Free UPSC Mathematics Optional Paper-1 2020 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 12, 2025ByAbstract Classes

UPSC M2020-1 WS

खण्ड-A / SECTION-A

(a) माना समुच्चय $V$ में सभी $n \times n$ के वास्तविक मैजिक वर्ग हैं। दिखाइए कि समुच्चय $V, R$ पर एक सदिश समष्टि है। दो भिन्न-भिन्न $2 \times 2$ मैजिक वर्ग के उदाहरण दीजिए।

Consider the set

V

of all

n \times n

real magic squares. Show that

V

is a vector space over

R

. Give examples of two distinct

2 \times 2

magic squares.

Answer:

Introduction

The problem asks us to prove that the set

V

of all

n \times n

real magic squares is a vector space over

R

. A magic square is a square grid of numbers such that the sums of the numbers in each row, each column, and both main diagonals are the same. We will use the properties of vector spaces to prove this.

To make the proof more explicit, let’s assume

X, Y,

and

Z

are

n \times n

magic squares with general entries as follows:

X = [\begin{array}{ccccc} x_{11} & x_{12} & x_{13} & \dots & x_{1 n} \\ x_{21} & x_{22} & x_{23} & \dots & x_{2 n} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ x_{n 1} & x_{n 2} & x_{n 3} & \dots & x_{n n} \end{array}]

Y = [\begin{array}{ccccc} y_{11} & y_{12} & y_{13} & \dots & y_{1 n} \\ y_{21} & y_{22} & y_{23} & \dots & y_{2 n} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ y_{n 1} & y_{n 2} & y_{n 3} & \dots & y_{n n} \end{array}]

Z = [\begin{array}{ccccc} z_{11} & z_{12} & z_{13} & \dots & z_{1 n} \\ z_{21} & z_{22} & z_{23} & \dots & z_{2 n} \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ z_{n 1} & z_{n 2} & z_{n 3} & \dots & z_{n n} \end{array}]

And let

a

and

b

be real numbers.

Work/Calculations

Property 1: Closure under Addition and Scalar Multiplication

$X + Y \in MS (n)$

Let’s substitute the values:

$(X + Y) = [x_{i j} + y_{i j}]$

The sum of each row, column, and diagonal in $X$ and $Y$ is the same constant $k$ . Therefore, the sum of each corresponding row, column, and diagonal in $X + Y$ will be $2 k$ , which means $X + Y$ is also a magic square.
$a X \in MS (n)$

Let’s substitute the values:

$a X = [a x_{i j}]$

If we multiply $X$ by a scalar $a$ , each row, column, and diagonal sum becomes $a k$ , which means $a X$ is also a magic square.

After calculating, we find that both

X + Y

and

a X

are in

MS (n)

, proving closure under addition and scalar multiplication.

Property 2: Commutativity of Addition

X + Y = Y + X

This is straightforward because matrix addition is commutative.

Property 3: Associativity of Addition

X + (Y + Z) = (X + Y) + Z

Matrix addition is associative, so this property holds.

Property 4: Existence of Zero Vector

Let

0

be the

n \times n

magic square where every entry is zero. Then,

X + 0 = 0 + X = X

Property 5: Existence of Additive Inverse

Let

X^{'} = - X

. Then,

X + X^{'} = X^{'} + X = 0

Property 6: Distributive Law 1

a (X + Y) = a X + a Y

This is a property of matrices, so it holds.

Property 7: Distributive Law 2

(a + b) X = a X + b X

This is also a property of matrices.

Property 8: Associativity of Scalar Multiplication

(a b) X = a (b X)

This is true for matrices.

Property 9: Multiplication by Identity

1 X = X

This is true for any matrix

X

We have shown that the set

V

of all

n \times n

real magic squares satisfies all the properties required for it to be a vector space over

R

. Therefore,

V

is indeed a vector space over

R

For a

2 \times 2

matrix to be a magic square, the sum of each row, each column, and both diagonals must be the same. Let’s consider a general

2 \times 2

magic square

M

with entries

a, b, c,

and

d

M = [\begin{array}{cc} a & b \\ c & d \end{array}]

For

M

to be a magic square, the following conditions must be met:

The sum of each row must be the same: $a + b = c + d$
The sum of each column must be the same: $a + c = b + d$
The sum of the diagonals must be the same: $a + d = b + c$

Example 1

Let’s choose

a = 1, b = 1, c = 1,

and

d = 1

. All the sums are

1 + 1 = 2

, so it’s a magic square.

M_{1} = [\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}]

Example 2

Let’s choose

a = 2, b = 2, c = 2,

and

d = 2

. All the sums are

2 + 2 = 4

, so it’s a magic square.

M_{2} = [\begin{array}{cc} 2 & 2 \\ 2 & 2 \end{array}]

Conclusion

Therefore, we conclude that the set

V

of all

n \times n

real magic squares is a valid vector space over

R

. This conclusion is underpinned by the rigorous application of vector space properties and the concrete examples provided for

2 \times 2

magic squares, affirming the validity of this mathematical concept.

(b) माना

M_{2} (R)

सभी

2 \times 2

वास्तविक आव्यूहों का सदिश समष्टि है। माना

B = [\begin{array}{cc} 1 & - 1 \\ - 4 & 4 \end{array}]

. माना

T : M_{2} (R) \to M_{2} (R)

एक रैखिक रूपांतरण है, जो

T (A) = B A

द्वारा परिभाषित है।

T

की कोटि (रिक) व शून्यता (नलिटि) ज्ञात कीजिए। आव्यूह

A

ज्ञात कीजिए, जो शून्य आव्यूह को प्रतिचित्रित करता है।

Let

M_{2} (R)

be the vector space of all

2 \times 2

real matrices. Let

B = [\begin{array}{cc} 1 & - 1 \\ - 4 & 4 \end{array}]

. Suppose

T : M_{2} (R) \to M_{2} (R)

is a linear transformation defined by

T (A) = B A

. Find the rank and nullity of

T

. Find a matrix

A

which maps to the null matrix.

Answer:

Introduction

We are given a vector space

M_{2} (R)

of all

2 \times 2

real matrices and a specific matrix

B

. A linear transformation

T : M_{2} (R) \to M_{2} (R)

is defined as

T (A) = B A

. We are asked to find the rank and nullity of

T

and to find a matrix

A

that maps to the null matrix under

T

Work/Calculations

Finding a Matrix $A$ that Maps to the Null Matrix

To find a matrix

A

that maps to the null matrix under

T

, we need to find

A

such that

B A = 0

Let

A = [\begin{array}{cc} a & b \\ c & d \end{array}]

Then

B A

is:

B A = [\begin{array}{cc} 1 & - 1 \\ - 4 & 4 \end{array}] [\begin{array}{cc} a & b \\ c & d \end{array}] = [\begin{array}{cc} a - c & b - d \\ - 4 a + 4 c & - 4 b + 4 d \end{array}]

For

B A

to be the null matrix, we need

a - c = 0

b - d = 0

- 4 a + 4 c = 0

, and

- 4 b + 4 d = 0

Solving these equations, we find

a = c

and

b = d

Therefore, any matrix

A

of the form

[\begin{array}{cc} a & b \\ a & b \end{array}]

will map to the null matrix under

T

Nullity of $T$

The nullity of

T

is the dimension of the null space of

T

, denoted as

N (T)

. The null space consists of all matrices

A

such that

B A = 0

Any matrix

A

of the form

[\begin{array}{cc} a & b \\ a & b \end{array}]

will map to the null matrix under

T

We can express this as:

[\begin{array}{ll} x_{1} & x_{2} \\ x_{3} & x_{4} \end{array}] = [\begin{array}{ll} p & q \\ p & q \end{array}] = p [\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}] + q [\begin{array}{ll} 0 & 1 \\ 0 & 1 \end{array}]

Thus, the null space is spanned by the matrices

[\begin{array}{ll} 1 & 0 \\ 1 & 0 \end{array}]

and

[\begin{array}{ll} 0 & 1 \\ 0 & 1 \end{array}]

, and its dimension is 2. Hence,

Nullity (T) = 2

Rank of $T$

By the rank-nullity theorem, we have:

ρ (T) + N (T) = 4 ⟹ ρ (T) = 4 - 2 = 2

So, the rank of

T

is 2.

Finding Matrix $A$ that Maps to the Null Matrix

Any matrix

A

of the form

[\begin{array}{cc} a & b \\ a & b \end{array}]

will map to the null matrix under

T

Lets take an example :

To find out if

A

maps to the null matrix under

T

, we need to compute

B A

Given

B = [\begin{array}{cc} 1 & - 1 \\ - 4 & 4 \end{array}]

and

A = [\begin{array}{ll} 2 & 3 \\ 2 & 3 \end{array}]

B A = [\begin{array}{cc} 1 & - 1 \\ - 4 & 4 \end{array}] [\begin{array}{ll} 2 & 3 \\ 2 & 3 \end{array}] = [\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}]

After calculating, we find that

B A

is indeed the null matrix.

Conclusion

The rank of $T$ is 1.
The nullity of $T$ is 1.
A matrix $A$ that maps to the null matrix under $T$ is of the form $[\begin{array}{cc} a & b \\ a & b \end{array}]$ Example is $A = [\begin{array}{ll} 2 & 3 \\ 2 & 3 \end{array}]$ .

(c)

lim_{x \to \frac{π}{4}} (\tan x)^{\tan 2 x}

का मान निकालिए।

Evaluate

lim_{x \to \frac{π}{4}} (\tan x)^{\tan 2 x}

Answer:

Equations/Concepts:
We’ll use the concept of limits, logarithms, and trigonometric identities.

Step-by-Step Explanation:

Let $y = (\tan x)^{\tan 2 x}$ .
Take the natural logarithm of both sides:

\begin{aligned} \log y & = \tan 2 x \log \tan x \end{aligned}

Now, let’s find the limit:

\begin{aligned} lim_{x \to \frac{π}{4}} \log y & = lim_{x \to \frac{π}{4}} \tan 2 x \log \tan x \end{aligned}

We can apply L’Hôpital’s rule here because it’s an indeterminate form ( $\frac{0}{0}$ ). So, differentiate the numerator and denominator with respect to $x$ :

\begin{aligned} = lim_{x \to \frac{π}{4}} \frac{\log \tan x}{\cot 2 x} \\ = lim_{x \to \frac{π}{4}} \frac{\frac{1}{\tan x} \sec^{2} x}{- 2 \csc^{2} 2 x} \end{aligned}

Now, substitute $x = \frac{π}{4}$ into the equation:

\begin{aligned} = \frac{\frac{1}{\tan \frac{π}{4}} \sec^{2} \frac{π}{4}}{- 2 \csc^{2} 2 \frac{π}{4}} \\ = \frac{1 \cdot 2}{- 2} \\ = - 1 \end{aligned}

Now, we’ve found the limit of $\log y$ as $x$ approaches $\frac{π}{4}$ to be $- 1$ . To find $y$ , we take the exponential of both sides:

\begin{aligned} y & = e^{- 1} \end{aligned}

Conclusion:
The limit of

(\tan x)^{\tan 2 x}

x

approaches

\frac{π}{4}

e^{- 1}

(d) वक्र

(2 x + 3) y = (x - 1)^{2}

के सभी अनंतस्पर्शी निकालिए।

Find all the asymptotes of the curve

(2 x + 3) y = (x - 1)^{2}

Answer:

Step-by-Step Solution:

Start with the given equation of the curve: $(2 x + 3) y = (x - 1)^{2}$ .
Rewrite the equation in standard form:

\begin{aligned} 2 x y + 3 y & = x^{2} - 2 x + 1 \\ x^{2} - 2 x y - 2 x - 3 y + 1 & = 0 \end{aligned}

To find the asymptotes, begin by assuming $y = m x + c$ .
Substitute $y = m x + c$ into the equation:

\begin{aligned} x^{2} - 2 x (m x + c) - 2 x - 3 (m x + c) + 1 & = 0 \\ x^{2} - 2 x^{2} m - 2 c x - 2 x - 3 m x - 3 c + 1 & = 0 \end{aligned}

Group the terms with $x^{2}$ and $x$ separately:

(1 - 2 m) x^{2} - (2 c + 2 + 3 m) x - 3 c + 1 = 0

To have asymptotes, the coefficient of $x^{2}$ should be zero and the constant term should not be zero. Equate them to zero:

\begin{aligned} 1 - 2 m & = 0 ⟹ m = \frac{1}{2} \\ 2 c + 2 + 3 m & = 0 ⟹ 2 c + 2 + 3 (\frac{1}{2}) = 0 \\ 2 c + 2 + \frac{3}{2} & = 0 \\ 2 c + \frac{7}{2} & = 0 \\ 2 c & = - \frac{7}{2} \\ c & = - \frac{7}{4} \end{aligned}

Thus, one asymptote is $y = \frac{x}{2} - \frac{7}{4}$ .
To find the vertical asymptotes, set the denominator equal to zero:

2 x + 3 = 0 ⟹ x = - \frac{3}{2}

Conclusion:
The curve

(2 x + 3) y = (x - 1)^{2}

has one slant asymptote given by

y = \frac{x}{2} - \frac{7}{4}

and one vertical asymptote at

x = - \frac{3}{2}

(e) दीर्घवृत्तज

2 x^{2} + 6 y^{2} + 3 z^{2} = 27

के स्पर्श समतल का समीकरण निकालिए, जो रेखा

x - y - z = 0 = x - y + 2 z - 9

से होकर गुजरता है।

Find the equations of the tangent plane to the ellipsoid

2 x^{2} + 6 y^{2} + 3 z^{2} = 27

which passes through the line

x - y - z = 0 = x - y + 2 z - 9

Answer:

Work/Calculations:

Equation of the Given Line:
We start with the equation of the given line:

$x - y - z = 0 (1)$
Equation of a Plane Through the Given Line:
The equation of any plane through the given line (1) can be written as:

$x - y - z + λ (x - y + 2 z - 9) = 0$

Simplifying:

$x (1 + λ) + y (- 1 - λ) + z (- 1 + 2 λ) = 9 λ (2)$
Ellipsoid Equation:
The equation of the given ellipsoid is $2 x^{2} + 6 y^{2} + 3 z^{2} = 27$ . We can normalize this equation as:

$\frac{2}{27} x^{2} + \frac{2}{9} y^{2} + \frac{1}{9} z^{2} = 1$

Here, $a = \frac{2}{27}$ , $b = \frac{2}{9}$ , and $c = \frac{1}{9}$ .
Using the Tangent Plane Equation:
If a plane (2) touches the given ellipsoid, we can use the equation for a tangent plane:

$\frac{l^{2}}{a} + \frac{m^{2}}{b} + \frac{n^{2}}{c} = p^{2} (3)$

Here, $(l, m, n)$ is the normal vector to the plane, and $p$ is the distance from the origin.
Substituting into (3):
Now, we substitute the values $a$ , $b$ , and $c$ into equation (3):

$\frac{27}{2} (1 + λ)^{2} + \frac{9}{2} (- 1 - λ)^{2} + 18 (2 λ - 1)^{2} = (9 λ)^{2}$
Solving for λ:
Simplifying and solving for λ:

$- 54 λ^{2} + 54 = 0$

$λ^{2} = 1$

This gives two solutions: $λ = \pm 1$ .

Conclusion:

Now, we can find the required equations of the tangent planes:

If $λ = 1$ :
Substituting $λ = 1$ into equation (2):

$2 x - 2 y + z = 9$

So, the equation of one of the tangent planes is $2 x - 2 y + z = 9$ .
If $λ = - 1$ :
Substituting $λ = - 1$ into equation (2):

$z = 3$

So, the equation of the other tangent plane is $z = 3$ .

These are the equations of the tangent planes to the ellipsoid that pass through the given line.

(a) $\int_{0}^{1} \tan^{- 1} (1 - \frac{1}{x}) d x$ का मान निकालिए।

Evaluate

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

Answer:

Introduction:
We are tasked with evaluating the integral

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

Work/Calculations:

Initial Setup:
We start by defining $I$ as the integral:

$I = \int_{0}^{1} \tan^{- 1} (1 - \frac{1}{x}) d x$
Rewriting the Integral:
We can express $\tan^{- 1} (1 - \frac{1}{x})$ as $\tan^{- 1} (\frac{x - 1}{x})$ :

$I = \int_{0}^{1} \tan^{- 1} (\frac{x - 1}{x}) d x (1)$
Applying Integral Property:
Utilizing the property $\int_{a}^{b} f^{'} (x) d x = \int_{a}^{b} f (a + b - x) d x$ , we obtain:

$I = \int_{0}^{1} \tan^{- 1} (- \frac{x}{1 - x}) d x$
Further Simplification:
We can simplify the integral by changing the sign inside the arctan function:

$I = \int_{0}^{1} \tan^{- 1} (\frac{x}{x - 1}) d x (2)$
Combining Equations (1) and (2):
Adding equations (1) and (2), we get:

$2 I = \int_{0}^{1} \tan^{- 1} (\frac{x - 1}{x}) d x + \int_{0}^{1} \tan^{- 1} (\frac{x}{x - 1}) d x$
Simplifying the Combined Integral:
Notice that $\tan^{- 1} x + \tan^{- 1} \frac{1}{x}$ has a specific property:

$\tan^{- 1} x + \tan^{- 1} \frac{1}{x} = {\begin{cases} \frac{π}{2} & if x > 0 \\ - \frac{π}{2} & if x < 0 \end{cases}$

Since $x \in (0, 1)$ , we have $x - 1 < 0$ and $\frac{x}{x - 1} < 0$ . Therefore, combining the integrals (3) results in:

$2 I = \int_{0}^{1} - \frac{π}{2} d x$
Evaluating the Integral:
Calculating the integral on the right-hand side:

$2 I = - \frac{π}{2} (x) |_{0}^{1} = - \frac{π}{2} (1 - 0) = - \frac{π}{2}$
Solving for $I$ :
Finally, solving for $I$ :

$I = - \frac{π}{4}$

Conclusion:
The corrected value of the given integral

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

- \frac{π}{4}

(b) एक

n \times n

आव्यूह

A

को परिभाषित कीजिए, जबकि

A = I - 2 u \cdot u^{T}

, जहाँ

u

एक इकाई स्तंभ सदिश है।
(i) परीक्षण कीजिए कि

A

सममित है।
(ii) परीक्षण कीजिए कि

A

लांबिक है।
(iii) दिखाइए कि आव्यूह

A

का अनुरेख

(n - 2)

है।
(iv) आव्यूह

A_{3 \times 3}

निकालिए, जबकि

u = [\begin{matrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{matrix}]

है।

Define an

n \times n

matrix as

A = I - 2 u \cdot u^{T}

, where

u

is a unit column vector.
(i) Examine if

A

is symmetric.
(ii) Examine if

A

is orthogonal.
(iii) Show that trace

(A) = n - 2

.
(iv) Find

A_{3 \times 3}

, when

u = [\begin{matrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{matrix}]

Answer:

Introduction

We are given an

n \times n

matrix

A

defined as

A = I - 2 u \cdot u^{T}

, where

u

is a unit column vector. We are tasked with examining the following properties of

A

Whether $A$ is symmetric.
Whether $A$ is orthogonal.
Examine the trace of $A$ .
Find $A_{3 \times 3}$ when $u = [\begin{matrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{matrix}]$ .

Work/Calculations

(i) Examine if $A$ is Symmetric

A matrix

A

is symmetric if

A = A^{T}

The transpose of

A

is given by

(I - 2 u \cdot u^{T})^{T}

After calculating, we get

A^{T} = I^{T} - (2 u \cdot u^{T})^{T}

Since

I^{T} = I

and

(2 u \cdot u^{T})^{T} = 2 u \cdot u^{T}

, we find that

A^{T} = I - 2 u \cdot u^{T}

Therefore,

A = A^{T}

, which means

A

is symmetric.

(ii) Examine if $A$ is Orthogonal

A matrix

A

is orthogonal if

A^{T} A = A A^{T} = I

Let’s substitute the values into the formula:

(I - 2 u \cdot u^{T})^{T} (I - 2 u \cdot u^{T})

After calculating, we get

I - 2 u \cdot u^{T} - 2 u \cdot u^{T} + 4 u \cdot u^{T} \cdot u \cdot u^{T}

Since

u

is a unit vector,

u \cdot u^{T} = 1

Therefore,

A^{T} A = I

and

A A^{T} = I

, which means

A

is orthogonal.

(iii) Show that Trace $(A) = n - 2$

The trace of

A

is the sum of its diagonal elements.

Trace

(A) = Trace (I) - 2 Trace (u \cdot u^{T})

Since

u

is a unit vector,

Trace (u \cdot u^{T}) = 1

Therefore, Trace

(A) = n - 2

(iv) Find $A_{3 \times 3}$ , when $u = [\begin{matrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{matrix}]$

A = I - 2 u \cdot u^{T}

After calculating, we get

A = (\begin{matrix} 1 - \frac{2}{9} & - \frac{4}{9} & - \frac{4}{9} \\ - \frac{4}{9} & 1 - \frac{8}{9} & - \frac{8}{9} \\ - \frac{4}{9} & - \frac{8}{9} & 1 - \frac{8}{9} \end{matrix})

Simplifying,

A = (\begin{matrix} \frac{7}{9} & - \frac{4}{9} & - \frac{4}{9} \\ - \frac{4}{9} & \frac{1}{9} & - \frac{8}{9} \\ - \frac{4}{9} & - \frac{8}{9} & \frac{1}{9} \end{matrix})

Conclusion

The matrix $A$ is symmetric.
The matrix $A$ is orthogonal.
The trace of $A$ is $n - 2$ .
When $u = [\begin{matrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{matrix}]$ , $A_{3 \times 3} = (\begin{matrix} \frac{7}{9} & - \frac{4}{9} & - \frac{4}{9} \\ - \frac{4}{9} & \frac{1}{9} & - \frac{8}{9} \\ - \frac{4}{9} & - \frac{8}{9} & \frac{1}{9} \end{matrix})$ .

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

के समांतर हैं तथा जिसका मार्गदर्शक वक्र

x^{2} + y^{2} = 4, z = 2

है।

Find the equation of the cylinder whose generators are parallel to the line

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

and whose guiding curve is

x^{2} + y^{2} = 4, z = 2

Answer:

Equation of a Generator:
Let $P (x_{1}, y_{1}, z_{1})$ be any point on the cylinder. The equation of a generator passing through $P$ can be expressed as:

$\frac{x - x_{1}}{1} = \frac{y - y_{1}}{- 2} = \frac{z - z_{1}}{3}$
Intersection with Plane $z = 2$ :
The generator intersects the plane $z = 2$ at a point given by:

$\begin{aligned} x - x_{1} & = \frac{y - y_{1}}{- 2} = \frac{z - z_{1}}{3} (1) \\ \Rightarrow (x_{1} + \frac{2 - z_{1}}{3}, - \frac{2 (2 - z_{1})}{3} + y_{1}, 2) \end{aligned}$
Intersection with Guiding Curve:
The generator also intersects the guiding curve, given by $x^{2} + y^{2} = 4$ and $z = 2$ . Substituting $z = 2$ into equation (1), we get:

$\begin{aligned} {(x_{1} + \frac{2 - z_{1}}{3})}^{2} + {(y_{1} - \frac{2 (2 - z_{1})}{3})}^{2} & = 4 \\ \Rightarrow {(3 x_{1} + 2 - z_{1})}^{2} + {(3 y_{1} - 4 + 2 z_{1})}^{2} & = 36 \\ \Rightarrow 9 x_{1}^{2} + 9 y_{1}^{2} + 5 z_{1}^{2} - 6 x_{1} z_{1} + 12 y_{1} z_{1} - 20 z_{1} + 12 x_{1} - 24 y_{1} - 16 & = 0 \end{aligned}$
Equation of the Cylinder:
The equation of the required cylinder is:

$9 x^{2} + 9 y^{2} + 5 z^{2} - 6 x z + 12 y z - 20 z + 12 x - 24 y - 16 = 0$

Conclusion:
The equation

9 x^{2} + 9 y^{2} + 5 z^{2} - 6 x z + 12 y z - 20 z + 12 x - 24 y - 16 = 0

represents the cylinder with generators parallel to the line

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

and a guiding curve defined by

x^{2} + y^{2} = 4

and

z = 2

3(a) निम्न फलन पर विचार कीजिए :

f (x) = \int_{0}^{x} (t^{2} - 5 t + 4) (t^{2} - 5 t + 6) d t

(i) फलन

f (x)

के क्रांतिक बिंदु निकालिए।
(ii) वे बिंदु निकालिए, जहाँ

f (x)

का स्थानीय न्यूनतम होगा।
(iii) वे बिंदु निकालिए, जहाँ

f (x)

का स्थानीय अधिकतम होगा।
(iv) फलन

f (x)

के

[0, 5]

में कितने शून्यक होंगे, निकालिए।

Consider the function

f (x) = \int_{0}^{x} (t^{2} - 5 t + 4) (t^{2} - 5 t + 6) d t

.
(i) Find the critical points of the function

f (x)

.
(ii) Find the points at which local minimum occurs.
(iii) Find the points at which local maximum occurs.
(iv) Find the number of zeros of the function

f (x)

[0, 5]

Answer:

Introduction

We are given the function

f (x) = \int_{0}^{x} (t^{2} - 5 t + 4) (t^{2} - 5 t + 6) d t

. We are tasked with:

Finding the critical points of $f (x)$ .
Identifying the points at which a local minimum occurs.
Identifying the points at which a local maximum occurs.
Determining the number of zeros of $f (x)$ in the interval $[0, 5]$ .

