Free UPSC Mathematics Optional Paper-2 2020 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 13, 2025ByAbstract Classes

खण्ड ‘

A^{'}

SECTION ‘

A^{'}

1.(a) मान लीजिए कि

S_{3}

व

Z_{3}

क्रमशः 3 प्रतीकों का क्रमचय समूह एवं मॉड्यूल 3 अवशिष्ट वर्गों के समूह हैं। दर्शाइए कि

S_{3}

का

Z_{3}

में तुच्छ समाकारिता के अतिरिक्त कोई भी समाकारिता नहीं है ।

Let

S_{3}

and

Z_{3}

be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of

S_{3}

Z_{3}

except the trivial homomorphism.

Answer:

To show that there is no homomorphism of

S_{3}

into

Z_{3}

except the trivial homomorphism, we can use the following properties of homomorphisms:

A homomorphism $ϕ : G \to H$ preserves the identity element, i.e., $ϕ (e_{G}) = e_{H}$ .
A homomorphism $ϕ : G \to H$ preserves the group operation, i.e., $ϕ (a * b) = ϕ (a) * ϕ (b)$ .
A homomorphism $ϕ : G \to H$ preserves the order of elements, i.e., if $a$ has order $n$ in $G$ , then $ϕ (a)$ has order dividing $n$ in $H$ .

Properties of $S_{3}$ and $Z_{3}$

$S_{3}$ is the permutation group on 3 symbols, and it has $3! = 6$ elements.
$Z_{3}$ is the group of residue classes modulo 3, and it has 3 elements: $[0], [1], [2]$ .

Steps to Show No Non-Trivial Homomorphism Exists

Identity Element: Any homomorphism $ϕ : S_{3} \to Z_{3}$ must map the identity element of $S_{3}$ (the identity permutation $e$ ) to the identity element of $Z_{3}$ ( $[0]$ ).

$ϕ (e) = [0]$
Order of Elements: The order of any element in $Z_{3}$ divides 3. In $S_{3}$ , we have elements of order 2 (e.g., transpositions) and elements of order 3 (e.g., 3-cycles). If there exists a non-trivial homomorphism $ϕ$ , then it must map elements of $S_{3}$ to elements of $Z_{3}$ in such a way that the order of the image divides the order of the original element.

However, $Z_{3}$ only has elements of order 1 ( $[0]$ ) and order 3 ( $[1], [2]$ ). There are no elements of order 2 in $Z_{3}$ .
Contradiction: $S_{3}$ contains elements of order 2 (transpositions). Any homomorphism $ϕ$ would have to map these elements to an element in $Z_{3}$ whose order divides 2. Since $Z_{3}$ contains no such elements (other than the identity), we reach a contradiction.

Therefore, the only homomorphism that can exist from

S_{3}

Z_{3}

is the trivial homomorphism that maps all elements of

S_{3}

to the identity element

[0]

Z_{3}

1.(b) मान लीजिए

R

मुख्य गुणजावली प्रान्त है । दर्शाइए कि

R

के विभाग-वलय की प्रत्येक गुणजावली, मुख्य गुणजावली है तथा

R / P, R

के अभाज्यगुणजावली

P

के लिए मुख्य गुणजावली प्रान्त है ।

Let

R

be a principal ideal domain. Show that every ideal of a quotient ring of

R

is principal ideal and

R / P

is a principal ideal domain for a prime ideal

P

R

Answer:

Introduction

The problem asks us to prove two things:

Every ideal of a quotient ring $R / P$ is a principal ideal.
If $P$ is a prime ideal of $R$ , then $R / P$ is a principal ideal domain (PID).

To prove these statements, we’ll use the properties of principal ideal domains and quotient rings.

Work/Calculations

Part 1: Every ideal of $R / P$ is a principal ideal

Let

I / P

be an ideal of

R / P

, where

I

is an ideal of

R

containing

P

Step 1: Show that $I$ is a principal ideal in $R$

Since

R

is a PID,

I

is generated by a single element

a

R

. That is,

I = (a)

Step 2: Show that $I / P$ is generated by $a + P$ in $R / P$

Let’s substitute the values:

I / P = (a) + P

After substituting, we can see that

I / P

is generated by

a + P

R / P

Therefore,

I / P

is a principal ideal in

R / P

Part 2: $R / P$ is a PID for a prime ideal $P$ of $R$

Step 1: Show that $R / P$ is an integral domain

Since

P

is a prime ideal,

R / P

is an integral domain.

Step 2: Show that every ideal in $R / P$ is principal

From Part 1, we know that every ideal in

R / P

is principal.

Step 3: Conclude that $R / P$ is a PID

Since

R / P

is an integral domain and every ideal in

R / P

is principal,

R / P

is a PID.

Conclusion

We have shown that every ideal of a quotient ring

R / P

is a principal ideal. Additionally, if

P

is a prime ideal of

R

, then

R / P

is a principal ideal domain. Both of these statements hold true when

R

is a principal ideal domain.

1.(c) सिद्ध कीजिए कि शर्त

| a_{n + 1} - a_{n} | ⩽ α | a_{n} - a_{n - 1} |

, जहाँ पर

0 < α < 1

को सभी प्राकृतिक संख्याओं

n ⩾ 2

के लिए सन्तुष्ट करने वाला अनुक्रम

(a_{n}^{'})

, कॉशी-अनुक्रम होता है ।

Prove that the sequence

(a_{n})

satisfying the condition

| a_{n + 1} - a_{n} | ⩽ α | a_{n} - a_{n - 1} |, 0 < α < 1

for all natural numbers

n ⩾ 2

, is a Cauchy sequence.

Answer:

Introduction

The problem asks us to prove that a sequence

(a_{n})

satisfying the condition

| a_{n + 1} - a_{n} | ⩽ α | a_{n} - a_{n - 1} |, 0 < α < 1

for all natural numbers

n ⩾ 2

, is a Cauchy sequence. A sequence is said to be Cauchy if for every

ϵ > 0

, there exists an

N

such that for all

m, n > N

| a_{m} - a_{n} | < ϵ

Work/Calculations

Step 1: Prove that $| a_{n + 1} - a_{n} |$ becomes arbitrarily small

We are given that

| a_{n + 1} - a_{n} | ⩽ α | a_{n} - a_{n - 1} |

Let’s substitute the values:

| a_{n + 1} - a_{n} | ⩽ α^{n} | a_{2} - a_{1} |

After substituting, we see that as

n

becomes large,

α^{n}

approaches zero (since

0 < α < 1

), making

| a_{n + 1} - a_{n} |

arbitrarily small.

Step 2: Prove that $(a_{n})$ is a Cauchy sequence

To prove that

(a_{n})

is a Cauchy sequence, we need to show that for any

ϵ > 0

, there exists an

N

such that for all

m, n > N

| a_{m} - a_{n} | < ϵ

Consider

m > n

and

m, n > N

. We have:

\begin{aligned} | a_{m} - a_{n} | & = | a_{m} - a_{m - 1} + a_{m - 1} - a_{m - 2} + \dots + a_{n + 1} - a_{n} | \\ \leq | a_{m} - a_{m - 1} | + | a_{m - 1} - a_{m - 2} | + \dots + | a_{n + 1} - a_{n} | \\ \leq α^{m - 1} | a_{2} - a_{1} | + α^{m - 2} | a_{2} - a_{1} | + \dots + α^{n} | a_{2} - a_{1} | \\ = α^{n} | a_{2} - a_{1} | (1 + α + α^{2} + \dots + α^{m - n - 1}) \\ = α^{n} | a_{2} - a_{1} | \frac{1 - α^{m - n}}{1 - α} \\ < α^{n} | a_{2} - a_{1} | \frac{1}{1 - α} \end{aligned}

We can make this less than

ϵ

by choosing

N

large enough so that

α^{N} | a_{2} - a_{1} | \frac{1}{1 - α} < ϵ

Conclusion

We have shown that for any

ϵ > 0

, there exists an

N

such that for all

m, n > N

| a_{m} - a_{n} | < ϵ

. Therefore, the sequence

(a_{n})

is a Cauchy sequence, as required.

1.(d) समाकल

\int_{C} (z^{2} + 3 z) d z

का,

(2, 0)

से

(0, 2)

तक वक्र

C

के वामावर्त अनुगत जहाँ पर

C

वृत्त

| z | = 2

है, मान निकालिए ।

Evaluate the integral

\int_{C} (z^{2} + 3 z) d z

counterclockwise from

(2, 0)

(0, 2)

along the curve

C

, where

C

is the circle

| z | = 2

Answer:

Introduction

The problem asks us to evaluate the integral

\int_{C} (z^{2} + 3 z) d z

counterclockwise from

(2, 0)

(0, 2)

along the curve

C

, where

C

is the circle

| z | = 2

Given Parameters and Parametrization

Given

| z | = 2

, i.e.

z = 2 e^{i θ}

and

0 \leq θ \leq \frac{π}{2}

Integral Setup

∴ \int_{C} (z^{2} + 3 z) d z = \int_{| z | = 2} (z^{2} + 3 z) d z

Substitution and Simplification

= \int_{0}^{π / 2} [{(2 e^{i θ})}^{2} + 3 (2 e^{i θ})] 2 e^{i θ} \cdot i d θ

Evaluation of the Integral

= {[8 i \frac{e^{3 i θ}}{3 i} + 12 i \frac{e^{2 i θ}}{2 i}]}_{0}^{\frac{π}{2}}

Final Simplification

= \frac{8}{3} (e^{i 3} π^{2} - 1) + 6 (e^{i π} - 1)

= \frac{8}{3} (- i - 1) + 6 (- 2)

= \frac{- 44}{3} - \frac{8}{3} i

Conclusion

The integral

\int_{C} (z^{2} + 3 z) d z

along the curve

C

, where

C

is the circle

| z | = 2

, is

\frac{- 44}{3} - \frac{8}{3} i

1.(e) यू.पी.एस.सी. के रखरखाव विभाग ने भवन में पर्दों की आवश्यकता-पूर्ति हेतु पर्दा-कपडे के पर्याप्त संख्या में टुकड़े खरीदे हैं। प्रत्येक टुकड़े की लम्बाई 17 फुट है। पर्दों की लम्बाई के अनुसार आवश्यकता निम्नलिखित है :

\begin{array}{cc} पर्दे की लम्बाई (फुटों में) आवश्यक संख्या \\ 5 & 700 \\ 9 & 400 \\ 7 & 300 \end{array}

टुकडों एवं सभी पर्दों की चौड़ाइयाँ समान हैं। विभिन्न रूप से काटे गये टुकड़ों की संख्या का निर्णय इस प्रकार करने हेतु कि कुल कटान-हानि न्यूनतम हो, एक रैखिक प्रोग्रामन समस्या का प्रामाणिक इस प्रकार करने हेतु कि कुल कटान-हानि न्यूनतम हो, एक रैखिक प्रोग्रामन समस्या का प्रामाणिक रूप में निर्धारण कीजिए । इसका एक आधारी सुसंगत हल भी दीजिए ।

UPSC maintenance section has purchased sufficient number of curtain cloth pieces to meet the curtain requirement of its building. The length of each piece is 17 feet. The requirement according to curtain length is as follows:

\begin{array}{cc} Curtain length (in feet) & Number required \\ 5 & 700 \\ 9 & 400 \\ 7 & 300 \end{array}

The width of all curtains is same as that of available pieces. Form a linear programming problem in standard form that decides the number of pieces cut in different ways so that the total trim loss is minimum. Also give a basic feasible solution to it.

Answer:

Introduction

The UPSC maintenance section has purchased curtain cloth pieces, each of length 17 feet, to meet the curtain requirements of its building. The goal is to cut these 17-foot pieces into smaller lengths of 5, 9, and 7 feet to meet the specific requirements while minimizing the total trim loss. We will formulate this as a linear programming problem (LPP) in standard form.

Variables

Let

x_{1}

be the number of 17-foot pieces cut into one 5-foot piece and one 12-foot piece.
Let

x_{2}

be the number of 17-foot pieces cut into one 9-foot piece and one 8-foot piece.
Let

x_{3}

be the number of 17-foot pieces cut into one 7-foot piece and one 10-foot piece.

Objective Function

The objective is to minimize the total trim loss, which is the sum of the remaining lengths after cutting the 17-foot pieces. The total trim loss

T

can be represented as:

Minimize T = 12 x_{1} + 8 x_{2} + 10 x_{3}

Constraints

The number of 5-foot pieces should be at least 700:
$x_{1} \geq 700$
The number of 9-foot pieces should be at least 400:
$x_{2} \geq 400$
The number of 7-foot pieces should be at least 300:
$x_{3} \geq 300$
All variables must be non-negative:
$x_{1}, x_{2}, x_{3} \geq 0$

Linear Programming Problem in Standard Form

Minimize T = 12 x_{1} + 8 x_{2} + 10 x_{3}

subject to:

\begin{aligned} x_{1} & \geq 700 \\ x_{2} & \geq 400 \\ x_{3} & \geq 300 \\ x_{1}, x_{2}, x_{3} & \geq 0 \end{aligned}

Basic Feasible Solution

A basic feasible solution can be obtained by setting the slack variables to zero and solving the constraints for

x_{1}, x_{2}, x_{3},

and

T

$x_{1} = 700$
$x_{2} = 400$
$x_{3} = 300$
$T = 12 \times 700 + 8 \times 400 + 10 \times 300$

After Calculating, we get:

T = 8400 + 3200 + 3000 = 14600

Conclusion

The linear programming problem to minimize the total trim loss while meeting the curtain requirements is formulated as above. A basic feasible solution suggests cutting 700 pieces into 5-foot lengths, 400 pieces into 9-foot lengths, and 300 pieces into 7-foot lengths, resulting in a total trim loss of 14,600 feet.

2.(a) मान लीजिए

G, n

समूहांक का परिमित चक्रीय समूह है। तब सिद्ध कीजिए कि

G

के

ϕ (n)

जनक हैं (जहाँ पर

ϕ

ऑयलर

ϕ

-फलन है) ।

Let

G

be a finite cyclic group of order

n

. Then prove that

G

has

ϕ (n)

generators (where

ϕ

is Euler’s

ϕ

-function).

Answer:

Introduction

The problem asks us to prove that a finite cyclic group

G

of order

n

has

ϕ (n)

generators, where

ϕ

is Euler’s

ϕ

-function. Euler’s

ϕ

-function

ϕ (n)

is defined as the number of positive integers less than

n

that are relatively prime to

n

Preliminaries

Let

G

be a finite cyclic group of order

n

generated by

a

. That is,

G = {a^{0}, a^{1}, a^{2}, \dots, a^{n - 1}}

Generators of $G$

A generator

g

G

is an element such that every element in

G

can be written as a power of

g

. In other words,

G = {g^{0}, g^{1}, g^{2}, \dots, g^{n - 1}}

Euler’s $ϕ$ -Function

Euler’s

ϕ

-function

ϕ (n)

counts the number of positive integers less than

n

that are relatively prime to

n

Proof

Claim: An element $a^{k}$ generates $G$ if and only if $gcd (k, n) = 1$ .

Proof of Claim:
- Forward Direction: Suppose $a^{k}$ generates $G$ . Then $⟨ a^{k} ⟩ = G$ , which means $a^{k}$ has order $n$ . By Lagrange’s theorem, the order of $a^{k}$ must divide $n$ . Since $a^{k}$ has order $n$ , $k$ and $n$ must be relatively prime.
- Backward Direction: Suppose $gcd (k, n) = 1$ . We want to show that $a^{k}$ generates $G$ . To do this, we need to show that every element $a^{m}$ in $G$ can be written as $(a^{k})^{r}$ for some integer $r$ .
Since $gcd (k, n) = 1$ , there exist integers $p$ and $q$ such that $p k + q n = 1$ . For any $a^{m}$ in $G$ , we have:

$a^{m} = a^{m p k} = (a^{k})^{m p} = (a^{k})^{r}$

where $r = m p$ . This shows that $a^{k}$ generates $G$ .
Counting Generators: The number of generators of $G$ is the same as the number of integers $k$ such that $0 < k < n$ and $gcd (k, n) = 1$ . This is precisely $ϕ (n)$ .

