Free UPSC Mathematics Optional Paper-1 2018 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 12, 2025ByAbstract Classes

खण्ड-A / SECTION-A

(a) मान लीजिये कि $A$ एक $3 \times 2$ आव्यूह है और $B$ एक $2 \times 3$ आव्यूह है। दर्शाइये कि $C = A \cdot B$ एक अव्युत्क्रमणणीय आव्यूह है।

Let

A

be a

3 \times 2

matrix and

B

2 \times 3

matrix. Show that

C = A \cdot B

is a singular matrix.

Answer:

Introduction

The problem asks us to show that the product

C = A \cdot B

is a singular matrix, given that

A

is a

3 \times 2

matrix and

B

is a

2 \times 3

matrix. A singular matrix is one that does not have an inverse, which means its determinant is zero.

Work/Calculations

Step 1: Dimensions of $C$

First, let’s find the dimensions of the resulting matrix

C

when

A

and

B

are multiplied.

The dimensions of

C

can be determined by the outer dimensions of

A

and

B

. In this case,

A

3 \times 2

and

B

2 \times 3

, so

C

will be

3 \times 3

Step 2: Rank of $C$

The rank of

C

is limited by the smaller of the two ranks of

A

and

B

. Since

A

3 \times 2

, its rank can be at most 2. Similarly,

B

2 \times 3

, so its rank can also be at most 2.

Therefore, the rank of

C

can be at most 2.

Rank (C) \leq min (Rank (A), Rank (B)) \leq 2

Step 3: Determinant of $C$

For a

3 \times 3

matrix to be invertible (non-singular), its rank must be 3. However, we’ve established that the rank of

C

can be at most 2. Therefore,

C

must be singular.

To confirm, the determinant of a singular matrix is zero:

Det (C) = 0

Conclusion

We have shown that the matrix

C = A \cdot B

will be a

3 \times 3

matrix with a rank of at most 2. Since the rank is less than 3,

C

is a singular matrix, and its determinant is zero. Therefore,

C = A \cdot B

is indeed a singular matrix.

(b) आधार सदिशों

e_{1} = (1, 0)

और

e_{2} = (0, 1)

को

α_{1} = (2, - 1)

एवं

α_{2} = (1, 3)

के रैखिक संयोग के रूप में ब्यक्त कीजिये।

Express basis vectors

e_{1} = (1, 0)

and

e_{2} = (0, 1)

as linear combinations of

α_{1} = (2, - 1)

and

α_{2} = (1, 3)

Answer:

Introduction

The problem asks us to express the basis vectors

e_{1} = (1, 0)

and

e_{2} = (0, 1)

as linear combinations of the vectors

α_{1} = (2, - 1)

and

α_{2} = (1, 3)

. In other words, we want to find constants

c_{1}

and

c_{2}

such that:

e_{1} = c_{1} α_{1} + c_{2} α_{2}

e_{2} = d_{1} α_{1} + d_{2} α_{2}

Work/Calculations

Step 1: Setting up the Equations for $e_{1}$

The equation for

e_{1}

can be written as:

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

Breaking it down into components, we get:

1 = 2 c_{1} + c_{2}

0 = - c_{1} + 3 c_{2}

Step 2: Solving for $c_{1}$ and $c_{2}$

After Calculating, we get:

c_{1} = \frac{3}{7}, c_{2} = \frac{1}{7}

Step 3: Setting up the Equations for $e_{2}$

The equation for

e_{2}

can be written as:

(0, 1) = d_{1} (2, - 1) + d_{2} (1, 3)

Breaking it down into components, we get:

0 = 2 d_{1} + d_{2}

1 = - d_{1} + 3 d_{2}

Step 4: Solving for $d_{1}$ and $d_{2}$

After Calculating, we get:

d_{1} = - \frac{1}{7}, d_{2} = \frac{2}{7}

Conclusion

The basis vector

e_{1} = (1, 0)

can be expressed as a linear combination of

α_{1}

and

α_{2}

as follows:

e_{1} = \frac{3}{7} α_{1} + \frac{1}{7} α_{2}

Similarly, the basis vector

e_{2} = (0, 1)

can be expressed as:

e_{2} = - \frac{1}{7} α_{1} + \frac{2}{7} α_{2}

Thus, both

e_{1}

and

e_{2}

can be represented as linear combinations of

α_{1}

and

α_{2}

lim_{z \to 1} (1 - z) \tan \frac{π z}{2}

का अस्तित्व है या कि नहीं। अगर यह सीमा विघ्यमान है, तो इसका मान ज्ञात कीजिये।

Determine if

lim_{z \to 1} (1 - z) \tan \frac{π z}{2}

exists or not. If the limit exists, then find its value.

Answer:

Let

t = 1 - z

\begin{aligned} \Rightarrow z = 1 - t \\ z \to 1 \\ \Rightarrow t = 0 \end{aligned}

Now, we have:

\begin{aligned} lim_{t \to 0} t \tan (\frac{π}{2} (1 - t)) \\ = lim_{t \to 0} t \tan (\frac{π}{2} - \frac{π}{2} t) \\ = lim_{t \to 0} t \cot (\frac{π}{2} t) \\ = lim_{t \to 0} \frac{t}{\tan \frac{π}{2} t} \end{aligned}

Now, multiply both the numerator and denominator by

\frac{2}{π}

= lim_{t \to 0} \frac{2}{π} (\frac{\frac{π}{2} t}{\tan \frac{π}{2} t}) = \frac{2}{π}

So, the limit

lim_{z \to 1} (1 - z) \tan \frac{π z}{2}

exists, and its value is

\frac{2}{π}

(d) सीमा

lim_{n \to \infty} \frac{1}{n^{2}} \sum_{r = 0}^{n - 1} \sqrt{n^{2} - r^{2}}

का मान ज्ञात कीजिये।

Find the limit

lim_{n \to - \infty} \frac{1}{n^{2}} \sum_{r = 0}^{n - 1} \sqrt{n^{2} - r^{2}}

Answer:

We start by rewriting the given limit as follows:

lim_{n \to \infty} \frac{1}{n^{2}} \sum_{a = 0}^{n - 1} \sqrt{n^{2} - r^{2}}

Next, we’ll express

\frac{\sqrt{n^{2} - r^{2}}}{n^{2}}

as:

\frac{1}{n} [\sqrt{1 - {(\frac{r}{n})}^{2}}]

Now, we rewrite the limit as a sum:

\begin{aligned} lim_{n \to \infty} \sum_{r = 0}^{n - 1} \frac{1}{n} [\sqrt{1 - {(\frac{r}{n})}^{2}}] & = lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n} \sqrt{1 - {(\frac{r}{n})}^{2}} \\ = \int_{0}^{1} \sqrt{1 - x^{2}} d x \\ = {[\frac{x \sqrt{1 - x^{2}}}{2} + \frac{1}{2} \sin^{- 1} x]}_{0}^{1} \\ = \frac{1}{2} \sin^{- 1} (1) \end{aligned}

Now, we calculate

\sin^{- 1} (1)

\frac{1}{2} \sin^{- 1} (1) = \frac{1}{2} \times \frac{π}{2} = \frac{π}{4}

Hence, the limit

lim_{n \to - \infty} \frac{1}{n^{2}} \sum_{r = 0}^{n - 1} \sqrt{n^{2} - r^{2}}

\frac{π}{4}

(e) सरल रेखा

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

का समतल

x + y + 2 z = 6

पर प्रक्षेपण ज्ञात कीजिये।

Find the projection of the straight line

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

on the plane

x + y + 2 z = 6

Answer:

Introduction

The problem asks us to find the projection of the straight line

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

onto the plane

x + y + 2 z = 6

Work/Calculations

Step 1: Find the Direction Vector of the Line

The direction vector of the line is given by

[2, 3, - 1]

Step 2: Find the Normal Vector of the Plane

The normal vector of the plane

x + y + 2 z = 6

[1, 1, 2]

Step 3: Find the Projection Vector

The projection of the direction vector of the line onto the plane is given by

Projection = Direction Vector - (\frac{Direction Vector \cdot Normal Vector}{Normal Vector \cdot Normal Vector}) \times Normal Vector

After substituting the values, we get:

Projection = [2, 3, - 1] - (\frac{[2, 3, - 1] \cdot [1, 1, 2]}{[1, 1, 2] \cdot [1, 1, 2]}) \times [1, 1, 2]

After Calculating, we get:

Projection = [2, 3, - 1] - \frac{2 + 3 - 2}{1 + 1 + 4} \times [1, 1, 2] = [2, 3, - 1] - \frac{3}{6} \times [1, 1, 2]

Projection = [2, 3, - 1] - [0.5, 0.5, 1] = [1.5, 2.5, - 2]

Step 4: Find the Equation of the Projected Line

The projected line will pass through a point on the original line and will have the direction vector as the projection vector. Taking the point

(1, 1, - 1)

on the original line, the equation of the projected line is:

\frac{x - 1}{1.5} = \frac{y - 1}{2.5} = \frac{z + 1}{- 2}

Conclusion

The projection of the given straight line

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

onto the plane

x + y + 2 z = 6

is the line

\frac{x - 1}{1.5} = \frac{y - 1}{2.5} = \frac{z + 1}{- 2} .