Work/Calculations

(i) Find the Critical Points of $f (x)$

By Leibnitz’s theorem, the derivative of

f (x)

is given by:

f^{'} (x) = (x^{2} - 5 x + 4) (x^{2} - 5 x + 6)

To find the critical points, we set

f^{'} (x) = 0

f^{'} (x) = (x + 1) (x - 4) (x - 2) (x - 3)

This gives us critical points at

x = 1, 2, 3, 4

(ii) Find the Points at Which Local Minimum Occurs

To find the local minima, we need to examine

f^{″} (x)

f^{″} (x) = (2 x - 5) [x^{2} - 5 x + 6] + [x^{2} - 5 x + 4] (2 x - 5)

After simplifying, we get:

f^{″} (x) = (2 x - 5) [x^{2} - 10 x + 10]

x = 2

f^{″} (x) = (4 - 5) (4 - 20 + 10) = 6

, which means

x = 2

is a local minimum.

(iii) Find the Points at Which Local Maximum Occurs

Using the same

f^{″} (x)

as above, we find:

x = 1

f^{″} (x) = (2 - 5) (1 - 10 + 10) = - 3

, indicating

x = 1

is a local maximum.

x = 3

f^{″} (x) = (6 - 5) (9 - 30 + 10) = - 11

, indicating

x = 3

is a local maximum.

x = 4

f^{″} (x) = (8 - 5) (16 - 40 + 10) = - 42

, indicating

x = 4

is a local maximum.

(iv) Find the Number of Zeros of $f (x)$ in $[0, 5]$

The function

f (x)

is an integral of a polynomial, and its graph shows that

f (0) = 0

. Therefore,

x = 0

is the only zero of the function

f (x)

[0, 5]

\begin{aligned} f (x) = [\frac{x^{5}}{5} - \frac{10 x^{4}}{4} + \frac{35 x^{3}}{3} - \frac{50 x^{2}}{2} + 24 x] \end{aligned}

\begin{aligned} f (1) = 8.36 (max) \\ f (2) = 7.73 (min) \\ f (3) = 8.1 (max) \\ f (4) = 7.46 (min) \\ f (0) = 0 \end{aligned}

Conclusion

The critical points of $f (x)$ are $x = 1, 2, 3, 4$ .
A local minimum occurs at $x = 2$ .
Local maxima occur at $x = 1, 3, 4$ .
The function $f (x)$ has only one zero in the interval $[0, 5]$ , which is $x = 0$ .

(b) माना

F

सम्मिश्र संख्याओं का एक उपक्षेत्र है व

T : F^{3} \to F^{3}

एक ऐसा फलन है, जो निम्न रूप से परिभाषित है :

T (x_{1}, x_{2}, x_{3}) = (x_{1} + x_{2} + 3 x_{3}, 2 x_{1} - x_{2}, - 3 x_{1} + x_{2} - x_{3})

a, b, c

पर क्या शर्तें हैं कि

(a, b, c), T

के शुन्य समष्टि में है?

T

की शून्यता निकालिए।

Let

F

be a subfield of complex numbers and

T

a function from

F^{3} \to F^{3}

defined by

T (x_{1}, x_{2}, x_{3}) = (x_{1} + x_{2} + 3 x_{3}, 2 x_{1} - x_{2}, - 3 x_{1} + x_{2} - x_{3})

. What are the conditions on

a, b, c

such that

(a, b, c)

be in the null space of

T

? Find the nullity of

T

Answer:

Introduction

We are given a function

T : F^{3} \to F^{3}

defined by

T (x_{1}, x_{2}, x_{3}) = (x_{1} + x_{2} + 3 x_{3}, 2 x_{1} - x_{2}, - 3 x_{1} + x_{2} - x_{3})

. The field

F

is a subfield of the complex numbers. We are tasked with finding the conditions on

a, b, c

such that

(a, b, c)

is in the null space of

T

, and to find the nullity of

T

Work/Calculations

Step 1: Define the Null Space

The null space of

T

, denoted

Null (T)

, is the set of all vectors

v

F^{3}

such that

T (v) = 0

Step 2: Find the Conditions for $(a, b, c)$

To find

(a, b, c)

in the null space of

T

, we set

T (a, b, c) = (0, 0, 0)

T (a, b, c) = (a + b + 3 c, 2 a - b, - 3 a + b - c) = (0, 0, 0)

This leads to the following system of equations:

$a + b + 3 c = 0$
$2 a - b = 0$
$- 3 a + b - c = 0$

Step 3: Row Reduction to Echelon Form

We can represent the system of equations as an augmented matrix and reduce it to row-echelon form:

\begin{array}{cccc} 1 & 1 & 3 & 0 \\ 2 & - 1 & 0 & 0 \\ - 3 & 1 & - 1 & 0 \end{array}

After row reduction, we get:

\begin{array}{cccc} 1 & 1 & 3 & 0 \\ 0 & - 3 & - 6 & 0 \\ 0 & 0 & 0 & 0 \end{array}

This leads to:

$a + b + 3 c = 0$
$- 3 b - 6 c = 0$
$b + 2 c = 0$

Step 4: Solve the System

Solving these equations, we find:

$b = - 2 c$
$a = - c$

Step 5: Find the Nullity of $T$

Null space is given by let

C = K

\begin{aligned} (a, b, c) = [- K, - 2 K, K] \\ = K [- 1, - 2, 1] \end{aligned}

The null space is spanned by the vector

[- 1, - 2, 1]

, which means the nullity of

T

is 1.

Conclusion

The conditions on $a, b, c$ such that $(a, b, c)$ is in the null space of $T$ are $a = - c$ and $b = - 2 c$ .
The nullity of $T$ is 1.

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

शंकु

5 y z - 8 z x - 3 x y = 0

के तीन परस्पर लांबिक जनकों के समुच्चय में से एक है, तब अन्य दो जनकों के समीकरण निकालिए।

If the straight line

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

represents one of a set of three mutually perpendicular generators of the cone

5 y z - 8 z x - 3 x y = 0

, then find the equations of the other two generators.

Answer:

Introduction

We are given a straight line

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

and a cone

5 y z - 8 z x - 3 x y = 0

. The straight line is one of the three mutually perpendicular generators of the cone. Our task is to find the equations of the other two generators.

Work/Calculations

Step 1: Equation of the Plane

Since the given line is one of the three mutually perpendicular generators, it is normal to the plane through the vertex that cuts the cone in two perpendicular generators. Therefore, the equation of the plane is:

x + 2 y + 3 z = 0 (Equation 1)

Step 2: Equation of the Line of Intersection

Let one of the lines of intersection between the plane and the cone be:

\frac{x}{l} = \frac{y}{m} = \frac{z}{n}

Then, we have:

$l + 2 m + 3 n = 0$
$5 m n - 8 n l - 3 l m = 0 (Equation 2)$

Step 3: Eliminate $l$ Between Equations

To find

m

and

n

, we eliminate

l

between the two equations. This gives us:

5 m n - (8 n + 3 m) [- (2 m + 3 n)] = 0

\Rightarrow 24 n^{2} + 30 m n + 6 m^{2} = 0

\Rightarrow m^{2} + 5 m n + 4 n^{2} = 0

\Rightarrow (m + n) (m + 4 n) = 0

Step 4: Solve for $m$ and $n$

From the above equation, we have two cases:

$m = - n$
$m = - 4 n$

For the first case

m = - n

, from Equation 2, we get

l + n = 0 \Rightarrow l = - n

Step 5: Equations of the Other Two Generators

Using the values of

l, m, n

, we find the equations of the other two generators:

When $m = - n$ , the equation becomes $\frac{x}{1} = \frac{y}{1} = \frac{z}{- 1}$
When $m = - 4 n$ , the equation becomes $\frac{x}{5} = \frac{y}{- 4} = \frac{z}{1}$

Conclusion

The equations of the other two mutually perpendicular generators of the cone are:

$\frac{x}{1} = \frac{y}{1} = \frac{z}{- 1}$
$\frac{x}{5} = \frac{y}{- 4} = \frac{z}{1}$

These generators are indeed perpendicular to each other and to the given generator, fulfilling the conditions of the problem.

4(a) माना

A = [\begin{array}{rrr} 1 & 0 & 2 \\ 2 & - 1 & 3 \\ 4 & 1 & 8 \end{array}] और B = [\begin{array}{rrr} - 11 & 2 & 2 \\ - 4 & 0 & 1 \\ 6 & - 1 & - 1 \end{array}]

(i)

A B

ज्ञात कीजिए।
(ii) सारणिक

(A)

व सारणिक

(B)

ज्ञात कीजिए।
(iii) निम्न रैखिक समीकरणों के निकाय का हल निकालिए :

x + 2 z = 3, 2 x - y + 3 z = 3, 4 x + y + 8 z = 14

Let

A = [\begin{array}{rrr} 1 & 0 & 2 \\ 2 & - 1 & 3 \\ 4 & 1 & 8 \end{array}] and B = [\begin{array}{rrr} - 11 & 2 & 2 \\ - 4 & 0 & 1 \\ 6 & - 1 & - 1 \end{array}]

(i) Find

A B

.
(ii) Find

\det (A)

and

\det (B)

.
(iii) Solve the following system of linear equations :

x + 2 z = 3, 2 x - y + 3 z = 3, 4 x + y + 8 z = 14

Answer:

Part (i): Find $A B$

To find the product

A B

, we’ll use the formula for matrix multiplication:

(A B)_{i j} = \sum_{k = 1}^{n} A_{i k} \times B_{k j}

Let’s substitute the values:

A B = [\begin{array}{rrr} 1 & 0 & 2 \\ 2 & - 1 & 3 \\ 4 & 1 & 8 \end{array}] \times [\begin{array}{rrr} - 11 & 2 & 2 \\ - 4 & 0 & 1 \\ 6 & - 1 & - 1 \end{array}]

After calculating, we get:

A B = [\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]

Part (ii): Find $det (A)$ and $det (B)$

To find the determinant of

A

and

B

, we’ll use the formula for a 3×3 matrix:

det (A) = a (e i - f h) - b (d i - f g) + c (d h - e g)

For

A

det (A) = 1 (- 1 \times 8 - 3 \times 1) - 0 + 2 (2 \times 1 - 4 \times - 1) = 1 (- 8 - 3) + 0 + 2 (2 + 4) = - 11 + 0 + 12 = 1

For

B

\begin{aligned} | B | = - 11 \times (0 \times (- 1) - 1 \times (- 1)) - 2 \times (- 4 \times (- 1) - 1 \times 6) + 2 \times (- 4 \times (- 1) - 0 \times 6) \\ | B | = - 11 \times (0 + 1) - 2 \times (4 - 6) + 2 \times (4 + 0) \\ | B | = - 11 \times (1) - 2 \times (- 2) + 2 \times (4) \\ | B | = - 11 + 4 + 8 \\ | B | = 1 \end{aligned}

Part (iii): Solve the system of linear equations

The system of equations is:

x + 2 z = 3, 2 x - y + 3 z = 3, 4 x + y + 8 z = 14

We can write this system as

A X = B

, where

A

is the coefficient matrix,

X

is the variable matrix, and

B

is the constant matrix.

A = [\begin{array}{rrr} 1 & 0 & 2 \\ 2 & - 1 & 3 \\ 4 & 1 & 8 \end{array}], X = [\begin{array}{r} x \\ y \\ z \end{array}], B = [\begin{array}{r} 3 \\ 3 \\ 14 \end{array}]

Converting given equations into matrix form

[\begin{array}{cccc} 1 & 0 & 2 & 3 \\ 2 & - 1 & 3 & 3 \\ 4 & 1 & 8 & 14 \end{array}]

\begin{aligned} R_{2} \leftarrow R_{2} - 2 \times R_{1} \\ = [\begin{array}{cccc} 1 & 0 & 2 & 3 \\ 0 & - 1 & - 1 & - 3 \\ 4 & 1 & 8 & 14 \end{array}] \end{aligned}

\begin{aligned} R_{3} \leftarrow R_{3} - 4 \times R_{1} \\ = [\begin{array}{cccc} 1 & 0 & 2 & 3 \\ 0 & - 1 & - 1 & - 3 \\ 0 & 1 & 0 & 2 \end{array}] \end{aligned}

\begin{aligned} R_{3} \leftarrow R_{3} + R_{2} \\ = [\begin{array}{rrrc} 1 & 0 & 2 & 3 \\ 0 & - 1 & - 1 & - 3 \\ 0 & 0 & - 1 & - 1 \end{array}] \end{aligned}

\begin{aligned} x + 2 z = 3 \to (1) \\ - y - z = - 3 \to (2) \\ - z = - 1 \to (3) \end{aligned}

Now use back substitution method From (3)

\begin{aligned} - z = - 1 \\ \Rightarrow z = 1 \end{aligned}

From (2)

\begin{aligned} - y - z = - 3 \\ \Rightarrow - y - (1) = - 3 \\ \Rightarrow - y - 1 = - 3 \\ \Rightarrow - y = - 3 + 1 \\ \Rightarrow - y = - 2 \\ \Rightarrow y = 2 \end{aligned}

From (1)

\begin{aligned} x + 2 z = 3 \\ \Rightarrow x + 2 (1) = 3 \\ \Rightarrow x + 2 = 3 \\ \Rightarrow x = 3 - 2 \\ \Rightarrow x = 1 \end{aligned}

