a} \cdot \mathbf{r})} \cdot \frac{d r}{d t}-\frac{r}{(\mathbf{a} \cdot \mathbf{r})^2}\left(\mathbf{a} \cdot \frac{d \mathbf{r}}{d t}\right)\right]=\)

⇒ \(=\frac{\mathbf{r}^{\prime}}{r^2}-\frac{2 r^{\prime} \mathbf{r}}{r^3}+\frac{\mathbf{b} r^{\prime}}{(\mathbf{a} \cdot \mathbf{r})}-\frac{\mathbf{b} r\left(\mathbf{a} \cdot \mathbf{r}^{\prime}\right)}{(\mathbf{a} \cdot \mathbf{r})^2}\)

(2) \(\frac{d}{d t}\left[r^3 \mathbf{r}+\mathbf{a} \times \mathbf{r}^{\prime}\right]=\frac{d}{d t}\left[r^3 \mathbf{r}\right]+\frac{d}{d t}\left(\mathbf{a} \times \mathbf{r}^{\prime}\right)\)

⇒ \(=r^3 \frac{d \mathbf{r}}{d t}+3 r^2 \frac{d t}{d t} \mathbf{r}+\mathbf{a} \times \frac{d(\mathbf{r})}{d t}=r^3 \mathbf{r}^{\prime}+3 r^2 r^{\prime} \mathbf{r}+\mathbf{a} \times \mathbf{r}^{\prime \prime}\)

Continuity Of Vector Functions Explained

Example. 4. If r = ae^λt + be^−λt where a and b are constant vectors show that\(\frac{d^2 \mathrm{r}}{d t^2}=\lambda^2 \mathrm{r}\)

Solution.

Given, r = ae^λt + be^−λt

Diffirentiating w.r.t. t, \(\frac{d r}{d t}=\lambda\left(a e^{\lambda t}-b e^{-\lambda t}\right) \cdots\)

⇒ \(\frac{d^2 r}{d t^2}=\lambda^2\left(a e^{i t}+b e^{-\lambda t}\right)=\lambda^2 r\)

Derivative of a Vector Function Solved Problems

Example. 5. Show that if r = acos wt + b sin wt, where a, b are constant vectors and w is a constant scalar, then \(\frac{d^2 r}{d d^2}=-w^2 \mathbf{r} \text { and } \mathbf{r} \times \frac{d r}{d t}=w(\mathbf{a} \times \mathbf{b}) \)

Solution.

Given r = acos wt + b sin wt, a, b are constant vectors and w is a constant scalar.

D. w.r.t t we get dr/dt = w(−asin wt + b cos wt)

⇒ \(\frac{d^2 \mathbf{r}}{d t^2}=\lambda^2\left(a e^{i t}+b e^{-\lambda t}\right)=\lambda^2 \mathbf{r}\)

⇒ \(\mathbf{r} \times \frac{d \mathbf{r}}{d t}=(\mathbf{a} \cos w t+\mathbf{b} \sin w t) \times w(-\mathbf{a} \sin w t+\mathbf{b} \cos w t)\)

= w (a × b cos² wt − b × asin² wt)

= w(a × b) (cos² wt + sin² wt) = w(a × b)

Example. 6. A particle moves so that its position vector is given by r = (cos ωt)i+(sin ωt)j, where ω is a constant scalar.

(1) Find the velocity of the particle and show that it is perpendicular to r.
(2) Find the acceleration and show that it is directed opposite to the direction of r.
(3) Show that the magnitude of the acceleration is proportional to the magnitude of r.

Solution.

Given r = (cos ωt)i + (sin ωt)j is the position vector of the particle.

D. W.r to t, we get dtr = [(− sin ωt)i + (cos ωt)j]ω

v = velocity = ω(− sin ωti + cos ωtj)

(1) r. v = ω[(cos ωti + sin ωtj) · (− sin ωti + cos νtj)

= ω(− sin ωt cos ωt + sin ωt cos ωt) = 0

(2) \(\frac{d^2 \mathrm{r}}{d t^2}=\omega^2(-\cos \omega t \mathrm{i}-\sin \omega t \mathrm{j})=-\omega^2 \bar{r}\)

acceleration = a = −ω²r, which has that it is the direction of opposite the direction of r.

(3) |a| = w²|r|

Example. 7. Suppose a particle P moves along a curve where parametric equations are x = t − sin t, y = 1 − cos t, z = 4sin(t/2), where t is time, (1) find the velocity and acceleration at time t = 0 (2) Find the magnitude of velocity and acceleration t = 0

Solution.

The position vector of the particle P is r = xi + yj + zk = (t − sin t)i + (1 − cos t)j + (4sin(t/2))k

D. w. r to t \(\frac{d \mathbf{r}}{d t}=(1-\cos t) \mathbf{i}+(\sin t) \mathbf{j}+\left(2 \cos \frac{t}{2}\right) \mathbf{k} \text { and } \frac{d^2 \mathbf{r}}{d t^2}=(\sin t) \mathbf{i}+(\cos t) \mathbf{j}+\left(-\sin \frac{t}{2}\right) \mathbf{k}\)

(1) The velocity v and acceleration a of P are, \(\mathbf{v}=\frac{d r}{d t}=(1-\cos t) \mathbf{i}+(\sin t) \mathbf{j}+\left(2 \cos \frac{t}{2}\right) \mathbf{k} ; \quad \mathbf{a}=\frac{d^2 \mathbf{r}}{d t^2}=(\sin t) \mathbf{i}+\cos t \mathbf{j}+ \)

(− sin t/2)k (b) at, t = 0, v = i + 2k, a = j

|v| = √1 + 4 = √5. | a| = 1

Examples Of Vector Function Scalar Variable Derivatives 

Example. 8. A particle moves along a curve whose parametric equations are x = e −t, y = 2cos³t, z = 2sin³t, where t is the time, determine its velocity and acceleration at any time.

Solution.

Given

A particle moves along a curve whose parametric equations are x = e −t, y = 2cos³t, z = 2sin³t, where t is the time

The position vector of the particle is r = xi + yj + zk = e −ti + (2cos³t)j + (2sin³t)k

D.W.R. to t

⇒ \(\frac{d \mathbf{r}}{d t}=\text { Velocity }=\mathbf{v}=-e^{-t^2} \mathbf{i}+(-6 \sin 3 t) \mathbf{j}+(6 \cos 3 t) \mathbf{k}\)

⇒ \(\mathbf{a}=\text { acceleration }=\frac{d^2 \mathbf{r}}{d t^2}=e^{-t} \mathbf{i}+(-18 \sin 3 t) \mathbf{j}+(-18 \cos 3 t) \mathbf{k} \)

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