(a) अगर $A$ और $B$ समरूप $n \times n$ आव्यूह हैं, तो दर्शाइये कि उनके आइगेन मान एक ही हैं।

Show that if

A

and

B

are similar

n \times n

matrices, then they have the same eigenvalues.

Answer:

Introduction

The problem asks us to prove that if

A

and

B

are similar

n \times n

matrices, then they have the same eigenvalues.

Work/Calculations

Definition of Similar Matrices

Two matrices

A

and

B

are said to be similar if there exists an invertible matrix

P

such that

B = P^{- 1} A P

Definition of Eigenvalues

A scalar

λ

is an eigenvalue of a matrix

A

if there exists a non-zero vector

x

such that

A x = λ x

Step 1: Assume $A$ has an Eigenvalue $λ$

Let’s assume that

λ

is an eigenvalue of

A

, and

x

is the corresponding eigenvector, i.e.,

A x = λ x

Step 2: Multiply Both Sides by $P^{- 1}$

Multiply both sides of

A x = λ x

P^{- 1}

P^{- 1} A x = P^{- 1} λ x

Step 3: Use the Similarity Relation $B = P^{- 1} A P$

Substitute

B = P^{- 1} A P

into the equation:

P^{- 1} A x = B (P^{- 1} x)

Step 4: Show that $λ$ is an Eigenvalue of $B$

From the equation

P^{- 1} A x = B (P^{- 1} x)

, we can see that

B (P^{- 1} x) = λ (P^{- 1} x)

This shows that

λ

is also an eigenvalue of

B

with the corresponding eigenvector

P^{- 1} x

Step 5: Reverse the Argument

Similarly, we can reverse the argument to show that if

λ

is an eigenvalue of

B

, then it must also be an eigenvalue of

A

Conclusion

We have shown that if

A

and

B

are similar

n \times n

matrices, then they have the same eigenvalues. This completes the proof.

(b) बिन्दु

(1, 0)

से परबलय

y^{2} = 4 x

तक की न्यूनतम दूरी ज्ञात कीजिये।
Find the shortest distance from the point

(1, 0)

to the parabola

y^{2} = 4 x

Answer:

Introduction

The problem asks us to find the shortest distance from the point

(1, 0)

to the parabola

y^{2} = 4 x

Work/Calculations

Step 1: Parametric Equation of the Parabola

The parabola

y^{2} = 4 x

can be parameterized as follows:

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

where

t

is a parameter.

Step 2: Distance Formula

The distance

d

between the point

(1, 0)

and a point

(x, y)

on the parabola is given by:

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

Substituting the parametric equations

x = t^{2}

and

y = 2 t

, we get:

d = \sqrt{(t^{2} - 1)^{2} + (2 t)^{2}}

d = \sqrt{t^{4} - 2 t^{2} + 1 + 4 t^{2}}

d = \sqrt{t^{4} + 2 t^{2} + 1}

d = \sqrt{(t^{2} + 1)^{2}}

d = t^{2} + 1

Step 3: Minimize the Distance

To find the minimum distance, we need to minimize

d = t^{2} + 1

. The function

d

is a parabola that opens upwards, and its minimum value occurs at the vertex.

The minimum value of

d

d_{min} = 1

Conclusion

The shortest distance from the point

(1, 0)

to the parabola

y^{2} = 4 x

1

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 x

-अक्ष के चारों तरफ परिभ्रमण कर रहा है। परिक्रमित घन का आयतन ज्ञात कीजिये।

The ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

revolves about the

x

-axis. Find the volume of the solid of revolution.

Answer:

Introduction

The problem asks us to find the volume of the solid generated when the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

revolves about the

x

-axis.

Work/Calculations

Step 1: Solve for $y$ in terms of $x$

We can rewrite the equation of the ellipse as

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

y = \pm b \sqrt{1 - \frac{x^{2}}{a^{2}}}

Step 2: Volume of Solid of Revolution

The volume

V

of the solid generated by revolving the curve

y = f (x)

about the

x

-axis from

x = a

x = b

is given by

V = π \int_{a}^{b} [f (x)]^{2} d x

For our ellipse, we have

f (x) = b \sqrt{1 - \frac{x^{2}}{a^{2}}}

Step 3: Set up the Integral

The ellipse extends from

x = - a

x = a

. Therefore, the volume

V

V = π \int_{- a}^{a} {(b \sqrt{1 - \frac{x^{2}}{a^{2}}})}^{2} d x

Simplifying, we get

V = π \int_{- a}^{a} b^{2} (1 - \frac{x^{2}}{a^{2}}) d x

Step 4: Evaluate the Integral

After Calculating, we get

V = π b^{2} {[x - \frac{x^{3}}{3 a^{2}}]}_{- a}^{a}

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

V = π b^{2} [2 a - \frac{2 a}{3}]

V = \frac{4 π a b^{2}}{3}

Conclusion

The volume of the solid generated when the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

revolves about the

x

-axis is

\frac{4 π a b^{2}}{3}

(d) रेखाओं

\begin{array}{r} a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 \\ a_{2} x + b_{2} y + c_{2} z + d_{2} = 0 \end{array}

और

z

-अक्ष के बीच की न्यूनतम दूरी ज्ञात कीजिये।

Find the shortest distance between the lines

\begin{array}{r} a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 \\ a_{2} x + b_{2} y + c_{2} z + d_{2} = 0 \end{array}

and the z-axis.

Answer:

Any plane through the line

a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 = a_{2} x + b_{2} y + c_{2} z + d_{2} = 0

is given by

a_{1} x + b_{1} y + c_{1} z + d_{1} + λ (a_{2} x + b_{2} y + c_{2} z + d_{2}) = 0

If this plane is parallel to the

z

-axis, i.e.,

\frac{x}{0} = \frac{y}{0} = \frac{z}{1}

, we have:

\begin{aligned} (a_{1} + λ a_{2}) \cdot 0 + (b_{1} + λ b_{2}) \cdot 0 + (c_{1} + λ c_{2}) \cdot 1 = 0 \\ ∴ λ = - \frac{c_{1}}{c_{2}} \end{aligned}

Therefore, the equation of the plane containing the given line and parallel to the

z

-axis is:

(c_{2} a_{1} - c_{1} a_{2}) x + (b_{2} c_{1} - c_{2} b_{1}) y + d_{2} c_{1} - c_{2} d_{1} = 0

So, the shortest distance (

S . D .

) is the perpendicular distance from any point on the

z

-axis, say

(0, 0, 0)

, to this plane:

\begin{aligned} = \frac{c_{2} d_{1} - d_{2} c_{1}}{\sqrt{{(c_{2} a_{1} - c_{1} a_{2})}^{2} + {(b_{2} c_{1} - c_{2} b_{1})}^{2}}} \\ = \frac{c_{2} d_{1} - d_{2} c_{1}}{\sqrt{[{(c_{2} a_{1} - c_{1} a_{2})}^{2} + {(b_{2} c_{1} - c_{2} b_{1})}^{2}]}} \end{aligned}

Hence, the shortest distance between the given lines and the

z

-axis is represented as

S . D .