Conclusion

We have proven that a finite cyclic group

G

of order

n

has

ϕ (n)

generators. The proof relies on the properties of Euler’s

ϕ

-function and the definition of a cyclic group.

2.(b) सिद्ध कीजिए कि फलन

f (x) = \sin x^{2}

अंतराल

[0, \infty

[ पर एकसमान संतत नहीं है।

Prove that the function

f (x) = \sin x^{2}

is not uniformly continuous on the interval

[0, \infty [

Answer:

Introduction

The problem asks us to prove that the function

f (x) = \sin (x^{2})

is not uniformly continuous on the interval

[0, \infty)

Definition of Uniform Continuity

A function

f (x)

is said to be uniformly continuous on an interval

I

if for every

ϵ > 0

, there exists a

δ > 0

such that for all

x, y

I

, if

| x - y | < δ

, then

| f (x) - f (y) | < ϵ

Proof by Contradiction

To prove that

f (x) = \sin (x^{2})

is not uniformly continuous on

[0, \infty)

, we’ll use a proof by contradiction. Assume that

f (x)

is uniformly continuous on

[0, \infty)

Assumption: Assume $f (x) = \sin (x^{2})$ is uniformly continuous on $[0, \infty)$ .
Choose $ϵ$ : Let $ϵ = \frac{1}{2}$ .
Find $δ$ : According to the definition of uniform continuity, there must exist a $δ > 0$ such that for all $x, y$ in $[0, \infty)$ , if $| x - y | < δ$ , then $| \sin (x^{2}) - \sin (y^{2}) | < \frac{1}{2}$ .
Construct Counterexample: Consider the sequence $x_{n} = \sqrt{n π}$ and $y_{n} = \sqrt{(n + 1) π}$ . We have:

$| x_{n} - y_{n} | = | \sqrt{n π} - \sqrt{(n + 1) π} | = | \sqrt{π} (\sqrt{n} - \sqrt{n + 1}) |$

As $n$ approaches infinity, $| x_{n} - y_{n} |$ approaches 0, which means for sufficiently large $n$ , $| x_{n} - y_{n} | < δ$ .
Evaluate $f (x_{n}) - f (y_{n})$ : We have $f (x_{n}) = \sin (n π) = 0$ and $f (y_{n}) = \sin ((n + 1) π) = 0$ . Therefore, $| f (x_{n}) - f (y_{n}) | = 0$ .
Contradiction: The function $f (x) = \sin (x^{2})$ oscillates infinitely many times as $x$ approaches infinity. This means that for any $δ > 0$ , we can find points $x, y$ such that $| x - y | < δ$ but $| \sin (x^{2}) - \sin (y^{2}) |$ is close to 1, contradicting the assumption that $f (x)$ is uniformly continuous.

Conclusion

We have shown that assuming

f (x) = \sin (x^{2})

is uniformly continuous leads to a contradiction. Therefore,

f (x) = \sin (x^{2})

is not uniformly continuous on the interval

[0, \infty)

2.(c) कन्टूर समाकलन का उपयोग कर, समाकल

\int_{0}^{2 π} \frac{1}{3 + 2 \sin θ} d θ

का मान ज्ञात कीजिए ।

Using contour integration, evaluate the integral

\int_{0}^{2 π} \frac{1}{3 + 2 \sin θ} d θ

Answer:

Introduction:
Using contour integration, evaluate the integral

\int_{0}^{2 π} \frac{1}{3 + 2 \sin θ} d θ .

Answer:

I = \int_{0}^{2 π} \frac{d θ}{3 + 2 \sin θ}

Step 1: Change of Variable
Consider

z = e^{i θ} \Rightarrow d z = i e^{i θ} d θ

\Rightarrow d θ = \frac{d z}{i z}, \sin θ = \frac{z - z^{- 1}}{2 i} = \frac{z^{2} - 1}{2 i z}

Step 2: Substitution
Then

I = \int \frac{1}{3 + \frac{2 (z^{2} - 1)}{2 i z}} \frac{d z}{i z} = \int \frac{\frac{1}{3 i z + z^{2} - 1}}{i z} \frac{d z}{i z} = \int_{c} \frac{d z}{z^{2} + 3 i z - 1} = \int_{c} f (z) d z

Step 3: Finding Poles

f (z)

has poles when

z^{2} + 3 i z - 1 = 0

\begin{aligned} z = \frac{- 3 i \pm \sqrt{9 i^{2} - 4 (- 1) (1)}}{2} \\ z = \frac{- 3 i \pm \sqrt{- 9 + 4}}{2} = \frac{- 3 i \pm \sqrt{5} i}{2} \end{aligned}

Thus

z_{1} = \frac{(- 3 + \sqrt{5}) i}{2}, z_{2} = \frac{(- 3 - \sqrt{5}) i}{2}

are poles.

Step 4: Residue Calculation
But only

z_{1}

lies inside

C

The residue at

z = z_{1}

\begin{aligned} lim_{z \to z_{1}} (z - z_{1}) \frac{1}{(z - z_{1}) (z - z_{2})} \\ lim_{z \to z_{1}} \frac{1}{z - z_{2}} = \frac{1}{z_{1} - z_{2}} = \frac{1}{\sqrt{5} i} \end{aligned}

Step 5: Applying Residue Theorem
By residue theorem

\int_{c} f (z) d z = 2 π i \times \frac{1}{\sqrt{5} i} = \frac{2 π}{\sqrt{5}}

Conclusion:
Therefore,

\int_{0}^{2 π} \frac{d θ}{3 + 2 \sin θ} = \frac{2 π}{\sqrt{5}}

(a) मान लीजिए $R, p (> 0)$ अभिलक्षण का एक परिमित क्षेत्र है । दर्शाइए कि $f (a) = a^{p}, \forall a \in R$ द्वारा परिभाषित प्रतिचित्रण $f : R \to R$ एकैक समाकारी है ।

Let

R

be a finite field of characteristic

p (> 0)

. Show that the mapping

f : R \to R

defined by

f (a) = a^{p}, \forall a \in R

is an isomorphism.

Answer:

Introduction

The problem asks us to prove that the mapping

f : R \to R

defined by

f (a) = a^{p}

for all

a \in R

is an isomorphism, where

R

is a finite field of characteristic

p > 0

Definitions

Field: A set $R$ with two operations $+$ and $\times$ that satisfy the field axioms.
Characteristic: A field $R$ has characteristic $p$ if $p$ is the smallest positive integer such that $p \cdot a = 0$ for all $a \in R$ . If no such $p$ exists, the characteristic is zero.
Isomorphism: A bijective map $f : R \to R$ that preserves the field operations.

Properties Needed

Frobenius Endomorphism: In a field of characteristic $p$ , $(a + b)^{p} = a^{p} + b^{p}$ and $(a b)^{p} = a^{p} b^{p}$ .

Proof

To prove that

f

is an isomorphism, we need to show that

f

is a bijective map that preserves addition and multiplication.

1. $f$ is a Homomorphism

Preservation of Addition:

$f (a + b) = (a + b)^{p} = a^{p} + b^{p} = f (a) + f (b)$
Preservation of Multiplication:

$f (a b) = (a b)^{p} = a^{p} b^{p} = f (a) f (b)$

2. $f$ is Injective (One-to-One)

Suppose

f (a) = f (b)

. Then

a^{p} = b^{p}

which implies

a^{p} - b^{p} = 0

. Since

R

is a field, it has no zero divisors, and we can factor

a^{p} - b^{p}

(a - b)^{p}

. This means

a - b = 0

a = b

, proving that

f

is injective.

3. $f$ is Surjective (Onto)

Since

R

is finite and

f

is injective,

f

must also be surjective. Alternatively, for any

b \in R

b = b^{p^{2}}

(by Fermat’s Little Theorem or the fact that

R^{*}

, the multiplicative group of

R

, has order

p^{n} - 1

). Thus,

f (b^{p - 1}) = (b^{p - 1})^{p} = b^{p} = b

, showing that

f

is surjective.

Conclusion

We have shown that

f

preserves both addition and multiplication, and is both injective and surjective. Therefore,

f : R \to R

defined by

f (a) = a^{p}

is an isomorphism.

3.(b) एकधा विधि के द्वारा निम्नलिखित रैखिक प्रोग्रामन समस्या को हल कीजिए :
न्यूनतमीकरण कीजिए

z = - 6 X_{1} - 2 X_{2} - 5 X_{3}

बशर्ते कि

\begin{matrix} 2 X_{1} - 3 X_{2} + X_{3} ⩽ 14 \\ - 4 X_{1} + 4 X_{2} + 10 X_{3} ⩽ 46 \\ 2 X_{1} + 2 X_{2} - 4 X_{3} ⩽ 37 \\ X_{1} ⩾ 2, X_{2} ⩾ 1, X_{3} ⩾ 3 \end{matrix}

Solve the linear programming problem using simplex method:

Minimize

z = - 6 X_{1} - 2 X_{2} - 5 X_{3}

subject to

2 X_{1} - 3 X_{2} + X_{3} ⩽ 14

\begin{matrix} - 4 X_{1} + 4 X_{2} + 10 X_{3} ⩽ 46 \\ 2 X_{1} + 2 X_{2} - 4 X_{3} ⩽ 37 \\ X_{1} ⩾ 2, X_{2} ⩾ 1, X_{3} ⩾ 3 \end{matrix}

Answer:

Introduction

The given Linear Programming Problem (LPP) aims to minimize the objective function

z = - 6 X_{1} - 2 X_{2} - 5 X_{3}

subject to certain constraints and variable bounds. The problem is transformed to a standard form by introducing slack variables and then solved using the Simplex method.

Problem Transformation

Let

x_{1} + 2 = X_{1}, x_{2} + 1 = X_{2}

and

x_{3} + 1 = X_{3}

Then, the objective function becomes:

\begin{aligned} min Z^{*} = - 6 (x_{1} + 2) - 2 (x_{2} + 1) - 5 (x_{3} + 3) \\ = - 6 x_{1} - 12 - 12 x_{2} - 2 - 5 x_{3} - 15 \\ = - 6 x_{1} - 2 x_{2} - 5 x_{3} - 29 \end{aligned}

Subject to the transformed constraints:

2 (x_{1} + 2) - 3 (x_{2} + 1) + (x_{3} + 3) ⩽ 14 \Rightarrow 2 x_{1} - 3 x_{2} + x_{3} ⩽ 10

- 4 (x_{1} + 2) + 4 (x_{2} + 1) + 10 (x_{3} + 3) ⩽ 46 \Rightarrow - 4 x_{1} + 4 x_{2} + 10 x_{3} ⩽ 20

2 (x_{1} + 2) + 2 (x_{2} + 1) - 4 (x_{3} + 3) ⩽ 37 \Rightarrow 2 x_{1} + 2 x_{2} - 4 x_{3} ⩽ 43

x_{1}, x_{2}, x_{3} ⩾ 0

After introducing slack variables

\begin{aligned} Min Z^{*} = - 6 x_{1} - 2 x_{2} - 5 x_{3} + 0 S_{1} + 0 S_{2} + 0 S_{3} \\ subject to \\ \begin{aligned} 2 x_{1} - 3 x_{2} + x_{3} + S_{1} & = 10 \\ - 4 x_{1} + 4 x_{2} + 10 x_{3} + S_{2} & = 20 \\ 2 x_{1} + 2 x_{2} - 4 x_{3} & + S_{3} = 43 \end{aligned} \\ and x_{1}, x_{2}, x_{3}, S_{1}, S_{2}, S_{3} \geq 0 \end{aligned}

Iteration-1		$C_{j}$	-6	-2	-5	0	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$S_{1}$	$S_{2}$	$S_{3}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{1}} \end{matrix}$
$S_{1}$	0	10	(2)	-3	1	1	0	0	$\frac{10}{2} = 5 \to$
$S_{2}$	0	20	-4	4	10	0	1	0	—
$S_{3}$	0	43	2	2	-4	0	0	1	$\frac{43}{2} = 21.5$
$Z = 0$		$Z_{j}$	0	0	$0$	0	0	0
$Z = 0$		$C_{j} - Z_{j}$	$- 6 ↑$	-2	-5	0	0	0

Negative minimum

C_{j} - Z_{j}

is -6 and its column index is 1 . So, the entering variable is

x_{1}

.
Minimum ratio is 5 and its row index is 1 . So, the leaving basis variable is

S_{1}

∴

The pivot element is 2 .
Entering

= x_{1}

, Departing

= S_{1}

, Key Element

= 2

\begin{aligned} R_{1} (new) = R_{1} (old) \div 2 \\ R_{2} (new) = R_{2} (old) + 4 R_{1} (new) \\ R_{3} (new) = R_{3} (old) - 2 R_{1} (new) \end{aligned}

Iteration-2		$C_{j}$	-6	-2	-5	0	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$S_{1}$	$S_{2}$	$S_{3}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{2}} \end{matrix}$
$x_{1}$	-6	5	1	-1.5	0.5	0.5	0	0	—
$S_{2}$	0	40	0	-2	12	2	1	0	—
$S_{3}$	0	33	0	(5)	-5	-1	0	1	$\frac{33}{5} = 6.6 \to$
$Z = - 30$		$Z_{j}$	-6	9	-3	-3	0	0
$Z = - 30$		$C_{j} - Z_{j}$	0	$- 11 ↑$	-2	3	0	0

Negative minimum

C_{j} - Z_{j}

is -11 and its column index is 2 . So, the entering variable is

x_{2}

.
Minimum ratio is 6.6 and its row index is 3 . So, the leaving basis variable is

S_{3}

∴

The pivot element is 5 .
Entering

= x_{2}

, Departing

= S_{3}

, Key Element

= 5

\begin{aligned} R_{3} (new) = R_{3} (old) \div 5 \\ R_{1} (new) = R_{1} (old) + 1.5 R_{3} (new) \\ R_{2} (new) = R_{2} (old) + 2 R_{3} (new) \end{aligned}

Iteration-3		$C_{j}$	-6	-2	-5	0	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$s_{1}$	$S_{2}$	$S_{3}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{3}} \end{matrix}$
$x_{1}$	-6	14.9	1	0	-1	0.2	0	0.3	—
$s_{2}$	0	53.2	0	0	(10)	1.6	1	0.4	$\frac{53.2}{10} = 5.32 \to$
$x_{2}$	-2	6.6	0	1	-1	-0.2	0	0.2	—
$Z = - 102.6$		$Z_{j}$	-6	-2	8	-0.8	0	-2.2
$Z = - 102.6$		$C_{j} - Z_{j}$	0	0	$- 13 ↑$	0.8	0	2.2

Negative minimum

C_{j} - Z_{j}

is -13 and its column index is 3 . So, the entering variable is

x_{3}

.
Minimum ratio is 5.32 and its row index is 2 . So, the leaving basis variable is

S_{2}

∴

The pivot element is 10 .
Entering

= x_{3}

, Departing

= S_{2}

, Key Element

= 10

\begin{aligned} R_{2} (new) = R_{2} (old) \div 10 \\ R_{1} (new) = R_{1} (old) + R_{2} (new) \\ R_{3} (new) = R_{3} (old) + R_{2} (new) \end{aligned}

Iteration-4		$C_{j}$	-6	-2	-5	0	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$S_{1}$	$S_{2}$	$S_{3}$	MinRatio
$x_{1}$	-6	20.22	1	0	0	0.36	0.1	0.34
$x_{3}$	-5	5.32	0	0	1	0.16	0.1	0.04
$x_{2}$	-2	11.92	0	1	0	-0.04	0.1	0.24
$Z = - 171.76$		$Z_{j}$	$- 6$	$- 2$	$- 5$	$- 2.88$	$- 1.3$	$- 2.72$
		$C_{j} - Z_{j}$	0	0	0	2.88	1.3	2.72

Since all

C_{j} - Z_{j} \geq 0

Hence, optimal solution is arrived with value of variables as :

x_{1} = 20.22, x_{2} = 11.92, x_{3} = 5.32

Min Z^{*} = - 171.76

Hence,

Min Z = Min Z^{*} - 29

Min Z = - 171.76 - 29 = - 200.76

Conclusion

The optimal solution to the given LPP is

x_{1} = 20.22, x_{2} = 11.92, x_{3} = 5.32

with a minimum value of

Z = - 200.76

. All

C_{j} - Z_{j}

values are non-negative, confirming that the solution is optimal.

