Free UPSC Mathematics Optional Paper-1 2021 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 12, 2025ByAbstract Classes

1.(a) यदि

A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 1 & 0 & 0 \end{array}]

है, तो

A^{- 1}

को ज्ञात किए बिना दर्शाइए कि

A^{2} = A^{- 1}

A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 1 & 0 & 0 \end{array}]

, then show that

A^{2} = A^{- 1}

(without finding

A^{- 1}

Answer:

Introduction

We are given a 3×3 matrix

A

and asked to prove that the square of this matrix,

A^{2}

, is equal to its inverse,

A^{- 1}

, without explicitly finding the inverse of

A

Assumptions

The problem assumes that the matrix

A

is invertible, i.e., it has an inverse

A^{- 1}

Definition

The inverse of a matrix

A

, denoted

A^{- 1}

, is a unique matrix such that when it is multiplied by

A

, the result is the identity matrix

I

. The identity matrix is a special square matrix with ones on the diagonal and zeros elsewhere.

Method/Approach

We will use the definition of the inverse of a matrix to solve this problem. If

A^{2}

is indeed the inverse of

A

, then the product

A A^{2}

should be the identity matrix

I

. We will calculate

A^{2}

and

A A^{2}

to verify this.

Work/Calculations

First, let’s calculate

A^{2}

A^{2} = A \cdot A = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 1 & 0 & 0 \end{array}] \cdot [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 1 & 0 & 0 \end{array}]

The calculation yields:

A^{2} = [\begin{array}{ccc} 0 & 0 & 1 \\ 0 & - 1 & 2 \\ 1 & - 1 & 1 \end{array}]

Next, we calculate

A A^{2}

A A^{2} = A \cdot A^{2} = [\begin{array}{ccc} 1 & - 1 & 1 \\ 2 & - 1 & 0 \\ 1 & 0 & 0 \end{array}] \cdot [\begin{array}{ccc} 0 & 0 & 1 \\ 0 & - 1 & 2 \\ 1 & - 1 & 1 \end{array}]

The calculation yields:

A A^{2} = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]

Conclusion

The product of

A

and

A^{2}

is the identity matrix

I

. Therefore, we have shown that

A^{2}

is indeed the inverse of

A

, i.e.,

A^{2} = A^{- 1}

, without explicitly finding

A^{- 1}

1.(b) क्रमित आधारक

B = {(0, 1, 1), (1, 0, 1), (1, 1, 0)}

के सापेक्ष

V_{3} (R)

पर परिभाषित रैखिक संकारक :

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

से संबन्धित आव्यूह ज्ञात कीजिए ।

Find the matrix associated with the linear operator on

V_{3} (R)

defined by

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

with respect to the ordered basis

B = {(0, 1, 1), (1, 0, 1), (1, 1, 0)}

Answer:

Introduction

We are tasked with finding the matrix representation of the linear operator

T : V_{3} (R) \to V_{3} (R)

defined by

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

. The matrix representation will be with respect to the ordered basis

B = {(0, 1, 1), (1, 0, 1), (1, 1, 0)}

Assumptions

$T$ is a linear operator.
$V_{3} (R)$ is a vector space over the real numbers $R$ .
The ordered basis $B$ consists of vectors in $V_{3} (R)$ .

Method/Approach

To find the matrix representation of

T

with respect to the basis

B

, we will:

Apply $T$ to each vector in the basis $B$ .
Express the resulting vectors in terms of the basis $B$ .
Use the coefficients as columns of the matrix representation of $T$ .

Work/Calculations

Step 1: Apply $T$ to each vector in $B$

Let’s apply

T

to the vectors

(0, 1, 1)

(1, 0, 1)

, and

(1, 1, 0)

$T (0, 1, 1) = (0 + 1, 0 - 1, 2 \times 1)$
$T (1, 0, 1) = (1 + 0, 1 - 0, 2 \times 1)$
$T (1, 1, 0) = (1 + 1, 1 - 1, 2 \times 0)$

After calculating, we get the following transformed vectors:

$T (0, 1, 1) = (1, - 1, 2)$
$T (1, 0, 1) = (1, 1, 2)$
$T (1, 1, 0) = (2, 0, 0)$

Step 2: Express the resulting vectors in terms of $B$

We need to express each of these vectors as a linear combination of the basis vectors in

B

$(1, - 1, 2) = a_{1} (0, 1, 1) + a_{2} (1, 0, 1) + a_{3} (1, 1, 0)$
$(1, 1, 2) = b_{1} (0, 1, 1) + b_{2} (1, 0, 1) + b_{3} (1, 1, 0)$
$(2, 0, 0) = c_{1} (0, 1, 1) + c_{2} (1, 0, 1) + c_{3} (1, 1, 0)$

After calculating, we find the coefficients as follows:

$a_{1} = 0, a_{2} = 2, a_{3} = - 1$
$b_{1} = 1, b_{2} = 1, b_{3} = 0$
$c_{1} = - 1, c_{2} = 1, c_{3} = 1$

Thus, the vectors

(1, - 1, 2)

(1, 1, 2)

, and

(2, 0, 0)

can be expressed in terms of the basis

B

as:

$(1, - 1, 2) = 0 \cdot (0, 1, 1) + 2 \cdot (1, 0, 1) + (- 1) \cdot (1, 1, 0)$
$(1, 1, 2) = 1 \cdot (0, 1, 1) + 1 \cdot (1, 0, 1) + 0 \cdot (1, 1, 0)$
$(2, 0, 0) = - 1 \cdot (0, 1, 1) + 1 \cdot (1, 0, 1) + 1 \cdot (1, 1, 0)$

Step 3: Form the Matrix Representation

The matrix representation

[T]_{B}

T

with respect to the basis

B

is formed by using these coefficients as columns:

[T]_{B} = (\begin{matrix} 0 & 1 & - 1 \\ 2 & 1 & 1 \\ - 1 & 0 & 1 \end{matrix})

Conclusion

The matrix representation of the linear operator

T

with respect to the ordered basis

B = {(0, 1, 1), (1, 0, 1), (1, 1, 0)}

is:

[T]_{B} = (\begin{matrix} 0 & 1 & - 1 \\ 2 & 1 & 1 \\ - 1 & 0 & 1 \end{matrix})

1.(c) दिया गया है :

Δ (x) = | \begin{array}{ccc} f (x + α) & f (x + 2 α) & f (x + 3 α) \\ f (α) & f (2 α) & f (3 α) \\ f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \end{array} |

जहाँ

f

एक वास्तविक-मान अवकलनीय फलन है तथा

α

एक अचर है।

lim_{x \to 0} \frac{Δ (x)}{x}

को ज्ञात कीजिए ।

Given :

Δ (x) = | \begin{array}{ccc} f (x + α) & f (x + 2 α) & f (x + 3 α) \\ f (α) & f (2 α) & f (3 α) \\ f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \end{array} |

where

f

is a real valued differentiable function and

α

is a constant. Find

lim_{x \to 0} \frac{Δ (x)}{x}

Answer:

Introduction

The problem asks us to find the limit of

\frac{Δ (x)}{x}

x \to 0

, where

Δ (x)

is a determinant involving a real-valued differentiable function

f (x)

and a constant

α

Assumptions

$f (x)$ is a real-valued differentiable function.
$α$ is a constant.
$x$ is a real number.

Method/Approach

To find the limit, we will:

Examine the determinant $Δ (x)$ as $x \to 0$ to see if it’s an indeterminate form.
Apply L’Hôpital’s Rule to $\frac{Δ (x)}{x}$ as $x \to 0$ .
Use determinant properties to simplify the expression.

Work/Calculations

Step 1: Examine $Δ (x)$ as $x \to 0$

x \to 0

, the determinant becomes:

lim_{x \to 0} \frac{Δ x}{x} = lim_{x \to 0} \frac{| \begin{array}{lll} f (x + α) & f (x + 2 α) & f (x + 3 α) \\ f (α) & f (2 α) & f (3 α) \\ f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \end{array} |}{x}

If we apply direct

x \to 0

to Numerator :

Δ = | \begin{array}{ccc} f (α) & f (2 α) & f (3 α) \\ f (α) & f (2 α) & f (3 α) \\ f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \end{array} |

Notice that the first and second rows are identical. Due to the determinant property, if any two rows (or columns) are identical, the determinant is zero. Therefore,

Δ = 0

The denominator

x

also becomes zero as

x \to 0

lim_{x \to 0} \frac{Δ x}{x} = lim_{x \to 0} \frac{| \begin{array}{lll} f (x + α) & f (x + 2 α) & f (x + 3 α) \\ f (α) & f (2 α) & f (3 α) \\ f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \end{array} |}{x} (\frac{0}{0})

So, we have an indeterminate form

\frac{0}{0}

, and we can apply L’Hôpital’s Rule.

Step 2: Apply L’Hôpital’s Rule

Differentiating the numerator

Δ (x)

with respect to

x

gives:

\frac{d}{d x} Δ (x) = | \begin{array}{ccc} f^{'} (x + α) & f^{'} (x + 2 α) & f^{'} (x + 3 α) \\ f (α) & f (2 α) & f (3 α) \\ f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \end{array} |

The derivative of the denominator

x

1

lim_{x \to 0} \frac{Δ x}{x} = lim_{x \to 0} \frac{| \begin{array}{lll} f^{'} (x + α) & f^{'} (x + 2 α) & f^{'} (x + 3 α) \\ f (α) & f (2 α) & f (3 α) \\ f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \end{array} |}{1}

Step 3: Use Determinant Properties

x \to 0

, the differentiated determinant becomes:

| \begin{array}{ccc} f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \\ f (α) & f (2 α) & f (3 α) \\ f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \end{array} |

Notice that the first and third rows are identical. Therefore, the determinant is zero.

\begin{aligned} lim_{x \to 0} \frac{Δ x}{x} = | \begin{array}{lll} f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \\ f (α) & f (2 α) & f (3 α) \\ f^{'} (α) & f^{'} (2 α) & f^{'} (3 α) \end{array} | \\ lim_{x \to 0} \frac{Δ x}{x} = 0 \end{aligned}

Conclusion

The limit

lim_{x \to 0} \frac{Δ (x)}{x} = 0

, as confirmed by L’Hôpital’s Rule and determinant properties.

1.(d) दर्शाइए कि

e^{x} \cos x = 1

के किन्हीं दो मूलों के बीच में

e^{x} \sin x - 1 = 0

का कम से कम एक मूल विद्यमान है ।

Show that between any two roots of

e^{x} \cos x = 1

, there exists at least one root of

e^{x} \sin x - 1 = 0

Answer:

Introduction

The problem asks us to prove that between any two roots of

e^{x} \cos x = 1

, there exists at least one root of

e^{x} \sin x - 1 = 0

Assumptions

$e^{x} \cos x = 1$ and $e^{x} \sin x - 1 = 0$ are continuous and differentiable functions.
$x$ is a real number.
$a$ and $b$ are two distinct roots of $e^{x} \cos x = 1$ .

Method/Approach

To prove the statement, we will:

Define a function $f (x)$ and examine its properties.
Apply Rolle’s Theorem to show that a root of $e^{x} \cos x + 1 = 0$ exists between $a$ and $b$ .

Work/Calculations

Step 1: Define $f (x)$ and Examine Its Properties

Consider

f (x) = e^{x} \sin x - 1 = 0

\begin{aligned} f (x) & \equiv e^{x} \sin x - 1 = 0 \\ f (x) & \equiv e^{x} \sin x = 1 \\ f (x) & \equiv \sin x = e^{- x} \\ f (x) & \equiv e^{- x} - \sin x = 0 \end{aligned}

Step 2: Apply Rolle’s Theorem

Rolle’s Theorem states that if a function

f (x)

is continuous on

[a, b]

and differentiable on

(a, b)

, and

f (a) = f (b)

, then there exists at least one

c

(a, b)

such that

f^{'} (c) = 0

In our Case

f (x) = e^{- x} - \sin x \forall x \in [a, b]

f^{'} (x) = - e^{- x} - \cos x

Let a and

b, a \neq b

, be any two roots of f(x) then

\begin{aligned} f (a) \equiv e^{- a} - \sin a = 0 \\ f (b) \equiv e^{- b} - \sin b = 0 \end{aligned}

Clearly

f

is continuous in

[a, b]

and derivable in

] a, b [

. Also

f (a) = f (b) = 0

Therefore the hypothesis of Rolle’s Theorem is satisfied by [a, b].

Therefore there exists

c \in] a, b [

such that

f^{'} (c) = 0

which implies

\begin{aligned} f^{'} (c) = 0 \\ - e^{- c} - \cos c = 0 \\ - e^{- c} = \cos c \\ - 1 = e^{c} \cos c \\ e^{c} \cos c + 1 = 0 \end{aligned}

e^{x} \cos x + 1 = 0

has a root

c

for some

c \in] a, b [

Conclusion

Between any two roots

a

and

b

e^{x} \cos x = 1

, there exists at least one root

c

e^{x} \cos x + 1 = 0

, as proven by Rolle’s Theorem. This also implies that between any two roots of

e^{x} \cos x = 1

, there exists at least one root of

e^{x} \sin x - 1 = 0

1.(e) उस बेलन का समीकरण ज्ञात कीजिए जिसके जनक, रेखा :

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

के समानान्तर हैं
तथा जिसका निर्देशक-वक्र

x^{2} + 2 y^{2} = 1, z = 0

है।

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

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

and whose guiding curve is

x^{2} + 2 y^{2} = 1, z = 0

Answer:

Introduction

The problem asks us to find the equation of a cylinder whose generators are parallel to the line

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

and whose guiding curve is

x^{2} + 2 y^{2} = 1

z = 0

Assumptions

The cylinder exists in a 3D Cartesian coordinate system.
The guiding curve of the cylinder lies in the plane $z = 0$ .

Definitions

Generator: A line that, when moved parallel to itself along a curve, defines a surface.
Guiding Curve: The curve along which the generator moves to form the surface.

Method/Approach

Write the equation of a generator parallel to the given line, passing through a point $(x_{1}, y_{1}, z_{1})$ .
Find the point where this generator intersects the plane $z = 0$ .
Use the equation of the guiding curve to find the locus of points $(x_{1}, y_{1}, z_{1})$ that form the cylinder.

Work/Calculations

Step 1: Equation of the Generator

Given the equation of the line as

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

, and a point

(x_{1}, y_{1}, z_{1})

on the generator, the equation of the generator can be written as:

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

Step 2: Intersection with Plane $z = 0$

This generator intersects the plane

z = 0

, yielding:

\frac{x - x_{1}}{1} = \frac{y - y_{1}}{- 2} = \frac{0 - z_{1}}{3}

Simplifying, we get:

x = x_{1} - \frac{z_{1}}{3}, y = y_{1} + \frac{2 z_{1}}{3}

Step 3: Locus of Points on the Generator

The point

(x_{1}, y_{1}, z_{1})

lies on the ellipse

x^{2} + 2 y^{2} = 1

. Substituting the values of

x

and

y

into this equation, we get:

{(x_{1} - \frac{z_{1}}{3})}^{2} + 2 {(y_{1} + \frac{2 z_{1}}{3})}^{2} = 1

\begin{aligned} ∴ {(3 x_{1} - z_{1})}^{2} + 2 {(3 y_{1} + 2 z_{1})}^{2} = 9 \\ ∴ 9 x_{1}^{2} + z_{1}^{2} - 6 x_{1} z_{1} + 2 (9 y_{1}^{2} + 4 z_{1}^{2} + 12 y_{1} z_{1}) = 9 \\ ∴ 9 x_{1}^{2} + 18 y_{1}^{2} + 9 z_{1}^{2} - 6 x_{1} z_{1} + 24 y_{1} z_{1} - 9 = 0 \\ ∴ locus of (x_{1}, y_{1}, z_{1}) is 3 x^{2} + 6 y^{2} + 3 z^{2} - 2 x z + 8 y z - 3 = 0 \end{aligned}

Simplifying, the locus of

(x_{1}, y_{1}, z_{1})

becomes:

3 x^{2} + 6 y^{2} + 3 z^{2} - 2 x z + 8 y z - 3 = 0

Conclusion

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

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

and whose guiding curve is

x^{2} + 2 y^{2} = 1

z = 0

3 x^{2} + 6 y^{2} + 3 z^{2} - 2 x z + 8 y z - 3 = 0

2.(a) दर्शाइए कि वे समतल, जो कि शंकु

a x^{2} + b y^{2} + c z^{2} = 0

को लंब जनकों में काटते हैं, शंकु

\frac{x^{2}}{b + c} + \frac{y^{2}}{c + a} + \frac{z^{2}}{a + b} = 0

को स्पर्श करते हैं ।

Show that the planes, which cut the cone

a x^{2} + b y^{2} + c z^{2} = 0

in perpendicular generators, touch the cone

\frac{x^{2}}{b + c} + \frac{y^{2}}{c + a} + \frac{z^{2}}{a + b} = 0

Answer:

Introduction

We are tasked with proving that any plane cutting the cone

a x^{2} + b y^{2} + c z^{2} = 0

along perpendicular generators also touches the cone

\frac{x^{2}}{b + c} + \frac{y^{2}}{c + a} + \frac{z^{2}}{a + b} = 0

Assumptions

The vertex of the given cone is at $(0, 0, 0)$ .
The plane passes through this vertex.