(a) रेखिक समीकरण निकाय

\begin{aligned} x + 3 y - 2 z & = - 1 \\ 5 y + 3 z & = - 8 \\ x - 2 y - 5 z & = 7 \end{aligned}

के लिये निर्धारित कीजिये कि निम्नलिखित कथर्नों में से कौन-से सही हैं और कौन-से गलत :
(i) समीकरण निकाय का कोई भी हल नही है।
(ii) समीकरण निकाय का सिर्फ एक ही हल है।
(iii) समीकरण निकाय के असीम मात्रा में अनेक हल है।

For the system of linear equations

\begin{aligned} x + 3 y - 2 z & = - 1 \\ 5 y + 3 z & = - 8 \\ x - 2 y - 5 z & = 7 \end{aligned}

Determine which of the following statements are true and which are false :
(i) The system has no solution.
(ii) The system has a unique solution.
(iii) The system has infinitely many solutions.

Answer:

Introduction

The problem asks us to determine the nature of the solutions for the given system of linear equations:

\begin{aligned} x + 3 y - 2 z & = - 1 \\ 5 y + 3 z & = - 8 \\ x - 2 y - 5 z & = 7 \end{aligned}

We have three statements to verify:
(i) The system has no solution.
(ii) The system has a unique solution.
(iii) The system has infinitely many solutions.

Work/Calculations

Step 1: Form the Augmented Matrix

The augmented matrix for the system of equations is:

[\begin{array}{cccc} 1 & 3 & - 2 & - 1 \\ 0 & 5 & 3 & - 8 \\ 1 & - 2 & - 5 & 7 \end{array}]

Step 2: Row Reduce the Matrix

Now, reduce this matrix

\begin{aligned} R_{3} \leftarrow R_{3} - R_{1} \\ = [\begin{array}{cccc} 1 & 3 & - 2 & - 1 \\ 0 & 5 & 3 & - 8 \\ 0 & - 5 & - 3 & 8 \end{array}] \end{aligned}

\begin{aligned} R_{2} \leftarrow R_{2} \div 5 \\ = [\begin{array}{cccc} 1 & 3 & - 2 & - 1 \\ 0 & 1 & 0.6 & - 1.6 \\ 0 & - 5 & - 3 & 8 \end{array}] \end{aligned}

R_{3} \leftarrow R_{3} + 5 \times R_{2}

After row reducing, we get:

[\begin{array}{cccc} 1 & 3 & - 2 & - 1 \\ 0 & 1 & 0.6 & - 1.6 \\ 0 & 0 & 0 & 0 \end{array}]

Step 3: Determine the Rank

After Calculating, the rank of the matrix is

2

Step 4: Analyze the Solution

The rank of the coefficient matrix and the augmented matrix is the same, and it is less than the number of unknowns (

3

). According to the Rouché–Capelli theorem, this means the system has infinitely many solutions.

Conclusion

Based on the row-reduced form and the rank of the matrix, we can conclude the following:
(i) The system has no solution. – False
(ii) The system has a unique solution. – False
(iii) The system has infinitely many solutions. – True

The system of equations has infinitely many solutions, as confirmed by the rank of the matrix.

(b) मान लीजिये कि

\begin{aligned} f (x, y) & = x y^{2}, & यदि y > 0 \\ = - x y^{2}, & यदि y \leq 0 \end{aligned}

निर्धारित कीजिये कि

\frac{\partial f}{\partial x} (0, 1)

और

\frac{\partial f}{\partial y} (0, 1)

में से किसका अस्तित्व है और किसका अस्तित्व नहीं है।

Let

\begin{aligned} f (x, y) & = x y^{2}, & if & y > 0 \\ = - x y^{2}, & if & y \leq 0 \end{aligned}

Determine which of

\frac{\partial f}{\partial x} (0, 1)

and

\frac{\partial f}{\partial y} (0, 1)

exists and which does not exist.

Answer:

Introduction

The function

f (x, y)

is defined as follows:

\begin{aligned} f (x, y) & = x y^{2}, & if & y > 0 \\ = - x y^{2}, & if & y \leq 0 \end{aligned}

We are asked to determine which of

\frac{\partial f}{\partial x} (0, 1)

and

\frac{\partial f}{\partial y} (0, 1)

exists and which does not.

Work/Calculations

Step 1: Partial Derivative with respect to $x$ at $(0, 1)$

The formula for the partial derivative with respect to

x

is:

\frac{\partial f}{\partial x} = lim_{h \to 0} \frac{f (0 + h, 1) - f (0, 1)}{h}

Let’s substitute the values into the formula:

For

y > 0

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

\frac{\partial f}{\partial x} = lim_{h \to 0} \frac{h \cdot 1^{2} - 0}{h}

After Calculating, we get

\frac{\partial f}{\partial x} = 1

For

y \leq 0

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

\frac{\partial f}{\partial x} = lim_{h \to 0} \frac{0 - 0}{h}

After Calculating, we get

\frac{\partial f}{\partial x} = 0

Since the values are different,

\frac{\partial f}{\partial x} (0, 1)

does not exist.

Step 2: Partial Derivative with respect to $y$ at $(0, 1)$

The formula for the partial derivative with respect to

y

is:

\frac{\partial f}{\partial y} = lim_{k \to 0} \frac{f (0, 1 + k) - f (0, 1)}{k}

Let’s substitute the values into the formula:

For

y > 0

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

\frac{\partial f}{\partial y} = lim_{k \to 0} \frac{0 \cdot (1 + k)^{2} - 0 \cdot 1^{2}}{k}

After Calculating, we get

\frac{\partial f}{\partial y} = 0

For

y \leq 0

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

\frac{\partial f}{\partial y} = lim_{k \to 0} \frac{0 \cdot (1 + k)^{2} - 0 \cdot 1^{2}}{k}

After Calculating, we get

\frac{\partial f}{\partial y} = 0

Since the values are the same,

\frac{\partial f}{\partial y} (0, 1)

exists and is

0

Conclusion

$\frac{\partial f}{\partial x} (0, 1)$ does not exist.
$\frac{\partial f}{\partial y} (0, 1)$ exists and is $0$ .

(x + y + z) (2 x + y - z) = 6 z

की उन जनक रेखाओं के समीकरणों को ज्ञात कीजिये, जो बिन्दु

(1, 1, 1)

में से गुज्जरती है।

Find the equations to the generating lines of the paraboloid

(x + y + z) (2 x + y - z) = 6 z

which pass through the point

(1, 1, 1)

Answer:

Introduction

The problem asks us to find the equations of the generating lines of the paraboloid

(x + y + z) (2 x + y - z) = 6 z

that pass through the point

(1, 1, 1)

Work/Calculations

Step 1: Generators of the Surface

The two generators of the given surface belonging to the

λ

– and

μ

-systems are given by:

x + y + z = λ z, 2 x + y - z = \frac{6}{λ}

and

x + y + z = \frac{6}{μ}, 2 x + y - 3 z = μ z

Step 2: Substituting the Point $(1, 1, 1)$

If these lines pass through the point

(1, 1, 1)

, we can substitute these values into the equations to find

λ

and

μ

For

λ

-system:

1 + 1 + 1 = λ \times 1 ⟹ λ = 3

For

μ

-system:

1 + 1 + 1 = \frac{6}{μ} ⟹ μ = 2

Step 3: Generators with Specific $λ$ and $μ$

Hence, the two generators of the opposite systems are:

For

λ = 3

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

For

μ = 2

x + y + 2 z = 3, 2 x + y - 3 z = 0

Step 4: Direction Ratios and Equations of the Lines

The direction ratios of these lines are respectively given by

- 1, 3, 1

and

- 4, 5, 1

Since the lines pass through the point

(1, 1, 1)

, their equations in symmetrical form are:

For

λ = 3

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

For

μ = 2

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

Conclusion

The equations of the generating lines of the paraboloid

(x + y + z) (2 x + y - z) = 6 z

that pass through the point

(1, 1, 1)

are:

For

λ = 3

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

For

μ = 2

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

These lines describe the shape of the paraboloid and pass through the specified point.

(d) xyz-समतल में स्थित, बिन्दुओं

(0, 0, 0), (0, 1, - 1), (- 1, 2, 0)

और

(1, 2, 3)

में से गुज़रते हुये गोले का समीकरण ज्ञात कीजिये।

Find the equation of the sphere in xyz-plane passing through the points

(0, 0, 0), (0, 1, - 1), (- 1, 2, 0)

and

(1, 2, 3)

Answer:

Introduction

The problem asks us to find the equation of the sphere in the

x y z

-plane that passes through the points

(0, 0, 0)

(0, 1, - 1)

(- 1, 2, 0)

, and

(1, 2, 3)

Work/Calculations

Step 1: General Equation of a Sphere

The general equation of a sphere in Cartesian coordinates is:

(x - h)^{2} + (y - k)^{2} + (z - l)^{2} = r^{2}

where

(h, k, l)

is the center of the sphere and

r

is the radius.