Solution using back substitution method.

x = 1, y = 2 and z = 1

(b) अतिपरवलयिक परवलयज

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 2 z

के लांबिक जनकों के प्रतिच्छेद बिंदु का बिंदुपथ निकालिए।

Find the locus of the point of intersection of the perpendicular generators of the hyperbolic paraboloid

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 2 z

Answer:

To find the locus, we need to consider the enveloping cone of the given hyperbolic paraboloid with a vertex at the point

(α, β, γ)

The equation of the enveloping cone can be represented as

S S_{1} = T^{2}

, where:

S = \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} - 2 z

S_{1} = \frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} - 2 γ

T = \frac{α x}{a^{2}} - \frac{β y}{b^{2}} - (z + γ)

Substituting these into

S S_{1} = T^{2}

, we get:

(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} - 2 z) (\frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} - 2 γ) = {(\frac{α x}{a^{2}} - \frac{β y}{b^{2}} - z - γ)}^{2}

For the enveloping cone to have three mutually perpendicular generators, the sum of the coefficients of

x^{2}, y^{2},

and

z^{2}

must be zero. That is:

[\frac{1}{a^{2}} (\frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} - 2 γ) - \frac{α^{2}}{a^{4}}] + [- \frac{1}{b^{2}} (\frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} - 2 γ) - \frac{β^{2}}{b^{4}}] + [- 1] = 0

Simplifying, we find:

α^{2} + β^{2} - 2 γ (a^{2} - b^{2}) + a^{2} b^{2} = 0

Conclusion

The locus of the point

(α, β, γ)

where the perpendicular generators of the hyperbolic paraboloid intersect is given by:

x^{2} + y^{2} - 2 (a^{2} - b^{2}) z + a^{2} b^{2} = 0

This equation satisfies all the conditions and constraints of the problem.

u = x^{2} + y^{2} + z^{2}

का चरम मान ज्ञात कीजिए, जो

2 x + 3 y + 5 z = 30

शर्त द्वारा प्रतिबंधित है।

Find an extreme value of the function

u = x^{2} + y^{2} + z^{2}

, subject to the condition

2 x + 3 y + 5 z = 30

, by using Lagrange’s method of undetermined multiplier.

Answer:

We start by defining the Lagrangian function

F

as follows:

F = x^{2} + y^{2} + z^{2} + λ (2 x + 3 y + 5 z - 30)

To find the extreme values, we set

d F = 0

. The differential

d F

is given by:

d F = (2 x + 2 λ) d x + (2 y + 3 λ) d y + (2 z + 5 λ) d z

From this, we get the following equations:

$2 x + 2 λ = 0$ (Equation 1)
$2 y + 3 λ = 0$ (Equation 2)
$2 z + 5 λ = 0$ (Equation 3)

Multiplying Equation 1 by

x

, Equation 2 by

y

, and Equation 3 by

z

, and then adding them, we get:

2 (x^{2} + y^{2} + z^{2}) + λ (2 x + 3 y + 5 z) = 0

Since

2 x + 3 y + 5 z = 30

, we can rewrite the above equation as:

U = - 15 λ (Equation 4)

Next, we solve for

x, y,

and

z

in terms of

λ

x = - \frac{2 λ}{2}, y = - \frac{3 λ}{2}, z = - \frac{5 λ}{2}

Substituting these into the constraint

2 x + 3 y + 5 z = 30

, we get:

- 4 λ - 9 λ - 25 λ = 60

- 38 λ = 60 \Rightarrow λ = - \frac{30}{19}

Using Equation 4, we find the extreme value

U

U_{extreme} = - 15 (- \frac{30}{19}) = \frac{450}{19}

Finally, we check the second derivative

d^{2} F

d^{2} F = 2 (d x^{2} + d y^{2} + d z^{2}) > 0

This confirms that the extreme value is a minimum.

Conclusion

The minimum value of

U

subject to the given constraint is

U_{min} = \frac{450}{19}

खण्ड-B / SECTION-B

5(a) निम्न अवकल समीकरण को हल कीजिए :

x \cos (\frac{y}{x}) (y d x + x d y) = y \sin (\frac{y}{x}) (x d y - y d x)

Solve the following differential equation :

x \cos (\frac{y}{x}) (y d x + x d y) = y \sin (\frac{y}{x}) (x d y - y d x)

Answer:

We start with the given differential equation:

x \cos (\frac{y}{x}) (y d x + x d y) = y \sin (\frac{y}{x}) (x d y - y d x) (Equation 1)

Rewriting Equation 1, we get:

(x \cos \frac{y}{x} + y \sin \frac{y}{x}) y - (y \sin \frac{y}{x} - x \cos \frac{y}{x}) x \frac{d y}{d x} = 0

Simplifying, we find:

\frac{d y}{d x} = \frac{{x \cos \frac{y}{x} + y \sin \frac{y}{x}} y}{{y \sin \frac{y}{x} - x \cos \frac{y}{x}} x} (Equation 2)

Further simplification yields:

\frac{d y}{d x} = \frac{[\cos \frac{y}{x} + \frac{y}{x} \sin \frac{y}{x}] (\frac{y}{x})}{[\frac{y}{x} \sin \frac{y}{x} - \cos \frac{y}{x}]} (Equation 3)

Let’s make a substitution:

\frac{y}{x} = v \Rightarrow y = v x

. Differentiating, we get:

\frac{d y}{d x} = v + x \frac{d v}{d x} (Equation 4)

Using Equations 3 and 4, we find:

\begin{aligned} v + x \frac{d v}{d x} & = \frac{v (\cos v + v \sin v)}{v \sin v - \cos v} \\ \Rightarrow x \frac{d v}{d x} & = \frac{v (\cos v + v \sin v)}{v \sin v - \cos v} - v = \frac{2 v \cos v}{v \sin v - \cos v} \end{aligned}

x \frac{d v}{d x} = \frac{2 v \cos v}{v \sin v - \cos v}

2 \frac{d x}{x} = \frac{v \sin v - \cos v}{v \cos v} d v = [\frac{\sin v}{\cos v} - \frac{1}{v}] d v

Integrating both sides, we get:

2 \log x = - \log \cos v - \log v + \log c

\log x^{2} = \log (\frac{c}{v} \cos v)

Simplifying, we find:

x^{2} v \cos v = c

Finally, substituting back

v = \frac{y}{x}

, we get:

x y \cos \frac{y}{x} = c

Or equivalently,

x y = c \sec (\frac{y}{x})

Conclusion

The solution to the given differential equation is

x y = c \sec (\frac{y}{x})

(b) वृत्त-कुल, जो बिंदु

(0, 2)

एवं

(0, - 2)

से गुजरता है, का लंबकोणीय संछेदी ज्ञात कीजिए।

Find the orthogonal trajectories of the family of circles passing through the points

(0, 2)

and

(0, - 2)

Answer:

We start with the general equation for a family of circles:

x^{2} + y^{2} + 2 g x + 2 f y + d = 0 (Equation 1)

Given that the circle passes through

(0, 2)

and

(0, - 2)

, we can substitute these points into Equation 1 to get:

4 + 4 f + d = 0 (Equation 2, from (0,2)) 4 - 4 f + d = 0 (Equation 3, from (0,-2))

Solving Equations 2 and 3, we find

f = 0

and

d = - 4

. Substituting these into Equation 1, we get:

x^{2} + y^{2} + 2 g x - 4 = 0 (Equation 4)

Differentiating Equation 4 with respect to

x

, we get:

2 x + 2 y (\frac{d y}{d x}) + 2 g = 0

Simplifying, we find:

g = - [x + y (\frac{d y}{d x})] (Equation 5)

Substituting the value of

g

into Equation 4, we get:

x^{2} + y^{2} - 2 x [x + y (\frac{d y}{d x})] - 4 = 0 (Equation 6)

Equation 6 is the differential equation for the family of circles.

For orthogonal trajectories, we replace

\frac{d y}{d x}

with

- \frac{d x}{d y}

. Doing so, we get:

\begin{aligned} Then we get x^{2} + y^{2} - 2 x [x - y (\frac{d x}{d y})] - 4 = 0 \\ \Rightarrow x^{2} + y^{2} - 4 = x - y (\frac{d x}{d y}) \\ \frac{x^{2} + y^{2} - 4}{2 x} - x = - y (\frac{d x}{d y}) \\ \Rightarrow \frac{- x^{2} + y^{2} - 4}{2 x} = - y (\frac{d x}{d y}) \\ \frac{x^{2} - y^{2} + 4}{2 x y} = (\frac{d x}{d y}) \\ \Rightarrow 2 x y d x - (x^{2} - y^{2} + 4) d y = 0 \end{aligned}

2 x y d x + (y^{2} - x^{2} - 4) d y = 0 (Equation 7)

Where

M = 2

xy and

N = y^{2} - x^{2} - 4

\frac{\partial M}{\partial y} = 2 x, \frac{\partial N}{\partial x} = - 2 x

This equation is not exact, but we can make it exact by multiplying by an integrating factor

I F

, which is

\frac{1}{y^{2}}

. Doing so, we get an exact equation:

\begin{aligned} \frac{1}{M} (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) = \frac{1}{2 x y} (- 2 x - 2 x) \\ = - \frac{4 x}{2 x y} \\ = - \frac{2}{y} \\ I . F = e^{- \int \frac{2}{y} d y} = e^{- 2 \log y} = e^{\log y^{- 2}} = \frac{1}{y^{2}} \end{aligned}

From (7)

\frac{2 x y}{y} d x + \frac{y^{2} - x^{2} - y}{y^{2}} d y = 0

\frac{2 x}{y} d x + (1 - \frac{4}{y^{2}}) d y = 0

Which is an exact equation

\int \frac{2 x}{y} d x + (1 - \frac{4}{y^{2}}) d y = 0

Which is an exact equation

\begin{aligned} \int \frac{2 x}{y} d x + \int 1 - \frac{4}{y^{2}} d y = c \\ \frac{x^{2}}{y} + y + \frac{4}{y} = c \end{aligned}

\frac{x^{2}}{y} + (y + \frac{4}{y}) = c

Conclusion

The orthogonal trajectories of the family of circles that pass through the points

(0, 2)

and

(0, - 2)

are described by the equation

\frac{x^{2}}{y} + (y + \frac{4}{y}) = c

(c)

a, b, c

के किस मान के लिए सदिश क्षेत्र

\bar{V} = (- 4 x - 3 y + a z) \hat{i} + (b x + 3 y + 5 z) \hat{j} + (4 x + c y + 3 z) \hat{k}

अघूर्णी है? तब

\bar{V}

को अदिश फलन

ϕ

की प्रवणता के रूप में व्यक्त कीजिए।

ϕ

को ज्ञात कीजिए।

For what value of

a, b, c

is the vector field

\bar{V} = (- 4 x - 3 y + a z) \hat{i} + (b x + 3 y + 5 z) \hat{j} + (4 x + c y + 3 z) \hat{k}

irrotational? Hence, express

\bar{V}

as the gradient of a scalar function

ϕ

. Determine

ϕ

Answer:

Introduction:
We need to determine the values of

a

b

, and

c

for which the vector field

\vec{V}

is irrotational. Additionally, we want to express

\vec{V}

as the gradient of a scalar function

ϕ

and find the expression for

ϕ

Work/Calculations:
A vector field

\vec{v}

is irrotational if its curl,

\nabla \times \vec{v}

, is equal to zero. The curl of

\vec{V}

is calculated as follows:

\nabla \times \vec{V} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ - 4 x - 3 y + a z & b x + 3 y + 5 z & 4 x + c y + 3 z \end{matrix} | = i (c - 5) - j (4 - a) + k (b + 3)

For

\vec{V}

to be irrotational, this expression must equal zero. Therefore, we have:

c - 5 = 0 ⟹ c = 5

4 - a = 0 ⟹ a = 4

b + 3 = 0 ⟹ b = - 3

To express

\vec{V}

as the gradient of a scalar function

ϕ

, we have:

\vec{V} = \nabla ϕ

Now, we have our

\vec{V}

as:

\vec{V} = (- 4 x - 3 y + 4 z) i + (9 - 3 x + 3 y + 5 z) j + (4 x + 5 y + 3 z) k = \frac{\partial ϕ}{\partial x} i + \frac{\partial ϕ}{\partial y} j + \frac{\partial ϕ}{\partial z} k (1)

Equating components:

\frac{\partial ϕ}{\partial x} = - 4 x - 3 y + 4 z ⟹ ϕ = - 2 x^{2} - 3 x y + 4 x z + f_{1} (y, z) (2)

\frac{\partial ϕ}{\partial y} = - 3 x + 3 y + 5 z ⟹ ϕ = - 3 x y + \frac{3}{2} y^{2} + 5 y z + f_{2} (x, y) (3)

\frac{\partial ϕ}{\partial z} = 4 x + 5 y + 3 z ⟹ ϕ = 4 x z + 5 y z + \frac{3}{2} z^{2} + f_{3} (x, y) (4)

These expressions for

ϕ

can be equated since they represent the same function

ϕ

. This gives us:

f_{1} (y, z) = \frac{3}{2} y^{2} + \frac{3}{2} z^{2} + 5 y z

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

f_{3} (x, y) = - 2 x^{2} + \frac{3}{2} y^{2} - 3 x y

So, our final expression for

ϕ

is:

ϕ = - 2 x^{2} + \frac{3}{2} y^{2} + \frac{3}{2} z^{2} - 3 x y + 4 x z + 5 y z + c

Where

c

is an arbitrary constant.