3.(c) यदि

u = \tan^{- 1} \frac{x^{3} + y^{3}}{x - y}, x \neq y

तब दर्शाइए कि

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = (1 - 4 \sin^{2} u) \sin 2 u

u = \tan^{- 1} \frac{x^{3} + y^{3}}{x - y}, x \neq y

then show that

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = (1 - 4 \sin^{2} u) \sin 2 u

Answer:

Introduction:
Here, we explore the nature of a function

u = \tan^{- 1} \frac{x^{3} + y^{3}}{x - y}

and investigate its homogeneity.

Homogeneity of $u$ :
We start by considering whether

u

is a homogeneous function.

However, we express

\tan u = \frac{x^{3} + y^{3}}{x - y}

z

, where

z

is a new variable. Thus, we obtain:

\begin{aligned} \tan u = \frac{x^{3} + y^{3}}{x - y} = z \to (1) \\ \Rightarrow z = x^{2} [\frac{1 + {(\frac{y}{x})}^{3}}{1 - (\frac{y}{x})}] \end{aligned}

Now,

z

is shown to be a homogeneous function of

x

and

y

of degree 2:

x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 2 z \to (2)

Derivatives of $z$ and $u$ :
Next, we calculate the derivatives of

z

and

u

with respect to

x

and

y

. These derivatives play a crucial role in our analysis. From equation (1), we have:

\begin{aligned} \frac{\partial z}{\partial x} = \sec^{2} u \frac{\partial u}{\partial x} \\ \frac{\partial z}{\partial y} = \sec^{2} u \frac{\partial u}{\partial y} \to (3) \end{aligned}

Simplification and Relationships:
Now, we use the derivatives to simplify and establish relationships between

x, y, z, u

From equation (2), we have:

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{2 \sin u}{\cos u} \cdot \cos^{2} u = \sin 2 u \to (4)

From the derivatives in equation (3), we further derive:

\begin{aligned} \frac{\partial^{2} z}{\partial x^{2}} = \sec^{2} u \frac{\partial^{2} u}{\partial x^{2}} + 2 \sec^{2} u \tan u {(\frac{\partial u}{\partial x})}^{2} \\ \frac{\partial^{2} z}{\partial y^{2}} = \sec^{2} u \frac{\partial^{2} u}{\partial y^{2}} + 2 \sec^{2} u \tan u {(\frac{\partial u}{\partial y})}^{2} \end{aligned}

And the mixed partial derivative:

\frac{\partial^{2} z}{\partial x \partial y} = \sec^{2} u \frac{\partial^{2} u}{\partial x \partial y} + 2 \sec^{2} u \tan u \frac{\partial u}{\partial x} \frac{\partial u}{\partial y}

Euler’s Theorem and Further Simplification:
By a corollary of Euler’s theorem, we reach the following relationships:

\begin{aligned} x^{2} \frac{\partial^{2} z}{\partial x^{2}} + 2 x y \frac{\partial^{2} z}{\partial x \partial y} + y^{2} \frac{\partial^{2} z}{\partial y^{2}} = 2 (2 - 1) z \\ \Rightarrow \sec^{2} u (x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}}) + 2 \sec^{2} u \tan u [x^{2} {(\frac{\partial u}{\partial x})}^{2} + 2 x y \frac{\partial u}{\partial x} \frac{\partial u}{\partial y} + y^{2} {(\frac{\partial u}{\partial y})}^{2}] = 2 \tan u \end{aligned}

divide by

s e c^{2} u

\begin{aligned} \Rightarrow x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} + 2 \tan u {(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y})}^{2} = 2 \sin u \cos u \\ \Rightarrow x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = \sin 2 u - 2 \tan u \sin^{2} 2 u (b y (4)) \\ = (1 - 4 \sin^{2} u) \sin 2 u \end{aligned}

Final Relationship:
Finally, we obtain:

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = (1 - 4 \sin^{2} u) \sin 2 u

This equation represents a significant relationship in our analysis.

4.(a) यदि

v (r, θ) = (r - \frac{1}{r}) \sin θ, r \neq 0

,
तब विश्लेषिक फलन

f (z) = u (r, θ) + i v (r, θ)

ज्ञात कीजिए ।

v (r, θ) = (r - \frac{1}{r}) \sin θ, r \neq 0

,
then find an analytic function

f (z) = u (r, θ) + i v (r, θ)

Answer:

Introduction:
The problem involves finding an analytic function

f (z)

given a complex-valued function

v (r, θ)

in polar coordinates.

Solution:

Step 1: Cauchy-Riemann Equations in Polar Coordinates
The Cauchy-Riemann equations in polar coordinates are as follows:

\begin{aligned} \frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial θ} (1) \\ \frac{1}{r} \frac{\partial u}{\partial θ} = - \frac{\partial v}{\partial r} (2) \end{aligned}

Step 2: Expressions for Partial Derivatives of $v$
From the given function

v (r, θ)

, we have the following expressions for the partial derivatives:

\begin{aligned} \frac{\partial v}{\partial r} = (1 + \frac{1}{r^{2}}) \sin θ (3a) \\ \frac{\partial v}{\partial θ} = (r - \frac{1}{r}) \cos θ (3b) \end{aligned}

Step 3: Solve for Partial Derivatives of $u$
Using equations (1) and (3), we can calculate the partial derivatives of

u

as follows:

\begin{aligned} \frac{\partial u}{\partial r} = \frac{1}{r} [r - \frac{1}{r}] \cos θ = (1 - \frac{1}{r^{2}}) \cos θ (4) \\ \frac{\partial u}{\partial θ} = - r \frac{\partial v}{\partial r} = - r [1 + \frac{1}{r^{2}}] \sin θ = [- r - \frac{1}{r}] \sin θ (5) \end{aligned}

Step 4: Integrate Partial Derivatives
Integrating equation (4) with respect to

r

and equation (5) with respect to

θ

, we obtain:

\begin{aligned} u (r, θ) = (r + \frac{1}{r}) \cos θ + f (θ) (6) \\ u (r, θ) = (r + \frac{1}{r}) \cos θ + f (r) (7) \end{aligned}

Step 5: Determine $f (θ)$ and $f (r)$
Equating equations (6) and (7), we find:

\begin{aligned} f (θ) = 0 = f (r) \\ u (r, θ) = (r + \frac{1}{r}) \cos θ \end{aligned}

Conclusion:
Therefore, the analytic function

f (z)

is given by:

f (z) = (r + \frac{1}{r}) \cos θ + i [r - \frac{1}{r}] \sin θ

4.(b) दर्शाइए कि

\int_{0}^{π / 2} \frac{\sin^{2} x}{\sin x + \cos x} d x = \frac{1}{\sqrt{2}} \log_{e} (1 + \sqrt{2})

Show that

\int_{0}^{π / 2} \frac{\sin^{2} x}{\sin x + \cos x} d x = \frac{1}{\sqrt{2}} \log_{e} (1 + \sqrt{2})

Answer:

Step 1: Define the Integral
We start by defining the integral

I

as:

I = \int_{0}^{π / 2} \frac{\sin^{2} x}{\sin x + \cos x} d x

Step 2: Apply Property of Definite Integrals
By using the property

\int_{0}^{a} f (x) = \int_{0}^{a} f (a - x)

, we can rewrite the integral as follows:

I = \int_{0}^{π / 2} \frac{\sin^{2} (\frac{π}{2} - x)}{\sin (\frac{π}{2} - x) + \cos (\frac{π}{2} - x)} d x

Step 3: Simplify the Integral
Simplifying the integrand further, we get:

I = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{\cos x + \sin x} d x

Step 4: Express the Integral in a Different Form
Next, we write the integral as:

2 I = \int_{0}^{π / 2} \frac{1}{\sin x + \cos x} d x = \frac{1}{\sqrt{2}} \int_{0}^{π / 2} \frac{1}{\cos (x - \frac{π}{4})} d x

Step 5: Evaluate the Integral
Now, we evaluate the integral:

2 I = \frac{1}{\sqrt{2}} {[\ln | \sec (x - \frac{π}{4}) + \tan (x - \frac{π}{4}) |]}_{0}^{π / 2}

Step 6: Calculate the Values
Calculate the values of the integral:

2 I = \frac{1}{\sqrt{2}} [\ln | \sqrt{2} + 1 | - \ln | \sqrt{2} - 1 |]

Step 7: Simplify
Simplify the expression:

2 I = \frac{1}{\sqrt{2}} \ln | \frac{\sqrt{2} + 1}{\sqrt{2} - 1} |

Step 8: Solve for $I$
Solve for

I

I = \frac{2}{2 \sqrt{2}} \ln | \sqrt{2} + 1 |

Step 9: Final Result
Finally, we arrive at the result:

I = \frac{1}{\sqrt{2}} \ln | \sqrt{2} + 1 |

Conclusion:
The integral

\int_{0}^{π / 2} \frac{\sin^{2} x}{\sin x + \cos x} d x

is equal to

\frac{1}{\sqrt{2}} \ln (1 + \sqrt{2})

, as shown.

(c) वोगेल की सम्निकटन विधि से निम्नलिखित परिवहन समस्या का आरंभिक आधारिक सुसंगत हल ज्ञात कीजिए । इस हल का उपयोग कर समस्या का इष्टतम हल एवं परिवहन लागत ज्ञात कीजिए ।

\begin{array}{llllll} D_{1} & D_{2} & D_{3} & D_{4} & Supply \\ S_{1} & 10 & 0 & 20 & 11 & 15 \\ S_{2} & 12 & 8 & 9 & 20 & 25 \\ S_{3} & 0 & 14 & 16 & 18 & 10 \\ Demand & 5 & 20 & 15 & 10 \end{array}

Find the initial basic feasible solution of the following transportation problem by Vogel’s approximation method and use it to find the optimal solution and the transportation cost of the problem.

\begin{array}{llllll} D_{1} & D_{2} & D_{3} & D_{4} & Supply \\ S_{1} & 10 & 0 & 20 & 11 & 15 \\ S_{2} & 12 & 8 & 9 & 20 & 25 \\ S_{3} & 0 & 14 & 16 & 18 & 10 \\ Demand & 5 & 20 & 15 & 10 \end{array}

Answer:

Initial feasible solution is

	$D_{1}$	$D_{2}$	$D_{3}$	$D_{4}$	Supply	Row Penalty
$S_{1}$	10	$0 (15)$	20	11	15	$10 \| 11 \| - \| - \| - \| - \|$
$S_{2}$	12	$8 (5)$	$9 (15)$	$20 (5)$	25	$1 \| 1 \| 1 \| 12 \| 20 \| - - \|$
$S_{3}$	$0 (5)$	14	16	$18 (5)$	10	$14 \| 2 \| 2 \| 4 \| 18 \| 18 \|$
Demand	5	20	15	10
	10	8	7	7
Column	—	8	7	7
Penalty	—	6	7	2
	—	—	—	2
	—	—	–	18

The minimum total transportation cost

= 0 \times 15 + 8 \times 5 + 9 \times 15 + 20 \times 5 + 0 \times 5 + 18 \times 5 = 365

Here, the number of allocated cells

= 6

is equal to

m + n - 1 = 3 + 4 - 1 = 6

∴

This solution is non-degenerate

Optimality test using modi method…
Allocation Table is

	$D_{1}$	$D_{2}$	$D_{3}$	$D_{4}$	Supply
$S_{1}$	10	$0 (15)$	20	11	15
$S_{2}$	12	$8 (5)$	$9 (15)$	$20 (5)$	25
$S_{3}$	$0 (5)$	14	16	$18 (5)$	10
Demand	5	20	15	10

Iteration-1 of optimality test

Find $u_{i}$ and $v_{j}$ for all occupied cells $(i, j)$ , where $c_{i j} = u_{i} + v_{j}$
Substituting, $u_{2} = 0$ , we get
$c_{22} = u_{2} + v_{2} \Rightarrow v_{2} = c_{22} - u_{2} \Rightarrow v_{2} = 8 - 0 \Rightarrow v_{2} = 8$
$c_{12} = u_{1} + v_{2} \Rightarrow u_{1} = c_{12} - v_{2} \Rightarrow u_{1} = 0 - 8 \Rightarrow u_{1} = - 8$
$c_{23} = u_{2} + v_{3} \Rightarrow v_{3} = c_{23} - u_{2} \Rightarrow v_{3} = 9 - 0 \Rightarrow v_{3} = 9$
$c_{24} = u_{2} + v_{4} \Rightarrow v_{4} = c_{24} - u_{2} \Rightarrow v_{4} = 20 - 0 \Rightarrow v_{4} = 20$
$c_{34} = u_{3} + v_{4} \Rightarrow u_{3} = c_{34} - v_{4} \Rightarrow u_{3} = 18 - 20 \Rightarrow u_{3} = - 2$
$c_{31} = u_{3} + v_{1} \Rightarrow v_{1} = c_{31} - u_{3} \Rightarrow v_{1} = 0 + 2 \Rightarrow v_{1} = 2$

\begin{array}{lllllll} D_{1} & D_{2} & D_{3} & D_{4} & Supply & u_{i} \\ S_{1} & 10 & 0 (15) & 20 & 11 & 15 & u_{1} = - 8 \\ S_{2} & 12 & 8 (5) & 9 (15) & 20 (5) & 25 & u_{2} = 0 \\ S_{3} & 0 (5) & 14 & 16 & 18 (5) & 10 & u_{3} = - 2 \\ Demand & 5 & 20 & 15 & 10 \\ v_{j} & v_{1} = 2 & v_{2} = 8 & v_{3} = 9 & v_{4} = 20 \end{array}

Find $d_{i j}$ for all unoccupied cells(i,j), where $d_{i j} = c_{i j} - (u_{i} + v_{j})$
$d_{11} = c_{11} - (u_{1} + v_{1}) = 10 - (- 8 + 2) = 16$
$d_{13} = c_{13} - (u_{1} + v_{3}) = 20 - (- 8 + 9) = 19$
$d_{14} = c_{14} - (u_{1} + v_{4}) = 11 - (- 8 + 20) = - 1$
$d_{21} = c_{21} - (u_{2} + v_{1}) = 12 - (0 + 2) = 10$
$d_{32} = c_{32} - (u_{3} + v_{2}) = 14 - (- 2 + 8) = 8$
$d_{33} = c_{33} - (u_{3} + v_{3}) = 16 - (- 2 + 9) = 9$

	$D_{1}$	$D_{2}$	$D_{3}$	$D_{4}$	Supply	$u_{i}$
$S_{1}$	$10 [16]$	$0 (15)$	$20 [19]$	$11 [- 1]$	15	$u_{1} = - 8$
$S_{2}$	$12 [10]$	$8 (5)$	$9 (15)$	$20 (5)$	25	$u_{2} = 0$
$S_{3}$	$0 (5)$	$14 [8]$	$16 [9]$	$18 (5)$	10	$u_{3} = - 2$
Demand	5	20	15	10
$v_{j}$	$v_{1} = 2$	$v_{2} = 8$	$v_{3} = 9$	$v_{4} = 20$

Now choose the minimum negative value from all $d_{i j}$ (opportunity cost) $= d_{14} = [- 1]$ and draw a closed path from $S_{1} D_{4}$ .
Closed path is $S_{1} D_{4} \to S_{1} D_{2} \to S_{2} D_{2} \to S_{2} D_{4}$
Closed path and plus/minus sign allocation…

	$D_{1}$	$D_{2}$	$D_{3}$	$D_{4}$	Supply	$u_{i}$
$S_{1}$	$10 [16]$	$0 (15) (-)$	$20 [19]$	$11 [- 1] (+)$	15	$u_{1} = - 8$
$S_{2}$	$12 [10]$	$8 (5) (+)$	$9 (15)$	$20 (5) (-)$	25	$u_{2} = 0$
$S_{3}$	$0 (5)$	$14 [8]$	$16 [9]$	$18 (5)$	10	$u_{3} = - 2$
Demand	5	20	15	10
$v_{j}$	$v_{1} = 2$	$v_{2} = 8$	$v_{3} = 9$	$v_{4} = 20$

Minimum allocated value among all negative position (-) on closed path $= 5$ Substract 5 from all (-) and Add it to all (+)

	$D_{1}$	$D_{2}$	$D_{3}$	$D_{4}$	Supply
$S_{1}$	10	$0 (10)$	20	$11 (5)$	15
$S_{2}$	12	$8 (10)$	$9 (15)$	20	25
$S_{3}$	$0 (5)$	14	16	$18 (5)$	10
Demand	5	20	15	10

Repeat the step 1 to 4 , until an optimal solution is obtained.