Definitions

$a, b, c$ : Coefficients of the original cone equation.
$u, v, w$ : Coefficients of the plane equation.
$l, m, n$ : Direction cosines of the intersecting lines between the cone and the plane.

Step 1: Establish Cone and Plane Equations

The given cone equation is:

a x^{2} + b y^{2} + c z^{2} = 0 (Equation 1)

Let the equation of any plane through the vertex of the cone be:

u x + v y + w z = 0 (Equation 2)

Step 2: Equations of Intersecting Lines

Assume that the lines formed by the intersection of the cone and the plane are:

\frac{x}{l} = \frac{y}{m} = \frac{z}{n} (Equation 3)

Applying this to Equation 1, we get:

a l^{2} + b m^{2} + c n^{2} = 0 (Equation 4)

And combining with Equation 2, we have:

u l + v m + w n = 0 (Equation 5)

Step 3: Elaborating the Condition for Perpendicular Generators

Eliminating $n$

Eliminating

n

between Equation 4 and Equation 5 gives:

(a w^{2} + c u^{2}) l^{2} + 2 c u v l m + (b w^{2} + c v^{2}) m^{2} = 0 (Equation 6)

Roots and Products

Since Equation 6 is quadratic in

\frac{l}{m}

, it shows that the plane cuts the cone in two lines. The product of the roots

\frac{l_{1}}{m_{1}}

and

\frac{l_{2}}{m_{2}}

is:

\frac{l_{1}}{m_{1}} \cdot \frac{l_{2}}{m_{2}} = \frac{b w^{2} + c v^{2}}{a w^{2} + c u^{2}} (Equation 7)

By symmetry, we write:

\frac{l_{1} l_{2}}{b w^{2} + c v^{2}} = \frac{m_{1} m_{2}}{c u^{2} + a w^{2}} = \frac{n_{1} n_{2}}{a v^{2} + b u^{2}} (Equation 8)

Step 4: Condition for Perpendicular Generators

The lines will be perpendicular if

l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0

, which simplifies to:

(b w^{2} + c v^{2}) + (c u^{2} + a w^{2}) + (a v^{2} + b u^{2}) = 0 (Equation 9)

Rearranging, this yields:

(b + c) u^{2} + (c + a) v^{2} + (a + b) w^{2} = 0 (Equation 10)

In general

(b + c) x^{2} + (c + a) y^{2} + (a + b) z^{2} = 0 (Equation 11)

Step 4: Reciprocal Cone

Now we will find reciprocal cone of

(b + c) x^{2} + (c + a) y^{2} + (a + b) z^{2} = 0 (Equation 11)

Let the cone reciprocal to the cone (11) be

A x^{2} + B y^{2} + C z^{2} + 2 F y z + 2 G z x + 2 H x y = 0 (Equation 12)

On Comparing equation (11) with

a x^{2} + b y^{2} + c z^{2} + 2 f y z + 2 g z x + 2 h x y = 0 .

we have

\begin{matrix} a = b + c, b = c + a, c = a + b, f = 0, g = 0, h = 0 . \end{matrix}

\begin{aligned} A = b c - f^{2} \\ A = (c + a) (a + b) - 0 \\ A = (c + a) (a + b) \\ B = c a - g^{2} \\ B = (a + b) (b + c) - 0 \\ B = (a + b) (b + c) \\ C = a b - h^{2} \\ C = (b + c) (c + a) - 0 \\ C = (b + c) (c + a) \\ F = g h - a f \\ F = 0 \\ G = h b - a g \\ G = 0 \\ H = b g - a h \\ H = 0 \end{aligned}

Putting these values of

A, B, C, F, G, H

in the equation (12), the required equation of the reciprocal cone is given by

\begin{aligned} A x^{2} + B y^{2} + c z^{2} + 2 F y z + 2 G z x + 2 H x y = 0 \\ (c + a) (a + b) x^{2} + (a + b) (b + c) y^{2} + (b + c) (c + a) z^{2} = 0 \end{aligned}

Divide by

(a + b) (b + c) (c + a)

\frac{x^{2}}{b + c} + \frac{y^{2}}{c + a} + \frac{z^{2}}{a + b} = 0

Conclusion

We have therefore proved that if a plane cuts the cone described by

a x^{2} + b y^{2} + c z^{2} = 0

in perpendicular generators, it must touch the cone described by

\frac{x^{2}}{b + c} + \frac{y^{2}}{c + a} + \frac{z^{2}}{a + b} = 0

2.(b) दिया गया है :

f (x, y) = | x^{2} - y^{2} |

, तब

f_{x y} (0, 0)

तथा

f_{y x} (0, 0)

ज्ञात कीजिए । अतः दर्शाइए कि

f_{x y} (0, 0) = f_{y x} (0, 0)

।

Given that

f (x, y) = | x^{2} - y^{2} |

. Find

f_{x y} (0, 0)

and

f_{y x} (0, 0)

.
Hence show that

f_{x y} (0, 0) = f_{y x} (0, 0)

Answer:

Introduction

The problem asks us to find

f_{x y} (0, 0)

and

f_{y x} (0, 0)

for the function

f (x, y) = | x^{2} - y^{2} |

and show that

f_{x y} (0, 0) = f_{y x} (0, 0)

Assumptions

The function $f (x, y) = | x^{2} - y^{2} |$ is defined for all $x, y \in R$ .

Definitions

$f_{x}$ and $f_{y}$ are the partial derivatives of $f$ with respect to $x$ and $y$ , respectively.
$f_{x y}$ and $f_{y x}$ are the second-order mixed partial derivatives of $f$ .

Method/Approach

We will use the limit definitions of partial derivatives to find

f_{x y} (0, 0)

and

f_{y x} (0, 0)

Work/Calculations

Step 1: Find $f_{x}$ and $f_{y}$

The function

f (x, y) = | x^{2} - y^{2} |

can be written as a piecewise function:

f (x, y) = {\begin{cases} x^{2} - y^{2}, & if x^{2} \geq y^{2} \\ y^{2} - x^{2}, & if x^{2} < y^{2} \end{cases}

Using the limit definition, we find:

$f_{x} (x, y)$

f_{x} (x, y) = lim_{h \to 0} \frac{f (x + h, y) - f (x, y)}{h}

For

x^{2} \geq y^{2}

f_{x} (x, y) = lim_{h \to 0} \frac{(x + h)^{2} - y^{2} - (x^{2} - y^{2})}{h} = 2 x

For

x^{2} < y^{2}

f_{x} (x, y) = lim_{h \to 0} \frac{y^{2} - (x + h)^{2} - (y^{2} - x^{2})}{h} = - 2 x

So,

f_{x} (x, y)

is:

f_{x} (x, y) = {\begin{cases} 2 x, & if x^{2} \geq y^{2} \\ - 2 x, & if x^{2} < y^{2} \end{cases}

$f_{y} (x, y)$

f_{y} (x, y) = lim_{k \to 0} \frac{f (x, y + k) - f (x, y)}{k}

For

x^{2} \geq y^{2}

f_{y} (x, y) = lim_{k \to 0} \frac{x^{2} - (y + k)^{2} - (x^{2} - y^{2})}{k} = - 2 y

For

x^{2} < y^{2}

f_{y} (x, y) = lim_{k \to 0} \frac{(y + k)^{2} - x^{2} - (y^{2} - x^{2})}{k} = 2 y

So,

f_{y} (x, y)

is:

f_{y} (x, y) = {\begin{cases} - 2 y, & if x^{2} \geq y^{2} \\ 2 y, & if x^{2} < y^{2} \end{cases}

Step 2: Find $f_{x y} (0, 0)$ and $f_{y x} (0, 0)$

Using the limit definition for

f_{x y}

and

f_{y x}

f_{x y} (0, 0) = lim_{k \to 0} \frac{f_{x} (0, 0 + k) - f_{x} (0, 0)}{k}

f_{y x} (0, 0) = lim_{h \to 0} \frac{f_{y} (0 + h, 0) - f_{y} (0, 0)}{h}

Both

f_{x y} (0, 0)

and

f_{y x} (0, 0)

are 0 because

f_{x} (0, 0) = f_{y} (0, 0) = 0

Conclusion

After performing the calculations, we find that

f_{x y} (0, 0) = f_{y x} (0, 0) = 0

, confirming that the mixed partial derivatives are equal at the point

(0, 0)

2.(c) दर्शाइए कि

S = {(x, 2 y, 3 x) : x, y

वास्तविक संख्याऐं हैं |

R^{3} (R)

का एक उपसमष्टि है ।

S

के दो आधार ज्ञात कीजिए ।

S

की विमा भी ज्ञात कीजिए ।

Show that

S = {(x, 2 y, 3 x) : x, y

are real numbers is a subspace of

R^{3} (R)

. Find two bases of

S

. Also find the dimension of

S

Answer:

Introduction

The problem asks us to show that the set

S = {(x, 2 y, 3 x) : x, y are real numbers}

is a subspace of

R^{3}

. Additionally, we are required to find two bases for

S

and determine its dimension.

Assumptions

$S$ is a set of vectors in $R^{3}$ defined as $S = {(x, 2 y, 3 x) : x, y are real numbers}$ .
$R^{3}$ is the vector space of all 3-tuples of real numbers.

Definitions

A subspace is a subset of a vector space that is itself a vector space.
A basis is a set of linearly independent vectors that span the vector space.
The dimension of a vector space is the number of vectors in any basis for the vector space.

Method/Approach

To show that

S

is a subspace of

R^{3}

, we need to verify the following properties:

The zero vector is in $S$ .
$S$ is closed under vector addition.
$S$ is closed under scalar multiplication.

Step 1: Zero Vector in $S$

The zero vector in

R^{3}

(0, 0, 0)

. This vector is in

S

when

x = 0

and

y = 0

Step 2: Closed Under Vector Addition

Let

u = (x_{1}, 2 y_{1}, 3 x_{1})

and

v = (x_{2}, 2 y_{2}, 3 x_{2})

be any vectors in

S

u + v = (x_{1} + x_{2}, 2 y_{1} + 2 y_{2}, 3 x_{1} + 3 x_{2})

Simplifying, we get

u + v = (x_{1} + x_{2}, 2 (y_{1} + y_{2}), 3 (x_{1} + x_{2}))

, which is also in

S

Step 3: Closed Under Scalar Multiplication

Let

u = (x, 2 y, 3 x)

be any vector in

S

and let

c

be any scalar.

c u = c (x, 2 y, 3 x) = (c x, 2 (c y), 3 (c x))

Simplifying, we get

c u = (c x, 2 c y, 3 c x)

, which is also in

S

Step 4: Finding Bases and Dimension

Any vector in

S

can be written as

(x, 2 y, 3 x)

, which can be rewritten as

x (1, 0, 3) + y (0, 2, 0)

Therefore, one basis for

S

{(1, 0, 3), (0, 2, 0)}

Another basis could be obtained by scaling these vectors, for example,

{(2, 0, 6), (0, 4, 0)}

Step 5: Dimension of $S$

The dimension of

S

is the number of vectors in any basis for

S

. In this case, the dimension is 2.

Conclusion

$S$ is a subspace of $R^{3}$ as it contains the zero vector and is closed under both vector addition and scalar multiplication.
Two bases for $S$ are ${(1, 0, 3), (0, 2, 0)}$ and ${(2, 0, 6), (0, 4, 0)}$ .
The dimension of $S$ is 2.

(a)(i) यदि $u = x^{2} + y^{2}, v = x^{2} - y^{2}$ , जहाँ पर $x = r \cos θ, y = r \sin θ$ हैं, तब $\frac{\partial (u, v)}{\partial (r, θ)}$ ज्ञात कीजिए ।

u = x^{2} + y^{2}, v = x^{2} - y^{2}

, where

x = r \cos θ, y = r \sin θ

, then find

\frac{\partial (u, v)}{\partial (r, θ)}

Answer:

Introduction

The question asks to find the Jacobian matrix

\frac{\partial (u, v)}{\partial (r, θ)}

for the functions

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

and

v = x^{2} - y^{2}

, where

x = r \cos θ

and

y = r \sin θ

Assumptions

$u, v, x, y, r, θ$ are all real numbers.
$r \geq 0$ and $0 \leq θ < 2 π$ .

Method/Approach

To find the Jacobian matrix, we’ll first find the partial derivatives of

u

and

v

with respect to

r

and

θ

. Then we’ll arrange these partial derivatives into a matrix.

Step 1: Express $u$ and $v$ in terms of $r$ and $θ$

Let’s substitute the values

x = r \cos θ

and

y = r \sin θ

into

u

and

v

u = r^{2} \cos^{2} θ + r^{2} \sin^{2} θ = r^{2} (\cos^{2} θ + \sin^{2} θ) = r^{2}

v = r^{2} \cos^{2} θ - r^{2} \sin^{2} θ = r^{2} (\cos^{2} θ - \sin^{2} θ) = r^{2} c o s (2 θ)

Step 2: Find the Partial Derivatives

We’ll find

\frac{\partial u}{\partial r}

\frac{\partial u}{\partial θ}

\frac{\partial v}{\partial r}

, and

\frac{\partial v}{\partial θ}

$\frac{\partial u}{\partial r} = 2 r$
$\frac{\partial u}{\partial θ} = 0$
$\frac{\partial v}{\partial r} = 2 r \cos 2 θ$
$\frac{\partial v}{\partial θ} = - 2 r^{2} \sin 2 θ$

Step 3: Form the Jacobian Matrix

The Jacobian matrix

\frac{\partial (u, v)}{\partial (r, θ)}

is given by:

(\begin{array}{ll} \frac{\partial u}{\partial r} & \frac{\partial u}{\partial θ} \\ \frac{\partial v}{\partial r} & \frac{\partial v}{\partial θ} \end{array}) = (\begin{array}{cc} 2 r & 0 \\ 2 r \cos 2 θ & - 2 r^{2} \sin 2 θ \end{array})

Conclusion

The Jacobian matrix

\frac{\partial (u, v)}{\partial (r, θ)}

for the given functions is:

(\begin{matrix} 2 r & 0 \\ 2 r \cos 2 θ & - 2 r^{2} \sin 2 θ \end{matrix})

This matrix contains all the partial derivatives of

u

and

v

with respect to

r

and

θ

3.(a)(ii) यदि

\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t

है, तो

f (1)

का मान ज्ञात कीजिए ।

\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t

, then find the value of

f (1)

Answer:

Introduction

The problem asks to find the value of

f (1)

given that

\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t

We’ll use the properties of definite integrals and differentiation to solve this problem.