(d) एक एकसमान छड़, जो ऊर्ध्वर्धर दशा में है, अपने एक सिरे पर स्वतंत्र रूप से वर्तन कर सकती है तथा दूसरे सिरे पर लगाए गए एक क्षैतिज बल, जिसका मान छड़ के भार का आधा है, द्वारा ऊर्ध्वाधर से एक तरफ खींची जाती है। बताइए कि ऊर्ध्वाधर से किस कोण पर छड़ विश्राम करेगी।

A uniform rod, in vertical position, can turn freely about one of its ends and is pulled aside from the vertical by a horizontal force acting at the other end of the rod and equal to half its weight. At what inclination to the vertical will the rod rest?

Answer:

Forces in the System

The forces acting on the system are:

The weight $W$ of the rod $A B$ , acting vertically downwards.
The force $P$ , which is $\frac{W}{2}$ , pulling the rod aside from the vertical.
The reaction $R$ at the end $A$ of the rod.

In equilibrium, these forces must intersect at a single point. Therefore, the reaction

R

should pass through

O

, the point of intersection of the directions of the forces

W

and

\frac{W}{2}

Geometry and Trigonometry

Let the length of the rod be

2 l

, so

A G = B G = l

△ A B C

, we have:

\cos θ = \frac{A C}{2 l} ⟹ A C = 2 l \cos θ

△ A O^{'} G

, we have:

\sin θ = \frac{A O^{'}}{l} ⟹ A O^{'} = l \sin θ

Equilibrium Condition

For the rod to be in equilibrium, the algebraic sum of the moments about point

A

must be zero. Therefore, we have:

P \cdot A C - W \cdot A O^{'} = 0

Substituting

P = \frac{W}{2}

and the expressions for

A C

and

A O^{'}

, we get:

\frac{W}{2} \cdot 2 l \cos θ - W \cdot l \sin θ = 0

Simplifying, we find:

\cos θ = \sin θ ⟹ \tan θ = 1

θ = \tan^{- 1} (1) = 45^{\circ}

Conclusion

The rod will come to rest at an inclination of

45^{\circ}

to the vertical.

(e) एक हल्की दृढ़ छड़

A B C

से तीन कण, जिनमें से हरेक का द्रव्यमान

m

है,

A, B

तथा

C

पर बंधे हुए हैं। उस छड़ को बिंदु

A

से

B C

दूरी के बराबर स्थित बिंदु पर एक बल

P

के द्वारा लम्बवत् मारा जाता है। सिद्ध कीजिए कि पैदा हुई गतिज ऊर्जा का मान

\frac{1}{2} \frac{p^{2}}{m} \frac{a^{2} - a b + b^{2}}{a^{2} + a b + b^{2}}

है, जहाँ

A B = a

तथा

B C = b

A light rigid rod

A B C

has three particles each of mass

m

attached to it at

A, B

and

C

. The rod is struck by a blow

P

at right angles to it at a point distant from

A

equal to

B C

. Prove that the kinetic energy set up is

\frac{1}{2} \frac{p^{2}}{m} \frac{a^{2} - a b + b^{2}}{a^{2} + a b + b^{2}}

, where

A B = a

and

B C = b

Answer:

Let the three particles each of mass

m

be placed at

A, B, C

of a light rod

A B C

, and let the impulse

P

be applied at

O

such that

A O = B C = b

, where

A B = a, B C = b

Initial Conditions and Variables

$A B = a$
$B C = b$
$A O = b$
$u^{'}$ is the velocity of $C$
$ω$ is the angular velocity of the rod just after the blow

Velocities of Points

The velocities of the points

A, B, C

and

O

just after the blow are:

Velocity of $C$ : $u^{'}$
Velocity of $B$ : $u + b ω$
Velocity of $A$ : $u + (a + b) ω$
Velocity of $O$ : $u + a ω$

Conservation of Momentum

Since the system was initially at rest, the total momentum perpendicular to the rod must be equal to the impulse

P

m u + m (u + b ω) + m {u + (a + b) ω} = P

3 u + ω (a + 2 b) = \frac{P}{m} (1)

Moments About Point $O$

Taking moments about

O

, we get:

\begin{aligned} {u + (a + b)} b - m (u + b ω) (a - b) - m u a = 0 \Rightarrow u (a - b) = b^{2} ω . \\ ∴ \frac{u}{b^{2}} = \frac{ω}{a - b} \Rightarrow \frac{3 u + (a + 2 b) ω}{3 b^{2} + (a + 2 b) (a - b)} = \frac{(P / m)}{(a^{2} + a b + b^{2})} \end{aligned}

Solving for $u$ and $ω$

From equations (1) and (2), we can solve for

u

and

ω

u = \frac{b^{2}}{a^{2} + a b + b^{2}} \frac{P}{m} and ω = \frac{(a - b)}{a^{2} + a b + b^{2}} \frac{P}{m}

Velocity of Point $O$

The velocity of point

O

is:

u + a ω = \frac{P}{m} \frac{1}{m (a^{2} + b^{2} + a b)} [b^{2} + a (a - b)] = \frac{a^{2} + b^{2} - a b}{a^{2} + b^{2} + a b} \cdot \frac{P}{m}

Kinetic Energy

Finally, the kinetic energy

K E

set up in the system is:

K E = \frac{1}{2} P \times (velocity of point O) = \frac{1}{2} \frac{P^{2} (a^{2} + b^{2} - a b)}{m (a^{2} + a b + b^{2})}

Conclusion

The kinetic energy set up in the system is

\frac{1}{2} \frac{p^{2}}{m} \frac{a^{2} - a b + b^{2}}{a^{2} + a b + b^{2}}

, as required.

(a) प्राचल विचरण विधि का प्रयोग करके, निम्न अवकल समीकरण का हल निकालिए, यदि $y = e^{- x}$ , पूरक फलन (CF) का एक हल है :

y^{''} + (1 - \cot x) y^{'} - y \cot x = \sin^{2} x

Using the method of variation of parameters, solve the differential equation

y^{''} + (1 - \cot x) y^{'} - y \cot x = \sin^{2} x

, if

y = e^{- x}

is one solution of CF.

Answer:

Introduction:
We are tasked with solving the differential equation

y^{''} + (1 - \cot x) y^{'} - y \cot x = \sin^{2} x

, given that

y = e^{- x}

is one of its complementary function (CF) solutions. We’ll use the method of variation of parameters to find the particular solution.

Work/Calculations:
Let’s begin by considering the homogeneous form of the equation:

y^{''} + (1 - \cot x) y^{'} - \cot x \cdot y = 0 (2)

Comparing this with the standard form

y^{''} + P y^{'} + Q y = 0

, we have

P = 1 - \cot x

and

Q = - \cot x

. We also observe that

1 - P + Q = 1 - (1 - \cot x) - \cot x = 0

, which indicates that

u = e^{- x}

is a part of the CF of equation (2).

Now, we assume the complete solution of (2) as

y = u \cdot v

. To find

v

, we use the formula for the second derivative of a product:

\frac{d^{2} v}{d x^{2}} + (P + \frac{2}{u} \frac{d u}{d x}) \frac{d v}{d x} = \frac{R}{y}

In our case,

P = 1 - \cot x

and

u = e^{- x}

, so:

\frac{d^{2} v}{d x^{2}} + (1 - \cot x + \frac{2}{e^{- x}} (- e^{- x})) \frac{d v}{d x} = 0

This simplifies to:

\frac{d^{2} v}{d x^{2}} - (1 + \cot x) \frac{d v}{d x} = 0 (3)

Let

\frac{d v}{d x} = q

, which transforms equation (3) into:

\frac{d q}{d x} - (1 + \cot x) q = 0

Solving this first-order linear differential equation:

\frac{1}{q} d q = (1 + \cot x) d x

By Integrating

\begin{aligned} \log q - \log c_{1} = x + \log \sin x \\ \Rightarrow q = c_{1} e^{x} \sin x \\ \frac{d q}{d x} = c_{1} \int e^{x} \sin x d x + c_{2} \\ v = \frac{1}{2} c_{1} e^{x} (\sin x - \cos x) + c_{2} \\ v = c_{1}^{'} e^{x} (\sin x - \cos x) + c_{2} where c_{1}^{'} = \frac{c_{1}}{2} \end{aligned}

Therefore, C. F of (2) is given by

\begin{aligned} y_{c} = e^{- x} {c_{1}^{'} e^{x} (\sin x - \cos x) + c_{2}} \\ y_{c} = c_{1}^{'} (\sin x - \cos x) + c_{2} e^{- x} \end{aligned}

Particular integral of (1)

y_{p} = U f (x) + V g (x) \to

(4)

Next, we proceed to find the particular integral

y_{p}

for the original equation (1). We’ve already determined

U

and

V

\sin x - \cos x

and

e^{- x}

, respectively. Now, we need to calculate the Wronskian

W

W = | \begin{matrix} \sin x - \cos x & e^{- x} \\ \cos x + \sin x & - e^{- x} \end{matrix} | = - 2 e^{- x} \sin x

We use

W

to find

f (x)

\begin{aligned} Where f (x) = - \int \frac{V}{W} d x = - \int \frac{e^{- x} \sin^{2} x}{- 2 e^{- x} \sin x} d x (from R = \sin^{2} x) \\ = \frac{1}{2} \int \sin^{2} x d x = - \frac{1}{2} \cos x \\ g (x) = \int \frac{U R}{W} = \int \frac{(\sin x - \cos x) \sin^{2} x}{- 2 e^{- x} \sin x} d x \\ = - \frac{1}{2} \int e^{x} (\sin x - \cos x) \sin x d x \\ = - \frac{1}{2} \int e^{x} (\sin^{2} x - \cos x \sin x) d x \\ = - \frac{1}{2} [\int e^{x} (\frac{1 - \cos 2 x}{2}) d x - \int e^{x} \frac{\sin 2 x}{2} d x] \\ = - \frac{1}{4} [e^{x} - \int e^{x} \cos 2 x d x - \int e^{x} \sin 2 x d x] \\ = - \frac{1}{4} [e^{x} - \frac{e^{x}}{5} (\cos 2 x + 2 \sin 2 x) - \frac{e^{x}}{5} (\sin x - 2 \cos 2 x)] \\ = - \frac{1}{4} e^{x} [1 + \frac{1}{5} (\cos 2 x - 3 \sin 2 x)] \end{aligned}

Now, let’s find

y_{p}

from equation (4):

\begin{aligned} y_{p} = - \frac{1}{2} \cos x (\sin x - \cos x) + e^{- x} [- \frac{1}{4} e^{x} (1 + \frac{1}{5} (\cos 2 x - 3 \sin 2 x))] \\ = - \frac{1}{2} \sin x \cos x + \frac{1}{2} \cos^{2} x - \frac{1}{4} - \frac{1}{20} (\cos 2 x - 3 \sin 2 x) \\ = - \frac{1}{4} \sin 2 x + \frac{1}{4} (1 + \cos 2 x) - \frac{1}{4} - \frac{1}{20} (\cos 2 x - 3 \sin 2 x) \\ = \frac{1}{20} (4 \cos 2 x - 2 \sin 2 x) \\ = \frac{1}{10} (2 \cos 2 x - \sin 2 x) \\ y = y_{c} + y_{p} \\ y = c_{1} (\sin x - \cos x) + c_{2} e^{- x} + \frac{1}{10} (2 \cos 2 x - \sin 2 x) \end{aligned}

Which is the required solution of given differential equation.

(b) दिए गए सदिश फलन

\bar{A}

, जहाँ

\bar{A} = (3 x^{2} + 6 y) \hat{i} - 14 y z \hat{j} + 20 x z^{2} \hat{k}

, के लिए

\int_{C} \bar{A} \cdot d \bar{r}

का मान निकालिए, जहाँ

C

बिंदु

(0, 0, 0)

से

(1, 1, 1)

तक निम्न पर्थों से निर्देशित है :
(i)

x = t, y = t^{2}, z = t^{3}

(ii) सरल रेखा

(0, 0, 0)

से

(1, 0, 0)

तक जोड़ने पर, फिर

(1, 1, 0)

तक तथा फिर

(1, 1, 1)

तक
(iii) सरल रेखा

(0, 0, 0)

से

(1, 1, 1)

तक जोड़ने पर
क्या सभी स्थितियों में परिणाम समान हैं? कारण की व्याख्या कीजिए।

For the vector function

\bar{A}

, where

\bar{A} = (3 x^{2} + 6 y) \hat{i} - 14 y z \hat{j} + 20 x z^{2} \hat{k}

, calculate

\int_{C} \bar{A} \cdot d \bar{r}

from

(0, 0, 0)

(1, 1, 1)

along the following paths :
(i)

x = t, y = t^{2}, z = t^{3}

(ii) Straight lines joining

(0, 0, 0)

(1, 0, 0)

, then to

(1, 1, 0)

and then to

(1, 1, 1)

(iii) Straight line joining

(0, 0, 0)

(1, 1, 1)

Is the result same in all the cases? Explain the reason.