Iteration-2 of optimality test

Find

u_{i}

and

v_{j}

for all occupied cells

(i, j)

, where

c_{i j} = u_{i} + v_{j}

Substituting, $u_{1} = 0$ , we get
$c_{12} = u_{1} + v_{2} \Rightarrow v_{2} = c_{12} - u_{1} \Rightarrow v_{2} = 0 - 0 \Rightarrow v_{2} = 0$
$c_{22} = u_{2} + v_{2} \Rightarrow u_{2} = c_{22} - v_{2} \Rightarrow u_{2} = 8 - 0 \Rightarrow u_{2} = 8$
$c_{23} = u_{2} + v_{3} \Rightarrow v_{3} = c_{23} - u_{2} \Rightarrow v_{3} = 9 - 8 \Rightarrow v_{3} = 1$
$c_{14} = u_{1} + v_{4} \Rightarrow v_{4} = c_{14} - u_{1} \Rightarrow v_{4} = 11 - 0 \Rightarrow v_{4} = 11$
$c_{34} = u_{3} + v_{4} \Rightarrow u_{3} = c_{34} - v_{4} \Rightarrow u_{3} = 18 - 11 \Rightarrow u_{3} = 7$
$c_{31} = u_{3} + v_{1} \Rightarrow v_{1} = c_{31} - u_{3} \Rightarrow v_{1} = 0 - 7 \Rightarrow v_{1} = - 7$

\begin{array}{lllllll} D_{1} & D_{2} & D_{3} & D_{4} & Supply & u_{i} \\ S_{1} & 10 & 0 (10) & 20 & 11 (5) & 15 & u_{1} = 0 \\ S_{2} & 12 & 8 (10) & 9 (15) & 20 & 25 & u_{2} = 8 \\ S_{3} & 0 (5) & 14 & 16 & 18 (5) & 10 & u_{3} = 7 \\ Demand & 5 & 20 & 15 & 10 \\ v_{j} & v_{1} = - 7 & v_{2} = 0 & v_{3} = 1 & v_{4} = 11 \end{array}

Find $d_{i j}$ for all unoccupied cells $(i, j)$ , where $d_{i j} = c_{i j} - (u_{i} + v_{j})$

$d_{11} = c_{11} - (u_{1} + v_{1}) = 10 - (0 - 7) = 17$
$d_{13} = c_{13} - (u_{1} + v_{3}) = 20 - (0 + 1) = 19$
$d_{21} = c_{21} - (u_{2} + v_{1}) = 12 - (8 - 7) = 11$
$d_{24} = c_{24} - (u_{2} + v_{4}) = 20 - (8 + 11) = 1$
$d_{32} = c_{32} - (u_{3} + v_{2}) = 14 - (7 + 0) = 7$
$d_{33} = c_{33} - (u_{3} + v_{3}) = 16 - (7 + 1) = 8$

	$D_{1}$	$D_{2}$	$D_{3}$	$D_{4}$	Supply	$u_{i}$
$S_{1}$	$10 [17]$	$0 (10)$	$20 [19]$	$11 (5)$	15	$u_{1} = 0$
$S_{2}$	$12 [11]$	$8 (10)$	$9 (15)$	$20 [1]$	25	$u_{2} = 8$
$S_{3}$	$0 (5)$	$14 [7]$	$16 [8]$	$18 (5)$	10	$u_{3} = 7$
Demand	5	20	15	10
$v_{j}$	$v_{1} = - 7$	$v_{2} = 0$	$v_{3} = 1$	$v_{4} = 11$

Since all

d_{i j} \geq 0

.
So final optimal solution is arrived.

	$D_{1}$	$D_{2}$	$D_{3}$	$D_{4}$	Supply
$S_{1}$	10	$0 (10)$	20	$11 (5)$	15
$S_{2}$	12	$8 (10)$	$9 (15)$	20	25
$S_{3}$	$0 (5)$	14	16	$18 (5)$	10
Demand	5	20	15	10

The minimum total transportation cost = 0 \times 10 + 11 \times 5 + 8 \times 10 + 9 \times 15 + 0 \times 5 + 18 \times 5 = 360

खण्ड ‘B’ SECTION ‘B’

5.(a)

z = y f (x) + x g (y)

से स्वैच्छिक फलनों

f (x)

व

g (y)

का विलोपन कर आंशिक अवकल समीकरण बनाइए तथा इसकी प्रकृति (दीर्घवृत्तीय, अतिपरवलीय या परवलीय)

x > 0, y > 0

क्षेत्र में इंगित कीजिए ।

Form a partial differential equation by eliminating the arbitrary functions

f (x)

and

g (y)

from

z = y f (x) + x g (y)

and specify its nature (elliptic, hyperbolic or parabolic) in the region

x > 0, y > 0

Answer:

Introduction:
The problem requires forming a partial differential equation by eliminating the arbitrary functions

f (x)

and

g (y)

from the given equation

z = y f (x) + x g (y)

. Additionally, it asks for specifying the nature of this equation in the region

x > 0, y > 0

Solution:

Step 1: Given Equation
We are given the equation:

z = y f (x) + x g (y) (1)

Step 2: Partial Differentiation
We differentiate equation (1) partially with respect to

x

and

y

\begin{aligned} \frac{\partial z}{\partial x} = y f^{'} (x) + g (y) (2) \\ \frac{\partial z}{\partial y} = f (x) + x g^{'} (y) (3) \end{aligned}

Step 3: Second Partial Differentiation
We differentiate equation (3) with respect to

x

\frac{\partial^{2} z}{\partial x \partial y} = f^{'} (x) + g^{'} (y) (4)

Step 4: Expressing $f^{'} (x)$ and $g^{'} (y)$
From equations (2) and (3), we express

f^{'} (x)

and

g^{'} (y)

as follows:

\begin{aligned} f^{'} (x) & = \frac{1}{y} (\frac{\partial z}{\partial x} - g (y)) \\ g^{'} (y) & = \frac{1}{x} (\frac{\partial z}{\partial y} - f (x)) \end{aligned}

Step 5: Substituting into Equation (4)
Substituting the expressions for

f^{'} (x)

and

g^{'} (y)

into equation (4), we obtain:

\begin{aligned} \frac{\partial^{2} z}{\partial x \partial y} = \frac{1}{y} (\frac{\partial z}{\partial x} - g (y)) + \frac{1}{x} (\frac{\partial z}{\partial y} - f (x)) \\ \Rightarrow x y \frac{\partial^{2} z}{\partial x \partial y} = x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} - {x g (y) + y f (x)} \\ \Rightarrow x y \frac{\partial^{2} z}{\partial x \partial y} = x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} - z \\ x y \frac{\partial^{2} z}{\partial x \partial y} - x \frac{\partial z}{\partial x} - y \frac{\partial z}{\partial y} + z = 0 \end{aligned}

Step 6: Compare with General Form
Comparing the obtained equation with the general form

R_{r} + S_{s} + T_{t} + f (x, y, z) = 0

, we have:

\begin{aligned} R & = 0 \\ S & = x y \\ T & = 0 \end{aligned}

Step 7: Nature of the Equation
The discriminant of the equation is given by

S^{2} - 4 R T = x^{2} y^{2} > 0

, indicating that it is a hyperbolic partial differential equation.

Conclusion:
The partial differential equation obtained from eliminating the arbitrary functions

f (x)

and

g (y)

from

z = y f (x) + x g (y)

is hyperbolic in the region

x > 0, y > 0

5.(b) दर्शाइए कि समीकरण :

f (x) = \cos \frac{π (x + 1)}{8} + 0 \cdot 148 x - 0 \cdot 9062 = 0

का एक मूल अन्तराल

(- 1, 0)

में तथा एक मूल

(0, 1)

में है । ऋणात्मक मूल की न्यूटन-रॉफसन विधि से दशमलव के चार स्थान तक सही गणना कीजिए।

Show that the equation:

f (x) = \cos \frac{π (x + 1)}{8} + 0 \cdot 148 x - 0 \cdot 9062 = 0

has one root in the interval

(- 1, 0)

and one in

(0, 1)

. Calculate the negative root correct to four decimal places using Newton-Raphson method.

Answer:

Introduction:
The problem involves showing that the equation

f (x) = \cos \frac{π (x + 1)}{8} + 0.148 x - 0.9062 = 0

has one root in the interval

(- 1, 0)

and one in

(0, 1)

. The task is to calculate the negative root accurate to four decimal places using the Newton-Raphson method.

Step 1: Given Equation
We are given the equation:

f (x) = \cos \frac{π (x + 1)}{8} + 0.148 x - 0.9062

Step 2: Calculate Derivative
Calculate the derivative of

f (x)

with respect to

x

f^{'} (x) = - \frac{π}{8} \sin (\frac{π (x + 1)}{8}) + 0.148

Step 3: Continuity of $f (x)$ and $f^{'} (x)$
It’s evident that

f (x)

and

f^{'} (x)

are continuous everywhere.

Step 4: Evaluate $f (- 1)$ , $f (0)$ , and $f (1)$
Calculate the values of

f

x = - 1

x = 0

, and

x = 1

\begin{aligned} f (- 1) = - 0.0542 \\ f (0) = 0.01768 \\ f (1) = - 0.05109 \end{aligned}

Since

f (- 1) \cdot f (0) < 0

, it implies that a root lies between

(- 1, 0)

. Similarly,

f (0) \cdot f (1) < 0

, indicating that another root lies between

(0, 1)

Step 5: Newton-Raphson Iteration
To calculate the negative root using the Newton-Raphson method, we start with an initial approximation

x_{0} = - 0.5

. The Newton’s iteration formula is given by:

x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})} (1), n = 0, 1, 2, \dots

Now, let’s calculate the iterations:

$n = 0$ :

$x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})} = - 0.5 - \frac{f (- 0.5)}{f^{'} (- 0.5)} = - 0.5138$
$n = 1$ :

$x_{2} = - 0.5138 - \frac{f (- 0.5138)}{f^{'} (- 0.5138)} = - 0.5046$
$n = 2$ :

$x_{3} = - 0.5022$

Therefore,

x = - 0.5022

is the required root, accurate to four decimal places.

Conclusion:
The equation

f (x) = \cos \frac{π (x + 1)}{8} + 0.148 x - 0.9062 = 0

has one root in the interval

(- 1, 0)

and one in

(0, 1)

. The negative root has been calculated to be

x = - 0.5022

with an accuracy of four decimal places using the Newton-Raphson method.

(c) मान लीजिए $g (w, x, y, z) = (w + x + y) (x + \bar{y} + z) (w + \bar{y})$ एक बूलीय-फलन है । $g (w, x, y, z)$ का योगात्मक प्रसामान्य स्वरूप (कन्जंकटिव नार्मल फॉर्म) प्राप्त कीजिए । $g (w, x, y, z)$ को उच्च-पदों (मैक्स टर्म्स) के गुणन के रूप में भी व्यक्त कीजिए ।

Let

g (w, x, y, z) = (w + x + y) (x + \bar{y} + z) (w + \bar{y})

be a Boolean function. Obtain the conjunctive normal form for

g (w, x, y, z)

. Also express

g (w, x, y, z)

as a product of maxterms.

Answer:

Introduction:
In this problem, we are given the conjunctive normal form of a boolean function

g (w, x, y, z)

and asked to create its truth table. The given function is represented as a product of its max terms.

Step 1: Expressing $g (w, x, y, z)$
We are given the conjunctive normal form of

g (w, x, y, z)

as follows:

\begin{aligned} g (w, x, y, z) = (w + x + y) (x + \bar{y} + z) (w + \bar{y}) \\ = (w + x + y + z \bar{z}) (w a + x + \bar{y} + z) (w + x \bar{x} + \bar{y} + z \bar{z}) \\ = (w + x + y + z) (w + x + y + \bar{z}) (w + x + \bar{y} + z) (w + x + \bar{y} + z) (w + x + \bar{y} + z) (w + x + \bar{y} + \bar{z}) (w + \bar{x} + \bar{y} + z) (w + \bar{x} + \bar{y} + \bar{z}) \\ g = (w + x + y + z) (w + x + y + \bar{z}) (w + x + \bar{y} + z) (w + x + \bar{y} + z) (w + x + \bar{y} + \bar{z}) (w + \bar{x} + \bar{y} + z) (w + \bar{x} + \bar{y} + \bar{z}) \end{aligned}

Step 2: Finding the Maxterms

A maxterm is a term that is an OR of all the variables or their complements. The maxterm is 1 when the function is 0. To express

g (w, x, y, z)

as a product of maxterms, we need to find the terms that make

g (w, x, y, z) = 0

From the expanded form, we can see that

g (w, x, y, z)

will be zero when all the terms are zero. This happens when:

$w = 0, x = 0, y = 0, z = 0$
$w = 0, x = 0, y = 0, z = 1$
$w = 0, x = 0, y = 1, z = 0$
$w = 0, x = 0, y = 1, z = 1$
$w = 0, x = 1, y = 0, z = 0$
$w = 0, x = 1, y = 0, z = 1$
$w = 0, x = 1, y = 1, z = 0$

So, the maxterms corresponding to these conditions are:

$(w + x + y + z)$
$(w + x + y + \bar{z})$
$(w + x + \bar{y} + z)$
$(w + x + \bar{y} + \bar{z})$
$(w + \bar{x} + y + z)$
$(w + \bar{x} + y + \bar{z})$
$(w + \bar{x} + \bar{y} + z)$
$(w + \bar{x} + \bar{y} + \bar{z})$

Conclusion

The Conjunctive Normal Form (CNF) of

g (w, x, y, z)

(w + x + y + z) (w + x + y + \bar{z}) (w + x + \bar{y} + z) (w + x + \bar{y} + \bar{z}) (w + \bar{x} + \bar{y} + z) (w + \bar{x} + \bar{y} + \bar{z})