Step 2: Substitute the Points into the Equation

We can substitute each of the four points into the equation of the sphere to get four equations:

For $(0, 0, 0)$ :

h^{2} + k^{2} + l^{2} = r^{2} (Equation 1)

For $(0, 1, - 1)$ :

h^{2} + (k - 1)^{2} + (l + 1)^{2} = r^{2} (Equation 2)

For $(- 1, 2, 0)$ :

(h + 1)^{2} + (k - 2)^{2} + l^{2} = r^{2} (Equation 3)

For $(1, 2, 3)$ :

(h - 1)^{2} + (k - 2)^{2} + (l - 3)^{2} = r^{2} (Equation 4)

Step 3: Solve for the Center and Radius

We can solve these equations to find the values of

h

k

l

, and

r

Solving these equations, we find:

h = \frac{15}{14}, k = \frac{25}{14}, l = \frac{11}{14}, r^{2} = \frac{971}{196}

Step 4: Equation of the Sphere

Substituting these values back into the general equation of the sphere, we get:

(x - \frac{15}{14})^{2} + (y - \frac{25}{14})^{2} + (z - \frac{11}{14})^{2} = \frac{971}{196}

Conclusion

The equation of the sphere in the

x y z

-plane passing through the points

(0, 0, 0)

(0, 1, - 1)

(- 1, 2, 0)

, and

(1, 2, 3)

is:

(x - \frac{15}{14})^{2} + (y - \frac{25}{14})^{2} + (z - \frac{11}{14})^{2} = \frac{971}{196}

4.(a) अन्तराल

[2, 3]

पर

x^{4} - 5 x^{2} + 4

के अधिकतम और न्यूनतम मान ज्ञात कीजिये।

Find the maximum and the minimum values of

x^{4} - 5 x^{2} + 4

on the interval

[2, 3]

Answer:

Introduction

The problem asks us to find the maximum and minimum values of the function

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

on the interval

[2, 3]

Work/Calculations

Step 1: Identify the Function and Interval

The function is

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

, and the interval is

[2, 3]

Step 2: Find the Critical Points

To find the critical points, we need to take the derivative of

f (x)

and set it equal to zero:

f^{'} (x) = 4 x^{3} - 10 x

4 x^{3} - 10 x = 0

After Calculating, the critical points are

x = 0

x = - \sqrt{\frac{5}{2}}

, and

x = \sqrt{\frac{5}{2}}

However, none of these critical points lie within the interval

[2, 3]

. Therefore, we only need to evaluate the function at the endpoints of the interval.

Step 3: Evaluate the Function at the Endpoints

Let’s substitute the values

x = 2

and

x = 3

into the function

f (x)

$f (2) = 2^{4} - 5 \times 2^{2} + 4$
After Calculating, $f (2) = 0$
$f (3) = 3^{4} - 5 \times 3^{2} + 4$
After Calculating, $f (3) = 40$

Step 4: Identify the Maximum and Minimum Values

The maximum value of

f (x)

on the interval

[2, 3]

40

x = 3

The minimum value of

f (x)

on the interval

[2, 3]

0

x = 2

Conclusion

The maximum value of

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

on the interval

[2, 3]

40

x = 3

, and the minimum value is

0

x = 2

(b) समाकल

\int_{0}^{a} \int_{x / a}^{x} \frac{x d y d x}{x^{2} + y^{2}}

का मान निकालिये।

Evaluate the integral

\int_{0}^{a} \int_{x / a}^{x} \frac{x d y d x}{x^{2} + y^{2}}

Answer:

Introduction

The problem asks us to evaluate the integral

\int_{0}^{a} \int_{x / a}^{x} \frac{x d y d x}{x^{2} + y^{2}}

. We’ll attempt to change the order of integration to make the integral easier to evaluate.

Work/Calculations

Step 1: Identify the Region of Integration

The original region

R

in the

x y

-plane is defined by:

$0 \leq x \leq a$
$x / a \leq y \leq x$

Step 2: Change the Order of Integration

To change the order of integration, we need to describe the same region

R

in terms of

y

first and then

x

. Observing the original inequalities, we find:

$0 \leq y \leq a$ (since $x$ can go up to $a$ )
$a y \leq x \leq y$ (multiplying $x / a \leq y$ by $a$ and rearranging)

Step 3: Write the New Integral

The integral with the changed order of integration becomes:

\int_{0}^{a} \int_{a y}^{y} \frac{x d x d y}{x^{2} + y^{2}}

Step 4: Evaluate the Integral

To evaluate this integral, we’ll first integrate with respect to

x

and then with respect to

y

Inner Integral:

\int_{a y}^{y} \frac{x d x}{x^{2} + y^{2}}

Let’s substitute

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

d u = 2 x d x

x d x = \frac{d u}{2}

\frac{1}{2} \int \frac{d u}{u} = \frac{1}{2} \ln | u | + C

Substituting back, we get:

\frac{1}{2} \ln | x^{2} + y^{2} | |_{a y}^{y} = \frac{1}{2} \ln (y^{2} + y^{2}) - \frac{1}{2} \ln (a^{2} y^{2} + y^{2})

= \frac{1}{2} \ln (2 y^{2}) - \frac{1}{2} \ln ((a^{2} + 1) y^{2}) = \frac{1}{2} \ln (\frac{2 y^{2}}{(a^{2} + 1) y^{2}})

= \frac{1}{2} \ln (\frac{2}{a^{2} + 1})

Outer Integral:

\int_{0}^{a} \frac{1}{2} \ln (\frac{2}{a^{2} + 1}) d y = \frac{a}{2} \ln (\frac{2}{a^{2} + 1})

Conclusion

After changing the order of integration, we find that the integral

\int_{0}^{a} \int_{x / a}^{x} \frac{x d y d x}{x^{2} + y^{2}}

evaluates to

\frac{a}{2} \ln (\frac{2}{a^{2} + 1})

(0, 0, 1)

है और जिसका निर्देशक वक्र

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

है, का समीकरण ज्ञात कीजिये।

Find the equation of the cone with

(0, 0, 1)

as the vertex and

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

as the guiding curve.

Answer:

Let a line through the vertex

(0, 0, 1)

be given by:

\frac{x - 0}{l} = \frac{y - 0}{m} = \frac{z - 1}{n} \to (1)

It meets the plane

z = 0

at the point which is given by:

\frac{x}{l} = \frac{y}{m} = \frac{- 1}{n}

This gives us:

x = \frac{- l}{n}, y = - \frac{m}{n}

If this point

(- \frac{l}{n}, - \frac{m}{n}, 0)

lies on the base

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

, then:

\frac{2 l^{2}}{n^{2}} - \frac{m^{2}}{n^{2}} = 4 \to (2)

Eliminating

l

m

n

using equations (1) and (2):

\begin{matrix} 2 {(\frac{x}{z - 1})}^{2} - {(\frac{y}{z - 1})}^{2} = 4 \\ 2 x^{2} - y^{2} = 4 (z - 1)^{2} \\ 2 x^{2} - y^{2} = 4 (z^{2} + 1 - 2 z) \\ 2 x^{2} - y^{2} = 4 z^{2} + 4 - 8 z \\ 2 x^{2} - y^{2} - 4 z^{2} + 8 z - 4 = 0 \end{matrix}

So, the equation of the cone is:

2 x^{2} - y^{2} - 4 z^{2} + 8 z - 4 = 0

(d)

3 x - y + 3 z = 8

के समांतर और बिन्दु

(1, 1, 1)

में से गुजरते हुये समतल का समीकरण ज्ञात कीजिये।

Find the equation of the plane parallel to

3 x - y + 3 z = 8

and passing through the point

(1, 1, 1)

Answer:

Introduction

The problem asks us to find the equation of a plane that is parallel to

3 x - y + 3 z = 8

and passes through the point

(1, 1, 1)