Assumptions

$f (t)$ is a continuous function.
$x$ and $t$ are real numbers.

Method/Approach

We’ll start by differentiating both sides of the equation with respect to

x

Leibnitz ‘s Rule :

\frac{d}{d x} \int_{u (x)}^{v (x)} f (t) d t = f (v (x)) \frac{d v}{d x} - f (u (x)) \frac{d u}{d x}

Step 1: Differentiate both sides

Differentiating both sides of the equation

\int_{0}^{x} f (t) d t = x + \int_{x}^{1} t f (t) d t

with respect to

x

, we get:

\begin{aligned} \Rightarrow \frac{d}{dx} (\int_{0}^{x} f (t) dt) = \frac{d}{dx} (x + \int_{x}^{1} tf (t) dt) \end{aligned}

\begin{aligned} f (x) = \frac{d}{d x} (x) + \frac{d}{d x} \int_{x}^{1} t f (t) d t \\ f (x) = 1 + [1 \cdot f (1) \frac{d}{d x} (1) - x \cdot f (x) \frac{d}{d x} (x)] using Leibnitz’s Rule. \\ f (x) = 1 + [0 - x f (x)] \end{aligned}

\begin{aligned} f (x) = 1 - x f (x) \\ f (x) [1 + x] = 1 \\ f (x) = \frac{1}{1 + x} \end{aligned}

Step 2: Find $f (1)$

Let’s substitute the values to find

f (1)

After substituting

x = 1

into

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

, we get:

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

Conclusion

After calculating, we find that

f (1) = \frac{1}{2}

3.(a)(iii)

\int_{a}^{b} (x - a)^{m} (b - x)^{n} d x

को बीटा-फलन के रूप में व्यक्त कीजिए ।

Express

\int_{a}^{b} (x - a)^{m} (b - x)^{n} d x

in terms of Beta function.

Answer:

Introduction

The problem asks us to express the integral

\int_{a}^{b} (x - a)^{m} (b - x)^{n} d x

in terms of the Beta function.

Definitions

The Beta function, $B (x, y)$ , is defined as $B (x, y) = \int_{0}^{1} t^{x - 1} (1 - t)^{y - 1} d t$

Method/Approach

Start with the integral $I_{m, n} = \int_{a}^{b} (x - a)^{m} (b - x)^{n} d x$ .
Use a change of variables to transform the integral into a form involving the Beta function.

Work/Calculations

Step 1: Define the Integral

I_{m, n} = \int_{a}^{b} (x - a)^{m} (b - x)^{n} d x

Step 2: Change of Variables

Let

t = \frac{x - a}{b - a}

. Then

x = a + (b - a) t

and

d x = (b - a) d t

The limits of integration also change:

When $x = a$ , $t = 0$
When $x = b$ , $t = 1$

The integral becomes:

\begin{aligned} I_{m, n} = \int_{0}^{1} [a + (b - a) t - a]^{m} [b - a - (b - a) t]^{n} (b - a) d t \\ I_{m, n} = \int_{0}^{1} (b - a)^{m} t^{m} (b - a)^{n} [1 - t]^{n} (b - a) d t \\ I_{m, n} = (b - a)^{m + n + 1} \int_{0}^{1} t^{m} (1 - t)^{n} d t \end{aligned}

Step 3: Relate to Beta Function

The integral

\int_{0}^{1} t^{m} (1 - t)^{n} d t

is precisely the definition of the Beta function

B (m + 1, n + 1)

So, we can write:

I_{m, n} = (b - a)^{m + n + 1} B (m + 1, n + 1)

Conclusion

The integral

\int_{a}^{b} (x - a)^{m} (b - x)^{n} d x

can be expressed in terms of the Beta function as

(b - a)^{m + n + 1} B (m + 1, n + 1)

. This relationship is established by using a change of variables to transform the integral into the form of the Beta function.

3.(b) अचर त्रिज्या

r

का एक गोला मूल-बिंदु

O

से गुजरता है तथा अक्षों को

A, B, C

बिन्दुओं पर काटता है ।

O

से समतल

A B C

पर खींचे गए लंब-पाद का बिन्दुपथ ज्ञात कीजिए ।

A sphere of constant radius

r

passes through the origin

O

and cuts the axes at the points

A, B

and

C

. Find, the locus of the foot of the perpendicular drawn from

O

to the plane

A B C

Answer:

Introduction

The problem asks us to find the locus of the foot of the perpendicular drawn from the origin

O

to the plane

A B C

, where

A

B

, and

C

are the points where a sphere of constant radius

r

cuts the axes.

Assumptions

The sphere has a constant radius $r$ .
The sphere passes through the origin $O$ .

Definitions

Locus: A set of points satisfying a certain condition.
Perpendicular: A line segment that is at a right angle to a given line or plane.

Work/Calculations

Let the equation of the plane

A B C

\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 - - - - (1)

Then the coordinates of

A, B, C

are respectively

(a, 0, 0), (0, b, 0), (0, 0, c)

Hence the equation of the sphere

O A B C

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

The radius

R

of the sphere (2) is given

\sqrt{{(\frac{a}{2})}^{2} + {(\frac{b}{2})}^{2} + {(\frac{c}{2})}^{2}} = R

i.e.,

a^{2} + b^{2} + c^{2} = 4 R^{2} - - - - (3)

The direction cosine to the normal to the plane (1) are proportional to

\frac{1}{a} + \frac{1}{b} + \frac{1}{c}

.
Therefore the equation of the perpendicular from

O

to the plane (1) are given by

\frac{x}{1 / a} + \frac{y}{1 / b} + \frac{z}{1 / c} = r (say) - - - - (4)

Any point on the line (4) is

(\frac{r}{a} + \frac{r}{b} + \frac{r}{c})

. This point will lie on the plane (1) if

(\frac{r}{a^{2}} + \frac{r}{b^{2}} + \frac{r}{c^{2}}) = 1 \Rightarrow r = \frac{1}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} = \frac{1}{a^{- 2} + b^{- 2} + c^{- 2}}

Let

(x^{'}, y^{'}, z^{'})

be the foot of the perpendicular drawn from

O

to plane (1). Then

\begin{aligned} x^{'} = \frac{r}{a} = \frac{a^{- 1}}{a^{- 2} + b^{- 2} + c^{- 2}} \\ y^{'} = \frac{r}{b} = \frac{b^{- 1}}{a^{- 2} + b^{- 2} + c^{- 2}} \\ z^{'} = \frac{r}{c} = \frac{c^{- 1}}{a^{- 2} + b^{- 2} + c^{- 2}} \end{aligned}

Now

\begin{array}{l} x^{' 2} + y^{' 2} + z^{' 2} = \frac{1}{a^{- 2} + b^{- 2} + c^{- 2}} \\ x^{' - 2} + y^{' - 2} + z^{' - 2} = \frac{a^{2} + b^{2} + c^{2}}{a^{- 2} + b^{- 2} + c^{- 2}} \end{array}} - - - - (5)

Using equation (3) and (5), we get

{(x^{' 2} + y^{' 2} + z^{' 2})}^{2} (x^{' - 2} + y^{' - 2} + z^{' - 2}) = 4 R^{2}

Hence, the required locus of

(x^{'}, y^{'}, z^{'})

{(x^{' 2} + y^{' 2} + z^{' 2})}^{2} (x^{' - 2} + y^{' - 2} + z^{' - 2}) = 4 R^{2}

Conclusion

The problem aimed to find the locus of the foot of the perpendicular drawn from the origin

O

to the plane

A B C

, where

A

B

, and

C

are the points where a sphere of constant radius

r

cuts the axes.

After a series of calculations involving the equations of the sphere and the plane, as well as the properties of perpendiculars and direction cosines, the locus of the foot of the perpendicular was found to be:

{(x^{' 2} + y^{' 2} + z^{' 2})}^{2} (x^{' - 2} + y^{' - 2} + z^{' - 2}) = 4 R^{2}

This equation describes the set of points that the foot of the perpendicular can occupy, given the conditions specified in the problem.

3.(c)(i) सिद्ध कीजिए कि एक वास्तविक सममित आव्यूह के दो भिन्न अभिलक्षणिक मानों के संगत अभिलक्षणिक सदिश, लांबिक हैं।

Prove that the eigen vectors, corresponding to two distinct eigen values of a real symmetric matrix, are orthogonal.

Answer:

Introduction

The problem asks us to prove that the eigenvectors corresponding to two distinct eigenvalues of a real symmetric matrix are orthogonal.

Assumptions

The matrix is real and symmetric.
We have two distinct eigenvalues.

Definitions

Eigenvector: A non-zero vector that only gets scaled when a matrix is applied to it.
Eigenvalue: The scalar by which an eigenvector gets scaled when a matrix is applied to it.
Orthogonal: Two vectors are orthogonal if their dot product is zero.

Theorem

For a real symmetric matrix

A

, the eigenvectors corresponding to distinct eigenvalues are orthogonal.

Method/Approach

Let

A

be a real symmetric matrix. Let

λ_{1}

and

λ_{2}

be two distinct eigenvalues of

A

, and let

x

and

y

be their corresponding eigenvectors. Then we have:

A x = λ_{1} x and A y = λ_{2} y

We aim to show that

x

and

y

are orthogonal, i.e.,

x \cdot y = 0

Work/Calculations

Step 1: Multiply the first equation by

y^{T}

from the left:

y^{T} A x = λ_{1} y^{T} x

Step 2: Multiply the second equation by

x^{T}

from the left:

x^{T} A y = λ_{2} x^{T} y

Step 3: Since

A

is symmetric,

A^{T} = A

. Therefore,

y^{T} A x = x^{T} A y

Step 4: Subtract the two equations:

λ_{1} y^{T} x - λ_{2} x^{T} y = 0

(λ_{1} - λ_{2}) y^{T} x = 0

Step 5: Since

λ_{1} \neq λ_{2}

, we must have

y^{T} x = 0

Conclusion

We have successfully proven that the eigenvectors corresponding to two distinct eigenvalues of a real symmetric matrix are orthogonal, as their dot product

y^{T} x

is zero. This result is fundamental in the theory of real symmetric matrices and has applications in various fields like physics, computer science, and engineering.

3.(c)(ii) दो वर्ग आव्यूह

A

तथा

B

जिनकी कोटि, 2 है के लिए दर्शाइए कि अनुरेख

(A B) =

अनुरेख

(B A)

। अतैव दर्शाइए कि

A B - B A \neq I_{2}

जहाँ

I_{2}

एक 2 -कोटि का तत्समक आव्यूह है ।

For two square matrices

A

and

B

of order 2 , show that trace

(A B) = trace (B A)

. Hence show that

A B - B A \neq I_{2}

, where

I_{2}

is an identity matrix of order 2.

Answer:

Introduction

We are tasked with proving two mathematical statements related to square matrices of order 2:

$trace (A B) = trace (B A)$
$A B - B A \neq I_{2}$ , where $I_{2}$ is the identity matrix of order 2.

Assumptions

$A$ and $B$ are square matrices of order 2.
$I_{2}$ is the identity matrix of order 2.

Definitions

Trace of a Matrix: The trace of a square matrix is the sum of the elements on its main diagonal.
Identity Matrix: A square matrix in which all the elements of the principal diagonal are ones and all other elements are zeros.

Method/Approach

We will use basic matrix multiplication rules and properties of the trace function to prove these statements.

Proof for $trace (A B) = trace (B A)$

Let

A = (\begin{matrix} a & b \\ c & d \end{matrix})

and

B = (\begin{matrix} w & x \\ y & z \end{matrix})

Calculate $A B$ and $B A$

A B = (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} w & x \\ y & z \end{matrix}) = (\begin{matrix} a w + b y & a x + b z \\ c w + d y & c x + d z \end{matrix})

B A = (\begin{matrix} w & x \\ y & z \end{matrix}) (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} w a + x c & w b + x d \\ y a + z c & y b + z d \end{matrix})

Calculate the trace of $A B$ and $B A$

trace (A B) = a w + b y + c x + d z

trace (B A) = w a + x c + y b + z d

Show that the traces are equal

trace (A B) = a w + b y + c x + d z = w a + x c + y b + z d = trace (B A)

Proof for $A B - B A \neq I_{2}$

Calculate $A B - B A$

A B - B A = (\begin{matrix} a w + b y & a x + b z \\ c w + d y & c x + d z \end{matrix}) - (\begin{matrix} w a + x c & w b + x d \\ y a + z c & y b + z d \end{matrix}) = (\begin{matrix} (a w + b y) - (w a + x c) & (a x + b z) - (w b + x d) \\ (c w + d y) - (y a + z c) & (c x + d z) - (y b + z d) \end{matrix})

Compare with $I_{2}$

The identity matrix

I_{2}

(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

. For

A B - B A

to be equal to

I_{2}

, the diagonal elements must be 1, and the off-diagonal elements must be 0.

However, the diagonal elements of

A B - B A

are

(a w + b y) - (w a + x c)

and

(c x + d z) - (y b + z d)

, which are not guaranteed to be 1. Therefore,

A B - B A \neq I_{2}

Conclusion

We have successfully proven that

trace (A B) = trace (B A)

for square matrices

A

and

B

of order 2. We have also shown that

A B - B A \neq I_{2}

, where

I_{2}

is the identity matrix of order 2.

4.(a)(i) निम्नलिखित आव्यूह का पंक्ति-समानीत सोपानक रूप में समानयन कीजिए एवं अतैव इसकी कोटि भी ज्ञात कीजिए ।

A = [\begin{array}{ccccc} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 2 & 6 & 2 & 6 & 2 \\ 3 & 9 & 1 & 10 & 6 \end{array}]

Reduce the following matrix to a row-reduced echelon form and hence also, find its rank:

A = [\begin{array}{ccccc} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 2 & 6 & 2 & 6 & 2 \\ 3 & 9 & 1 & 10 & 6 \end{array}]

Answer:

Introduction

We are given a matrix

A

and are tasked with reducing it to its row-reduced echelon form (RREF). Additionally, we need to find the rank of the matrix.

Definitions

Row-Reduced Echelon Form (RREF): A matrix is in RREF if it satisfies the following conditions:
1. All zero rows are at the bottom of the matrix.
2. The leading entry of each nonzero row occurs to the right of the leading entry of the previous row.
3. The leading entry in any nonzero row is 1.
4. All entries in the column above and below a leading 1 are zero.
Rank of a Matrix: The rank of a matrix is the dimension of the column space (or equivalently, the row space), which is the maximum number of linearly independent columns (or rows).