The function

g (w, x, y, z)

can be expressed as a product of the maxterms

(w + x + y + z) (w + x + y + \bar{z}) (w + x + \bar{y} + z) (w + x + \bar{y} + \bar{z}) (w + \bar{x} + y + z) (w + \bar{x} + y + \bar{z}) (w + \bar{x} + \bar{y} + z) (w + \bar{x} + \bar{y} + \bar{z})

5.(d) आंशिक अवकल समीकरण :

(D^{3} - 2 D^{2} D^{'} - D D^{' 2} + 2 D^{' 3}) z = e^{2 x + y^{'}} + \sin (x - 2 y)

D \equiv \frac{\partial}{\partial x}, D^{'} \equiv \frac{\partial}{\partial y}

को हल कीजिए ।

Solve the partial differential equation :

(D^{3} - 2 D^{2} D^{'} - D D^{' 2} + 2 D^{' 3}) z = e^{2 x + y} + \sin (x - 2 y)

D \equiv \frac{\partial}{\partial x}, D^{'} \equiv \frac{\partial}{\partial y}

Answer:

Introduction:
The problem involves solving the partial differential equation:

(D^{3} - 2 D^{2} D^{'} - D D^{' 2} + 2 D^{' 3}) z = e^{2 x + y} + \sin (x - 2 y)

where

D

represents the partial derivative with respect to

x

, and

D^{'}

represents the partial derivative with respect to

y

Step 1: Auxiliary Equation
To solve the differential equation, we first find the auxiliary equation. The auxiliary equation is given by:

m^{3} - 2 m^{2} - m + 2 = 0

Solving this equation, we find the roots

m = 2

m = 1

, and

m = - 1

Step 2: Complementary Function (C.F.)
The complementary function is given by:

C . F = ϕ_{1} (y + 2 x) + ϕ_{2} (y + x) + ϕ_{3} (y - x) (1)

Step 3: Particular Integral (P.I.)
To find the particular integral, we need to solve for

P . I

in the equation:

P . I = \frac{e^{2 x + y}}{(D^{3} - 2 D^{2} D^{'} - D D^{'} 2 + 2 D^{'} 3)} + \frac{\sin (x - 2 y)}{(D - D^{'}) (D + D^{'}) (D - 2 D^{'})}

Simplify this expression:

\begin{aligned} For P . I = \frac{e^{2 x + y}}{D^{3} - 2 D^{2} D^{'} - D D^{'} 2 + 2 D^{'} 3} + \frac{\sin (x - 2 y)}{(D - D^{'}) (D + D^{'}) (D - 2 D^{'})} \\ = \frac{e^{2 x + y}}{(D - D^{'}) (D + D^{'}) (D - 2 D^{'})} + \frac{\sin (x - 2 y)}{(D - D^{'}) (D + D^{'}) (D - 2 D^{'})} \\ = \frac{e^{2 x + y}}{(2 - 1) (2 + 1) (D - 2 D^{'})} + \frac{1}{D - 2 D^{'}} \times \frac{1}{D + D^{'}} \times \frac{1}{3} \int \sin v d v \\ = \frac{e^{2 x + y}}{3 (D - 2 D^{'})} - \frac{1}{3} \frac{1}{D - 2 D^{'}} \cdot \frac{\cos (x - 2 y)}{D + D^{'}} \\ = \frac{e^{2 x + y}}{3 (D - 2 D^{'})} - \frac{1}{3 (D - 2 D^{'})} \times \frac{1}{1 - 2} \int \cos v d v \\ = \frac{e^{2 x + y}}{3 (D - 2 D^{'})} + \frac{\sin (x - 2 y)}{3 (D - 2 D^{'})} \\ = \frac{e^{2 x + y}}{3 (D - 2 D^{'})} + \frac{1}{3} \times \frac{1}{1 + 4} \int \sin v d v \\ = \frac{e^{2 x + y}}{3 (D - 2 D^{'})} - \frac{\cos (x - 2 y)}{15} \\ = \frac{e^{2 x + y}}{3} \cdot \frac{x^{'}}{1.1!} - \frac{\cos (x - 2 y)}{15} \\ = \frac{x e^{2 x + y}}{3} - \frac{\cos (x - 2 y)}{15} \end{aligned}

Step 4: Final Solution
Now, we combine the Complementary Function (C.F.) and the Particular Integral (P.I.) to find the final solution:

\begin{aligned} z = C . F + P . I \\ z = ϕ_{1} (y + 2 x) + ϕ_{2} (y + x) + ϕ_{3} (y - x) + \frac{x}{3} e^{2 x + y} - \frac{\cos (x - 2 y)}{15} \end{aligned}

Conclusion:
The solution to the partial differential equation is given by:

z = ϕ_{1} (y + 2 x) + ϕ_{2} (y + x) + ϕ_{3} (y - x) + \frac{x}{3} e^{2 x + y} - \frac{\cos (x - 2 y)}{15}

5.(e) सिद्ध कीजिए कि त्रिभुजाकार पटल

A B C

का इसके तल में

A

से होकर जाने वाली किसी भी अक्ष के सापेक्ष जड़त्व-आघूर्ण

\frac{M}{6} (β^{2} + β γ + γ^{2})

है, जहाँ पर

M

पटल की संहति तथा

β

व

γ

क्रमशः

B

व

C

से अक्ष पर डाले गये लम्बों की लम्बाइयां हैं।

Prove that the moment of inertia of a triangular lamina

A B C

about any axis through

A

in its plane is

\frac{M}{6} (β^{2} + β γ + γ^{2})

where

M

is the mass of the lamina and

β, γ

are respectively the length of perpendiculars from

B

and

C

on the axis.

Answer:

Introduction:
In this problem, we are tasked with proving the moment of inertia of a triangular lamina

A B C

about any axis through point

A

in its plane. The formula to be proven is:

\frac{M}{6} (β^{2} + β γ + γ^{2})

Where

M

is the mass of the lamina, and

β

and

γ

represent the lengths of perpendiculars from points

B

and

C

to the axis, respectively.

Step 1: Mass Distribution
Let

m

be the mass of the triangle

A B C

. We consider the triangle as being equipotential to three particles, each with mass

\frac{m}{3}

, placed at the midpoints

D, E, F

of its sides.

Step 2: Perpendicular Distances
Now, let

L M

be any line through vertex

A

and within the plane of triangle

A B C

. We define

β

as the distance from vertex

B

to line

L M

(i.e.,

B T = β

), and

γ

as the distance from vertex

C

to line

L M

(i.e.,

C K = γ

). Therefore, the perpendicular distances of points

D, E, F

from line

L M

are as follows:

D M = \frac{1}{2} (β + γ), E N = \frac{1}{2} γ, and F P = \frac{1}{2} β

Step 3: Moment of Inertia Calculation
The moment of inertia (

M . I .

) of triangle

A B C

about line

L M

can be calculated as the sum of the moment of inertia of the masses

\frac{m}{3}

at points

D, E, F

about line

L M

. Using the formula for

M . I .

of a point mass about an axis, which is

I = m \cdot r^{2}

, we have:

\begin{aligned} M . I . & = Sum of M. I of the masses \frac{m}{3} each at D, E, F about LM \\ = \frac{m}{3} \cdot D M^{2} + \frac{m}{3} \cdot E N^{2} + \frac{m}{3} \cdot F P^{2} \\ = \frac{m}{3} [\frac{1}{4} (β + γ)^{2} + \frac{1}{4} γ^{2} + \frac{1}{4} β^{2}] \\ = \frac{m}{6} (β^{2} + γ^{2} + β γ) \end{aligned}

Conclusion:
The moment of inertia of the triangular lamina

A B C

about any axis passing through point

A

in its plane is given by the formula:

\frac{M}{6} (β^{2} + β γ + γ^{2})

Where

M

represents the mass of the lamina, and

β

and

γ

are the lengths of perpendiculars from points

B

and

C

to the axis, respectively.

6.(a) आंशिक अवकल समीकरण :

(x - y) y^{2} \frac{\partial z}{\partial x} + (y - x) x^{2} \frac{\partial z}{\partial y} = (x^{2} + y^{2}) z

के वक्र :

x z = a^{3}, y = 0

को अपने ऊपर समाहित करने वाले समाकल पृष्ठ को ज्ञात कीजिए ।

Find the integral surface of the partial differential equation :

(x - y) y^{2} \frac{\partial z}{\partial x} + (y - x) x^{2} \frac{\partial z}{\partial y} = (x^{2} + y^{2}) z

that contains the curve :

x z = a^{3}, y = 0

on it.

Answer:

Introduction:
We are given a partial differential equation:

(x - y) y^{2} \frac{\partial z}{\partial x} + (y - x) x^{2} \frac{\partial z}{\partial y} = (x^{2} + y^{2}) z

and we need to find the integral surface of this equation that contains the curve

x z = a^{3}

and

y = 0

Step 1: Initial Equation and Curve
Given equation (1) is:

(x - y) y^{2} \frac{\partial z}{\partial x} + (y - x) x^{2} \frac{\partial z}{\partial y} = (x^{2} + y^{2}) z

and the given curve (2) is

x z = a^{3}

and

y = 0

Step 2: Lagrange’s Auxiliary Equation
We apply Lagrange’s auxiliary equations to equation (1):

\frac{d x}{(x - y) y^{2}} = \frac{d y}{(y - x) x^{2}} = \frac{d z}{(x^{2} + y^{2}) z}

This results in equation (3).

Step 3: Integration of Auxiliary Equations
For each fraction of equation (3), we get:

\frac{d x - d y}{(x - y) (y^{2} + x^{2})} = \frac{d z}{(x^{2} + y^{2}) z}

This simplifies to

\frac{d (x - y)}{x - y} - \frac{d z}{z} = 0

Integrating this equation, we arrive at:

\frac{x - y}{z} = C_{1}

This is equation (4).

Step 4: Integration of First Two Fractions
Taking the first two fractions from equation (3), we obtain:

3 x^{2} d x + 3 y^{2} d y = 0

Integrating this equation leads to:

x^{3} + y^{3} = C_{2}

This is equation (5).

Step 5: Parametric Form of the Curve
The parametric form of the given curve (2) is:

z = t, x = \frac{a^{3}}{t}, y = 0

This is equation (6).

Step 6: Substituting Parametric Form into Equations
Substituting the values from equation (6) into equations (4) and (5), we get:

\frac{a^{3}}{t^{2}} = C_{1} and t^{3} = \frac{a^{9}}{C_{2}}

Squaring both sides of the second equation, we have:

t^{6} = \frac{a^{18}}{C_{2}^{2}}

Equating these expressions, we obtain:

\frac{a^{9}}{C_{1}^{3}} = \frac{a^{18}}{C_{2}^{2}} or C_{2}^{2} = a^{9} C_{1}^{3}

This is equation (9).

Step 7: Finding the Integral Surface
Substituting the values of

C_{1}

and

C_{2}

from equations (4) and (5) into equation (9), we arrive at the required integral surface

g (1)

{(x^{3} + y^{3})}^{2} = \frac{a^{9} (x - y)^{3}}{z^{3}}

Conclusion:
The integral surface of the partial differential equation:

(x - y) y^{2} \frac{\partial z}{\partial x} + (y - x) x^{2} \frac{\partial z}{\partial y} = (x^{2} + y^{2}) z

that contains the curve

x z = a^{3}

and

y = 0

is given by the equation:

{(x^{3} + y^{3})}^{2} = \frac{a^{9} (x - y)^{3}}{z^{3}}

6.(b) समीकरण निकाय :

4 x + y + 2 z = 4

3 x + 5 y + z = 7

x + y + 3 z = 3

के हल के लिए गाउस-सीडल पुनरावर्ती क्रिया-विधि निर्धारित कीजिए तथा आरंभिक सदिश

X^{(0)} = 0

से प्रारंभ करके तीन बार पुनरावर्त कीजिए । यथातथ (बिल्कुल ठीक) हल भी निकालिए और पुनरावर्त हलों से तुलना कीजिए।

For the solution of the system of equations :

4 x + y + 2 z = 4

\begin{aligned} 3 x + 5 y + z = 7 \\ x + y + 3 z = 3 \end{aligned}

set up the Gauss-Seidel iterative scheme and iterate three times starting with the initial vector

X^{(0)} = 0

. Also find the exact solutions and compare with the iterated solutions.

Answer:

Introduction:

We were given a system of three linear equations and tasked with solving it using the Gauss-Seidel iterative method. We started with an initial guess

X^{(0)} = (0, 0, 0)

and performed three iterations to find approximate solutions for

x

y

, and

z

. To check the accuracy of our approximations, we also solved the system exactly using Cramer’s Rule.

Total Equations are 3

\begin{aligned} 4 x + y + 2 z = 4 \\ 3 x + 5 y + z = 7 \\ x + y + 3 z = 3 \end{aligned}

From the above equations

x_{k + 1} = \frac{1}{4} (4 - y_{k} - 2 z_{k})

y_{k + 1} = \frac{1}{5} (7 - 3 x_{k + 1} - z_{k})

z_{k + 1} = \frac{1}{3} (3 - x_{k + 1} - y_{k + 1})

Initial gauss

(x, y, z) = (0, 0, 0)

Solution steps are

1^{st}

Approximation

x_{1} = \frac{1}{4} [4 - (0) - 2 (0)] = \frac{1}{4} [4] = 1

\begin{aligned} y_{1} = \frac{1}{5} [7 - 3 (1) - (0)] = \frac{1}{5} [4] = 0.8 \\ z_{1} = \frac{1}{3} [3 - (1) - (0.8)] = \frac{1}{3} [1.2] = 0.4 \end{aligned}

2^{nd}

Approximation

x_{2} = \frac{1}{4} [4 - (0.8) - 2 (0.4)] = \frac{1}{4} [2.4] = 0.6

y_{2} = \frac{1}{5} [7 - 3 (0.6) - (0.4)] = \frac{1}{5} [4.8] = 0.96

z_{2} = \frac{1}{3} [3 - (0.6) - (0.96)] = \frac{1}{3} [1.44] = 0.48

3^{r d}

Approximation

\begin{aligned} x_{3} = \frac{1}{4} [4 - (0.96) - 2 (0.48)] = \frac{1}{4} [2.08] = 0.52 \\ y_{3} = \frac{1}{5} [7 - 3 (0.52) - (0.48)] = \frac{1}{5} [4.96] = 0.992 \\ z_{3} = \frac{1}{3} [3 - (0.52) - (0.992)] = \frac{1}{3} [1.488] = 0.496 \end{aligned}

Iterations are tabulated as below

Iteration	$x$	$y$	$z$
1	1	0.8	0.4
2	0.6	0.96	0.48
3	0.52	0.992	0.496

Exact solutions using Cramer’s Rule:

Use Cramer’s Rule to find the values of

x, y, z

\begin{aligned} \frac{x}{D_{x}} = \frac{- y}{D_{y}} = \frac{z}{D_{z}} = \frac{- 1}{D} \\ D_{x} = | \begin{array}{lll} 1 & 2 & - 4 \\ 5 & 1 & - 7 \\ 1 & 3 & - 3 \end{array} | \\ = 1 \times | \begin{array}{ll} 1 & - 7 \\ 3 & - 3 \end{array} | - 2 \times | \begin{array}{cc} 5 & - 7 \\ 1 & - 3 \end{array} | - 4 \times | \begin{array}{cc} 5 & 1 \\ 1 & 3 \end{array} | \\ = 1 \times (1 \times (- 3) - (- 7) \times 3) - 2 \times (5 \times (- 3) - (- 7) \times 1) - 4 \times (5 \times 3 - 1 \times 1) \\ = 1 \times (- 3 + 21) - 2 \times (- 15 + 7) - 4 \times (15 - 1) \\ = 1 \times (18) - 2 \times (- 8) - 4 \times (14) \\ = 18 + 16 - 56 \\ = - 22 \end{aligned}

\begin{aligned} D_{y} = | \begin{array}{lll} 4 & 2 & - 4 \\ 3 & 1 & - 7 \\ 1 & 3 & - 3 \end{array} | \\ = 4 \times | \begin{array}{ll} 1 & - 7 \\ 3 & - 3 \end{array} | - 2 \times | \begin{array}{cc} 3 & - 7 \\ 1 & - 3 \end{array} | - 4 \times | \begin{array}{ll} 3 & 1 \\ 1 & 3 \end{array} | \\ = 4 \times (1 \times (- 3) - (- 7) \times 3) - 2 \times (3 \times (- 3) - (- 7) \times 1) - 4 \times (3 \times 3 - 1 \times 1) \\ = 4 \times (- 3 + 21) - 2 \times (- 9 + 7) - 4 \times (9 - 1) \\ = 4 \times (18) - 2 \times (- 2) - 4 \times (8) \\ = 72 + 4 - 32 \\ = 44 \end{aligned}

\begin{aligned} D_{z} = | \begin{array}{lll} 4 & 1 & - 4 \\ 3 & 5 & - 7 \\ 1 & 1 & - 3 \end{array} | \\ = 4 \times | \begin{array}{ll} 5 & - 7 \\ 1 & - 3 \end{array} | - 1 \times | \begin{array}{ll} 3 & - 7 \\ 1 & - 3 \end{array} | - 4 \times | \begin{array}{cc} 3 & 5 \\ 1 & 1 \end{array} | \\ = 4 \times (5 \times (- 3) - (- 7) \times 1) - 1 \times (3 \times (- 3) - (- 7) \times 1) - 4 \times (3 \times 1 - 5 \times 1) \\ = 4 \times (- 15 + 7) - 1 \times (- 9 + 7) - 4 \times (3 - 5) \\ = 4 \times (- 8) - 1 \times (- 2) - 4 \times (- 2) \\ = - 32 + 2 + 8 \\ = - 22 \end{aligned}

\begin{aligned} D = | \begin{array}{lll} 4 & 1 & 2 \\ 3 & 5 & 1 \\ 1 & 1 & 3 \end{array} | \\ = 4 \times | \begin{array}{ll} 5 & 1 \\ 1 & 3 \end{array} | - 1 \times | \begin{array}{ll} 3 & 1 \\ 1 & 3 \end{array} | + 2 \times | \begin{array}{ll} 3 & 5 \\ 1 & 1 \end{array} | \\ = 4 \times (5 \times 3 - 1 \times 1) - 1 \times (3 \times 3 - 1 \times 1) + 2 \times (3 \times 1 - 5 \times 1) \\ = 4 \times (15 - 1) - 1 \times (9 - 1) + 2 \times (3 - 5) \\ = 4 \times (14) - 1 \times (8) + 2 \times (- 2) \\ = 56 - 8 - 4 \\ = 44 \end{aligned}

\begin{aligned} \frac{x}{D_{x}} = \frac{- y}{D_{y}} = \frac{z}{D_{z}} = \frac{- 1}{D} \\ ∴ \frac{x}{- 22} = \frac{- y}{44} = \frac{z}{- 22} = \frac{- 1}{44} \\ ∴ \frac{x}{- 22} = \frac{- 1}{44}, \frac{- y}{44} = \frac{- 1}{44}, \frac{z}{- 22} = \frac{- 1}{44} \\ ∴ x = \frac{22}{44}, y = \frac{44}{44}, z = \frac{22}{44} \\ ∴ x = \frac{1}{2}, y = 1, z = \frac{1}{2} \end{aligned}

Comparison:

After doing three iterations with the Gauss-Seidel method, we got the approximate solutions as

x_{3} = 0.52

y_{3} = 0.992

, and

z_{3} = 0.496

Then, we used Cramer’s Rule to find the exact solutions, which turned out to be

x = \frac{1}{2}

y = 1

, and

z = \frac{1}{2}

When we compared the two sets of solutions, we noticed that the approximate solutions were really close to the exact ones. Specifically,

x_{3}

was almost

\frac{1}{2}

y_{3}

was nearly

1

, and

z_{3}

was close to

\frac{1}{2}

(c) एक कण जिसकी संहति $m$ है, $x^{2} + y^{2} = R^{2}$ , जहाँ पर $R$ अचर है, द्वारा परिभाषित बेलन पर गति के लिए व्यवरोधित है। कण मूल बिन्दु की ओर लगे बल जो कण की मूल बिन्दु से दूरी $r$ के अनुपाती है, से प्रतिबन्धित है । बल $\vec{F} = - k \vec{r}$ , जहाँ पर $k$ अचर है, से दिया गया है ।

By writing down the Hamiltonian, find the equations of motion of a particle of mass

m

constrained to move on the surface of a cylinder defined by

x^{2} + y^{2} = R^{2}

R

is a constant. The particle is subject to a force directed towards the origin and proportional to the distance

r

of the particle from the origin given by

\vec{F} = - k \vec{r}, k

is a constant.

Answer:

Introduction:
We are tasked with finding the equations of motion for a particle of mass

m

constrained to move on the surface of a cylinder defined by

x^{2} + y^{2} = R^{2}

, where

R

is a constant. The particle is subject to a force

\vec{F} = - k \vec{r}

, where

k

is a constant.

Step 1: Finding Potential Energy
Given the force

\vec{F} = - k \vec{r}

, we can find the potential energy

V

as follows:

F = - \frac{\partial V}{\partial r}, V = \frac{k r^{2}}{2}

Step 2: Coordinate Transformation
Let’s consider the coordinates

x = ρ \cos ϕ

y = ρ \sin ϕ

, and

z = z

. Now, differentiate these coordinates with respect to time:

\dot{x} = \dot{ρ} \cos ϕ - ρ \sin ϕ \dot{ϕ}; \dot{y} = \dot{ρ} \sin ϕ + ρ \cos ϕ \dot{ϕ}; \dot{z} = \dot{z}

We can also find the velocity squared

v^{2}

v^{2} = {\dot{x}}^{2} + {\dot{y}}^{2} + {\dot{z}}^{2} - ({\dot{ρ}}^{2} + ρ^{2} {\dot{ϕ}}^{2} + {\dot{z}}^{2})

Step 3: Kinetic Energy
Next, let’s find the kinetic energy

T

T = \frac{1}{2} m v^{2} = \frac{1}{2} m ({\dot{ρ}}^{2} + ρ^{2} {\dot{ϕ}}^{2} + {\dot{z}}^{2})

Step 4: Constraints
We have the constraint

x^{2} + y^{2} = R^{2}

, which implies

ρ = R

and

\dot{ρ} = 0

. Substituting these values, we get

v

and

T

in terms of

z

v = \frac{k}{2} (R^{2} + z^{2}); T = \frac{1}{2} m (R^{2} {\dot{ϕ}}^{2} + z^{2})

Step 5: Lagrangian
Now, let’s find the Lagrangian

L

, which is the difference between kinetic and potential energy:

L = T - V = \frac{1}{2} m (R^{2} {\dot{ϕ}}^{2} + z^{2}) - \frac{k}{2} (R^{2} + z^{2})

Importantly, the Lagrangian explicitly does not depend on time

t

Step 6: Hamiltonian
We need to find the Hamiltonian

H

with the help of the generalized coordinates and momenta:

H = T + V = \frac{1}{2} m (R^{2} {\dot{ϕ}}^{2} + z^{2}) + \frac{k}{2} (R^{2} + z^{2})

Step 7: Equations of Motion
To find the equations of motion, we differentiate the Hamiltonian with respect to the generalized coordinates and momenta:

\begin{aligned} H & = T + V = \frac{1}{2} m (R^{2} ϕ^{2} + y^{2}) + \frac{k}{2} (R^{2} + z^{2}) \\ p ϕ & = \frac{\partial L}{\partial ϕ} = m R^{2} ϕ; p z = \frac{\partial L}{\partial z} = m z \\ H & = \frac{1}{2} m (R^{2} {(\frac{p ϕ}{m R^{2}})}^{2} + {(\frac{p z}{M})}^{2}) + \frac{k}{2} (R^{2} + z^{2}) \\ H & = \frac{1}{2} \frac{p ϕ^{2}}{m R^{2}} + \frac{1}{2} \frac{p z^{2}}{m} + \frac{k}{2} (R^{2} + z^{2}) required Hamiltonian \\ ϕ = \frac{\partial H}{\partial p_{ϕ}} & = \frac{p ϕ}{m R^{2}}; z = \frac{\partial H}{\partial p_{z}} = \frac{p z}{m}; p ϕ = - \frac{\partial H}{\partial ϕ} = 0; p z = - \frac{\partial H}{\partial z} = - k z \\ ϕ = \frac{p ϕ}{m R^{2}} & = 0; z = \frac{p z}{m} = - \frac{k z}{m} this is the required equation of motion \end{aligned}

Conclusion:
The equations of motion for a particle of mass

m

constrained to move on the surface of a cylinder, subject to the force

\vec{F} = - k \vec{r}

, are given by:

ϕ = 0; z = - \frac{k z}{m}

These equations describe the motion of the particle in the

ϕ

and

z

directions.

7.(a) आंशिक अवकल समीकरण :

z = \frac{1}{2} (p^{2} + q^{2}) + (p - x) (q - y); p \equiv \frac{\partial z}{\partial x}, q \equiv \frac{\partial z}{\partial y}

का हल ज्ञात कीजिये जो कि

x

-अक्ष से गुजरता हो ।

Find the solution of the partial differential equation:

z = \frac{1}{2} (p^{2} + q^{2}) + (p - x) (q - y); p \equiv \frac{\partial z}{\partial x}, q \equiv \frac{\partial z}{\partial y}

which passes through the

x

-axis.

Answer:

Introduction:
In this problem, we are tasked with finding the solution to the partial differential equation:

z = \frac{1}{2} (p^{2} + q^{2}) + (p - x) (q - y)

where

p

is defined as

\frac{\partial z}{\partial x}

and

q

\frac{\partial z}{\partial y}

. We are looking for an integral surface of this equation that passes through the

x

-axis.

Solution:

Step 1: Given Equation
The given equation is:

z = \frac{1}{2} (p^{2} + q^{2}) + (p - x) (q - y) (1)

We want to find an integral surface that satisfies this equation and passes through the

x

-axis, which is defined by

y = 0

and

z = 0

Step 2: Parametric Form
Rewrite the conditions

y = 0

and

z = 0

in parametric form:

x = λ, y = 0, z = 0, where λ is the parameter (3)

Step 3: Initial Values
Let the initial values

x_{0}, y_{0}, z_{0}, p_{0}, q_{0}

x, y, z, p, q

be taken as:

x_{0} = x_{0} (λ) = λ, y_{0} = y_{0} (λ) = 0, z_{0} = z_{0} (λ) = 0 (4)

Let

p_{0}, q_{0}

be the initial values of

p

and

q

corresponding to the initial values

x_{0}, y_{0}, z_{0}

. Since the initial values

(x_{0}, y_{0}, z_{0}, p_{0}, q_{0})

satisfy equation (1), we have:

\begin{aligned} z_{0} & = \frac{1}{2} (p_{0}^{2} + q_{0}^{2}) + (p_{0} - λ) (q_{0} - 0) \\ 0 & = \frac{1}{2} (p_{0}^{2} + q_{0}^{2}) + q_{0} (p_{0} - λ) (5) \end{aligned}

Step 4: Solving for $p_{0}$ and $q_{0}$
We can solve equations (5) for

p_{0}

and

q_{0}

\begin{aligned} p_{0}^{2} + q_{0}^{2} + 2 q_{0} p_{0} - 2 q_{0} λ & = 0 (5) \\ p_{0} & = 0 (6) \end{aligned}

Solving (5) and (6)

p_{0} = 0

and

q_{0} = 2 λ \to (4 A)

Collecting relations (4) and (4A) together, initial value of

x_{0}, y_{0}, z_{0}, p_{0}, q_{0}

are given by

x_{0} = λ, y_{0} = 0, z_{0} = 0, p_{0} = 0, q_{0} = 2 λ where t = t_{0} = 0 \to (7)

Let

f (x, y, z, p, q) = \frac{1}{2} (p^{2} + q^{2}) - p q - p y - q x + x y - z = 0 \to

(8)
The usual characteristic equations of (8) are given by

\begin{aligned} \frac{d x}{d t} = \frac{d f}{d p} = p + q - y \to (9) \\ \frac{d y}{d t} = \frac{d f}{d q} = q + p - x \to (10) \\ \frac{d z}{d t} = p (\frac{\partial f}{\partial p}) + q (\frac{\partial f}{\partial q}) = p (p + q - y) + q (q + p - x) \to (11) \\ \frac{d p}{d t} = - (\frac{\partial f}{\partial x}) - p (\frac{\partial f}{\partial z}) = p + q - y \to (12) \end{aligned}

And

\frac{d q}{d t} = - (\frac{\partial f}{\partial y}) - q (\frac{\partial f}{\partial z}) = p + q - x \to

(3)
From (9) and (12), (

\frac{d x}{d t}) - (\frac{d p}{d t}) = 0

so that

x - p = C_{1} \to

(14)
Where

C_{1}

is an arbitrary constant. Using initial conditions (7), (14) gives

λ - 0 = C_{1} \Rightarrow C_{1} = λ .