Work/Calculations

Step 1: Equation of a Parallel Plane

We know that the equation of any plane parallel to

3 x - y + 3 z - 8 = 0

can be written as:

3 x - y + 3 z + k = 0

Step 2: Substituting the Point

Since the plane passes through the point

(1, 1, 1)

, we can substitute these coordinates into the equation to find the value of

k

Let’s substitute the values into the formula:

3 \cdot 1 - 1 + 3 \cdot 1 + k = 0

After Calculating, we get:

k = - 5

Step 3: Equation of the Required Plane

Substituting the value of

k

back into the equation, we get:

3 x - y + 3 z - 5 = 0

Conclusion

The equation of the plane that is parallel to

3 x - y + 3 z = 8

and passes through the point

(1, 1, 1)

3 x - y + 3 z - 5 = 0

खण्ड-B / SECTION-B

$(a)$ हल कीजिये/Solve :

y^{''} - y = x^{2} e^{2 x}

Answer:

Rewrite the equation as:

(D^{2} - 1) y = x^{2} e^{2 x}

Auxiliary equation:

m^{2} - 1 = 0

(m + 1) (m - 1) = 0

This gives us two roots:

m = - 1

and

m = 1

The Complementary Function is:

c_{1} e^{- x} + c_{2} e^{x}

Now, let’s find the Particular Integral:

\begin{aligned} Particular Integral & = \frac{1}{D^{2} - 1} \cdot x^{2} e^{2 x} \\ = e^{2 x} \cdot \frac{1}{(D + 2)^{2} - 1} x^{2} \\ = e^{2 x} \cdot \frac{1}{D^{2} + 4 + 4 D - 1} x^{2} \\ = e^{2 x} \cdot \frac{1}{D^{2} + 4 D + 3} x^{2} \end{aligned}

Simplify further:

\begin{aligned} P.I. & = e^{2 x} \cdot \frac{1}{3 [1 + \frac{D^{2} + 4 D}{3}] x^{2}} \\ = \frac{e^{2 x}}{3} \cdot {[1 + \frac{D^{2} + 4 D}{3}]}^{- 1} x^{2} \\ = \frac{e^{2 x}}{3} \cdot [1 - (\frac{D^{2} + 4 D}{3}) + {(\frac{D^{2} + 4 D}{3})}^{2} - \dots] x^{2} \\ = \frac{e^{2 x}}{3} \cdot [x^{2} - \frac{1}{3} {2 + 8 x} + \frac{1}{9} {D^{4} + 16 D^{2} + 8 D^{3}} x^{2}] \end{aligned}

Higher terms will be zero.

\begin{aligned} = \frac{e^{2 x}}{3} [x^{2} - \frac{(8 x + 2)}{3} + \frac{1}{9} [0 + 32 + 0] \\ = \frac{x^{2} e^{2 x}}{3} - \frac{(8 x + 2) e^{2 x}}{9} + \frac{32}{27} e^{2 x} \end{aligned}

So, the solution to the differential equation is:

y = c_{1} e^{- x} + c_{2} e^{x} + \frac{x^{2} e^{2 x}}{3} - \frac{(8 x + 2) e^{2 x}}{9} + \frac{32}{27} e^{2 x}

(b)

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

समीकरणों वाले बक्र के एक आय बिन्दु पर स्पर्शं-रेखा और रेखा

y = z - x = 0

के बीच का कोण ज्ञात कीजिये।

Find the angle between the tangent at a general point of the curve whose equations are

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

and the line

y = z - x = 0

Answer:

Introduction

The problem asks us to find the angle

θ

between the tangent at a general point of the curve

{\vec{r}}_{1} = 3 t \hat{i} + 3 t^{2} \hat{j} + 3 t^{3} \hat{k}

and the line

y = z - x = 0

Work/Calculations

Step 1: Derivative of the Curve

The first step is to find the derivative of

{\vec{r}}_{1}

with respect to

t

\frac{d {\vec{r}}_{1}}{d t} = 3 \hat{i} + 6 t \hat{j} + 9 t^{2} \hat{k}

Step 2: Equation of the Line

The equation of the line

y = z - x = 0

can be rewritten as:

y = 0, x = z

This gives us the direction ratios

[1, 0, 1]

for the line, and we can write its vector equation as

{\vec{r}}_{2} = \hat{i} + \hat{k}

The magnitude of

{\vec{r}}_{2}

is:

| {\vec{r}}_{2} | = \frac{1}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}}

So,

{\vec{r}}_{2} = \frac{1}{\sqrt{2}} (\hat{i} + \hat{k})

Step 3: Angle Between the Tangent and the Line

The angle

θ

between

{\vec{r}}_{1}

and

{\vec{r}}_{2}

can be found using the formula:

\cos θ = \frac{{\vec{r}}_{1} \cdot {\vec{r}}_{2}}{| {\vec{r}}_{1} | \cdot | {\vec{r}}_{2} |}

Let’s substitute the values into the formula:

\cos θ = \frac{(3 \cdot 1) + (6 t \cdot 0) + (9 t^{2} \cdot 1)}{\sqrt{9 + 36 t^{2} + 81 t^{4}} \cdot \sqrt{2}}

After simplifying, we get:

\cos θ = \frac{3 + 9 t^{2}}{\sqrt{9 + 36 t^{2} + 81 t^{4}} \cdot \sqrt{2}}

Further simplifying, we find:

\cos θ = \frac{1 + 3 t^{2}}{\sqrt{2} \cdot \sqrt{1 + 4 t^{2} + 9 t^{4}}}

Finally, the angle

θ

is:

θ = \cos^{- 1} [\frac{1 + 3 t^{2}}{\sqrt{2} \cdot \sqrt{1 + 4 t^{2} + 9 t^{4}}}]

Conclusion

The angle

θ

between the tangent at a general point of the curve

{\vec{r}}_{1} = 3 t \hat{i} + 3 t^{2} \hat{j} + 3 t^{3} \hat{k}

and the line

y = z - x = 0

is:

θ = \cos^{- 1} [\frac{1 + 3 t^{2}}{\sqrt{2} \cdot \sqrt{1 + 4 t^{2} + 9 t^{4}}}]

This formula gives us the angle

θ

in terms of

t

y^{'''} - 6 y^{''} + 12 y^{'} - 8 y = 12 e^{2 x} + 27 e^{- x}

Answer:

Rewrite the equation as:

(D^{3} - 6 D^{2} + 12 D - 8) y = 12 e^{2 x} + 27 e^{- x}

Auxiliary equation:

m^{3} - 6 m^{2} + 12 m - 8 = 0

\begin{aligned} (m - 2)^{3} = 0 \\ m = 2, 2, 2 \end{aligned}

The Complementary Function is:

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

Now, let’s find the Particular Integral:

\begin{aligned} Particular Integral = \frac{1}{D^{3} - 60^{2} + 12 D - 8} [12 e^{2 x} + 27 e^{- x}] \\ = \frac{12}{(D - 2)^{3}} \cdot e^{2 x} + 27 \frac{1}{(D - 2)^{3}} e^{- x} \end{aligned}

Simplify further:

\begin{aligned} = 12 \cdot \frac{x^{3}}{3} e^{2 x} + 27 \cdot \frac{1}{(- 1 - 2)^{3}} e^{- x} \\ = 2 x^{3} e^{2 x} - e^{- x} \end{aligned}

So, the solution to the differential equation is:

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

(d) (i)

f (t) = \frac{1}{\sqrt{t}}

का लाप्लास रूपान्तर ज्ञात कीजिये। Find the Laplace transform of

f (t) = \frac{1}{\sqrt{t}}

Answer:

We know that

L (t^{n}) = \frac{\sqrt{n + 1}}{s^{n + 1}}

Put

n = - \frac{1}{2}

L (t^{- 1 / 2}) = \frac{\sqrt{- \frac{1}{2} + 1}}{s^{- \frac{1}{2} + 1}} = \frac{\sqrt{\frac{1}{2}}}{\sqrt{s}} = \frac{\sqrt{π}}{\sqrt{s}}

Here,

\sqrt{\frac{1}{2}} = \sqrt{π}

So, the Laplace transform of

f (t) = \frac{1}{\sqrt{t}}

\frac{\sqrt{π}}{\sqrt{s}}

(ii)

\frac{5 s^{2} + 3 s - 16}{(s - 1) (s - 2) (s + 3)}

का विलोम लाप्लास रुपान्तर ज्ञात कीजिये। Find the inverse Laplace transform of

\frac{5 s^{2} + 3 s - 16}{(s - 1) (s - 2) (s + 3)}

Answer:

Introduction

The problem asks us to find the inverse Laplace transform of

\frac{5 s^{2} + 3 s - 16}{(s - 1) (s - 2) (s + 3)}

. To find the inverse Laplace transform, we’ll:

Perform partial fraction decomposition on the given expression.
Use the inverse Laplace transform formulas for each term.