Method/Approach

We will use the Wolfram plugin to calculate the RREF of the given matrix

A

Calculate the RREF of $A$

The given matrix

A

is:

A = [\begin{array}{ccccc} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 2 & 6 & 2 & 6 & 2 \\ 3 & 9 & 1 & 10 & 6 \end{array}]

\begin{aligned} R_{3} \leftarrow R_{3} - 2 \times R_{1} \\ = [\begin{array}{ccccc} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 0 & 0 & - 2 & - 2 & 0 \\ 3 & 9 & 1 & 10 & 6 \end{array}] \\ R_{4} \leftarrow R_{4} - 3 \times R_{1} \\ = [\begin{array}{ccccc} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 0 & 0 & - 2 & - 2 & 0 \\ 0 & 0 & - 5 & - 2 & 3 \end{array}] \\ R_{2} \leftarrow R_{2} \div 2 \\ = [\begin{array}{ccccc} 1 & 3 & 2 & 4 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & - 2 & - 2 & 0 \\ 0 & 0 & - 5 & - 2 & 3 \end{array}] \end{aligned}

\begin{aligned} R_{1} \leftarrow R_{1} - 2 \times R_{2} \\ = [\begin{array}{ccccc} 1 & 3 & 0 & 2 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & - 2 & - 2 & 0 \\ 0 & 0 & - 5 & - 2 & 3 \end{array}] \\ R_{3} \leftarrow R_{3} + 2 \times R_{2} \\ = [\begin{array}{ccccc} 1 & 3 & 0 & 2 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & - 5 & - 2 & 3 \end{array}] \\ R_{4} \leftarrow R_{4} + 5 \times R_{2} \\ = [\begin{array}{lllll} 1 & 3 & 0 & 2 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 3 & 3 \end{array}] \end{aligned}

After interchanging rows

R_{3} \leftrightarrow R_{4}

= [\begin{array}{lllll} 1 & 3 & 0 & 2 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 3 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{array}]

\begin{aligned} R_{3} \leftarrow R_{3} \div 3 \\ = [\begin{array}{lllll} 1 & 3 & 0 & 2 & 1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}] \end{aligned}

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

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

Find the Rank of $A$

The rank of a matrix is equal to the number of nonzero rows in its RREF. In this case, there are 3 nonzero rows, so the rank of

A

is 3.

Conclusion

The row-reduced echelon form of the given matrix

A

is:

[\begin{array}{ccccc} 1 & 3 & 0 & 0 & - 1 \\ 0 & 0 & 1 & 0 & - 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}]

The rank of the matrix

A

is 3.

4.(a)(ii) सम्मिश्र संख्या क्षेत्र पर आव्यूह

A = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array})

के अभिलक्षणिक मान तथा संगत अभिलक्षणिक सदिशों को ज्ञात कीजिए ।

Find the eigen values and the corresponding eigen vectors of the matrix

A = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array})

, over the complex-number field.

Answer:

Introduction

We are tasked with finding the eigenvalues and corresponding eigenvectors of the given

2 \times 2

matrix

A

over the complex-number field. The matrix

A

is:

A = (\begin{array}{cc} 0 & - i \\ i & 0 \end{array})

Definitions

Eigenvalue: A scalar $λ$ is an eigenvalue of a matrix $A$ if there exists a nonzero vector $x$ such that $A x = λ x$ .
Eigenvector: A vector $x$ is an eigenvector corresponding to an eigenvalue $λ$ if $A x = λ x$ and $x \neq 0$ .

Method/Approach

To find the eigenvalues and eigenvectors, we will solve the characteristic equation

det (A - λ I) = 0

and then find the eigenvectors corresponding to each eigenvalue.

Eigenvalues

\begin{aligned} | A - λ I | = 0 \\ | \begin{array}{cc} (- λ) & - i \\ i & (- λ) \end{array} | = 0 \\ ∴ (- λ) \times (- λ) - - i \times i = 0 \\ ∴ (λ^{2}) - 1 = 0 \\ ∴ (λ^{2} - 1) = 0 \\ ∴ (λ - 1) (λ + 1) = 0 \\ ∴ (λ - 1) = 0 or (λ + 1) = 0 \end{aligned}

∴

The eigenvalues of the matrix

A

are given by

λ = 1, - 1

Eigenvector for $λ_{1} = 1$

Substitute the Eigenvalue into the Equation

(A - λ I) x = (\begin{array}{cc} 0 - 1 & - i \\ i & 0 - 1 \end{array}) x = 0

\Rightarrow (\begin{array}{cc} - 1 & - i \\ i & - 1 \end{array}) (\begin{matrix} x \\ y \end{matrix}) = 0

Solve the System of Equations

- 1 x - i y = 0 (Equation 1)

i x - 1 y = 0 (Equation 2)

From Equation 1, we get

x = - i y

If we choose

y = 1

, then

x = - i

So, a possible eigenvector corresponding to

λ_{1} = 1

v_{1} = [\begin{matrix} - i \\ 1 \end{matrix}]

Eigenvector for $λ_{2} = - 1$

Substitute the Eigenvalue into the Equation

(A - λ I) x = (\begin{array}{cc} 0 + 1 & - i \\ i & 0 + 1 \end{array}) x = 0

\Rightarrow (\begin{array}{cc} 1 & - i \\ i & 1 \end{array}) (\begin{matrix} x \\ y \end{matrix}) = 0

Solve the System of Equations

1 x - i y = 0 (Equation 3)

i x + 1 y = 0 (Equation 4)

From Equation 3, we get

x = i y

If we choose

y = 1

, then

x = i

So, a possible eigenvector corresponding to

λ_{2} = - 1

v_{2} = [\begin{matrix} i \\ 1 \end{matrix}]

Conclusion

The eigenvalues of the given matrix

A

are

λ_{1} = 1

and

λ_{2} = - 1

. The corresponding eigenvectors are

v_{1} = [\begin{matrix} - i \\ 1 \end{matrix}]

and

v_{2} = [\begin{matrix} i \\ 1 \end{matrix}]

, respectively.

4.(b) दर्शाइए कि ऐस्ट्रॉइड :

x^{2 / 3} + y^{2 / 3} = a^{2 / 3}

का पूरा क्षेत्रफल

\frac{3}{8} π a^{2}

है।

Show that the entire area of the Astroid :

x^{2 / 3} + y^{2 / 3} = a^{2 / 3}

\frac{3}{8} π a^{2}

Answer:

Introduction

We are tasked with finding the area of the astroid curve defined by

x^{2 / 3} + y^{2 / 3} = a^{2 / 3}

Assumptions

$a > 0$
The curve is symmetric about both the x-axis and y-axis.

Method/Approach

To find the area of the astroid, we can integrate the function that defines the curve. The equation of the astroid is

x^{2 / 3} + y^{2 / 3} = a^{2 / 3}

. Solving for

y

, we get:

y = {(a^{2 / 3} - x^{2 / 3})}^{3 / 2}

Since the curve is symmetric about both axes, we can find the area of one quadrant and then multiply it by 4.

\begin{aligned} Required Area = 4 \int_{0}^{a} y d x \\ = 4 \int_{0}^{a} {(a^{2 / 3} - x^{2 / 3})}^{3 / 2} d x \\ = 4 \int_{\frac{π}{2}}^{0} (a \sin 3 θ) (- 3 a \cos^{2} θ \sin θ) d θ \end{aligned}

(Let

x = a \cos^{2} θ

)

\begin{aligned} = 12 a^{2} \int_{0}^{\frac{π}{2}} \sin^{4} θ \cos^{2} θ d θ \\ = 12 a^{2} \times \frac{3}{6} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \frac{π}{2} \\ = (\frac{3 π a^{2}}{8}) square unit \end{aligned}

Conclusion

The entire area of the astroid

x^{2 / 3} + y^{2 / 3} = a^{2 / 3}

\frac{3}{8} π a^{2}

\begin{aligned} \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}, \\ \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} \end{aligned}

को अंतर्विष्ट करने वाले समतल का समीकरण ज्ञात कीजिए । दी गई रेखाओं के प्रतिच्छेद बिंदु को भी ज्ञात कीजिए।

Find equation of the plane containing the lines

\begin{aligned} \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}, \\ \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} . \end{aligned}

Also find the point of intersection of the given lines.

Answer:

Introduction

The problem is to find the equation of the plane containing two given lines and to find their point of intersection. The lines are given as:

\begin{aligned} \frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}, \\ \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5} . \end{aligned}

Definitions

Line in 3D Space: A line in 3D space can be represented using parametric equations.
Plane in 3D Space: A plane in 3D space can be represented using the equation $A x + B y + C z + D = 0$ .

Method/Approach

Step 1: Find the Equation of the Plane

The equation of the plane containing the given lines can be found using the determinant method:

| \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{matrix} | = 0

For the given lines, this becomes:

| \begin{matrix} x + 1 & y + 3 & z + 5 \\ 3 & 5 & 7 \\ 1 & 3 & 5 \end{matrix} | = 0

After expanding the determinant, we get:

4 (x + 1) - 8 (y + 3) + 4 (z + 5) = 0

\Rightarrow 4 x - 8 y + 4 z = 0

\Rightarrow x - 2 y + z = 0

Step 2: Find the Point of Intersection

Express the Lines in Parametric Form

First, let’s express the lines in parametric form:

Line 1: $x = 3 t_{1} - 1$ , $y = 5 t_{1} - 3$ , $z = 7 t_{1} - 5$
Line 2: $x = t_{2} + 2$ , $y = 3 t_{2} + 4$ , $z = 5 t_{2} + 6$

Equate the Parametric Equations

To find the point of intersection, we equate the parametric equations of the two lines:

$3 t_{1} - 1 = t_{2} + 2$
$5 t_{1} - 3 = 3 t_{2} + 4$
$7 t_{1} - 5 = 5 t_{2} + 6$

Solve for $t_{1}$ and $t_{2}$

Solve these equations to find

t_{1}

and

t_{2}

\begin{aligned} 3 t_{1} - t_{2} = 3, \\ 5 t_{1} - 3 t_{2} = 7, \\ 7 t_{1} - 5 t_{2} = 11, \end{aligned}

We find

t_{1} = \frac{1}{2}

and

t_{2} = - \frac{3}{2}

Find the Point of Intersection

Now, let’s substitute

t_{1} = \frac{1}{2}

and

t_{2} = - \frac{3}{2}

back into the parametric equations of the lines to find the point of intersection.

For Line 1, the parametric equations are:

x = 3 t_{1} - 1, y = 5 t_{1} - 3, z = 7 t_{1} - 5

After substituting

t_{1} = \frac{1}{2}

, we get:

(x, y, z) = (3 \times \frac{1}{2} - 1, 5 \times \frac{1}{2} - 3, 7 \times \frac{1}{2} - 5)

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

Conclusion

The equation of the plane containing the given lines is

x - 2 y + z = 0

The point of intersection of the given lines is

(\frac{1}{2}, - \frac{1}{2}, - \frac{3}{2})

खण्ड ‘B’ SECTION ‘

B

‘

5.(a) अवकल समीकरण :

\frac{d^{2} y}{d x^{2}} + 2 y = x^{2} e^{3 x} + e^{x} \cos 2 x

को हल कीजिए ।

Solve the differential equation :

\frac{d^{2} y}{d x^{2}} + 2 y = x^{2} e^{3 x} + e^{x} \cos 2 x

Answer:

Introduction

The problem asks to solve the second-order linear differential equation:

\frac{d^{2} y}{d x^{2}} + 2 y = x^{2} e^{3 x} + e^{x} \cos 2 x

Method/Approach

To solve this differential equation, we use symbolic operators to find the general solution.

Solve the Differential Equation

The differential equation is:

\frac{d^{2} y}{d x^{2}} + 2 y = x^{2} e^{3 x} + e^{x} \cos 2 x

(D^{2} + 2) y = x^{2} e^{3 x} + e^{x} \cos (2 x)

The Auxiliary equation is

m^{2} + 2 = 0

\begin{aligned} m^{2} = - 2 \\ m = \pm \sqrt{- 2} \\ m = \pm i \sqrt{2} \end{aligned}

Complementary function

(C \cdot F \cdot) = c_{1} Cos (\sqrt{2} x) + c_{2} Sin (\sqrt{2} x)

Particular Integral (P.I.)

= \frac{1}{(D^{2} + 2)} [x^{2} e^{3 x} + e^{x} \cos (2 x)]

\begin{matrix} P . I . = \frac{1}{(D^{2} + 2)} x^{2} e^{3 x} + \frac{1}{(D^{2} + 2)} e^{x} \cos (2 x) \\ D \to D + 3 D \to D + 1 \end{matrix}

\begin{aligned} P . I . = e^{3 x} \cdot \frac{1}{(D + 3)^{2} + 2} \cdot x^{2} + \frac{e^{x} \cdot 1}{(D + 1)^{2} + 2} \cos 2 x \\ P . I . = e^{3 x} \cdot \frac{1}{D^{2} + 9 + 6 D + 2} \cdot x^{2} + e^{x} \cdot \frac{1}{D^{2} + 1 + 2 D + 2} \cos 2 x \\ P . I . = e^{3 x} \cdot \frac{1}{D^{2} + 6 D + 11} \cdot x^{2} + e^{x} \cdot \frac{1}{D^{2} + 2 D + 3} \cos 2 x \\ D^{2} ⟶ - 4 in Second Term \end{aligned}

\begin{aligned} P . I . = e^{3 x} \cdot \frac{1}{11 [1 + \frac{D^{2} + 6 D}{11}]} \cdot x^{2} + \frac{e^{x} \cdot 1}{- 4 + 2 D + 3} \cdot \cos (2 x) \\ P . I . = \frac{e^{3 x}}{11} \cdot {[1 + \frac{D^{2} + 6 D}{11}]}^{- 1} \cdot x^{2} + e^{x} \cdot \frac{1}{2 D - 1} \cdot \cos (2 x) \\ P . I . = \frac{e^{3 x}}{11} \cdot [1 - \frac{D^{2} + 6 D}{11} + {(\frac{D^{2} + 6 D}{11})}^{2} - \dots] \cdot x^{2} + e^{x} \cdot \frac{2 D + 1}{4 D^{2} - 1} \cdot \cos (2 x) \\ P . I . = \frac{e^{3 x}}{11} [x^{2} - \frac{1}{11} [D^{2} x^{2} + 6 D x^{2}] + \frac{1}{121} [D^{4} + 36 D^{2} + 12 D^{3}] x^{2} - \dots] + e^{x} \frac{(2 D + 1)}{- 17} \cos 2 x \\ P . I . = \frac{e^{3 x}}{11} [x^{2} - \frac{1}{11} [2 + 12 x] + \frac{1}{121} [0 + 72 + 0]] - \frac{e^{x}}{17} [2 D \cos (2 x) + \cos (2 x)] \end{aligned}

P . I . = \frac{e^{3 x}}{11} [x^{2} - \frac{12 x + 2}{11} + \frac{72}{121}] - \frac{e^{x}}{17} [\cos (2 x) - 4 \sin (2 x)]

P . I . = \frac{e^{3 x} (121 x^{2} - 132 x + 50)}{1331} - \frac{e^{x} (\cos (2 x) - 4 \sin (2 x))}{17}

General Solution = Complementary function + Particular Integral

y (x) = C_{1} \cos (\sqrt{2} x) + C_{2} \sin (\sqrt{2} x) + \frac{e^{3 x} (121 x^{2} - 132 x + 50)}{1331} - \frac{e^{x} (\cos (2 x) - 4 \sin (2 x))}{17}

Conclusion

The general solution of the given second-order linear differential equation is:

y (x) = C_{1} \cos (\sqrt{2} x) + C_{2} \sin (\sqrt{2} x) + \frac{e^{3 x} (121 x^{2} - 132 x + 50)}{1331} - \frac{e^{x} (\cos (2 x) - 4 \sin (2 x))}{17}

Here,

C_{1}

and

C_{2}

are arbitrary constants.

5.(b) लाप्लास रूपान्तर विधि का उपयोग करते हुए प्रारम्भिक मान समस्या :

\frac{d^{2} y}{d x^{2}} + 4 y = e^{- 2 x} \sin 2 x; y (0) = y^{'} (0) = 0

को हल कीजिए ।

Solve the initial value problem :

\frac{d^{2} y}{d x^{2}} + 4 y = e^{- 2 x} \sin 2 x; y (0) = y^{'} (0) = 0

using Laplace transform method.

Answer:

Introduction

The problem at hand is to solve the second-order ordinary differential equation (ODE) with initial conditions:

\frac{d^{2} y}{d x^{2}} + 4 y = e^{- 2 x} \sin (2 x), y (0) = 0, y^{'} (0) = 0

We will use the Laplace Transform method to find the solution.

Assumptions

The function $y (x)$ and its derivatives are continuous and well-defined.
The Laplace Transform exists for the given functions.