Hence (14) reduces to

x - p = λ \Rightarrow x = p + λ \to (15)

From (10) and (13),

\frac{d y}{d t} - \frac{d q}{d t} = 0 so that y - q = C_{2} \to (16)

Where

C_{2}

is an arbitrary constant
Using initial conditions (7), (16) gives

0 - 2 λ = C_{2} \Rightarrow C_{2} = 2 λ

Hence (16), reduces to

y - q = 2 λ \Rightarrow y = q - 2 λ \to (17)

Hence

\frac{d (p + q - x)}{d t} = \frac{d p}{d t} + \frac{d q}{d t} - \frac{d x}{d t} = p + q - y + p + q - x - (p + q - y)

[using (9) , (12) and (13) ]

\Rightarrow \frac{d (p + q - x)}{d t} = p + q - x \Rightarrow \frac{d (p + q - x)}{p + q - x} d t

Integrating,

\log (p + q - x) - \log C_{3} = t (

) p + q - x = c_{3} e^{t} \to (18)

Where

c_{3}

is an arbitrary constant. Using initial conditions

(7), (18)

gives

0 + 2 λ - λ = c_{3} \Rightarrow c_{3} = λ

Hence (18) reduces to

p + q - x = λ e^{t} \to

(19)

\frac{d (p + q - y)}{d t} = \frac{d p}{d t} + \frac{d q}{d t} - \frac{d y}{d t} = p + q - y + p + q - x - (q + p - x)

[using (10), (12) and (13) ]

\Rightarrow \frac{d (p + q - y)}{d t} = p + q - y \Rightarrow \frac{d (p + q - y)}{p + q - y} = d t

Integrating,

\log (p + q - y) - \log C_{4} = t \Rightarrow p + q - y = c_{4} e^{t} \to (20)

Where

c_{4}

is an arbitrary constant. Using initial conditions (7), (20) gives

0 + 2 λ - 0 = c_{4} \Rightarrow c_{4} = 2 λ,

Hence (20) reduces to

p + q - y = 2 λ e^{t} \to

(21)
From (9) and (21)

\frac{d x}{d t} = 2 λ e^{t} so that x = 2 λ e^{t} + c_{5} \to (22)

Where

c_{5}

is an arbitrary constant. Using initial conditions

(7), (22)

gives

λ = 2 λ + c_{5} \Rightarrow c_{5} = - λ

Hence (22) reduces to

x = 2 λ e^{t} - λ \Rightarrow x = λ (2 e^{t} - 1) \to (23)

From (10) and (7),

\frac{d y}{d t} = λ e^{t} so that y = λ e^{t} + c_{6} \to (24)

Where

c_{6}

is an arbitrary constant. Using initial conditions (7), (24) gives

0 = λ + c_{6} \Rightarrow C_{6} = - λ

Hence (24) reduces to

y = λ e^{t} - λ \Rightarrow y = λ (e^{t} - 1) \to (25)

Substituting value of y from (17) in (12), we get

\frac{d p}{d t} = p + q - (q - 2 λ) \Rightarrow \frac{d p}{d t} - p = 2 λ \to (26)

Whose integrating factor

= e^{\int (- 1) d t} = e^{- t}

and solution is

\begin{aligned} p e^{- t} = \int (2 λ) e^{- t} d t + c_{7} = - 2 λ e^{- t} + c_{3} \\ \Rightarrow p = - 2 λ + c_{3} e^{t} \to (27) \end{aligned}

Where

C_{7}

is an arbitrary constant. Using initial condition (7), (27) gives

0 = - 2 λ + c_{7} \Rightarrow c_{7} = 2 λ

Hence (27) reduces to

p = - 2 λ + 2 λ e^{t} \Rightarrow p = 2 λ (e^{t} - 1) \to (28)

Substituting value of

x

from (23) in (13), we get

\begin{aligned} \frac{d q}{d t} = p + q - (p + λ) \\ \Rightarrow \frac{d q}{d t} - q = - λ \to (29) \end{aligned}

Whose integrating factor

= e^{\int (- 1) d t} = e^{t}

and solution is

q e^{- t} = \int (- λ) e^{- t} d t + C_{8} = λ e^{- t} + c_{8} \Rightarrow q = λ + c_{8} e^{t} \to (30)

Where

C_{8}

is an arbitrary constant. Using initial condition (7), (30) gives

2 λ = λ + C_{8} \Rightarrow C_{8} = λ

Hence (30) reduces to

q = λ + λ e^{t}

\Rightarrow q = λ (1 + e^{t}) \to (31)

Substitute the values of

p + q - x

and

p + q - y

from (13) and (24) respectively in (1), we have

\frac{d z}{d t} = p (2 λ e^{t}) + q (λ e^{t}) = 2 λ (e^{t} - 1) (2 λ e^{t}) + λ (1 + e^{t}) (λ e^{t})

[on putting values of

p

and

q

with help of (28) and (31) ]

\begin{aligned} \Rightarrow \frac{d z}{d t} = 5 λ^{2} e^{2 t} - 3 λ^{2} e^{t} \Rightarrow d z = (5 λ^{2} e^{2 t} - 3 λ^{2} e^{t}) d t \\ Integrating z = \frac{5}{2} λ^{2} e^{2 t} - 3 λ^{2} e^{t} + C_{9} \to (32) \end{aligned}

Where

C_{9}

is an arbitrary constant. Using initial conditions (7), namely

z = 0

, where

t = 0

, (32) gives

0 = \frac{5}{2} λ^{2} - 3 λ^{2} + C_{9} \Rightarrow C_{9} = 3 λ^{2} - \frac{5}{2} λ^{2}

Hence (32) reduces to

z = \frac{5}{2} λ^{2} (e^{2 t} - 1) - 3 λ^{2} (e^{t} - 1) \to (33)

Solving (23) and (25) for

λ

and

e^{t} = \frac{x - y}{x - 2 y} \to

(34)
Eliminating

λ

and

e^{t}

from (33) and (34), we have

\begin{aligned} z = \frac{5}{2} (x - 2 y)^{2} {{(\frac{x - y}{x - 2 y})}^{2} - 1} - 3 (x - 2 y)^{2} (\frac{x - y}{x - 2 y} - 1) \\ z = \frac{5}{2} {(x - y)^{2} - (x - 2 y)^{2}} - 3 {(x - 2 y) (x - y) - (x - 2 y)^{2}} \\ z = \frac{1}{2} y (4 x - 3 y) \end{aligned}

Conclusion:
The solution to the given partial differential equation is:

z = \frac{1}{2} y (4 x - 3 y)

This represents the integral surface of the equation that passes through the

x

-axis.

7.(b) क्षेत्रकलन के लिए

\int_{0}^{1} f (x) \frac{d x}{\sqrt{x (1 - x)}} = α_{1} f (0) + α_{2} f (\frac{1}{2}) + α_{3} f (1)

द्वारा उस सूत्र को ज्ञात कीजिए जो अधिकतम सम्भव घात के बहुपद के लिए यथातथ (बिल्कुल ठीक) हो । सूत्र का उपयोग

\int_{0}^{1} \frac{d x}{\sqrt{x - x^{3}}}

का (दशमलव के तीन स्थानों तक सही) मूल्यांकन के लिए कीजिए ।

Find a quadrature formula

\int_{0}^{1} f (x) \frac{d x}{\sqrt{x (1 - x)}} = α_{1} f (0) + α_{2} f (\frac{1}{2}) + α_{3} f (1)

which is exact for polynomials of highest possible degree. Then use the formula to evaluate

\int_{0}^{1} \frac{d x}{\sqrt{x - x^{3}}}

(correct up to three decimal places).

Answer:

Introduction:

The problem involves finding a quadrature formula to approximate the integral

\int_{0}^{1} f (x) \frac{d x}{\sqrt{x (1 - x)}}

, such that it is exact for polynomials of the highest possible degree. Additionally, we will use this formula to evaluate

\int_{0}^{1} \frac{d x}{\sqrt{x - x^{3}}}

up to three decimal places.

Step 1: Exactness for Polynomials
To make the method exact for polynomials of degree up to 2, we derive the following expressions for different functions

f (x)

For

f (x) = 1

I_{1} = \int_{0}^{1} \frac{d x}{\sqrt{x (1 - x)}} = α_{1} + α_{2} + α_{3}

For

f (x) = x

I_{2} = \int_{0}^{1} \frac{x d x}{\sqrt{x (1 - x)}} = \frac{1}{2} α_{1} + α_{3}

For

f (x) = x^{2}

I_{3} = \int_{0}^{1} \frac{x^{2} d x}{\sqrt{x (1 - x)}} = \frac{1}{4} α_{2} + α_{3}

Step 2: Evaluating $I_{1}, I_{2}, I_{3}$
We calculate

I_{1}, I_{2}, I_{3}

as follows:

For

I_{1}

I_{1} = 2 \int_{0}^{d} \frac{d x}{\sqrt{1 - (2 x - 1)^{2}}} = \int_{- 1}^{1} \frac{d t}{\sqrt{1 - t^{2}}} = {[\sin^{- 1} t]}_{- 1}^{1} = π

For

I_{2}

I_{2} = 2 \int_{0}^{d} \frac{x d x}{\sqrt{1 - (2 x - 1)^{2}}} = \int_{- 1}^{1} \frac{(t + 1)}{2 \sqrt{1 - t^{2}}} d t = \frac{1}{2} \int_{- 1}^{1} \frac{t}{\sqrt{1 - t^{2}}} d t = \frac{π}{2}

For

I_{3}

I_{3} = 2 \int_{0}^{d} \frac{x^{2} d x}{\sqrt{1 - (2 x - 1)^{2}}} = \frac{1}{4} \int_{- 1}^{1} \frac{(t + 1)^{2}}{\sqrt{1 - t^{2}}} d t = \frac{1}{4} \int_{- 1}^{d} \frac{t^{2}}{\sqrt{1 - t^{2}}} d t + \frac{1}{4 α} \int_{- 1}^{d} \frac{t}{\sqrt{1 - t^{2}}} d t + \frac{1}{4} \int_{- 1}^{1} \frac{d t}{\sqrt{1 - t^{2}}} = \frac{3 π}{8}

Step 3: Quadrature Formula Equations
Now, we have the equations:

α_{1} + α_{2} + α_{3} = π

\frac{1}{2} α_{1} + α_{3} = \frac{π}{2}

\frac{1}{4} α_{2} + α_{3} = \frac{3 π}{8}

Solving these equations, we obtain the quadrature formula:

\int_{0}^{1} \frac{f (x) d x}{\sqrt{x (1 - x)}} = \frac{π}{4} [f (0) + 2 f (\frac{1}{2}) + f (1)]

Step 4: Evaluating $I$
Using this formula, we can evaluate:

I = \int_{0}^{1} \frac{d x}{\sqrt{x - x^{3}}} = \int_{0}^{1} \frac{d x}{\sqrt{1 + x} \sqrt{x (1 - x)}} = \int_{0}^{1} \frac{f (x) d x}{\sqrt{x (1 - x)}}

Where

f (x) = \frac{1}{\sqrt{1 + x}}

. Calculating

I

, we get:

I = \frac{π}{4} [1 + \frac{2 \sqrt{2}}{\sqrt{3}} + \frac{\sqrt{2}}{2}] \approx 2.62331

The exact value is

I = 2.622

Conclusion:

The quadrature formula is successfully used to approximate the integral, yielding an approximate value of

2.62331

(correct up to three decimal places).

(c) एक द्विविमीय द्रव्य-प्रवाह का वेग विभव $ϕ (x, y) = x y + x^{2} - {\dot{y}}^{2}$ द्वारा दिया गया है । इस प्रवाह का धारा-फलन ज्ञात कीजिए ।

A velocity potential in a two-dimensional fluid flow is given by

ϕ (x, y) = x y + x^{2} - y^{2}

. Find the stream function for this flow.

Answer:

Introduction:

A velocity potential in a two-dimensional fluid flow is given by

ϕ (x, y) = x y + x^{2} - y^{2}

. In this problem, we are tasked with finding the stream function for this flow.

Step 1: Velocity Components
As given, the velocity components are obtained from the negative gradient of the velocity potential

ϕ

u = - \frac{\partial ϕ}{\partial x} = - [y + 2 x]

v = - \frac{\partial ϕ}{\partial y} = - [x - 2 y]

Step 2: Complex Potential and Cauchy-Riemann Equations
Now, we consider the complex potential

w = ϕ + Ψ

and recognize that both

ϕ

and

Ψ

must satisfy the Cauchy-Riemann equations:

\frac{\partial ϕ}{\partial x} = \frac{\partial Ψ}{\partial y} (2a)

\frac{\partial ϕ}{\partial y} = - \frac{\partial Ψ}{\partial x} (2b)

Step 3: Solving for $Ψ$
Using equation (2a), we have:

\frac{\partial Ψ}{\partial y} = y + 2 x (3)

Integrating with respect to

y

, we obtain:

Ψ = \frac{y^{2}}{2} + 2 x y + f (x) (1)

Now, using equation (2b), we get:

\frac{\partial Ψ}{\partial x} = - x + 2 y (4)

Step 4: Solving for $f (x)$
Differentiating equation (1) with respect to

x

and comparing it with equation (4):

2 y + f^{'} (x) = - x + 2 y

This implies:

f^{'} (x) = - x

Integrating

f^{'} (x)

with respect to

x

, we find:

f (x) = - \frac{x^{2}}{2} + c (5)

Conclusion:

The stream function

Ψ

is determined as follows:

Ψ = \frac{y^{2}}{2} + 2 x y - \frac{x^{2}}{2} + c

This completes the solution.

8.(a) लम्बाई

l

की कसकर खींची गई लचीली-पतली डोरी का एक सिरा मूल बिन्दु पर तथा दूसरा

x = l

पर बंधा है। आरंभिक अवस्था में इसे

x = \frac{l}{3}

बिन्दु से ऐसे खींचकर छोड़ा जाता है ताकि यह

x - y

तल में

h

ऊँचाई के त्रिभुज का आकार लेता है। किसी भी दूरी

x

तथा समय

t

, डोरी को विरामावस्था से छोड़ने के बाद, पर विस्थापन

y

को ज्ञात कीजिए ।
डोरी में

\frac{क्षैतिज तनाव}{डोरी की इकाई लम्बाई की संहति} = c^{2}

लीजिए ।

One end of a tightly stretched flexible thin string of length

l

is fixed at the origin and the other at

x = l

. It is plucked at

x = \frac{l}{3}

so that it assumes initially the shape of a triangle of height

h

in the

x - y

plane. Find the displacement

y

at any distance

x

and at any time

t

after the string is released from rest. Take,

\frac{horizontal tension}{mass per unit length} = c^{2}

Answer:

Introduction:
One end of a tightly stretched flexible thin string of length

l

is fixed at the origin and the other at

x = l

. It is plucked at

x = \frac{l}{3}

so that it assumes initially the shape of a triangle of height

h

in the

x - y

plane. Find the displacement

y

at any distance

x

and at any time

t

after the string is released from rest. Take,

\frac{horizontal tension}{mass per unit length} = c^{2}

Step 1: The displacement function

y (x, t)

is the solution of the wave equation

\frac{\partial^{2} y}{\partial x^{2}} = \frac{1}{c^{2}} \frac{\partial^{2} y}{\partial t^{2}} \to (1)

Subject to the boundary conditions

y (0, t) = y (l, t) = 0 \forall t ⩾ 0 \to (2)

Step 2: Initial position of the string at

t = 0

is made up of two straight line segments OB and BA as shown in the figure and the string is released from rest. The equation of

OB

is given by

y - 0 = \frac{h - 0}{\frac{l}{3} - 0} (x - 0) \Rightarrow y = \frac{3 h x}{l}, 0 ⩽ x ⩽ \frac{l}{3}

The equation of BA is given by

y - 0 = \frac{h - 0}{\frac{l}{3} - l} (x - l) \Rightarrow y = \frac{3 h (l - x)}{2 l}, \frac{l}{3} ⩽ x ⩽ l

Hence the initial displacement is given by

\begin{aligned} u (x, 0) = f (x) = {\begin{array}{c} \frac{3 h x}{l}, 0 ⩽ x ⩽ \frac{l}{3} \\ \frac{3 h (l - x)}{2 l}, \frac{l}{3} ⩽ x ⩽ l \end{array} \to (3) \\ And the initial velocity = \frac{\partial u}{\partial t} I_{t = 0} = 0 \to (4) \end{aligned}

Step 3: Let

Y (x, t) = X (x) T (t) \to (5)

is the solution of (1). From (1), we have

\begin{aligned} X T = \frac{1}{c^{2}} \times T^{''} \Rightarrow \frac{X^{n}}{X} = \frac{1}{c^{2}} \frac{T^{4}}{T} = P (say) \\ \Rightarrow X - PX = 0 and T - P c^{2} = 0 \to (7) \end{aligned}

Using (2), (5) gives

X (0) T (t) = 0

and

X (l) T (t) = 0 (

because

T (t) = 0

leads

T = 0 \forall t)

X (0) = 0 and X (l) = 0 \to (8)

Which are boundary condition. We now solve (6) under boundary conditions (8). Three cases arise:

Case (i): Let

P = 0

The solution of (6) is

X (n) = A n + B

Using boundary condition, we get

A = 0, B = 0

\Rightarrow X (x) = 0

which leads to

T = 0

which does not satisfy

I . C

so reject

P = 0

Case (ii): Let

P = λ^{2}, λ \neq 0

The solution of (6) is

X (x) = A e^{λ x} + B e^{- λ x}

Using boundary condition (8) we get

A = 0, B = 0

\Rightarrow X (x) = 0

, which leads to

Y = 0

which does not satisfy (3) and (4)
So reject

P = λ^{2}

Case (iii): Let

P = - λ^{2}, λ \neq 0

The solution of (6) is

X (x) = A \cos λ x + B \sin λ x

Using boundary condition (8), we get

X (0) = 0 = A (1) + B (0) \Rightarrow A = 0

And

X (l) = 0 = 0 + B \sin λ l

\Rightarrow B \sin λ l = 0 \Rightarrow \sin λ l = 0 (B \neq 0) \Rightarrow λ l = n π \Rightarrow λ = \frac{n π}{l}, n = 1, 2, 3, \dots X (l) = B \sin \frac{n π x}{l}, n = 1, 2, 3, \dots