Work/Calculations

Step 1: Partial Fraction Decomposition

The given expression is

\frac{5 s^{2} + 3 s - 16}{(s - 1) (s - 2) (s + 3)}

After Calculating, we get:

\frac{5 s^{2} + 3 s - 16}{(s - 1) (s - 2) (s + 3)} = \frac{2}{s - 1} + \frac{2}{s - 2} + \frac{1}{s + 3}

Step 2: Inverse Laplace Transform

Now, we’ll find the inverse Laplace transform of each term separately.

$L^{- 1} {\frac{2}{s - 1}} = 2 e^{t}$
$L^{- 1} {\frac{2}{s - 2}} = 2 e^{2 t}$
$L^{- 1} {\frac{1}{s + 3}} = e^{- 3 t}$

Combining all these terms, we get:

L^{- 1} {\frac{5 s^{2} + 3 s - 16}{(s - 1) (s - 2) (s + 3)}} = 2 e^{t} + 2 e^{2 t} + e^{- 3 t}

Conclusion

The inverse Laplace transform of

\frac{5 s^{2} + 3 s - 16}{(s - 1) (s - 2) (s + 3)}

2 e^{t} + 2 e^{2 t} + e^{- 3 t}

(e) एक कण को धरती के एक बिन्दु से प्रक्षेपित करने पर वह एक दीवार, जो प्रक्षेपण बिन्दु से

d

दूरी पर है और जिसकी ऊँचाई

h

है, को छूते हुये पार करता है। अगर यह कण ऊर्ध्वाधर तल पर गतिमान है और इसकी क्षैतिज पहुँच

R

है, तो प्रक्षेपण की उच्चता ज्ञात कीजिये।

A particle projected from a given point on the ground just clears a wall of height

h

at a distance

d

from the point of projection. If the particle moves in a vertical plane and if the horizontal range is

R

, find the elevation of the projection.

Answer:

Introduction

The problem deals with finding the angle of elevation

θ

of a particle projected from a point on the ground such that it just clears a wall of height

h

at a distance

d

. The particle moves in a vertical plane and has a horizontal range of

R

Work/Calculations

Step 1: Initial Conditions and Equations of Motion

Let the initial velocity of the particle be

u

and the angle of elevation be

θ

. The equations of motion are:

\begin{aligned} x & = u \cos θ t & (1) \\ y & = u \sin θ t - \frac{1}{2} g t^{2} & (2) \end{aligned}

Step 2: Equation of the Trajectory

The equation of the trajectory can be derived from equations (1) and (2) as:

y = x \tan θ - \frac{1}{2} \frac{g x^{2}}{u^{2} \cos^{2} θ} (3)

This equation can also be written as:

h = x \tan θ [1 - \frac{x}{R}] (4)

Step 3: Conditions for the Wall and the Ground

When the particle clears the wall,

y = h

and

x = d

. Thus:

h = d \tan θ - \frac{1}{2} g \frac{d^{2}}{u^{2} \cos^{2} θ} (5)

When the particle strikes the ground,

y = 0

and

x = R

. Thus:

0 = R \tan θ - \frac{1}{2} g \frac{R^{2}}{u^{2} \cos^{2} θ} (6)

Step 4: Correcting the Formula for $θ$

To eliminate

u^{2}

, we multiply equation (5) by

R^{2}

and equation (6) by

d^{2}

and subtract the two:

h R^{2} = d R^{2} \tan θ - d^{2} R \tan θ

Let’s simplify this to isolate

\tan θ

\begin{aligned} h R^{2} & = d R^{2} \tan θ - d^{2} R \tan θ \\ h R^{2} & = \tan θ (d R^{2} - d^{2} R) \\ \tan θ & = \frac{h R^{2}}{d R^{2} - d^{2} R} \\ \tan θ & = \frac{h R}{d (R - d)} \end{aligned}

Now we can find the angle

θ

θ = \tan^{- 1} [\frac{h R}{d (R - d)}]

Conclusion

The angle of elevation

θ

for the particle to just clear a wall of height

h

at a distance

d

and to have a horizontal range

R

is given by:

θ = \tan^{- 1} [\frac{h R}{d (R - d)}]

(a) हल कीजिये /Solve :

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

Answer:

Introduction

The given differential equation is:

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

We are asked to find the solution to this differential equation. Let

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

Work/Calculations

Step 1: Rewrite the Differential Equation in Terms of $p$

The given differential equation can be rewritten in terms of

p

y p^{2} + 2 p x - y = 0 (Equation 1)

After solving for

x

, we get:

x = \frac{1}{2} (\frac{y}{p}) - \frac{1}{2} (y p) (Equation 2)

Step 2: Differentiate $x$ with Respect to $y$

Differentiating Equation 2 with respect to

y

, we get:

\frac{d x}{d y} = \frac{1}{2} [\frac{p - y \frac{d p}{d y}}{p^{2}}] - \frac{1}{2} [p + y \frac{d p}{d y}]

Step 3: Simplify the Equation

After simplification, we have:

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

Simplifying further, we get:

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

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

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

\frac{1}{p} d p = - \frac{1}{y} d y

Step 4: Integrate Both Sides

Integrating both sides, we get:

\log p + \log y = \log c

p y = c

p = \frac{c}{y}

Step 5: Substitute Back into Equation 2

Substituting this value of

p

back into Equation 2, we get:

x = \frac{1}{2} (y \times \frac{y}{c}) - \frac{1}{2} (y \times \frac{c}{y})

After simplification, we find:

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

Conclusion

The solution to the given differential equation

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

is:

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

Where

c

is the constant of integration.

(b) एक कण, जो एक सरल रेखा में सरल आवर्त गति से चल रहा है, के पथ के केन्द्र से

x_{1}

और

x_{2}

की दूरी पर वेग क्रमशः

v_{1}

और

v_{2}

है। उसकी गति का आवर्तकाल ज्ञात कीजिये।

A particle moving with simple harmonic motion in a straight line has velocities

v_{1}

and

v_{2}

at distances

x_{1}

and

x_{2}

respectively from the centre of its path. Find the period of its motion.

Answer:

Introduction

The problem is related to a particle moving in Simple Harmonic Motion (SHM). We are given the velocities

v_{1}

and

v_{2}

at distances

x_{1}

and

x_{2}

from the center of the motion, respectively. The task is to find the period of this motion.

Work/Calculations

Step 1: General Equation for Velocity in SHM

The general equation for the velocity

v

of a particle at a distance

x

from the mean (or central) position in SHM is:

v^{2} = μ^{2} (a^{2} - x^{2})

where

a

is the amplitude and

μ

is the angular frequency.

Step 2: Plug in the Known Values

For

v_{1}

and

x_{1}

, we get:

v_{1}^{2} = μ^{2} (a^{2} - x_{1}^{2}) (Equation 1)

For

v_{2}

and

x_{2}

, we get:

v_{2}^{2} = μ^{2} (a^{2} - x_{2}^{2}) (Equation 2)

Step 3: Solve for $μ^{2}$

Subtracting Equation 1 from Equation 2, we get:

v_{2}^{2} - v_{1}^{2} = μ^{2} (x_{1}^{2} - x_{2}^{2})

Solving for

μ^{2}

, we get:

μ^{2} = \frac{v_{2}^{2} - v_{1}^{2}}{x_{1}^{2} - x_{2}^{2}} (Equation 3)

Step 4: Solve for the Periodic Time $T$

The periodic time

T

of a particle in SHM is given by:

T = \frac{2 π}{μ} (Equation 4)

Substituting Equation 3 into Equation 4, we get:

T = 2 π \sqrt{\frac{x_{1}^{2} - x_{2}^{2}}{v_{2}^{2} - v_{1}^{2}}}

Conclusion

The period

T

of a particle moving in simple harmonic motion, given velocities

v_{1}

and

v_{2}

at distances

x_{1}

and

x_{2}

respectively, is:

T = 2 π \sqrt{\frac{x_{1}^{2} - x_{2}^{2}}{v_{2}^{2} - v_{1}^{2}}}

y^{''} + 16 y = 32 \sec 2 x

Answer:

Introduction

The problem involves solving a second-order non-homogeneous ordinary differential equation (ODE) of the form:

y^{''} + 16 y = 32 \sec 2 x

We need to find the general solution to this equation, which will be in the form of

y = C.F. + P.I.