Method/Approach

We will proceed with the following steps:

Take the Laplace Transform of both sides of the equation.
Solve for $Y (s)$ , the Laplace Transform of $y (x)$ .
Use the inverse Laplace Transform to find $y (x)$ .

Step 1: Laplace Transform of Both Sides

Taking the Laplace Transform of both sides of the equation, we get:

L {\frac{d^{2} y}{d x^{2}}} + 4 L {y} = L {e^{- 2 x} \sin (2 x)}

Using the properties of Laplace Transforms, this becomes:

s^{2} Y (s) - s y (0) - y^{'} (0) + 4 Y (s) = \frac{2}{(s + 2)^{2} + 4}

Substituting the initial conditions

y (0) = 0

and

y^{'} (0) = 0

, we get:

s^{2} Y (s) + 4 Y (s) = \frac{2}{(s + 2)^{2} + 4}

Step 2: Solve for $Y (s)$

Simplifying, we find:

Y (s) (s^{2} + 4) = \frac{2}{(s + 2)^{2} + 4}

Y (s) = \frac{2}{(s^{2} + 4) ((s + 2)^{2} + 4)}

Step 3: Inverse Laplace Transform

To find

y (x)

, we need to take the inverse Laplace Transform of

Y (s)

. This involves partial fraction decomposition.

\begin{aligned} \Rightarrow L (y) = \frac{2}{(s^{2} + 4) (s^{2} + 4 s + 8)} = \frac{1}{10} [\frac{1 - s}{s^{2} + 4} + \frac{s + 3}{s^{2} + 4 s + 8}] \\ = \frac{1}{10} [\frac{1}{s^{2} + 4} - \frac{s}{s^{2} + 4} + \frac{s + 2}{(s + 2)^{2} + 2^{2}} + \frac{2}{2 [(s + 2)^{2} + 2^{2}]}] \end{aligned}

By using inverse Laplace Transform Formulas

y (x) = \frac{1}{10} [\frac{\sin 2 t}{2} - \cos 2 t + e^{- 2 t} \cos 2 t + \frac{e^{- 2 t} \sin 2 t}{2}]

Conclusion

The solution to the initial value problem

\frac{d^{2} y}{d x^{2}} + 4 y = e^{- 2 x} \sin (2 x), y (0) = 0, y^{'} (0) = 0

is given by:

y (x) = \frac{1}{10} [\frac{\sin 2 t}{2} - \cos 2 t + e^{- 2 t} \cos 2 t + \frac{e^{- 2 t} \sin 2 t}{2}]

This solution was obtained using the Laplace Transform method.

(c) दो छड़े $L M$ व $M N$ बिन्दु $M$ पर दृढ़ता से इस प्रकार जुड़ी हैं कि $(L M)^{2} + (M N)^{2} = (L N)^{2}$ तथा वे स्वतन्त्र रूप से साम्यावस्था में स्थिर बिन्दु $L$ पर टँगी हैं। माना कि दोनों एकसमान छड़ों का प्रति एकांक लम्बाई, भार $ω$ है। छड़ $L M$ का ऊर्ध्वाधर दिशा के साथ बने कोण को छड़ों की लम्बाई के रूप में ज्ञात कीजिए ।

Two rods

L M

and

M N

are joined rigidly at the point

M

such that

(L M)^{2} + (M N)^{2} = (L N)^{2}

and they are hanged freely in equilibrium from a fixed point

L

. Let

ω

be the weight per unit length of both the rods which are uniform. Determine the angle, which the rod

L M

makes with the vertical direction, in terms of lengths of the rods.

Answer:

Introduction

The problem involves two uniform rods

L M

and

M N

joined rigidly at point

M

and hanging freely from a fixed point

L

. The rods satisfy the condition

(L M)^{2} + (M N)^{2} = (L N)^{2}

. The weight per unit length of both rods is

ω

. We are asked to find the angle

θ

that the rod

L M

makes with the vertical direction.

Assumptions

The rods are uniform, and their weight per unit length is $ω$ .
The system is in equilibrium, so the sum of the moments about point $L$ is zero.

Method/Approach

Calculate the weights of the rods $L M$ and $M N$ and identify their points of action.
Write the equation for the sum of the moments about point $L$ being zero.
Solve for $\tan θ$ in terms of the lengths of the rods.

Step 1: Calculate the Weights

The weights of the rods

L M

and

M N

are

ω \cdot L M

and

ω \cdot M N

respectively. These weights act at the midpoints

G_{1}

and

G_{2}

L M

and

M N

Step 2: Equation for Moments

In equilibrium, the sum of the moments of the weights about point

L

is zero. Let

G_{1} P

and

G_{2} Q

be the perpendiculars from

G_{1}

and

G_{2}

to the vertical line through

L

The equation for the moments about

L

is:

ω \cdot L M \cdot G_{1} P - ω \cdot M N \cdot G_{2} S = 0 - - - - (1)

We have

\sin θ = \frac{G_{1} P}{L G_{1}}

, so

G_{1} P = \frac{L M}{2} \sin θ

Similarly,

\cos θ = \frac{G_{2} Q}{M G_{2}}

, so

G_{2} Q = \frac{M N}{2} \cos θ

And, in

\begin{aligned} △ L M R Sin θ & = \frac{M R}{L M} \\ M R & = L M \sin θ \end{aligned}

\begin{aligned} G_{2} Q = \frac{M N}{2} \cos θ \\ G_{2} S + S Q = \frac{M N}{2} \cos θ \\ G_{2} S + M R = \frac{M N}{2} \cos θ \\ G_{2} S = \frac{M N}{2} \cos θ - M R \\ G_{2} S = \frac{M N}{2} \cos θ - L M \sin θ \end{aligned}

Put these values in equation (1)

ω \cdot L M \cdot G_{1} P - ω \cdot M N \cdot G_{2} S = 0

ω \cdot L M \cdot \frac{L M}{2} \sin θ - ω \cdot M N \cdot (\frac{M N}{2} \cos θ - L M \sin θ) = 0

\frac{ω}{2} \cdot L M^{2} \sin θ = ω \cdot M N [\frac{M N \cos θ - 2 L M \sin θ}{2}]

Step 3: Solve for $\tan θ$

Divide by

\frac{ω}{2} \cos θ

\begin{aligned} L M^{2} \tan θ = M N^{2} - 2 M N \cdot L M \tan θ \\ \tan θ [L M^{2} + 2 M N \cdot L M] = M N^{2} \\ \tan θ = \frac{M N^{2}}{L M^{2} + 2 M N \cdot L M} \end{aligned}

Conclusion

The angle

θ

that the rod

L M

makes with the vertical direction is given by:

\tan θ = \frac{M N^{2}}{L M^{2} + 2 M N \cdot L M}

This angle is determined solely by the lengths of the rods

L M

and

M N

(d) यदि एक ग्रह, जो सूर्य के परितः वृत्तीय कक्षा में परिभ्रमण करता है, अचानक अपनी कक्षा में रोक दिया जाता है, तो वह समय, जिसमें वह सूर्य में गिर जाएगा, ज्ञात कीजिए। इसके गिरने के समय का ग्रह के परिभ्रमण आवर्तकाल से अनुपात भी ज्ञात कीजिए ।

If a planet, which revolves around the Sun in a circular orbit, is suddenly stopped in its orbit, then find the time in which it would fall into the Sun. Also, find the ratio of its falling time to the period of revolution of the planet.

Answer:

Introduction

The problem asks us to find the time it would take for a planet to fall into the Sun if it is suddenly stopped in its circular orbit. We are also asked to find the ratio of this falling time to the period of revolution of the planet. The planet initially revolves in a circular orbit of radius

a

around the Sun and is suddenly stopped at point

P

. The planet then starts moving towards the Sun under a given acceleration.

Assumptions

The orbit of the planet is circular with radius $a$ .
The planet is suddenly stopped, making its velocity zero.
The planet moves towards the Sun under the acceleration $\frac{μ}{{(distance)}^{2}}$ .

Definitions

$μ$ : Gravitational parameter
$a$ : Radius of the circular orbit
$v$ : Velocity of the planet
$r$ : Distance from the Sun
$t$ : Time taken to fall into the Sun
$T$ : Periodic time of planet’s revolution

Method/Approach

We will use the given equation of motion and integrate it to find the time

t

it takes for the planet to fall into the Sun. We will also find the ratio

\frac{t}{T}

Work/Calculations

Step 1: Equation of Motion

The equation of motion for the planet is given as:

v \frac{d v}{d r} = - \frac{μ}{r^{2}}

Let’s integrate this equation to find

v

in terms of

r

Step 2: Integration and Finding $v$

Integrating, we get:

v^{2} = \frac{2 μ}{r} + A

When

r = a

and

v = 0

, we find

A = - \frac{2 μ}{a}

. Therefore, the equation becomes:

v^{2} = \frac{2 μ (a - r)}{a r}

We also have

v = \frac{d r}{d t}

, so:

v = - \sqrt{2 μ} \sqrt{\frac{a - r}{a r}}

Here,

r

is decreasing.

Step 3: Finding Time $t$

We have:

\int_{a}^{0} \sqrt{\frac{r}{a - r}} d r = - \sqrt{\frac{2 μ}{a}} \int_{0}^{t} d t

After Calculating, we get:

- \sqrt{\frac{2 μ}{a}} t = - \int_{0}^{\frac{π}{2}} 2 a \cos^{2} θ d θ

Therefore,

\sqrt{\frac{2 μ}{a}} t = a \frac{π}{2}

Finally, we find

t

as:

t = \frac{\sqrt{2} π a^{3 / 2}}{4 \sqrt{μ}}

Step 4: Finding $\frac{t}{T}$

The periodic time

T

is given as

T = \frac{2 π a^{3 / 2}}{\sqrt{μ}}

. Therefore, the ratio

\frac{t}{T}

is:

\frac{t}{T} = \frac{\sqrt{2}}{8}

Conclusion

The time taken by the planet to fall into the Sun is

t = \frac{\sqrt{2} π a^{3 / 2}}{4 \sqrt{μ}}

. The ratio of this falling time to the period of revolution of the planet is

\frac{t}{T} = \frac{\sqrt{2}}{8}

5.(e) दर्शाइए कि

\nabla^{2} [\nabla \cdot (\frac{\vec{r}}{r^{2}})] = \frac{2}{r^{4}}

, जहाँ

\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}

है ।

Show that

\nabla^{2} [\nabla \cdot (\frac{\vec{r}}{r^{2}})] = \frac{2}{r^{4}}

, where

\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}

Answer:

Introduction

The question asks for the Laplacian

\nabla^{2}

of the function

\frac{1}{r}

and to show that

\nabla^{2} [\nabla \cdot (\frac{\vec{r}}{r^{2}})] = \frac{2}{r^{4}}

, where

r = \sqrt{x^{2} + y^{2} + z^{2}}

is the radial distance from the origin in a three-dimensional Cartesian coordinate system.

Assumptions

$r$ is a positive real number representing the radial distance from the origin.
We are working in a three-dimensional Cartesian coordinate system.

Work/Calculations

We know that

\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}

\begin{aligned} \vec{r} = \sqrt{x^{2} + y^{2} + z^{2}} \\ r^{2} = x^{2} + y^{2} + z^{2} - - - - (1) \end{aligned}

Differentiate equation (1) with respect to ‘

x

‘

\begin{aligned} 2 π \frac{\partial r}{\partial x} & = 2 x \\ \frac{\partial r}{\partial x} & = \frac{x}{r} \end{aligned}

Differentiate equation (1) with respect to ‘

y

‘

\begin{aligned} 2 r \frac{\partial r}{\partial y} & = 2 y \\ \frac{\partial r}{\partial y} & = \frac{y}{r} \end{aligned}

Differentiate equation (1) with respect to ‘

z

‘

\begin{aligned} 2 r \frac{\partial r}{\partial z} & = 2 z \\ \frac{\partial r}{\partial z} & = \frac{z}{r} \end{aligned}

\begin{aligned} Consider \nabla \cdot (\frac{\vec{r}}{r^{2}}) = [(\frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k}) \cdot (\frac{x \hat{i} + y \hat{j} + z \hat{k}}{r^{2}})] \\ \nabla \cdot (\frac{\vec{r}}{r^{2}}) = [\frac{\partial}{\partial x} (\frac{x}{r^{2}}) + \frac{\partial}{\partial y} (\frac{y}{r^{2}}) + \frac{\partial}{\partial z} (\frac{z}{r^{2}})] \\ \nabla \cdot (\frac{\vec{r}}{r^{2}}) = [\frac{r^{2} - x \cdot 2 r \frac{\partial r}{\partial x}}{r^{4}} + \frac{r^{2} - y \cdot 2 r \frac{\partial r}{\partial y}}{r^{4}} + \frac{r^{2} - z \cdot 2 r \frac{\partial r}{\partial z}}{r^{4}}] \\ \nabla \cdot (\frac{\vec{r}}{r^{2}}) = [\frac{r^{2} - \frac{2 r x^{2}}{r}}{r^{4}} + \frac{r^{2} - \frac{2 r y^{2}}{r}}{r^{4}} + \frac{r^{2} - \frac{2 r z^{2}}{r}}{r^{4}}] \\ \nabla \cdot (\frac{\vec{r}}{r^{2}}) = [\frac{3 r^{2} - 2 (x^{2} + y^{2} + z^{2})}{r^{4}}] \\ \nabla \cdot (\frac{\vec{r}}{r^{2}}) = [\frac{3 r^{2} - 2 r^{2}}{r^{4}}] \\ \nabla \cdot (\frac{\vec{r}}{r^{2}}) = [\frac{r^{2}}{r^{4}}] \\ \nabla \cdot (\frac{\vec{r}}{r^{2}}) = \frac{1}{r^{2}} \end{aligned}

Now we will calculate

\nabla^{2} [\nabla \cdot (\frac{\vec{r}}{r^{2}})]

\begin{aligned} \nabla^{2} [\nabla \cdot (\frac{\vec{r}}{r^{2}})] & = \nabla^{2} [\frac{1}{r^{2}}] \\ = \nabla \cdot \nabla (\frac{1}{r^{2}}) \\ = \nabla \cdot [\nabla (\frac{1}{r^{2}})] \\ = \nabla \cdot [\hat{i} \frac{\partial}{\partial x} (\frac{1}{r^{2}}) + \hat{j} \frac{\partial}{\partial y} (\frac{1}{r^{2}}) + \hat{k} \frac{\partial}{\partial z} (\frac{1}{r^{2}})] \\ = \nabla \cdot [\hat{i} {- \frac{2}{r^{3}} \frac{\partial r}{\partial x}} + \hat{j} {- \frac{2}{r^{3}} \frac{\partial r}{\partial y}} + \hat{k} {\frac{- 2}{r^{4}} \frac{\partial r}{\partial z}}] \\ = \nabla \cdot [- \frac{2}{r^{3}} \cdot \frac{x}{r} \hat{i} - \frac{2}{r^{3}} \frac{y}{r} \hat{j} - \frac{2}{r^{3}} \frac{z}{r} \hat{k}] \\ = - 2 \nabla \cdot [\frac{x \hat{i} + y \hat{j} + z \hat{k}}{r^{4}}] \\ = - 2 (\hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}) \cdot (\frac{x}{r^{4}} \hat{i} + \frac{y}{r^{4}} \hat{j} + \frac{z}{r^{4}} \hat{k}) \\ = - 2 [\frac{\partial}{\partial x} (\frac{x}{r^{4}}) + \frac{\partial}{\partial y} (\frac{y}{r^{4}}) + \frac{\partial}{\partial z} (\frac{z}{r^{4}})] \end{aligned}

\begin{aligned} \nabla^{2} [\nabla \cdot (\frac{\vec{r}}{r^{2}})] & = - 2 [\frac{r^{4} - x \cdot 4 r^{3} \frac{\partial r}{\partial x}}{r^{8}} + \frac{r^{4} - y \cdot 4 r^{3} \frac{\partial r}{\partial y}}{r^{8}} + \frac{r^{4} - z \cdot 4 r^{3} \frac{\partial r}{\partial z}}{r^{8}}] \\ = - 2 [\frac{r^{4} - \frac{4 r^{3} x^{2}}{r}}{r^{8}} + \frac{r^{4} - \frac{4 r^{3} y^{2}}{r}}{r^{8}} + \frac{r^{4} - \frac{4 r^{3} z^{2}}{r}}{r^{8}}] \\ = - 2 [\frac{3 r^{4} - 4 r^{2} (x^{2} + y^{2} + z^{2})}{r^{8}}] \\ = - 2 [\frac{3 r^{4} - 4 r^{2} \cdot r^{2}}{r^{8}}] \\ = + \frac{2 r^{4}}{r^{8}} \\ = \frac{2}{r^{4}} \end{aligned}

Conclusion

The Laplacian $\nabla^{2} (\frac{1}{r})$ of the function $\frac{1}{r}$ where $r = \sqrt{x^{2} + y^{2} + z^{2}}$ is $\frac{1}{r^{4}}$ .
The expression $\nabla^{2} [\nabla \cdot (\frac{\vec{r}}{r^{2}})] = \frac{2}{r^{4}}$ is verified.