Answer:

Part (i): Path $x = t, y = t^{2}, z = t^{3}$

Vector $d \bar{r}$

For the curve

C

, we have

d \bar{r} = \frac{d \bar{r}}{d t} d t = (i + 2 t j + 3 t^{2} k) d t

Limits of Integration

The points

(0, 0, 0)

and

(1, 1, 1)

correspond to

t = 0

and

t = 1

, respectively.

Integral Calculation

\begin{aligned} \int_{C} \bar{A} \cdot d \bar{r} & = \int_{0}^{1} [(3 t^{2} + 6 t^{2}) i - 14 t^{2} t^{3} j + 20 t (t^{3})^{2} k] \cdot (i + 2 t j + 3 t^{2} k) d t \\ = \int_{0}^{1} (9 t^{2} - 28 t^{6} + 60 t^{9}) d t \\ = {[3 t^{3} - 4 t^{7} + 6 t^{10}]}_{0}^{1} \\ = 3 - 4 + 6 \\ = 5 \end{aligned}

The integral along this path is 5.

Part (ii): Straight Lines Joining Points

Path 1: $(0, 0, 0)$ to $(1, 0, 0)$

For this path,

x = t, y = 0, z = 0

and

0 \leq t \leq 1

\begin{aligned} \int_{C 1} \bar{A} \cdot d \bar{r} & = \int_{0}^{1} (3 t^{2} + 6 \cdot 0) d t \\ = \int_{0}^{1} 3 t^{2} d t \\ = {[t^{3}]}_{0}^{1} \\ = 1 - 0 \\ = 1 \end{aligned}

Path 2: $(1, 0, 0)$ to $(1, 1, 0)$

For this path,

x = 1, y = t, z = 0

and

0 \leq t \leq 1

\begin{aligned} \int_{C 2} \bar{A} \cdot d \bar{r} & = \int_{0}^{1} (3 \cdot 1^{2} + 6 t) d t \\ = \int_{0}^{1} (3 + 6 t) d t \\ = {[3 t + 3 t^{2}]}_{0}^{1} \\ = 3 + 3 \\ = 6 \end{aligned}

Path 3: $(1, 1, 0)$ to $(1, 1, 1)$

For this path,

x = 1, y = 1, z = t

and

0 \leq t \leq 1

\begin{aligned} \int_{C 3} \bar{A} \cdot d \bar{r} & = \int_{0}^{1} (3 \cdot 1^{2} + 6 \cdot 1) - 14 \cdot 1 \cdot t + 20 \cdot 1 \cdot t^{2} d t \\ = \int_{0}^{1} (9 - 14 t + 20 t^{2}) d t \\ = {[9 t - 7 t^{2} + \frac{20}{3} t^{3}]}_{0}^{1} \\ = 9 - 7 + \frac{20}{3} \\ = 2 + \frac{20}{3} \\ = \frac{6}{3} + \frac{20}{3} \\ = \frac{26}{3} \end{aligned}

Total Integral for Part (ii)

\int_{C} \bar{A} \cdot d \bar{r} = 1 + 6 + \frac{26}{3} = 7 + \frac{26}{3} = \frac{21}{3} + \frac{26}{3} = \frac{47}{3}

Part (iii): Straight Line from $(0, 0, 0)$ to $(1, 1, 1)$

For this path,

x = t, y = t, z = t

and

0 \leq t \leq 1

The vector field

\bar{A} = (3 x^{2} + 6 y) \hat{i} - 14 y z \hat{j} + 20 x z^{2} \hat{k}

The differential path element

d \bar{r} = d t \hat{i} + d t \hat{j} + d t \hat{k}

\begin{aligned} \int_{C} \bar{A} \cdot d \bar{r} & = \int_{0}^{1} [(3 t^{2} + 6 t) \hat{i} - 14 t^{2} \hat{j} + 20 t^{3} \hat{k}] \cdot (d t \hat{i} + d t \hat{j} + d t \hat{k}) \\ = \int_{0}^{1} (3 t^{2} + 6 t) d t - \int_{0}^{1} 14 t^{2} d t + \int_{0}^{1} 20 t^{3} d t \\ = {[\frac{3}{3} t^{3} + 3 t^{2}]}_{0}^{1} - {[\frac{14}{3} t^{3}]}_{0}^{1} + {[\frac{20}{4} t^{4}]}_{0}^{1} \\ = 1 + 3 - \frac{14}{3} + 5 \\ = 4 - \frac{14}{3} + 5 \\ = 9 - \frac{14}{3} \\ = \frac{27}{3} - \frac{14}{3} \\ = \frac{13}{3} \end{aligned}

Summary

For part (i), the integral is

5

For part (ii), the integral is

\frac{47}{3}

For part (iii), the integral is

\frac{13}{3}

The result is not the same for all paths, which indicates that the vector field

\bar{A}

is not conservative. In a conservative vector field, the line integral is path-independent, meaning it would be the same for all paths between two points. Here, that is not the case.

A D

दो आलंब

B

एवं

C

पर विश्राम करता है, जबकि

A B = B C = C D

. यह पाया गया कि दंड झुक जाएगा यदि एक भार

p kg

, बिंदु

A

से लटकाया जाए या एक भार

q kg

, बिंदु

D

से लटकाया जाए। दंड का भार बताइए।

A beam

A D

rests on two supports

B

and

C

, where

A B = B C = C D

. It is found that the beam will tilt when a weight of

p kg

is hung from

A

or when a weight of

q kg

is hung from

D

. Find the weight of the beam.

Answer:

Given Information

Lengths of the segments: $A B = B C = C D = a$
Weight hung from $A$ : $p$ kg
Weight hung from $D$ : $q$ kg

Diagram

Let

G

be the center of gravity (C.G.) of the beam, where its weight

W

acts. Let

B G = x

Case 1: Weight $p$ is hung from $A$

When a weight

p

is hung from

A

, the beam tilts about

B

, and the contact at

C

is just about to break. Therefore, the torque due to

p

must be balanced by the torque due to

W

p \cdot a = W \cdot x (Equation 1)

Case 2: Weight $q$ is hung from $D$

When a weight

q

is hung from

D

, the beam tilts about

C

, and the contact at

B

is just about to break. Therefore, the torque due to

q

must be balanced by the torque due to

W

q \cdot a = W \cdot (a - x) (Equation 2)

Finding the Weight of the Beam

Adding Equation 1 and Equation 2:

(p + q) \cdot a = W \cdot a

Simplifying, we find:

W = p + q

Conclusion

The weight of the beam is

W = p + q

kg.

(a) स्टोक्स प्रमेय को सत्यापित कीजिए, जबकि सदिश क्षेत्र $\bar{F} = x y \hat{i} + y z \hat{j} + x z \hat{k}$ एक सतह $S$ पर है जो कि एक बेलन $z = 1 - x^{2}, 0 \leq x \leq 1, - 2 \leq y \leq 2$ का हिस्सा है, जहाँ $S$ उपरिमुखी अभिविन्यस्त है।

Verify the Stokes’ theorem for the vector field

\bar{F} = x y \hat{i} + y z \hat{j} + x z \hat{k}

on the surface

S

which is the part of the cylinder

z = 1 - x^{2}

for

0 \leq x \leq 1, - 2 \leq y \leq 2

;

S

is oriented upwards.

Answer:

Stokes’ Theorem

Stokes’ theorem states that

∮_{C} \vec{F} \cdot d \vec{r} = \iint_{S} (\nabla \times \vec{F}) \cdot \hat{n} d S

Line Integral Calculation

The curve

C

is the boundary of the surface

S

, and it consists of four segments:

c_{1}, c_{2}, c_{3}, c_{4}

Along $c_{1}$ :

The curve

c_{1}

is defined by

x^{2} + z = 1

and

y = - 2

. The differential

d y = 0

∮_{c_{1}} \vec{F} \cdot d \vec{r} = \int_{0}^{1} (- 2 x d x + x (1 - x^{2}) (- 2 x d x)) = - \frac{19}{15}

Along $c_{2}$ :

The curve

c_{2}

is defined by

z = 0

x = 1

, and

y

varies from

- 2

2

. The differentials

d x = 0

and

d z = 0

∮_{c_{2}} \vec{F} \cdot d \vec{r} = \int_{- 2}^{2} 0 d y = 0

Along $c_{3}$ :

The curve

c_{3}

is defined by

x^{2} + z = 1

and

y = 2

. The differential

d y = 0

∮_{c_{3}} \vec{F} \cdot d \vec{r} = \int_{0}^{1} (2 x d x + x (1 - x^{2}) (- 2 x d x)) = - \frac{11}{15}

Along $c_{4}$ :

The curve

c_{4}

is defined by

z = 1

x = 0

, and

y

varies from

- 2

2

. The differentials

d x = 0

and

d z = 0

∮_{c_{4}} \vec{F} \cdot d \vec{r} = \int_{- 2}^{2} 0 d y = 0

Adding all these together, we get:

∮_{C} \vec{F} \cdot d \vec{r} = - \frac{19}{15} + 0 + - \frac{11}{15} + 0 = - \frac{30}{15} = - 2

Surface Integral Calculation

First, let’s find

\nabla \times \vec{F}

\nabla \times \vec{F} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x y & y z & x z \end{array} | = - (y \hat{i} + z \hat{j} + x \hat{k})

The unit normal vector to the surface

S

is given by

\hat{n} = \frac{\nabla ϕ}{| \nabla ϕ |} = \frac{(2 x \hat{i} + \hat{k})}{\sqrt{4 x^{2} + 1}}

The differential area element

d S

in this case is

d S = \frac{d x d y}{| \hat{n} \cdot \hat{k} |} = \frac{d x d y}{\sqrt{4 x^{2} + 1}}

Now, let’s find the surface integral:

\iint_{S} (\nabla \times \vec{F}) \cdot \hat{n} d S

= \iint_{S} - (y \hat{i} + z \hat{j} + x \hat{k}) \cdot \frac{(2 x \hat{i} + \hat{k})}{\sqrt{4 x^{2} + 1}} \times \frac{d x d y}{\sqrt{4 x^{2} + 1}}

After Calculating, we get:

\iint_{S} (- 2 x y - x) \frac{d x d y}{4 x^{2} + 1}

The limits for

x

are from 0 to 1, and for

y

are from -2 to 2.

z = 1 - x^{2}

After Calculating, we get:

\iint_{S} (- 2 x y - x) \frac{d x d y}{4 x^{2} + 1} = 0

Conclusion

The line integral is

- 2

and the surface integral is

0

. Since these two values are not equal, Stokes’ theorem is not verified for this particular vector field and surface.

This could be due to the specific conditions or constraints of the surface and vector field in question.

(b) लाप्लास रूपांतरण का प्रयोग करके प्रारंभिक मान समस्या

t y^{''} + 2 t y^{'} + 2 y = 2; y (0) = 1

तथा

y^{'} (0)

स्वेच्छ है, को हल कीजिए। क्या इस प्रश्न का हल अद्वितीय है?

Using Laplace transform, solve the initial value problem

t y^{''} + 2 t y^{'} + 2 y = 2

;

y (0) = 1

and

y^{'} (0)

is arbitrary. Does this problem have a unique solution?

Answer:

Using Laplace Transform to Solve the Initial Value Problem

Step 1: The Initial Differential Equation

Given the differential equation:

t y^{''} + 2 t y^{'} + 2 y = 2

We’ll use the Laplace transform to solve it.

Step 2: Applying the Laplace Transform Property

We can use the property of Laplace transforms:

L {t f (t)} = - \frac{d}{d s} [L {f (t)}]

Now apply the Laplace transform to the given equation:

L {t y^{''} + 2 t y^{'} + 2 y} = L (2)

Step 3: Laplace Transform Calculation

Using the Laplace transform property and the initial conditions, we get:

- \frac{d}{d s} [s^{2} \bar{y} - s y (0) - y^{'} (0)] - 2 \frac{d}{d s} [s \bar{y} - y (0)] + 2 \bar{y} = \frac{2}{s}

Given

y (0)

and that

y^{'} (0)

is an arbitrary constant

A

- \frac{d}{d s} [s^{2} \bar{y} - s (1) - A] - 2 \frac{d}{d s} [s \bar{y} - 1] + 2 \bar{y} = \frac{2}{s}

Step 4: Simplification

Simplify the equation:

- s^{2} \frac{d \bar{y}}{d s} - 2 s \bar{y} + 1 - 2 \bar{y} - 2 s \frac{d \bar{y}}{d s} + 2 \bar{y} = \frac{2}{s}

- s^{2} \frac{d \bar{y}}{d s} - 2 s \bar{y} - 2 s \frac{d \bar{y}}{d s} = \frac{2}{s} - 1

- (s^{2} + 2 s) \frac{d \bar{y}}{d s} - 2 s \bar{y} = \frac{2}{s} - 1

\frac{d \bar{y}}{d s} + \frac{2 s \bar{y}}{s (s + 2)} = (1 - \frac{2}{s}) \frac{1}{s (s + 2)} \to (1)

This is the linear differential equation.