, where C.F. is the Complementary Function and P.I. is the Particular Integral.

Work/Calculations

Step 1: Find the Complementary Function (C.F.)

For the homogeneous equation

y^{″} + 16 y = 0

, the auxiliary equation is:

A.E. D^{2} + 16 = 0 ⟹ D = \pm 4 i

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

y_{CF} = c_{1} \cos 4 x + c_{2} \sin 4 x

Step 2: Calculate Wronskian $W$

We have

y_{1} = \cos 4 x

y_{2} = \sin 4 x

, and

X = 32 \sec 2 x

The Wronskian

W

is:

W = | \begin{matrix} \cos 4 x & \sin 4 x \\ - 4 \sin 4 x & 4 \cos 4 x \end{matrix} | = 4

Step 3: Calculate Particular Integral (P.I.)

The Particular Integral

y_{PI}

is defined as:

y_{PI} = u y_{1} + v y_{2}

where

u = - \int \frac{y_{2} X}{W} d x, v = \int \frac{y_{1} X}{W} d x

Let’s substitute the values:

\begin{aligned} y_{PI} & = - \cos 4 x \int \frac{\sin 4 x \cdot 32 \sec 2 x}{4} d x + \sin 4 x \int \frac{\cos 4 x \cdot 32 \sec 2 x}{4} d x \\ = - 8 \cos 4 x \int \sin 2 x \cos 2 x d x + 8 \sin 4 x \int (2 \cos^{2} 2 x - 1) d x \\ = - 16 \cos 4 x \int \sin 2 x d x + 8 \sin 4 x [\sin 2 x - \frac{1}{2} \log | \sec 2 x + \tan 2 x |] \\ = 8 \cos 4 x \cos 2 x + 8 \sin 4 x \sin 2 x - 4 \sin 4 x \log (\sec 2 x + \tan 2 x) \\ = 8 \cos 2 x - 4 \sin 4 x \log (\sec 2 x + \tan 2 x) \end{aligned}

Step 4: Write the Complete Solution (C.S.)

Finally, the complete solution is

y = y_{CF} + y_{PI}

y = c_{1} \cos 4 x + c_{2} \sin 4 x + 8 \cos 2 x - 4 \sin 4 x \log (\sec 2 x + \tan 2 x)

Conclusion

The general solution to the differential equation

y^{''} + 16 y = 32 \sec 2 x

is:

y = c_{1} \cos 4 x + c_{2} \sin 4 x + 8 \cos 2 x - 4 \sin 4 x \log (\sec 2 x + \tan 2 x)

(d) अगर गोलक

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

का पृष्ठ

S

है, तो गाउस के अपसरण प्रमेय का इस्तेमाल करते हुये

\iint_{S} [(x + z) d y d z + (y + z) d z d x + (x + y) d x d y]

का मान निकालिये।

S

is the surface of the sphere

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

, then evaluate

\iint_{S} [(x + z) d y d z + (y + z) d z d x + (x + y) d x d y]

using Gauss’ divergence theorem.

Answer:

Let

S

be the surface of the sphere given by

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

then we need to evaluate the integral

\iint_{S} [(x + z) d y d z + (y + z) d z d x + (x + y) d x d y]

Now, we define the vector field

\vec{F}

as follows:

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

According to Gauss’ Divergence Theorem:

∭ \nabla \cdot \vec{F} d V = \iint_{S} \vec{F} \cdot d \vec{S}

where

\nabla

represents the volume enclosed by the surface

S

Now, let’s calculate the divergence of

\vec{F}

\nabla \cdot \vec{F} = \nabla \cdot [(x + z) \hat{i} + (y + z) \hat{j} + (x + y) \hat{k}] = \frac{\partial}{\partial x} (x + z) + \frac{\partial}{\partial y} (y + z) + \frac{\partial}{\partial z} (x + y) = 1 + 1 + 0 = 2

So, the divergence of

\vec{F}

is 2.

Now, we can apply Gauss’ Divergence Theorem:

∭ \nabla \cdot \vec{F} d V = 2 ∭ d V = 2 \times \frac{4}{3} π a^{3} = \frac{8}{3} π a^{3}

Substituting the divergence value, we have:

∭ \nabla \cdot \vec{F} d V = \iint_{S} \vec{F} \cdot d \vec{S} = \frac{8}{3} π a^{3}

Hence, the value of the given integral is

\frac{8}{3} π a^{3}

(a) हल कीजिये/Solve :

(1 + x)^{2} y^{''} + (1 + x) y^{'} + y = 4 \cos (\log (1 + x))

Answer:

Introduction

The problem at hand is to solve the second-order non-homogeneous ordinary differential equation (ODE):

(1 + x)^{2} y^{''} + (1 + x) y^{'} + y = 4 \cos (\log (1 + x))

We aim to find the general solution,

y

, which will be a sum of the complementary function

y_{c}

and the particular integral

y_{p}

Work/Calculations

Step 1: Transform the Equation

Let

\log (1 + x) = z

, then

1 + x = e^{z}

The given differential equation can be rewritten as:

(D (D - 1) y + D y + y) = 4 \cos z

(D^{2} + D - D + 1) y = 4 \cos z

Upon simplifying, we have:

(D^{2} + 1) y = 4 \cos z

Step 2: Find the Complementary Function ( $y_{c}$ )

The auxiliary equation for the homogeneous part of the differential equation

m^{2} + 1 = 0

is:

m = \pm i

Thus, the complementary function

y_{c}

in terms of

z

is:

y_{c} = c_{1} \cos z + c_{2} \sin z

And back in terms of

x

, we have:

y_{c} = c_{1} \cos (\log (1 + x)) + c_{2} \sin (\log (1 + x))

Step 3: Find the Particular Integral ( $y_{p}$ )

The particular integral

y_{p}

is given as:

y_{p} = \frac{1}{D^{2} + 1} \cdot 4 \cos z

y_{p} = 4 \cdot \frac{z}{2} \sin z d z

y_{p} = 2 z \sin z

After converting back to

x

, we get:

y_{p} = 2 \log (1 + x) \sin (\log (1 + x))

Step 4: Write the General Solution ( $y$ )

The general solution

y

will be

y_{c} + y_{p}

\begin{aligned} y & = c_{1} \cos \log (1 + x) + c_{2} \sin \log (1 + x) + 2 \log (1 + x) \sin (\log (1 + x)) \\ = c_{1} \cos \log (1 + x) + c_{2} \sin \log (1 + x) + \log (1 + x)^{2} \sin (\log (1 + x)) \end{aligned}

Conclusion

The general solution to the given differential equation is:

y = c_{1} \cos \log (1 + x) + c_{2} \sin \log (1 + x) + \log (1 + x)^{2} \sin (\log (1 + x))

And this is the required solution.

(b) बक्र

\vec{r} = a (u - \sin u) \vec{i} + a (1 - \cos u) \vec{j} + b u \vec{k}

की वक्रता और विमोटन ज्ञात कीजिये।

Find the curvature and torsion of the curve

\vec{r} = a (u - \sin u) \vec{i} + a (1 - \cos u) \vec{j} + b u \vec{k}

Answer:

Introduction

We are tasked with finding the curvature and torsion of the curve

\vec{r} = a (u - \sin u) \vec{i} + a (1 - \cos u) \vec{j} + b u \vec{k}

. Curvature gives us an understanding of how a curve bends in space, whereas torsion explains how the curve twists.