(a) एक भारी डोरी, जिसका घनत्व एक समान नहीं है, दो बिन्दुओं से टँगी हुई है। माना कि $T_{1}, T_{2}, T_{3}$ क्रमशः कैटिनरी के बीच के बिन्दुओं $A, B, C$ पर तनाव हैं, जिन पर इसके क्षैतिज के साथ आनति कोण, सार्व अंतर $β$ के साथ समांतर श्रेढ़ी में हैं। माना कि डोरी के $A B$ तथा $B C$ भागों के भार क्रमशः $ω_{1}$ तथा $ω_{2}$ हैं। सिद्ध कीजिए
(i) $T_{1}, T_{2}$ तथा $T_{3}$ का हरात्मक माध्य $= \frac{3 T_{2}}{1 + 2 \cos β}$
(ii) $\frac{T_{1}}{T_{3}} = \frac{ω_{1}}{ω_{2}}$

A heavy string, which is not of uniform density, is hung up from two points. Let

T_{1}, T_{2}, T_{3}

be the tensions at the intermediate points

A, B, C

of the catenary respectively where its inclinations to the horizontal are in arithmetic progression with common difference

β

. Let

ω_{1}

and

ω_{2}

be the weights of the parts

A B

and

B C

of the string respectively. Prove that
(i) Harmonic mean of

T_{1}, T_{2}

and

T_{3} = \frac{3 T_{2}}{1 + 2 \cos β}

(ii)

\frac{T_{1}}{T_{3}} = \frac{ω_{1}}{ω_{2}}

Answer:

Let

O

be the lowest point of the catenary and

T_{1}, T_{2}, T_{3}

be the tensions at

A, B

and

C

respectively.

T_{0}

be the tension at o , then consider the equilibrium of the portion

O A

of the chain.

T_{0} = T_{1} \cos (α - β) - - - - (1)

Similarly, considering the equilibrium of portions

O B

and

O C

we get,

T_{0} = T_{1} \cos (α - β) = T_{2} \cos α = T_{3} \cos (α + β) - - - - (2)

Again considering the equilibrium of

A B

in vertical direction,

T_{2} \sin α - T_{1} \sin (α - β) = ω_{1} - - - - (3)

Similarly the equilibrium of portion

B C

in vertical direction gives:

T_{3} \sin (α + β) - T_{2} \sin α = ω_{2} - - - - (4)

Part (i): Harmonic Mean of $T_{1}, T_{2}, T_{3}$

The harmonic mean

H

T_{1}, T_{2}, T_{3}

is given by:

H = \frac{3}{\frac{1}{T_{1}} + \frac{1}{T_{2}} + \frac{1}{T_{3}}}

We aim to prove that

H = \frac{3 T_{2}}{1 + 2 \cos β}

Step 1: Substitute $T_{1}, T_{2}, T_{3}$ in terms of $T_{0}$

From Above, we have:

T_{1} = \frac{T_{0}}{\cos (α - β)}

T_{2} = \frac{T_{0}}{\cos (α)}

T_{3} = \frac{T_{0}}{\cos (α + β)}

Step 2: Substitute into Harmonic Mean Formula

H = \frac{3}{\frac{\cos (α - β)}{T_{0}} + \frac{\cos (α)}{T_{0}} + \frac{\cos (α + β)}{T_{0}}}

Step 3: Simplify $H$

H = \frac{3 T_{0}}{\cos (α - β) + \cos (α) + \cos (α + β)}

Using trigonometric identities, we can simplify this to the desired form.

Using the trigonometric identity

\cos (A) + \cos (B) = 2 \cos (\frac{A + B}{2}) \cos (\frac{A - B}{2})

, we get:

H = \frac{3 T_{0}}{2 \cos (α) \cos (β) + \cos (α)} = \frac{3 T_{0}}{\cos (α) (1 + 2 \cos (β))}

Since

T_{2} = \frac{T_{0}}{\cos (α)}

, we have:

H = \frac{3 T_{2}}{1 + 2 \cos (β)}

This proves the first part.

Part (ii): Ratio of $T_{1}$ to $T_{3}$

We aim to prove that

\frac{T_{1}}{T_{3}} = \frac{ω_{1}}{ω_{2}}

Using the given equilibrium equations, we can express

ω_{1}

and

ω_{2}

in terms of

T_{1}, T_{2}, T_{3}

and

α, β

Multiply equation (3) by

T_{3}

& equation (4) by

T_{1}

and subtracting,
We get

ω_{1} T_{3} - ω_{2} T_{1} = T_{2} T_{3} \sin α - T_{1} T_{3} \sin (α - β) - T_{1} T_{3} \sin (α + β) + T_{1} T_{2} \sin α

ω_{1} T_{s} - ω_{2} T_{1} = T_{1} T_{s} \frac{\sin α \cos (α - β)}{\cos α} - T_{1} T_{s} \sin (α - β) - T_{1} T_{s} \sin (α + β) + T_{1} T_{s} \frac{\cos (α + β) \sin α}{\cos α}

ω_{1} T_{3} - ω_{2} T_{1} = T_{1} T_{3} [\frac{Sin α}{Cos α} {Cos (α - β) + Cos (α + β)} - {Sin (α - β) + Sin (α + β)}]

Using

\cos C + \cos D = 2 \cos \frac{C + D}{2} \cos \frac{C - D}{2}

and

\sin C + \sin D = 2 \sin \frac{C + D}{2} \cos \frac{C - D}{2}

\begin{aligned} ω_{1} T_{3} - ω_{2} T_{1} & = T_{1} T_{3} [\frac{\sin α}{\cos α} \cdot (2 \cos α \cos β) - (2 \sin α \cos β)] \\ ω_{1} T_{3} - ω_{2} T_{1} & = T_{1} T_{3} [2 \sin α \cos β - 2 \sin α \cos β] \\ ω_{1} T_{3} - ω_{2} T_{1} & = T_{1} T_{3} [0] \\ ω_{1} T_{3} - ω_{2} T_{1} & = 0 \end{aligned}

\begin{aligned} ω_{1} T_{3} = ω_{2} T_{1} \\ \frac{T_{1}}{T_{3}} = \frac{ω_{1}}{ω_{2}} \end{aligned}

This proves the second part.

6.(b) सभी अन्तर्त्रस्त (शामिल) चरणों को दर्शति हुए समीकरण :

\frac{d^{2} y}{d x^{2}} + (\tan x - 3 \cos x) \frac{d y}{d x} + 2 y \cos^{2} x = \cos^{4} x

को पूर्ण रूप से हल कीजिए ।

Solve the equation:

\frac{d^{2} y}{d x^{2}} + (\tan x - 3 \cos x) \frac{d y}{d x} + 2 y \cos^{2} x = \cos^{4} x

completely by demonstrating all the steps involved.

Answer:

Given equation

\frac{d^{2} y}{d x^{2}} + (\tan x - 3 \cos x) \frac{d y}{d x} + 2 y \cos^{2} x = \cos^{4} x

Compare this equation with

\frac{d^{2} y}{d x^{2}} + P \frac{d y}{d x} + Q y = R

We get

\begin{aligned} P = \tan x - 3 \cos x \\ Q = 2 \cos^{2} x \\ R = \cos^{4} x \end{aligned}

To solve the given differential equation we change the independent variable from

x

z

by choosing

z

such that

\frac{Q}{{(\frac{d z}{d x})}^{2}} = \frac{2 \cos^{2} x}{{(\frac{d z}{d x})}^{2}} = Constant = 2 (say)

Then

\begin{aligned} {(\frac{d z}{d x})}^{2} = \cos^{2} x \\ \frac{d z}{d x} = \cos x \\ z = \sin x \end{aligned}

Now by the substitution

z = \sin x

, the given differential equation is transformed into

\frac{d^{2} y}{d z^{2}} + P_{1} \frac{d y}{d z} + Q_{1} y = R_{1}

Where

P_{1} = \frac{\frac{d^{2} z}{d x^{2}} + P \cdot \frac{d z}{d x}}{{(\frac{d z}{d x})}^{2}}

\begin{aligned} P_{1} = \frac{- \sin x + (\tan x - 3 \cos x) \cdot \cos x}{\cos^{2} x} \\ P_{1} = - 3 \\ Q_{1} = \frac{Q}{{(\frac{d z}{d x})}^{2}} = \frac{2 \cos^{2} x}{\cos^{2} x} = 2 \\ R_{1} = \frac{R}{{(\frac{d z}{d x})}^{2}} = \frac{\cos^{4} x}{\cos^{2} x} = \cos^{2} x \\ P_{1} = 1 - \sin^{2} x \\ P_{1} = 1 - z^{2} \end{aligned}

So, The transformed equation is

\begin{aligned} \frac{d^{2} y}{d z^{2}} - \frac{3 d y}{d z} + 2 y = 1 - z^{2} \\ (D^{2} - 3 D + 2) y = 1 - z^{2} \end{aligned}

A.E.

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

\begin{aligned} (m - 1) (m - 2) = 0 \\ m = 1, 2 \end{aligned}

C.F.

y_{c} = c_{1} e^{z} + c_{2} e^{2 z}

P.I.

y_{p} = \frac{1}{D^{2} - 3 D + 2} (1 - z^{2})

\begin{aligned} y_{p} = \frac{1}{2 [1 + \frac{D^{2} - 3 D}{2}]} (1 - z^{2}) \\ y_{p} = \frac{1}{2} {[1 + \frac{D^{2} - 3 D}{2}]}^{- 1} (1 - z^{2}) \\ y_{p} = \frac{1}{2} [1 - (\frac{D^{2} - 3 D}{2}) + {(\frac{D^{2} - 3 D}{2})}^{2} \dots -] (1 - z^{2}) \\ y_{p} = \frac{1}{2} [1 - (\frac{D^{2} - 3 D}{2}) + \frac{1}{4} [D^{4} + 9 D^{2} - 6 D^{3}] \dots] (1 - z^{2}) \end{aligned}

\begin{aligned} D (1 - z^{2}) = - 2 z \\ D^{2} (1 - z^{2}) = - 2 \\ D^{3} (1 - z^{2}) = 0 \end{aligned}

y_{ρ} = \frac{1}{2} [(1 - z^{2}) - \frac{1}{2} [- 2 + 6 z] + \frac{1}{4} [0 - 18 - 0]]

Higher derivatives are zero.

\begin{aligned} y_{p} = \frac{1 - z^{2}}{2} - \frac{(- z + 6 z)}{4} - \frac{18}{8} \\ y_{p} = \frac{1}{2} - \frac{z^{2}}{2} + \frac{1}{2} - \frac{3 z}{2} - \frac{9}{4} \\ y_{p} = - \frac{z^{2}}{2} - \frac{3 z}{2} - \frac{5}{4} \end{aligned}

So,

G . S = C . F . + P . I .

\begin{aligned} y = y_{c} + y_{p} \\ y = c_{1} e^{z} + c_{2} e^{2 z} - \frac{z^{2}}{2} - \frac{3 z}{2} - \frac{5}{4} \end{aligned}

Put

z = \sin x

y = c_{1} e^{\sin x} + c_{2} e^{2 \sin x} - \frac{\sin^{2} x}{2} - \frac{3 \sin x}{2} - \frac{5}{4}

Which is our required solution.

6.(c)

\int_{C} \vec{F} \cdot d \vec{r}

का मान निकालिए,
जहाँ

C, x y

-समतल में एक स्वैच्छिक संवृत वक्र है तथा

\vec{F} = \frac{- y \hat{i} + x \hat{j}}{x^{2} + y^{2}}

है।

Evaluate

\int_{C} \vec{F} \cdot d \vec{r}

, where

C

is an arbitrary closed curve in the

x y

-plane and

\vec{F} = \frac{- y \hat{i} + x \hat{j}}{x^{2} + y^{2}}

Answer:

To evaluate the line integral

\int_{C} \vec{F} \cdot d \vec{r}

, where

C

is an arbitrary closed curve in the

x y

-plane and

\vec{F} = \frac{- y \hat{i} + x \hat{j}}{x^{2} + y^{2}}

, we can use the following steps:

Step 1: Check for Conservative Vector Field

First, let’s check if

\vec{F}

is a conservative vector field. A vector field

\vec{F} = P \hat{i} + Q \hat{j}

is conservative if

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

For

\vec{F} = \frac{- y \hat{i} + x \hat{j}}{x^{2} + y^{2}}

, we have:

P = \frac{- y}{x^{2} + y^{2}}, Q = \frac{x}{x^{2} + y^{2}}

Let’s find

\frac{\partial P}{\partial y}

and

\frac{\partial Q}{\partial x}

to check if the field is conservative.

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

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

Step 2: Evaluate the Partial Derivatives

To find the partial derivatives, we can use the quotient rule for differentiation, which is

\frac{d}{d x} (\frac{u}{v}) = \frac{v u^{'} - u v^{'}}{v^{2}}

Let’s evaluate these derivatives.

The partial derivatives are:

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

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

Since

\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}

, the vector field

\vec{F}

is conservative.

Step 3: Find the Potential Function

Since the vector field is conservative, there exists a scalar potential function

f (x, y)

such that:

\vec{F} = \nabla f (x, y)

This means:

\frac{\partial f}{\partial x} = P, \frac{\partial f}{\partial y} = Q

We can integrate these partial derivatives to find

f (x, y)

Step 4: Evaluate the Line Integral for a Conservative Field

For a conservative vector field, the line integral over a closed curve

C

is zero:

∮_{C} \vec{F} \cdot d \vec{r} = 0

Therefore, the line integral of

\vec{F}

over any arbitrary closed curve

C

in the

x y

-plane is zero.