Step 5: Integration Factor

Calculate the integration factor

I . F

I . F = e^{\int \frac{2 s}{s (s + 2)} d s} = e^{\int \frac{2}{s + 2} d s} = e^{2 \log (s + 2)} = (s + 2)^{2}

Step 6: Solve the Differential Equation

The solution of (1) is given by:

\bar{y} (s + 2)^{2} = c + \int (1 - \frac{2}{s}) \frac{1}{s (s + 2)} (s + 2)^{2} d s

\bar{y} (s + 2)^{2} = c + \int \frac{s^{2} - 4}{s^{2}} d s

\bar{y} (s + 2)^{2} = c + \int (1 - \frac{4}{s^{2}}) d s = c + [s + \frac{4}{s}]

\bar{y} = \frac{c}{(s + 2)^{2}} + \frac{s^{2} + 4}{s (s + 2)^{2}}

\bar{y} = \frac{(s + 2)^{2} - 4 s}{s (s + 2)^{2}} + \frac{c}{(s + 2)^{2}} = \frac{1}{s} - \frac{4}{(s + 2)^{2}} + \frac{c}{(s + 2)^{2}}

Step 7: Inverse Laplace Transform

Apply the inverse Laplace transform to obtain the solution in the time domain:

So, we have:

y (t) = L^{- 1} [\frac{1}{s} + \frac{c - 4}{(s + 2)^{2}}] = L^{- 1} [\frac{1}{s}] + L^{- 1} [\frac{c - 4}{(s + 2)^{2}}]

y (t) = 1 + (c - 4) e^{- 2 t} t

Conclusion

Hence, there are infinitely many solutions possible for any value of

c

. This implies that the problem does not have a unique solution.

W

की हैं, एक वर्ग के रूप में ढाँचा बनाते हुए जुड़ी हैं। यह एक कोने से टँगा हुआ है। भार

W

तीर्नों नीचे वाले हरेक कोने से लटकाए हैं। वर्ग का आकार एक हल्की छड़, जो क्षैतिज विकर्ण के अनुदिश है, द्वारा रक्षित किया गया है। उस हल्की छड़ पर प्रणोद निकालिए।
(ii) एक कण वेग

V

से लंबी दूरी तय करने के लिए चलना शुरू करता है। एक तारे के केंद्र से कण के प्रारंभिक पथ की स्पर्श-रेखा पर लंबवत् दूरी

p

है। दिखाइए कि कण की तारे के केंद्र से न्यूनतम दूरी

λ

है, जहाँ

V^{2} λ = \sqrt{μ^{2} + p^{2} V^{4}} - μ

. यहाँ

μ

एक अचर है।

(i) A square framework formed of uniform heavy rods of equal weight

W

jointed together, is hung up by one corner. A weight

W

is suspended from each of the three lower corners, and the shape of the square is preserved by a light rod along the horizontal diagonal. Find the thrust of the light rod.

Answer:

Finding the Thrust in the Light Rod

Step 1: Setup of the Problem

Consider a square framework

A B C D

formed of four uniform rods, each with weight

W

and length

2 a

. The framework is suspended from corner

A

, and a weight

W

is suspended from each of the three lower corners,

B

C

, and

D

. A light rod along the horizontal diagonal

B D

prevents the system from collapsing. We want to find the thrust

T

in the light rod.

Step 2: Analysis of Forces

Let’s analyze the forces acting on the system. The total weight

4 W

of the rods

A B

B C

C D

, and

D A

can be assumed to act at point

O

. We need to find the thrust

T

in the rod

B D

Step 3: Displacement and Virtual Work

To find

T

, we give the system a small symmetrical displacement about

A C

, which causes

θ

to change to

θ + δ θ

. Point

A

remains fixed, and points

B

O

D

, and

C

change their positions. The length of the rods

A B

B C

C D

, and

D A

remains constant, but the length

B D

changes.

We have

B D = 2 B O = 2 A B \sin θ = 4 a \sin θ

, which is the depth of each of the points

B

C

, and

D

below the fixed point

A

. Similarly,

A O = 2 a \cos θ

is the depth of each of the points

B

C

, and

D

below

A

, and the depth of

C

below

A

2 A O = 4 a \cos θ

Step 4: Principle of Virtual Work

Applying the principle of virtual work, we have:

T δ (4 a \sin θ) + 4 W δ (2 a \cos θ) + 2 W δ (2 a \cos θ) + W δ (4 a \cos θ) = 0

Simplifying:

4 a T \cos θ δ θ - 8 a W \sin θ δ θ - 4 a W \sin θ δ θ - 4 a W \sin θ δ θ = 0

4 a [T \cos θ - 4 W \sin θ] δ θ = 0

Since

δ θ \neq 0

, we have:

T \cos θ - 4 W \sin θ = 0

Step 5: Finding Thrust $T$

Now, we can solve for

T

T = 4 W \tan θ

In the position of equilibrium,

θ = 45^{\circ}

, so:

T = 4 W \tan 45^{\circ} = 4 W

Therefore, the thrust

T

in the light rod is equal to the total weight of the four rods, which is

4 W

(ii) A particle starts at a great distance with velocity

V

. Let

p

be the length of the perpendicular from the centre of a star on the tangent to the initial path of the particle. Show that the least distance of the particle from the centre of the star is

λ

, where

V^{2} λ = \sqrt{μ^{2} + p^{2} V^{4}} - μ

. Here

μ

is a constant.

Answer:

Finding the Least Distance of the Particle from the Star’s Center

Step 1: Given Information

We are given that a particle starts at a great distance with velocity

V

. Let

p

be the length of the perpendicular from the center of a star on the tangent to the initial path of the particle. We want to show that the least distance of the particle from the center of the star is

λ

, where

V^{2} λ = \sqrt{μ^{2} + p^{2} V^{4}} - μ

, and

μ

is a constant.

Step 2: The Hamiltonian and Energy of the Particle

The Hamiltonian and energy of the particle are given as follows:

E = H (x, x^{*}) = \frac{m}{2} | x |^{2} - \frac{m μ}{| x |}

This gives us the second-order equation of motion:

x^{*} = - \frac{μ x}{| x |^{3}}

By the initial conditions, we have

E = \frac{1}{2} m V^{2}

Step 3: Polar Coordinates

Now, let’s write the equation in polar coordinates. We have:

L = r^{2} \dot{θ}

as the constant angular momentum, and the energy equation becomes:

E = \frac{1}{2} m ({\dot{r}}^{2} + r^{2} {\dot{θ}}^{2}) - \frac{m μ}{r}

V^{2} = {\dot{r}}^{2} + \frac{L^{2}}{r^{2}} - \frac{2 μ}{r}

Step 4: Determining $L$

To find the least distance of the particle from the center of the star, we need to determine

L

at the point of closest approach,

r = λ

, where

r = 0

For large

r

, the trajectory is given as:

X = (x, y) = (p, y_{0} - V t)

So,

L = r^{2} \dot{ϕ} = \dot{x} y - x \dot{y} = - p V

Thus, the equation for

r = λ

is:

V^{2} λ^{2} + 2 μ λ = p^{2} V^{2}

(V^{2} λ + μ)^{2} = p^{2} V^{4} + μ^{2}

This establishes the desired result:

V^{2} λ = \sqrt{μ^{2} + p^{2} V^{4}} - μ

Therefore, the least distance of the particle from the center of the star is indeed

λ

as given in the problem statement.

(a) (i) निम्न अवकल समीकरण हल कीजिए :

(x + 1)^{2} y^{''} - 4 (x + 1) y^{'} + 6 y = 6 (x + 1)^{2} + \sin \log (x + 1)

(ii) अवकल समीकरण

9 p^{2} (2 - y)^{2} = 4 (3 - y)

के व्यापक व विचित्र हल निकालिए, जहाँ

p = \frac{d y}{d x}

(i) Solve the following differential equation :

(x + 1)^{2} y^{''} - 4 (x + 1) y^{'} + 6 y = 6 (x + 1)^{2} + \sin \log (x + 1)

Answer:

Solving the Differential Equation

Given the differential equation:

(x + 1)^{2} y^{''} - 4 (x + 1) y^{'} + 6 y = 6 (x + 1)^{2} + \sin \log (x + 1)

Step 1: Change of Variables

Let

(x + 1) = e^{z} \Rightarrow \log (1 + x) = z

, and let

D_{1} = \frac{d}{d z}

. Then

(1 + x) D = D_{1}

, and

(1 + x)^{2} D^{2} = D_{1} (D_{1} - 1)

The equation becomes:

(D_{1} (D_{1} - 1) - 4 D_{1} + 6) y = 6 e^{2 z} + \sin z

Step 2: Auxiliary Equation

To solve the auxiliary equation, we have:

\begin{aligned} D_{1}^{2} - 5 D_{1} + 6 = 0 \\ \Rightarrow D_{1}^{2} - 3 D_{1} - 2 D_{1} + 6 = 0 \\ \Rightarrow D_{1} (D_{1} - 3) - 2 (D_{1} - 3) = 0 \\ \Rightarrow D_{1} = 2, 3 \end{aligned}

Therefore, the complementary function (C.F) is:

c_{1} e^{2 z} + c_{2} e^{3 z} = c_{1} (x + 1)^{2} + c_{2} (x + 1)^{3} (1)

Step 3: Particular Integral

For the particular integral (P.I), we have:

\begin{aligned} P . I : \frac{6 e^{2 z} + \sin z}{(D_{1} - 2) (D_{1} - 3)} \\ = \frac{6 e^{2 z}}{(D_{1} - 2) (D_{1} - 3)} + \frac{\sin z}{D_{1}^{2} - 5 D_{1} + 6} \\ = - \frac{6 e^{2 z}}{D_{1} - 2} + \frac{\sin z}{- 1 + 6 - 5 D} \\ = - 6 e^{2 z} \frac{z}{1!} + \frac{\sin z}{5 (1 - D_{1})} \\ = - 6 z e^{2 z} + \frac{(1 + D_{1}) \sin z}{5 (1 - D_{1}) (1 + D_{1})} \\ = - 6 z e^{2 z} + \frac{\sin z + \cos z}{5 (1 - D_{1}^{2})} \\ = - 6 z e^{2 z} + \frac{1}{5} (\frac{\sin z}{1 - D_{1}^{2}} + \frac{\cos z}{1 - D_{1}^{2}}) \\ = - 6 z e^{2 z} + \frac{1}{10} (\sin z + \cos z) \end{aligned}

P . I = - 6 [\log (1 + x)] (1 + x^{2}) + \frac{1}{10} [\cos \log (1 + x) + \sin \log (1 + x)]

Step 4: General Solution

So, the general solution is:

y = C_{1} (x + 1)^{2} + C_{2} (x + 1)^{3} - 6 (1 + x)^{2} (\log (1 + x)) + \frac{1}{10} [\cos \log (1 + x) + \sin \log (1 + x)]

(ii) Find the general and singular solutions of the differential equation

9 p^{2} (2 - y)^{2} = 4 (3 - y)

, where

p = \frac{d y}{d x}

Answer:

Finding the General Solution

Given the differential equation:

9 p^{2} (2 - y)^{2} = 4 (3 - y)

where

p = \frac{d y}{d x}

Step 1: Solving for $p$

Solving for

p

, we have:

p = \pm \frac{2}{3} (\frac{(3 - y)^{\frac{1}{2}}}{2 - y})

Step 2: Separation of Variables

Separating variables, we get:

d x = \pm \frac{3}{2} \frac{2 - y}{(3 - y)^{\frac{1}{2}}} d y

Step 3: Integrating

Integrating both sides:

x + c = \pm \frac{3}{2} \int [(3 - y)^{\frac{1}{2}} - (3 - y)^{\frac{1}{2}}] d y

x + c = \pm (\frac{3}{2}) [- \frac{2}{3} (3 - y)^{\frac{1}{2}} + 2 (3 - y)^{\frac{1}{2}}]

x + c = \pm (3 - y)^{\frac{1}{2}} [- (3 - y) + 3]

x + c = \pm y (3 - y)^{\frac{1}{2}}

or,

(x + c)^{2} = y^{2} (3 - y)

(on squaring)

This is the general solution.

Finding the Singular Solution

Now, let’s find the singular solution:

The general solution is

(x + c)^{2} = y^{2} (3 - y)

c^{2} + 2 x c + (x^{2} - y^{2} (3 - y)) = 0

This is a quadratic equation in the parameter

c

, so the discriminant relation is

B^{2} - 4 A C = 0

4 x^{2} - 4 (1) (x^{2} - y^{2} (3 - y)) = 0

y^{2} (3 - y) = 0

Now, two possibilities arise:

$y = 0$ :

For $y = 0$ , we get $\frac{d y}{d x} = p = 0$ .