Hence non-zero solutions of

X_{n} (x)

of (6) are given by

X_{n} (x) = B_{n} \sin \frac{n π x}{l} \to

(9) From (7),

\begin{aligned} T^{μ} - P C^{2} T = 0 \\ \Rightarrow T^{*} + λ^{2} c^{2} T = 0 (P = - λ^{2}) \\ \Rightarrow T^{*} + \frac{n^{2} π^{2}}{l^{2}} c^{2} T = 0 (λ = \frac{n π}{l}) \end{aligned}

Whose general solution is

\begin{aligned} T_{n} (t) = C_{n} \cos (\frac{n π c t}{l}) + D_{n} \sin (\frac{n π c t}{l}) \\ y_{n} (x, t) = X_{n} (x) T_{n} (t) \\ = B_{n} \sin (\frac{n π x}{l}) [c_{n} \cos \frac{n π c t}{l} + D_{n} \sin \frac{n π c t}{l}] \\ = [E_{n} \cos \frac{n π c t}{l} + F_{n} \sin \frac{n π c t}{l}] \sin (\frac{n π x}{l}) \end{aligned}

Are solutions of (1) satisfying (2)
Where

E_{n} = B_{n} C_{n}

and

F_{n} = B_{n} D_{n}

In order to obtain a solution also satisfying (3) and (4)
We consider more general solution.

\begin{aligned} y (x, t) = \sum_{n = 1}^{\infty} y_{n} (x, t) \\ y (x, t) = \sum_{n = 1}^{\infty} [E_{n} \cos \frac{n π c t}{l} + F_{n} \sin \frac{n π c t}{l}] \sin (\frac{n π x}{l}) \to (10) \\ \frac{\partial y}{\partial t} = \sum_{n = 1}^{\infty} [E_{n} (\sin \frac{π c t}{l}) \frac{n π c}{l} + \frac{n π c}{l} F_{n} \cos \frac{n π c t}{l}] \sin \frac{n π r}{l} \to (11) \\ = \sum_{n = 1}^{\infty} \frac{n π c}{l} F_{n} \sin \frac{n π x}{l} b y (4) \end{aligned}

Where

{F_{n} = ⌋}_{0} 0 = 0

By putting

t = 0

in (10),

f (x) = \sum_{n = 1}^{\infty} E_{n} \sin \frac{n π x}{l}

Where

E_{n} = \frac{2}{l} \int_{0} f (x) \sin \frac{n π x}{l} d x

\begin{aligned} \Rightarrow E_{n} = \frac{2}{l} [\int_{0}^{\frac{l}{3}} f (x) \sin \frac{n π x}{l} d x + \int_{\frac{l}{3}}^{l} f (x) \sin \frac{n π x}{l} d x] \\ = \frac{2}{l} [\int_{0}^{\frac{1}{3}} \frac{3 h x}{l} \sin \frac{n π x}{l} d x + \int_{\frac{1}{3}}^{l} \frac{3 h (- x)}{2 l} \sin \frac{n π x}{l} d x] \\ = \frac{6 h}{l^{2}} \int_{0}^{\frac{1}{3}} x \sin \frac{n π x}{l} d x + \frac{3 h}{2 l^{2}} \int_{\frac{1}{3}} (l - x) \sin \frac{n π x}{l} d x \end{aligned}

= \frac{6 h}{l^{2}} {[(x) (- \frac{l}{n π} \cos \frac{n π x}{l}) - 1 (- \frac{l^{2}}{n^{2} π^{2}} \sin \frac{n π x}{l})]}_{0}^{\frac{1}{3}} + \frac{3 h}{l^{2}} [(l - x) (- \frac{l}{n π} \cos \frac{n π x}{l}) - (- 1) (- \frac{l^{2}}{n^{2} π^{2}} \sin \frac{π x}{l})]

= \frac{9 h}{n^{2} π^{2}} \sin \frac{n π}{3}

(10) implies

y (x, t) = \sum_{n = 1}^{\infty} \frac{9 h}{n^{2} π^{2}} \sin \frac{n π}{3} \cdot \sin \frac{n π x}{l} \cos \frac{n π c t}{l}

Which is the required displacement function

8.(b) बिन्दुओं

x_{0}, x_{0} + ε

तथा

x_{1}

के सापेक्ष तीन-बिन्दु लेगरान्ज-अन्तर्वेशन बहुपद को लिखिए । तदुपरान्त limit

ε \to 0

करने पर निम्नलिखित सम्बन्ध को स्थापित कीजिए :

f (x) = \frac{(x_{1} - x) (x + x_{1} - 2 x_{0})}{{(x_{1} - x_{0})}^{2}} f (x_{0}) + \frac{(x - x_{0}) (x_{1} - x)}{(x_{1} - x_{0})} f^{'} (x_{0}) + \frac{{(x - x_{0})}^{2}}{(x_{1} - x_{0})} f (x_{1}) + E (x)

जहाँ पर

E (x) = \frac{1}{6} {(x - x_{0})}^{2} (x - x_{1}) f^{'''} (ξ)

त्रुटि-फलन है और
न्यूनतम

(x_{0}, x_{0} + ε, x_{1}) < ξ <

उच्चतम

(x_{0}, x_{0} + ε, x_{1})

Write the three point Lagrangian interpolating polynomial relative to the points

x_{0}, x_{0} + ε

and

x_{1}

. Then by taking the limit

ε \to 0

, establish the relation

f (x) = \frac{(x_{1} - x) (x + x_{1} - 2 x_{0})}{{(x_{1} - x_{0})}^{2}} f (x_{0}) + \frac{(x - x_{0}) (x_{1} - x)}{(x_{1} - x_{0})} f^{'} (x_{0}) + \frac{{(x - x_{0})}^{2}}{(x_{1} - x_{0})} f (x_{1}) + E (x)

where

E (x) = \frac{1}{6} {(x - x_{0})}^{2} (x - x_{1}) f^{'''} (ξ)

is the error function and min.

(x_{0}, x_{0} + ε, x_{1}) < ξ < max . (x_{0}, x_{0} + ε, x_{1})

Answer:

Introduction:

The problem involves writing the three-point Lagrangian interpolating polynomial relative to the points

x_{0}, x_{0} + ε

and

x_{1}

. Then, by taking the limit

ε \to 0

, we establish the relation for

f (x)

as described in the provided equation.

Step 1: Lagrangian Three-Point Formula

We start with the Lagrangian three-point formula. Given the points

x_{0}, x_{0} + t, x_{1}

and corresponding function values

f (x_{0}), f (x_{0} + t), f (x_{1})

, the formula is as follows:

f (x) = \frac{(x - x_{0} - t) (x - x_{1})}{(- t) (x_{0} - x_{1})} f (x_{0}) + \frac{(x - x_{0}) (x - x_{1})}{(t) (x_{0} + t - x_{1})} f (x_{0} + t) + \frac{(x - x_{0}) (x - x_{0} - t)}{(x - x_{0}) (x_{1} - x_{0} - t)} f (x_{1}) + Error (E) \to (1)

Where

E = \frac{(x - x_{0}) (x - x_{0} - t) (x - x_{1})}{3!} f^{''} (ξ)

Step 2: Taylor’s Series

By Taylor’s series, we can express

f (x_{0} + t)

as follows:

f (x_{0} + t) = f (x_{0}) + t f^{'} (x_{0}) + \frac{t^{2}}{2!} f^{''} (x_{0}) + \dots

Step 3: Simplifying Equation (1)

Now, from Equation (1), we simplify it by neglecting higher-order terms of

t

and obtain:

\begin{aligned} f (x) = \frac{(x - x_{0} - t) (x - x_{1})}{(- t) (x_{0} - x_{1})} f (x_{0}) + [\frac{(x - x_{0}) (x - x_{1})}{t (x_{0} + t - x_{1})} [f (x_{0}) + t f^{'} (x_{0})]] + \frac{(x - x_{0}) (x - x_{0} + t)}{(x_{1} - x_{0}) (x_{1} - x_{0} - t)} f (x_{1}) + E \\ = (x - x_{1}) \frac{f (x_{0})}{t} [- \frac{x - x_{0} - t}{x_{0} - x_{1}} + \frac{x - x_{0}}{x_{0} + t - x_{1}}] + \frac{(x - x_{0}) (x - x_{1})}{x_{0} + t - x_{1}} f^{'} (x_{0}) + \frac{(x - x_{0}) (x - x_{0} + t)}{(x_{1} - x_{0}) (x_{1} - x_{0} - t)} f (x_{1}) + E \\ = (x - x_{1}) \frac{f (x_{0})}{t} [- \frac{x - x_{0}}{x_{0} - x_{1}} + \frac{t}{x_{0} - x_{1}} + \frac{x - x_{0}}{x_{0} + t - x_{1}}] + \frac{(x - x_{0}) (x - x_{1})}{x_{0} + t - x_{1}} f^{'} (x_{0}) + \frac{(x - x_{0}) (x - x_{0} + t)}{(x_{1} - x_{0}) (x_{1} - x_{0} - t)} f (x_{1}) + E \\ = (x - x_{1}) f (x_{0}) [\frac{x - x_{0}}{(x_{0} - x_{1}) (x_{0} + t - x_{1})} + \frac{1}{x - x_{1}}] + (x - x_{0}) (x - x_{1}) \frac{f^{'} (x_{0})}{x_{0} + t - x_{1}} + \frac{(x - x_{0}) (x - x_{0} + t)}{(x_{1} - x_{0}) (x_{1} - x_{0} - t)} f (x_{1}) + E \end{aligned}

Step 4: Taking the Limit $ε \to 0$

t

tends to

0

, we get the final relation:

f (x) = \frac{(x_{1} - x) (x + x_{1} - 2 x_{0})}{{(x_{1} - x_{0})}^{2}} f (x_{0}) + \frac{(x - x_{0}) (x_{1} - x)}{(x_{1} - x_{0})} f^{'} (x_{0}) + \frac{{(x - x_{0})}^{2}}{(x_{1} - x_{0})} f (x_{1}) + E

Where

E = \frac{(x - x_{0}) (x - x_{1})}{6} f^{'''} (ξ)

with

min (x_{0}, x_{0} + ε, x_{1}) < ξ < max (x_{0}, x_{0} + ε, x_{1})

Conclusion:

The desired relation for

f (x)

is established as described, and it includes the error function

E (x)

(c) $\frac{m}{2}$ शक्ति वाले दो स्रोत, बिन्दुओं $(\pm a, 0)$ पर स्थित हैं। दर्शाइए कि वृत $x^{2} + y^{2} = a^{2}$ के किसी भी बिन्दु पर वेग $y$ -अक्ष के समान्तर तथा $y$ के व्युत्क्रमानुपाती है।

Two sources of strength

\frac{m}{2}

are placed at the points

(\pm a, 0)

. Show that at any point on the circle

x^{2} + y^{2} = a^{2}

, the velocity is parallel to the

y

-axis and is inversely proportional to

y

Answer:

Introduction:

The problem involves two sources of strength

\frac{m}{2}

placed at the points

(\pm a, 0)

. The goal is to demonstrate that at any point on the circle

x^{2} + y^{2} = a^{2}

, the velocity is parallel to the

y

-axis and inversely proportional to

y

Step 1: Complex Potential Calculation

Given the strength of the source is

\frac{m}{2}

, we can calculate the complex potential as follows:

\begin{aligned} w = - \frac{m}{2} \ln (z - a) - \frac{m}{2} \ln (z + a) \\ q = \frac{d w}{d z} = - \frac{m}{2} [\frac{1}{z - a} + \frac{1}{z + a}] \end{aligned}

Step 2: Simplifying $q$

Simplify the expression for

q

as follows:

q = - \frac{m}{2} [\frac{z + a + z - a}{(z - a) (z + a)}] = - \frac{m}{2} [\frac{2 z}{(z - a) (z + a)}] = - \frac{m z}{z^{2} - a^{2}} \to (1)

Step 3: Parameterize Points on the Circle

We parameterize points on the circle

x^{2} + y^{2} = a^{2}

as follows:

z = x + iy on circle x^{2} + y^{2} = a^{2} \Rightarrow x = a \cos θ, y = a \sin θ

Step 4: Express $z$ in Terms of $θ$

Express

z

in terms of

θ

as follows:

z = a [\cos θ + i \sin θ] = a e^{i θ}

Step 5: Substituting $z$ into Equation (1)

Substitute

z

into Equation (1) as follows:

q = - \frac{m a e^{i θ}}{a^{2} e^{2 i θ} - a^{2}} = - \frac{m}{a [e^{i θ} - e^{- i θ}]} = - \frac{2 m}{a [e^{i θ} - e^{- i θ}]} = - \frac{2 m}{2 a [e^{i θ} - e^{- i θ}]} = - \frac{m}{2 a i \sin θ} (\sin θ = \frac{e^{i θ} - e^{i θ}}{2 i})

q = - \frac{m}{y} (

because

y = a \sin θ)

Step 6: Final Conclusion

Conclude that

q α \frac{1}{y}

, indicating that the velocity is parallel to the

y

-axis and inversely proportional to

y

Conclusion:

The analysis demonstrates that at any point on the circle

x^{2} + y^{2} = a^{2}

, the velocity is indeed parallel to the

y

-axis and inversely proportional to

y

Free UPSC Mathematics Optional Paper-2 2020 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Properties of S 3 S 3 S_(3)S_3S3 and Z 3 Z 3 Z_(3)Z_3Z3

Steps to Show No Non-Trivial Homomorphism Exists

Introduction

Work/Calculations

Part 1: Every ideal of R / P R / P R//PR/PR/P is a principal ideal

Part 2: R / P R / P R//PR/PR/P is a PID for a prime ideal P P PPP of R R RRR

Conclusion

Introduction

Work/Calculations

Step 1: Prove that | a n + 1 − a n | a n + 1 − a n |a_(n+1)-a_(n)|\left|a_{n+1} – a_n\right||an+1−an| becomes arbitrarily small

Step 2: Prove that ( a n ) a n (a_(n))\left(a_n\right)(an) is a Cauchy sequence

Conclusion

Introduction

Given Parameters and Parametrization

Integral Setup

Substitution and Simplification

Evaluation of the Integral

Final Simplification

Conclusion

Introduction

Variables

Objective Function

Constraints

Linear Programming Problem in Standard Form

Basic Feasible Solution

Conclusion

Introduction

Preliminaries

Generators of G G GGG

Euler’s ϕ ϕ phi\phiϕ-Function

Proof

Conclusion

Introduction

Definition of Uniform Continuity

Proof by Contradiction

Conclusion

Introduction

Definitions

Properties Needed

Proof

1. f f fff is a Homomorphism

2. f f fff is Injective (One-to-One)

3. f f fff is Surjective (Onto)

Conclusion

Introduction

Problem Transformation

Conclusion

Step 2: Finding the Maxterms

Conclusion

Introduction:

Comparison:

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Properties of $S_{3}$ and $Z_{3}$

Part 1: Every ideal of $R / P$ is a principal ideal

Part 2: $R / P$ is a PID for a prime ideal $P$ of $R$

Step 1: Prove that $| a_{n + 1} - a_{n} |$ becomes arbitrarily small

Step 2: Prove that $(a_{n})$ is a Cauchy sequence

Generators of $G$

Euler’s $ϕ$ -Function

1. $f$ is a Homomorphism

2. $f$ is Injective (One-to-One)

3. $f$ is Surjective (Onto)