(a) प्रथम अष्टांशक में $y^{2} + z^{2} = 9$ तथा $x = 2$ द्वारा परिबद्ध क्षेत्र पर $\vec{F} = 2 x^{2} y \hat{i} - y^{2} \hat{j} + 4 x z^{2} \hat{k}$ के लिए गाउस अपसरण प्रमेय को सत्यापित कीजिए ।

Verify Gauss divergence theorem for

\vec{F} = 2 x^{2} y \hat{i} - y^{2} \hat{j} + 4 x z^{2} \hat{k}

taken over the region in the first octant bounded by

y^{2} + z^{2} = 9

and

x = 2

Answer:

\overset{―}{F} = 2 x^{2} y \hat{i} - y^{2} \hat{j} + 4 x z^{2} \hat{k} - - - - (1) | given

\begin{aligned} div \overset{―}{F} & = (\hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y} + \hat{k} \frac{\partial}{\partial z}) \cdot (2 x^{2} y \hat{i} - y^{2} \hat{j} + 4 x z^{2} \hat{k}) = 4 x y - 2 y + 8 x z \\ ∴ ∭_{V} div \overset{―}{F} d V & = \int_{x = 0}^{2} \int_{y = 0}^{3} \int_{z = 0}^{\sqrt{9 - y^{2}}} (4 x y - 2 y + 8 x z) d z d y d x \\ = \int_{x = 0}^{2} \int_{y = 0}^{3} [4 x y \sqrt{9 - y^{2}} - 2 y \sqrt{9 - y^{2}} + 4 x (9 - y^{2})] d y d x \\ = \int_{x = 0}^{2} {[\frac{- 4 x}{2} \cdot \frac{2}{3} {(9 - y^{2})}^{3 / 2} + \frac{2}{3} {(9 - y^{2})}^{3 / 2} + 36 x y - \frac{4}{3} x y^{3}]}_{0}^{3} d x \\ = \int_{x = 0}^{2} (108 x - 18 x) d x = {(54 x^{2} - 18 x)}_{0}^{2} \\ = 216 - 36 = 180 - - - - (2) \end{aligned}

To evaluate

\iint_{S} \overset{―}{F} \cdot \hat{n} d S

The surface

S

consists of five parts.

\iint_{S} \overset{―}{F} \cdot \hat{n} d S = \iint_{OABC} \overset{―}{F} \cdot \hat{n} d S + \iint_{OCE} \overset{―}{F} \cdot \hat{n} d S + \iint_{OADE} \overset{―}{F} \cdot \hat{n} d S + \iint_{ABD} \overset{―}{F} \cdot \hat{n} d S + \iint_{BDEC} \overset{―}{F} \cdot \hat{n} d S - - - - (3)

Over the surface OABC,

z = 0, d z = 0, \hat{n} = - \hat{k}, d S = d x d y

\iint_{OABC} \overset{―}{F}, \hat{n} d S = \iint (2 x^{2} y \hat{i} - y^{2} \hat{j}) (- \hat{k}) d x d y = 0 - - - - (4)

Over the surface OCE,

x = 0, d x = 0, \hat{n} = - \hat{i}, d S = d y d z

\iint_{OCE} \overset{―}{F} \cdot \hat{n} d S = \iint (- y^{2} \hat{j}) \cdot (- \hat{i}) d y d z = 0 - - - - (5)

Over the surface OADE,

y = 0, d y = 0, \hat{n} = - \hat{j}, d 8 = d z d x

\iint_{ODE} \overset{―}{F}, \hat{n} d S = \iint (4 x z^{2} \hat{k}), (- \hat{j}) d z d x = 0 - - - - (6)

Over the surface ABD,

x = 2, d x = 0, \hat{A} = \hat{f}, d 8 = d y d z

\begin{aligned} \iint_{ABC} \overset{―}{F} \cdot \hat{n} d S & = \iint (8 y \hat{i} - y^{2} \hat{j} + 8 z^{2} \hat{\hat{h}}), \hat{ı} d y d z \\ = \int_{z = 0}^{3} \int_{y = 0}^{\sqrt{9 - z^{3}}} 8 y d y d z = \int_{z = 0}^{3} {(4 y^{2})}_{0}^{\sqrt{9 - z^{3}}} d z \\ = 4 \int_{z = 0}^{3} (9 - z^{2}) d z = 4 {(9 z - \frac{z^{3}}{3})}_{0}^{3} = 4 (27 - 9) = 72 - - - - (7) \end{aligned}

Over the surface BDEC,

\begin{aligned} \hat{n} & = \frac{grad ϕ}{| grad ϕ |} where ϕ = y^{2} + z^{2} - 9 = 0 \\ ∴ \hat{n} & = \frac{2 (y \hat{j} + z \hat{k})}{2 \sqrt{y^{2} + z^{2}}} = \frac{y \hat{j} + z \hat{k}}{3} \end{aligned}

\begin{aligned} \iint_{BDEC} \overset{―}{F} \cdot \hat{n} d S & = \iint_{BDEC} (2 x^{2} y \hat{i} - y^{2} \hat{j} + 4 x z^{2} \hat{k}) \cdot (\frac{y \hat{j} + z \hat{k}}{3}) \frac{d x d y}{(\hat{n} \cdot \hat{k})} \\ = \iint_{BDEC} \frac{(- y^{3} + 4 x z^{3})}{3} \frac{d x d y}{(\frac{z}{3})} = \int_{x = 0}^{2} \int_{y = 0}^{3} (- \frac{y^{3}}{z} + 4 x z^{2}) d x d y \\ = \int_{x = 0}^{2} \int_{y = 0}^{π / 2} [\frac{- 27 \sin^{3} θ}{3 \cos θ} + 4 x (9 \cos^{2} θ)] 3 \cos θ d θ d x \\ = \int_{x = 0}^{2} \int_{y = 0}^{π / 2} (- 27 \sin^{3} θ + 108 x \cos^{3} θ) d θ d x \\ = \int_{x = 0}^{2} (\frac{- 54}{3} + 108 x \times \frac{2}{3}) d x = \int_{0}^{2} (- 18 + 72 x) d x \\ = {(- 18 x + 36 x^{2})}_{0}^{2} = 108 - - - - (8) \end{aligned}

From eqn. (3),

\iint_{S} \vec{F} \cdot \hat{n} d S = 0 + 0 + 0 + 72 + 108 = 180 - - - - (9)

From results (2) and (8), it is clear that

\iint_{B} \vec{F} \cdot \hat{n} d S = ∭_{V} div \vec{F} d V

Hence, Gauss’ divergence theorem is verified.

7.(b) अवकल समीकरण :

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

के सभी सम्भव हल ज्ञात कीजिए ।

Find all possible solutions of the differential equation :

y^{2} \log y = x y \frac{d y}{d x} + {(\frac{d y}{d x})}^{2} .

Answer:

Given equation

y^{2} \log_{e} y = x y \frac{d y}{d x} + {(\frac{d y}{d x})}^{2}

\begin{array}{ll} {(\frac{d y}{d x})}^{2} + x y (\frac{d y}{d x}) - y^{2} \log_{e} y = 0 \\ p^{2} + x y p - y^{2} \log_{e} y = 0 & ∵ p = \frac{d y}{d x} \end{array}

Solving for

x

, this equation can be written as

\begin{aligned} x y p = y^{2} \log_{e} y - p^{2} \\ x = \frac{y \log_{e} y}{p} - \frac{p}{y} - - - - (1) \end{aligned}

Differentiate equation (1) with respect to ‘

y

‘ and writing

\frac{1}{p}

for

\frac{d x}{d y}

we get,

\frac{1}{p} = \frac{\log_{e} y}{p} + y \cdot \frac{1}{y} \cdot \frac{1}{p} - \frac{y \log_{e} y}{p^{2}} \frac{d p}{d y} + \frac{p}{y^{2}} - \frac{1}{y} \cdot \frac{d p}{d y}

\begin{aligned} \frac{p}{y^{2}} + \frac{\log_{e} y}{p} = \frac{1}{y} \frac{d p}{d y} (1 + \frac{y^{2}}{p^{2}} \log_{e} y) \\ \frac{p}{y^{2}} [1 + \frac{y^{2}}{p^{2}} \log_{e} y] = \frac{1}{y} \frac{d p}{d y} [1 + \frac{y^{2}}{p^{2}} \log_{e} y] \end{aligned}

\frac{p}{y^{2}} = \frac{1}{y} \frac{d p}{d y}

Cancelling the common factor on either side which will give us the singular solution of ( 1 )

\begin{aligned} \frac{p}{y} = \frac{d p}{d y} \\ \frac{d p}{p} = \frac{d y}{y} \end{aligned}

on integrating,

\begin{aligned} \log_{e} p & = \log_{e} y + \log_{e} c \\ \log_{e} p & = \log_{e} y c \\ p & = c y \end{aligned}

put this value in equation (1), we get

\begin{aligned} x = \frac{y \log_{e} y}{c y} - \frac{c y}{y} \\ x = \frac{\log_{e} y}{c} - c \end{aligned}

\log_{e} y = c x + c^{2}

which is our required solution.

(c) एक भारी कण $a$ लम्बाई की अवितान्य डोरी से एक स्थिर बिन्दु से टँगा है तथा $\sqrt{2 g h}$ वेग से क्षैतिज दिशा में प्रक्षेपित किया जाता है । यदि $\frac{5 a}{2} > h > a$ है, तो सिद्ध कीजिए कि प्रक्षेपण बिन्दु से $\frac{1}{3} (a + 2 h)$ ऊँचाई पहुँचने पर कण की वृत्तीय गति समाप्त हो जाती है । यह भी सिद्ध कीजिए कि उस कण द्वारा प्रक्षेपण बिंदु से ऊपर प्राप्य अधिकतम ऊँचाई $\frac{(4 a - h) (a + 2 h)^{2}}{27 a^{2}}$ है ।

A heavy particle hangs by an inextensible string of length

a

from a fixed point and is then projected horizontally with a velocity

\sqrt{2 g h}

\frac{5 a}{2} > h > a

, then prove that the circular motion ceases when the particle has reached the height

\frac{1}{3} (a + 2 h)

from the point of projection. Also, prove that the greatest height ever reached by the particle above the point of projection is

\frac{(4 a - h) (a + 2 h)^{2}}{27 a^{2}}

Answer:

Let a particle of mass

m

be attached to one end of a string of length a whole other end is fixed at

O

. The particle is projected horizontally with a velocity

u = \sqrt{2 g h}

from

A

P

is the portion of the particle at time t such that angle of

A O P = θ

and

arc A P = s

Then the equations of the motion of the particle are

m \frac{d^{2} s}{d t^{2}} = - m g \sin θ \to (1)

And

m \frac{v^{2}}{a} = T - m g \cos θ \to (2)

Also

s = a θ \to (3)

From (1) and (3), we have

a \frac{d^{2} θ}{d t^{2}} = - g \sin θ

Multiplying both sides by

2 a \frac{d θ}{d t}

and integrating, we have

v^{2} = {(a \frac{d θ}{d t})}^{2} = 2 a g \cos θ + A

But at the point A,

θ = 0

and

v = u = \sqrt{2 g h}

\begin{aligned} A = 2 g h - 2 a g \\ v^{2} = 2 a g \cos θ + 2 g h - 2 a g \to (4) \end{aligned}

From (2) and (4), we have

T = \frac{m}{a} (v^{2} + a g \cos θ) = \frac{m}{a} (3 a g \cos θ + 2 g h - 2 a g)

If the particle leaves the circular path at

Q

where

θ = θ_{1}

, then

T = 0

when

θ = θ_{1}

0 = \frac{m}{a} (3 a g \cos θ_{1} + 2 g h - 2 a g) \Rightarrow \cos θ_{1} = - \frac{2 h - 2 a}{3 a}

Since

\frac{5}{2} a > h > a

that is

5 a > 2 h > 2 a

, therefore

\cos θ_{1}

is negative and its absolute value is

< 1

. so

θ_{1}

is real and

\frac{1}{2} π < θ_{1} < π

Thus, the particle leaves the circular path at

Q

before arriving at the highest point

Height of the point

Q

above

A = A L = A O + O L = a + a \cos (π - θ_{1})

\begin{aligned} = a - a \cos θ_{1} \\ = a + a \frac{2 h - 2 a}{3 a} = \frac{1}{3} (a + 2 h) \end{aligned}

That is the particle leaves the circular path when it has reached a height

\frac{1}{3} (a + 2 h)

above the point of projection.

v_{1}

is the velocity of the particle at the point

Q

, then from (4), we have

\begin{aligned} v_{1}^{2} = 2 a g \cos θ_{1} + 2 g h - 2 a g \\ = - 2 a g \frac{2 h - 2 a}{3 a} + 2 g (h - a) \\ = 2 g (h - a) (1 - \frac{2}{3}) = \frac{2}{3} g (h - a) \end{aligned}

If angle of LOQ

= α

, then

α = π - θ_{1}

\cos α = \cos (π - θ_{1}) = - \cos θ_{1} = \frac{2 (h - a)}{3 a}

Thus the particle leaves the circular path at the point

Q

with velocity

v_{1} = \sqrt{{\frac{2}{3} g (h - a)}}

at an angle

α = \cos^{- 1} {\frac{2 (h - d)}{3 a}}

to the horizontal and will subsequently describe a parabolic path.

Maximum height of the particle above the point

Q

\begin{aligned} H = \frac{v_{1}^{2} \sin^{2} α}{2 g} = \frac{v_{1}^{2}}{2 g} (1 - \cos^{2} α) = \frac{1}{3} (h - a) [1 - \frac{4}{9 a^{2}} (h - a^{2}] \\ = \frac{1}{27 a^{2}} (h - a) [9 a^{2} - 4 (h^{2} - 2 a h + a^{2})] \\ = \frac{h - a}{27 a^{2}} [5 a^{2} + 8 a h - 4 h^{2}] = \frac{1}{27 a^{2}} (h - a) (a + 2 h) (5 a - 2 h) \end{aligned}

Greatest height ever reached by the particle above the point of projection

A

\begin{aligned} = A L + H = \frac{1}{2} (a + 2 h) + \frac{1}{27 a^{2}} (h - a) (a + 2 h) (5 a - 2 h) \\ = \frac{1}{27 a^{2}} (a + 2 h) [9 a^{2} + (h - a) (5 a - 2 h)] \\ = \frac{1}{27 a^{2}} (a + 2 h) [4 a^{2} + 7 a h - 2 h^{2}] \\ = \frac{1}{27 a^{2}} (a + 2 h) (a + 2 h) (4 a - h) \\ = \frac{1}{27 a^{2}} (4 a - h) (a + 2 h)^{2} \end{aligned}

(a)(i) संनाभि शांकव कुल
$\frac{x^{2}}{a^{2} + λ} + \frac{y^{2}}{b^{2} + λ} = 1; a > b > 0$ अचर हैं तथा $λ$ एक प्राचल है,
के लंबकोणीय संछेदी ज्ञात कीजिए । दर्शाइए कि दिया गया वक्र-कुल स्वलांबिक है।

Find the orthogonal trajectories of the family of confocal conics

\frac{x^{2}}{a^{2} + λ} + \frac{y^{2}}{b^{2} + λ} = 1; a > b > 0

are constants and

λ

is a parameter. Show that the given family of curves is self orthogonal.

Answer:

Given

\frac{x^{2}}{a^{2} + λ} + \frac{y^{2}}{b^{2} + λ} = 1 \to (1)

Differentiating,

\frac{2 x}{a^{2} + λ} + \frac{2 y}{b^{2} + λ} \frac{d y}{d x} = 0

λ (x + y \frac{d y}{d x}) = - (b^{2} x + a^{2} y \frac{d y}{d x})

\begin{aligned} λ = - \frac{{b^{2} x + a^{2} y (\frac{d y}{d x})}}{{x + y (\frac{d y}{d x})}} \\ a^{2} + λ = a^{2} - b^{2} x + \frac{a^{2} y (\frac{d y}{d x})}{x + y (\frac{d y}{d x})} = \frac{(a^{2} - b^{2}) x}{x + y (\frac{d y}{d x})} and \\ b^{2} + λ = b^{2} - \frac{b^{2} x + a^{2} y \frac{d y}{d x}}{x + y (\frac{d y}{d x})} = - \frac{(a^{2} - b^{2}) y (\frac{d y}{d x})}{x + y (\frac{d y}{d x})} \end{aligned}

Putting the above values of

(a^{2} + λ)

and

(b^{2} + λ)

in (1), we have

\frac{x^{2} {x + y (\frac{d y}{d x})}}{(a^{2} - b^{2}) x} - \frac{y^{2} {x + y (\frac{d y}{d x})}}{(a^{2} - b^{2}) y (\frac{d y}{d x})} = 1

{x + y (\frac{d y}{d x})} {x - y (\frac{d x}{d y})} = a^{2} - b^{2} \to (2)

Which is the differential equation of the family (1), Replacing

\frac{d y}{d x} b y (- \frac{d x}{d y})

in (2), the differential equation of the required orthogonal trajectories is

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

{x + y (\frac{d y}{d x})} {x - y (\frac{d x}{d y})} = a^{2} - b^{2} \to (3)

Which is the same as the differential equation (2) of the given family of curves (1).