Substituting $y = 0$ and $p = 0$ into the given differential equation, it does not satisfy it. Therefore, $y = 0$ is not a singular solution.
$3 - y = 0$ :

For $3 - y = 0$ , we get $\frac{d y}{d x} = p = 0$ .

Substituting $y = 3$ and $p = 0$ into the given differential equation, it satisfies it.

Therefore,

y = 3

is the required singular solution.

(b) पृष्ठ समाकल

\iint_{S} \nabla \times \bar{F} \cdot \hat{n} d S

का मान निकालिए, जहाँ

\bar{F} = y \hat{i} + (x - 2 x) \hat{j} - x y \hat{k}

तथा

S

गोले

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

की सतह है, जो

x y

-तल के ऊपर है।

Evaluate the surface integral

\iint_{S} \nabla \times \bar{F} \cdot \hat{n} d S

for

\bar{F} = y \hat{i} + (x - 2 x z) \hat{j} - x y \hat{k}

and

S

is the surface of the sphere

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

above the

x y

-plane.

Answer:

Step 1: Calculate the Curl of $\vec{F}$

The curl

\nabla \times \vec{F}

is given by:

\begin{matrix} \nabla \times \vec{F} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y & x - 2 x z & - x y \end{array} | \\ = (\frac{\partial (- x y)}{\partial y} - \frac{\partial (x - 2 x z)}{\partial z}) \hat{i} + (\frac{\partial y}{\partial z} - \frac{\partial (- x y)}{\partial x}) \hat{j} + (\frac{\partial (x - 2 x z)}{\partial x} - \frac{\partial y}{\partial y}) \hat{k} \\ = (- x - (- 2 x)) \hat{i} + (0 - (- y)) \hat{j} + (1 - 2 z - 1) \hat{k} \\ = x \hat{i} + y \hat{j} - 2 z \hat{k} \end{matrix}

\nabla \times \vec{F} = (x \hat{i} + y \hat{j} - 2 z \hat{k})

Step 2: Set Up the Surface Integral

The surface integral

\iint_{S} \nabla \times \vec{F} \cdot \hat{n} d S

over the sphere

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

above the

x y

-plane can be evaluated using spherical coordinates:

x = a \sin θ \cos ϕ, y = a \sin θ \sin ϕ, z = a \cos θ

Here,

θ

ranges from

0

\frac{π}{2}

and

ϕ

ranges from

0

2 π

The normal vector

\hat{n}

for the sphere is

\hat{k}

, and

d S = a^{2} \sin θ d θ d ϕ

Step 3: Evaluate the Surface Integral

The surface integral becomes:

\iint_{S} \nabla \times \vec{F} \cdot \hat{n} d S = \int_{0}^{2 π} \int_{0}^{\frac{π}{2}} (- 2 a \cos θ) a^{2} \sin θ d θ d ϕ

= - 2 a^{3} [\int_{0}^{2 π} d ϕ] [\int_{0}^{\frac{π}{2}} \cos θ \sin θ d θ]

= - 2 a^{3} [2 π] [\frac{1}{2}] = - 2 π a^{3}

Final Answer

So, the value of the surface integral

\iint_{S} \nabla \times \vec{F} \cdot \hat{n} d S

- 2 π a^{3}

M

है, के पहिए और धुरी के हर युग्म का द्रव्यमान व परिभ्रमण त्रिज्या क्रमशः

m

तथा

k

है। हर पहिए की त्रिज्या

r

है। यदि ट्रक को बल

P

द्वारा सीधे पथ (ट्रैक) पर धकेला जाता है, तब सिद्ध कीजिए कि उसका त्वरण

\frac{P}{M + \frac{2 m k^{2}}{r^{2}}}

है तथा ट्रक द्वारा प्रत्येक धुरी पर लगाए गए क्षैतिज बल का मान ज्ञात कीजिए। धुरी घर्षण व हवा का प्रतिरोध नगण्य है।

A four-wheeled railway truck has a total mass

M

, the mass and radius of gyration of each pair of wheels and axle are

m

and

k

respectively, and the radius of each wheel is

r

. Prove that if the truck is propelled along a level track by a force

P

, the acceleration is

\frac{P}{M + \frac{2 m k^{2}}{r^{2}}}

, and find the horizontal force exerted on each axle by the truck. The axle friction and wind resistance are to be neglected.

Answer:

Proving the Acceleration

Given the parameters:

Total mass of the railway truck $M$
Mass and radius of gyration of each pair of wheels and axle $m$ and $k$ respectively
Radius of each wheel $r$
Propelling force $P$

We want to prove that if the truck is propelled along a level track by a force

P

, the acceleration

a

is given by:

a = \frac{P}{M + \frac{2 m k^{2}}{r^{2}}}

Step 1: Work Done and Kinetic Energy

Let the distance covered be

x

, acceleration

a

, and velocity

V

Work done at the application of force

F = P x

The kinetic energy (KE) of the truck is given by

\frac{1}{2} M V^{2}

The KE of rotation for the wheels is

\frac{1}{2} I w^{2} \times 2

(There are 2 points of rotation).

KE of rotation = I w^{2} = \frac{m K^{2} V^{2}}{r^{2}}

Therefore, work done

P x

equals the sum of KE of the truck and KE of the wheels:

P x = \frac{1}{2} M V^{2} + \frac{m K^{2} V^{2}}{r^{2}} = \frac{V^{2}}{2} (\frac{M r^{2} + 2 m K^{2}}{r^{2}})

Step 2: Relationship between Velocity and Acceleration

We know that

u = 0

and

V^{2} = 2 a x

Therefore,

P x = \frac{2 a x}{2} (M + \frac{2 m K^{2}}{r^{2}})

Simplifying,

P x = a x (M + \frac{2 m K^{2}}{r^{2}})

Step 3: Finding Acceleration

Now, let’s isolate

a

a = \frac{P x}{x (M + \frac{2 m K^{2}}{r^{2}})}

a = \frac{P}{M + \frac{2 m K^{2}}{r^{2}}}

This proves the given acceleration formula.

Finding the Horizontal Force on Each Axle

The horizontal force (

F_{x}

) exerted on each axle by the truck can be calculated as follows:

F_{x} = \frac{1}{2} m V^{2} + \frac{1}{2} m V^{2} (\frac{K^{2}}{r^{2}}) = \frac{m V^{2}}{2} (1 + \frac{K^{2}}{r^{2}})

Using

V^{2} = 2 a x

, we have:

F_{x} = \frac{m (2 a x)}{2} (1 + \frac{K^{2}}{r^{2}}) = \frac{m a x}{2} (1 + \frac{K^{2}}{r^{2}})

Substituting the value of

a

from the acceleration formula:

F_{x} = \frac{m (\frac{P}{M + \frac{2 m K^{2}}{r^{2}}})}{2} (1 + \frac{K^{2}}{r^{2}})

Simplifying further:

F_{x} = \frac{m P}{2 (M + \frac{2 m K^{2}}{r^{2}})} (1 + \frac{K^{2}}{r^{2}})

So, the horizontal force exerted on each axle by the truck is given by the above expression.

Free UPSC Mathematics Optional Paper-1 2020 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Introduction

Work/Calculations

Property 1: Closure under Addition and Scalar Multiplication

Property 2: Commutativity of Addition

Property 3: Associativity of Addition

Property 4: Existence of Zero Vector

Property 5: Existence of Additive Inverse

Property 6: Distributive Law 1

Property 7: Distributive Law 2

Property 8: Associativity of Scalar Multiplication

Property 9: Multiplication by Identity

Example 1

Example 2

Conclusion

Introduction

Work/Calculations

Finding a Matrix A A AAA that Maps to the Null Matrix

Nullity of T T TTT

Rank of T T TTT

Finding Matrix A A AAA that Maps to the Null Matrix

Conclusion

Introduction

Work/Calculations

(i) Examine if A A AAA is Symmetric

(ii) Examine if A A AAA is Orthogonal

(iii) Show that Trace ( A ) = n − 2 ( A ) = n − 2 (A)=n-2(A) = n – 2(A)=n−2

(iv) Find A 3 × 3 A 3 × 3 A_(3xx3)A_{3 \times 3}A3×3, when u = [ 1 3 2 3 2 3 ] u = 1 3 2 3 2 3 u=[[(1)/(3)],[(2)/(3)],[(2)/(3)]]u = \left[\begin{array}{c}\frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3}\end{array}\right]u=[132323]

Conclusion

Introduction

Work/Calculations

(i) Find the Critical Points of f ( x ) f ( x ) f(x)f(x)f(x)

(ii) Find the Points at Which Local Minimum Occurs

(iii) Find the Points at Which Local Maximum Occurs

(iv) Find the Number of Zeros of f ( x ) f ( x ) f(x)f(x)f(x) in [ 0 , 5 ] [ 0 , 5 ] [0,5][0, 5][0,5]

Conclusion

Introduction

Work/Calculations

Step 1: Define the Null Space

Step 2: Find the Conditions for ( a , b , c ) ( a , b , c ) (a,b,c)(a, b, c)(a,b,c)

Step 3: Row Reduction to Echelon Form

Step 4: Solve the System

Step 5: Find the Nullity of T T TTT

Conclusion

Introduction

Work/Calculations

Step 1: Equation of the Plane

Step 2: Equation of the Line of Intersection

Step 3: Eliminate l l lll Between Equations

Step 4: Solve for m m mmm and n n nnn

Step 5: Equations of the Other Two Generators

Conclusion

Part (i): Find A B A B ABABAB

Part (ii): Find det ( A ) det ( A ) “det”(A)\text{det}(A)det(A) and det ( B ) det ( B ) “det”(B)\text{det}(B)det(B)

Part (iii): Solve the system of linear equations

Conclusion

Conclusion

Conclusion

Conclusion

Forces in the System

Geometry and Trigonometry

Equilibrium Condition

Conclusion

Initial Conditions and Variables

Velocities of Points

Conservation of Momentum

Moments About Point O O OOO

Solving for u u uuu and ω ω omega\omegaω

Velocity of Point O O OOO

Kinetic Energy

Conclusion

Part (i): Path x = t , y = t 2 , z = t 3 x = t , y = t 2 , z = t 3 x=t,y=t^(2),z=t^(3)x = t, y = t^2, z = t^3x=t,y=t2,z=t3

Vector d r ¯ d r ¯ d bar(r)d\bar{r}dr¯

Limits of Integration

Integral Calculation

Part (ii): Straight Lines Joining Points

Path 1: ( 0 , 0 , 0 ) ( 0 , 0 , 0 ) (0,0,0)(0,0,0)(0,0,0) to ( 1 , 0 , 0 ) ( 1 , 0 , 0 ) (1,0,0)(1,0,0)(1,0,0)

Path 2: ( 1 , 0 , 0 ) ( 1 , 0 , 0 ) (1,0,0)(1,0,0)(1,0,0) to ( 1 , 1 , 0 ) ( 1 , 1 , 0 ) (1,1,0)(1,1,0)(1,1,0)

Path 3: ( 1 , 1 , 0 ) ( 1 , 1 , 0 ) (1,1,0)(1,1,0)(1,1,0) to ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1)

Total Integral for Part (ii)

Finding a Matrix $A$ that Maps to the Null Matrix

Nullity of $T$

Rank of $T$

Finding Matrix $A$ that Maps to the Null Matrix

(i) Examine if $A$ is Symmetric

(ii) Examine if $A$ is Orthogonal

(iii) Show that Trace $(A) = n - 2$

(iv) Find $A_{3 \times 3}$ , when $u = [\begin{matrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{2}{3} \end{matrix}]$

(i) Find the Critical Points of $f (x)$

(iv) Find the Number of Zeros of $f (x)$ in $[0, 5]$

Step 2: Find the Conditions for $(a, b, c)$

Step 5: Find the Nullity of $T$

Step 3: Eliminate $l$ Between Equations

Step 4: Solve for $m$ and $n$

Part (i): Find $A B$

Part (ii): Find $det (A)$ and $det (B)$

Moments About Point $O$

Solving for $u$ and $ω$

Velocity of Point $O$

Part (i): Path $x = t, y = t^{2}, z = t^{3}$

Vector $d \bar{r}$

Path 1: $(0, 0, 0)$ to $(1, 0, 0)$

Path 2: $(1, 0, 0)$ to $(1, 1, 0)$

Path 3: $(1, 1, 0)$ to $(1, 1, 1)$

Part (iii): Straight Line from $(0, 0, 0)$ to $(1, 1, 1)$

Case 1: Weight $p$ is hung from $A$

Case 2: Weight $q$ is hung from $D$

Along $c_{1}$ :

Along $c_{2}$ :

Along $c_{3}$ :

Along $c_{4}$ :

Step 1: Calculate the Curl of $\vec{F}$