Hence, the system of the given curve (1) is self orthogonal, that is each member of the given family of curves intersects its own members orthogonally.

(a)(ii) अवकल समीकरण : $x^{2} \frac{d^{2} y}{d x^{2}} - 2 x (1 + x) \frac{d y}{d x} + 2 (1 + x) y = 0$ का व्यापक हल ज्ञात कीजिए । अतः अवकल समीकरण : $x^{2} \frac{d^{2} y}{d x^{2}} - 2 x (1 + x) \frac{d y}{d x} + 2 (1 + x) y = x^{3}$ को प्राचल विचरण विधि द्वारा हल कीजिए ।

Find the general solution of the differential equation :

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

.
Hence, solve the differential equation:

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

by the method of variation of parameters.

Answer:

Important Result:

(i)

y = x

is a solution if

P + Q x = 0

.
(ii)

y = x^{2}

is a solution if

2 + 2 Px + {Qx}^{2} = 0

.
(i)

y = e^{x}

is a part of C.F. if

1 + P + Q = 0

.
(ii)

y = e^{- x}

is a part of C.F. if

1 - P + Q = 0

.
(iii)

y = e^{m_{x}}

is a part of C.F. if

m^{2} + m P + Q = 0

.
(iv)

y = x

is a part of C.F. if

P + Q x = 0

.
(v)

y = x^{2}

is a pait of C.F. if

2 + 2 P x + Q x^{2} = 0

.
(vi)

y = x^{m}

is a part of C.F. if

m (m - 1) + P m x + Q x^{2} = 0

Very Important. If in a linear differential equation

A \frac{d^{2} y}{d x^{2}} + B \frac{d y}{d x} + C y = X,

the sum of the coefficients of

d^{2} y / d x^{2}, d y / d x

and

y

is zero i.e., if

A + B + C = 0

, then

y = e^{x}

is a part of the C.F. of the solution of the differential equation.
Solved Examples

Let’s Solve our question in parts:

Part – (a)

Given differential equation

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

Divide by

x^{2}

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

Compare with

\frac{d^{2} y}{d x^{2}} + P \frac{d y}{d x} + Q y = R

P = - \frac{2 (1 + x)}{x}, Q = \frac{2 (1 + x)}{x^{2}}, R = 0

Here

P + Q x = \frac{- 2 (1 + x)}{x} + \frac{2 (1 + x)}{x^{2}} \cdot x

\begin{aligned} P + Q x = \frac{- 2 (1 + x)}{x} + \frac{2 (1 + x)}{x} \\ P + Q x = 0 \end{aligned}

∵ P + Q x = 0

So,

y = x

is a solution of given differential equation.

Part – (b)

Given differential equation

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

Divide by

x^{2}

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

Compare with

\frac{d^{2} y}{d x^{2}} + P \frac{d y}{d x} + Q y = R

P = - \frac{2 (1 + x)}{x}, Q = \frac{2 (1 + x)}{x^{2}}, R = x

Let

y = v x

\begin{aligned} \frac{d y}{d x} & = v + x \frac{d v}{d x} \\ \frac{d^{2} y}{d x^{2}} & = 2 \frac{d v}{d x} + \frac{x d^{2} v}{d x^{2}} \end{aligned}

Put all there values in equation

\frac{d^{2} v}{d x^{2}} - \frac{2 d v}{d x} = 1

\frac{d p}{d x} - 2 p = 1 ∵ p = \frac{d v}{d x}

This equation is linear in

p

Integrating factor

(I . F .) = e^{\int - 2 d x} = e^{- 2 x}

General Solution:

p \cdot e^{- 2 x} = \int 1 \cdot e^{- 2 x} d x + c_{1}

\begin{matrix} p \cdot e^{- 2 x} = \frac{e^{- 2 x}}{- 2} + c_{1} \\ p = - \frac{1}{2} + c_{1} e^{2 x} \\ \frac{d v}{d x} = \frac{- 1}{2} + c_{1} e^{2 x} \end{matrix}

Integrate,

v = - \frac{x}{2} + \frac{c_{1}}{2} e^{2 x} + c_{2}

So, Complete solution of given equation is

\begin{aligned} y = v x \\ y = \frac{- x^{2}}{2} + \frac{c_{1} x e^{2 x}}{2} + x c_{2} \end{aligned}

8.(b) द्रव्यमान

m

का एक कण, जो की प्रक्षेपण बिन्दु से वेग

u

के साथ क्षैतिज दिशा के साथ

θ

कोण बनाने वाली दिशा में प्रक्षेपण बिन्दु से गुजरने वाले ऊर्ध्वाधर समतल में प्रक्षेपित किया जाता है, उसकी गति तथा पथ का वर्णन कीजिए । यदि कणों को उसी बिन्दु से उसी ऊर्ध्वाधर समतल में वेग

4 \sqrt{g}

के साथ प्रक्षेपित किया जाता है, तो उनके पथों के शीर्षों के बिन्दुपथ को भी निर्धारित कीजिए ।

Describe the motion and path of a particle of mass

m

which is projected in a vertical plane through a point of projection with velocity

u

in a direction making an angle

θ

with the horizontal direction. Further, if particles are projected from that point in the same vertical plane with velocity

4 \sqrt{g}

, then determine the locus of vertices of their paths.

Answer:

Given Problem

A particle of mass

m

is projected in a vertical plane with an initial velocity

u

at an angle

θ

with the horizontal direction.

Equations of Motion

Given the particle with the mass

m

, is projected from the point in the vertical plane with velocity

u

Which is projected in a vertical plane through a point of projection with velocity

u

in a direction making an angle

θ

with horizontal direction.

The initial velocity

θ

Then

m \ddot{x} = 0

and

m \ddot{y} = - m g \Rightarrow \ddot{y} = - g

\begin{aligned} \dot{y} = - g t + B when t = 0; u \sin θ = B \\ \dot{y} = - g t + u \sin θ \end{aligned}

By Integrating

\begin{aligned} y = - \frac{g t^{2}}{2} + u \sin θ t + B^{'} when t = 0, y = 0; B^{'} = 0 \\ y = u \sin θ t - \frac{g t^{2}}{2} \end{aligned}

Then

m \ddot{x} = 0, \ddot{x} = 0, \dot{x} = c

Path and Trajectory

\begin{aligned} x = u \cos θ t \end{aligned}

By eliminating t we get path and trajectory

Here

t = \frac{x}{u \cos θ}

Therefore, path of trajectory;

y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}

Multiply both sides by

- \frac{2 u^{2} \cos^{2} α}{g}

By rearranging we get

(x - \frac{u^{2} \sin α \cos α}{g}) = \frac{- 2 a^{2} \cos^{2} α}{g} (y - \frac{u^{2} \sin^{2} α}{2 g})

Above equation is in the form of

(x - h)^{2} = - \frac{4 a}{y - k}

with vertices

(\frac{u^{2} \sin θ \cos θ}{g}, \frac{u^{2} \sin^{2} θ}{2 g})

Vertices of the Path

Vertices is

(\frac{u^{2} \sin θ \cos θ}{g}, \frac{u^{2} \sin^{2} θ}{2 g})

Given that

u = 4 \sqrt{g}

Then the vertices is

(8 \sin 2 θ, 8 \sin^{2} θ)

Special Case: $u = 4 \sqrt{g}$

When

u = 4 \sqrt{g}

, the vertices become:

(8 \sin 2 θ, 8 \sin^{2} θ)

Locus

The vertices of the paths are given by:

(8 \sin 2 θ, 8 \sin^{2} θ)

To find the locus of these vertices, let’s eliminate the parameter

θ

. We can use trigonometric identities to do this.

We know that

\sin^{2} θ = \frac{1 - \cos (2 θ)}{2}

So, the coordinates of the vertices can be rewritten as:

x = 8 \sin 2 θ

y = 8 (\frac{1 - \cos (2 θ)}{2})

y = 4 - 4 \cos (2 θ)

Now, we can express

\cos (2 θ)

in terms of

x

\sin 2 θ = \frac{x}{8}

\cos^{2} (2 θ) + \sin^{2} (2 θ) = 1

\cos^{2} (2 θ) = 1 - {(\frac{x}{8})}^{2}

\cos (2 θ) = \pm \sqrt{1 - {(\frac{x}{8})}^{2}}

Substituting this into the equation for

y

y = 4 - 4 \cos (2 θ)

y = 4 \pm 4 \sqrt{1 - {(\frac{x}{8})}^{2}}

y = 4 \pm 4 \sqrt{1 - {(\frac{x}{8})}^{2}}

This equation describes the locus of the vertices of the paths when

u = 4 \sqrt{g}

(c) स्टोक्स प्रमेय का उपयोग करते हुए $\iint_{S} (\nabla \times \vec{F}) \cdot \hat{n} d S$ का मान निकालिए, जहाँ पर $\vec{F} = (x^{2} + y - 4) \hat{i} + 3 x y \hat{j} + (2 x y + z^{2}) \hat{k}$ तथा $S$ , परवलयज $z = 4 - (x^{2} + y^{2})$ का $x y$ -समतल से ऊपर का पृष्ठ है। यहाँ $\hat{n}, S$ पर एकक बहिर्मुखी अभिलंब सदिश है ।

Using Stokes’ theorem, evaluate

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

, where

\vec{F} = (x^{2} + y - 4) \hat{i} + 3 x y \hat{j} + (2 x y + z^{2}) \hat{k}

and

S

is the surface of the paraboloid

z = 4 - (x^{2} + y^{2})

above the

x y

-plane. Here,

\hat{n}

is the unit outward normal vector on

S

Answer:

Stoke’s theorem is

\int_{C} \overset{―}{F} \cdot d \bar{r} = \iint_{S} curl \overset{―}{F} \cdot \hat{n} d S

where

C

is the circle,

x^{2} + y^{2} = 4, z = 0

Also,

\overset{―}{F} \cdot d \bar{r} = [(x^{2} + y - 4) \hat{i} + 3 x y \hat{j} + (2 x z + z^{2}) \hat{k}] \cdot (d x \hat{i} + d y \hat{j} + d z \hat{k})

= (x^{2} + y - 4) d x + 3 x y d y + (2 x z + z^{2}) d z

On the circle

\begin{array}{ll} z = 0, & ∴ d z = 0 \\ x = 2 \cos ϕ, & ∴ d x = - 2 \sin ϕ d ϕ \\ y = 2 \sin ϕ, & ∴ d y = 2 \cos ϕ d ϕ \end{array}

\begin{aligned} ∴ \int_{C} \bar{F} \cdot d \bar{r} & = \int_{C} [x^{2} + y - 4) d x + 3 x y d y] \\ = \int_{0}^{2 π} (4 \cos^{2} ϕ + 2 \sin ϕ - 4) (- 2 \sin ϕ d ϕ) + 3 (2 c o s ϕ) (2 s i n ϕ) (2 c o s ϕ d ϕ) \\ = \int_{0}^{2 π} (- 8 \cos^{2} ϕ \sin ϕ - 4 \sin^{2} ϕ + 8 \sin^{2} ϕ + 24 \cos^{2} ϕ \sin ϕ) d ϕ \\ = \int_{0}^{2 π} (16 \cos^{2} ϕ \sin ϕ - 4 \sin^{2} ϕ + 8 \sin ϕ) d ϕ \\ = - 4 \int_{0}^{2 π} \sin^{2} ϕ d ϕ = - 4 \times 4 \int_{0}^{π / 2} \sin^{2} ϕ d ϕ = - 16 (\frac{1}{2} \frac{π}{2}) = - 4 π \end{aligned}

Free UPSC Mathematics Optional Paper-1 2021 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Introduction

Assumptions

Method/Approach

Work/Calculations

Step 1: Apply T T TTT to each vector in B B BBB

Step 2: Express the resulting vectors in terms of B B BBB

Step 3: Form the Matrix Representation

Conclusion

Introduction

Assumptions

Method/Approach

Work/Calculations

Step 1: Examine Δ ( x ) Δ ( x ) Delta(x)\Delta(x)Δ(x) as x → 0 x → 0 x rarr0x \rightarrow 0x→0

Step 2: Apply L’Hôpital’s Rule

Step 3: Use Determinant Properties

Conclusion

Introduction

Assumptions

Method/Approach

Work/Calculations

Step 1: Define f ( x ) f ( x ) f(x)f(x)f(x) and Examine Its Properties

Step 2: Apply Rolle’s Theorem

Conclusion

Introduction

Assumptions

Definitions

Method/Approach

Work/Calculations

Step 1: Equation of the Generator

Step 2: Intersection with Plane z = 0 z = 0 z=0z=0z=0

Step 3: Locus of Points on the Generator

Conclusion

Introduction

Assumptions

Definitions

Step 1: Establish Cone and Plane Equations

Step 2: Equations of Intersecting Lines

Step 3: Elaborating the Condition for Perpendicular Generators

Eliminating n n nnn

Roots and Products

Step 4: Condition for Perpendicular Generators

Step 4: Reciprocal Cone

Conclusion

Introduction

Assumptions

Definitions

Method/Approach

Work/Calculations

Step 1: Find f x f x f_(x)f_xfx and f y f y f_(y)f_yfy

Step 2: Find f x y ( 0 , 0 ) f x y ( 0 , 0 ) f_(xy)(0,0)f_{xy}(0,0)fxy(0,0) and f y x ( 0 , 0 ) f y x ( 0 , 0 ) f_(yx)(0,0)f_{yx}(0,0)fyx(0,0)

Conclusion

Introduction

Assumptions

Definitions

Method/Approach

Step 1: Zero Vector in S S SSS

Step 2: Closed Under Vector Addition

Step 3: Closed Under Scalar Multiplication

Step 4: Finding Bases and Dimension

Step 5: Dimension of S S SSS

Conclusion

Introduction

Assumptions

Method/Approach

Step 1: Express u u uuu and v v vvv in terms of r r rrr and θ θ theta\thetaθ

Step 2: Find the Partial Derivatives

Step 3: Form the Jacobian Matrix

Conclusion

Introduction

Assumptions

Method/Approach

Step 1: Differentiate both sides

Step 2: Find f ( 1 ) f ( 1 ) f(1)f(1)f(1)

Conclusion

Introduction

Definitions

Method/Approach

Work/Calculations

Step 1: Define the Integral

Step 1: Apply $T$ to each vector in $B$

Step 2: Express the resulting vectors in terms of $B$

Step 1: Examine $Δ (x)$ as $x \to 0$

Step 1: Define $f (x)$ and Examine Its Properties

Step 2: Intersection with Plane $z = 0$

Eliminating $n$

Step 1: Find $f_{x}$ and $f_{y}$

Step 2: Find $f_{x y} (0, 0)$ and $f_{y x} (0, 0)$

Step 1: Zero Vector in $S$

Step 5: Dimension of $S$

Step 1: Express $u$ and $v$ in terms of $r$ and $θ$

Step 2: Find $f (1)$

Proof for $trace (A B) = trace (B A)$

Proof for $A B - B A \neq I_{2}$

Calculate the RREF of $A$

Find the Rank of $A$

Eigenvector for $λ_{1} = 1$

Eigenvector for $λ_{2} = - 1$

Solve for $t_{1}$ and $t_{2}$

Step 2: Solve for $Y (s)$

Step 3: Solve for $\tan θ$