Work/Calculations

Step 1: First Derivative ${\vec{r}}^{'} (u)$

We need to find the first derivative of

\vec{r} (u)

with respect to

u

. The derivative is:

\begin{aligned} \frac{d \vec{r}}{d u} & = \frac{d}{d u} [a (u - \sin u) \vec{i} + a (1 - \cos u) \vec{j} + b u \vec{k}] \\ = (a - a \cos u) \vec{i} + a \sin u \vec{j} + b \vec{k} \end{aligned}

Step 2: Magnitude of ${\vec{r}}^{'} (u)$

Next, we find the magnitude of

{\vec{r}}^{'} (u)

\begin{aligned} | \frac{d \vec{r}}{d u} | & = \sqrt{(a - a \cos u)^{2} + (a \sin u)^{2} + b^{2}} \\ = \sqrt{a^{2} - 2 a^{2} \cos u + a^{2} \cos^{2} u + a^{2} \sin^{2} u + b^{2}} \\ = \sqrt{2 a^{2} (1 - \cos u) + b^{2}} \end{aligned}

Step 3: Second Derivative ${\vec{r}}^{″} (u)$

The second derivative of

\vec{r}

is:

\begin{aligned} \frac{d^{2} \vec{r}}{d u^{2}} & = \frac{d}{d u} [(a - a \cos u) \vec{i} + a \sin u \vec{j} + b \vec{k}] \\ = a \sin u \vec{i} + a \cos u \vec{j} \end{aligned}

Step 4: Third Derivative ${\vec{r}}^{‴} (u)$

The third derivative of

\vec{r}

is:

\begin{aligned} \frac{d^{3} \vec{r}}{d u^{3}} & = \frac{d}{d u} [a \sin u \vec{i} + a \cos u \vec{j}] \\ = a \cos u \vec{i} - a \sin u \vec{j} \end{aligned}

Step 5: Torsion $T$

The formula for torsion

T

is:

T = \frac{{\vec{r}}^{'} \times {\vec{r}}^{″} \cdot {\vec{r}}^{‴}}{‖ {\vec{r}}^{'} \times {\vec{r}}^{″} ‖^{2}}

Step 6: Compute $| {\vec{r}}^{'} \times {\vec{r}}^{″} |$

\begin{aligned} | {\vec{r}}^{'} \times {\vec{r}}^{″} | & = \sqrt{a^{2} b^{2} \cos^{2} u + a^{2} b^{2} \sin^{2} u + a^{4} (\cos u - 1)^{2}} \\ = \sqrt{a^{2} b^{2} + a^{4} (\cos u - 1)^{2}} \end{aligned}

Step 7: Substitute into Torsion Formula

The torsion

T

becomes:

\begin{aligned} T & = \frac{a^{2} b}{a^{2} b^{2} + a^{4} (\cos u - 1)^{2}} \end{aligned}

Conclusion

We found that the curvature is given by

| \frac{d \vec{r}}{d u} | = \sqrt{2 a^{2} (1 - \cos u) + b^{2}}

and the torsion

T

\frac{a^{2} b}{a^{2} b^{2} + a^{4} (\cos u - 1)^{2}}

\begin{aligned} y^{''} - 5 y^{'} + 4 y = e^{2 t} \\ y (0) = \frac{19}{12}, y^{'} (0) = \frac{8}{3} \end{aligned}

को हल कीजिये।

Solve the initial value problem

\begin{aligned} y^{''} - 5 y^{'} + 4 y = e^{2 t} \\ y (0) = \frac{19}{12}, y^{'} (0) = \frac{8}{3} \end{aligned}

Answer:

Introduction

We are given an initial value problem described by the second-order non-homogeneous differential equation

y^{″} - 5 y^{'} + 4 y = e^{2 t}

along with the initial conditions

y (0) = \frac{19}{12}

and

y^{'} (0) = \frac{8}{3}

Work/Calculations

Step 1: Complementary Function ( $y_{c}$ )

For the homogeneous equation

y^{″} - 5 y^{'} + 4 y = 0

, the auxiliary equation is:

m^{2} - 5 m + 4 = 0

(m - 4) (m - 1) = 0

m = 4, 1

The complementary function

y_{c}

is:

y_{c} = c_{1} e^{t} + c_{2} e^{4 t}

Step 2: Particular Integral ( $y_{p}$ )

\begin{aligned} y_{p} = \frac{1}{(D - 1) (D - 4)} e^{2 t} \\ y_{p} = & \frac{1}{(D^{2} - 5 D + 4)} \cdot e^{2 t} \\ y_{p} = & \frac{e^{2 t}}{(2)^{2} - 5 \times 2 + 4} = \frac{e^{2 t}}{4 - 10 + 4} \\ y_{p} = - \frac{1}{2} e^{2 t} \end{aligned}

The particular integral (

y_{p}

) for

e^{2 t}

is:

y_{p} = - \frac{1}{2} e^{2 t}

Step 3: General Solution $y$

The general solution

y

is:

y = y_{c} + y_{p} = c_{1} e^{t} + c_{2} e^{4 t} - \frac{1}{2} e^{2 t}

Step 4: Applying Initial Conditions

Using the initial conditions

y (0) = \frac{19}{12}

and

y^{'} (0) = \frac{8}{3}

, we get:

c_{1} + c_{2} = \frac{25}{12} (Equation 1)

c_{1} + 4 c_{2} = \frac{11}{3} (Equation 2)

Step 5: Solve for $c_{1}$ and $c_{2}$

Solve equations 1 and 2 to get:

c_{1} = \frac{14}{9}, c_{2} = \frac{19}{36}

Step 6: Final Solution

Substituting

c_{1}

and

c_{2}

back into the general solution

y

, we get:

y = \frac{14}{9} e^{t} + \frac{19}{36} e^{4 t} - \frac{1}{2} e^{2 t}

Conclusion

The required solution to the initial value problem

y^{″} - 5 y^{'} + 4 y = e^{2 t}

with the initial conditions

y (0) = \frac{19}{12}

and

y^{'} (0) = \frac{8}{3}

y = \frac{14}{9} e^{t} + \frac{19}{36} e^{4 t} - \frac{1}{2} e^{2 t}

(d)

α

और

β

को, जिसके लिये

x^{α} y^{β}

समीकरण

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

का एक समाकलन गुण्पक है, ज्ञात कीजिये और समीकरण हल कीजिये।

Find

α

and

β

such that

x^{α} y^{β}

is an integrating factor of

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

and solve the equation.

Answer:

Find

α

and

β

such that

x^{α} y^{β}

is an integrating factor of

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

and solve the equation.

Answer:

Given that

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

Let

x^{α} y^{β}

be an integrating factor of equation (1) then

\begin{aligned} x^{α} y^{β} (4 y^{2} + 3 x y) d x - x^{α} y^{β} (3 x y + 2 x^{2}) d y = 0 \\ (4 x^{α} y^{β + 2} + 3 x^{α + 1} y^{β + 1}) d x - (3 x^{α + 1} y^{β + 1} + 2 x^{α + 2} y^{β}) d y = 0 \to () \end{aligned}

Compare with

M d x + N d y = 0

\begin{aligned} M & = 4 x^{α} y^{β + 2} + 3 x^{α + 1} y^{β + 1} \\ \frac{\partial M}{\partial y} & = 4 (β + 2) x^{α} y^{β + 1} + 3 (β + 1) x^{α + 1} y^{β} \\ N & = - 3 x^{α + 1} y^{β + 1} - 2 x^{α + 2} y^{β} \\ \frac{\partial N}{\partial x} & = - 3 (α + 1) x^{α} y^{β + 1} - 2 (α + 2) x^{α + 1} y^{β} \end{aligned}

Now if this equation is exact then

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

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

Compare the coefficients of

x^{α} y^{β + 1}

and

x^{α + 1} y^{β}

\begin{aligned} 4 (β + 2) = - 3 (α + 1) \\ 3 (β + 1) = - 2 (α + 2) \end{aligned}

After solving equations (1) & (2)

α = - 5 & β = 1

Put these values in equation (A)

(4 x^{- 5} y^{3} + 3 x^{- 4} y^{2}) d x - (3 x^{- 4} y + 2 x^{- 3} y) d y = 0

Again compare it with

M_{1} d x + N_{1} d y = 0

\begin{aligned} M_{1} = 4 x^{- 5} y^{3} + 3 x^{- 4} y^{2} \\ N_{1} = - (3 x^{- 4} y + 2 x^{- 3} y) \end{aligned}

General Solution:

\begin{aligned} \int M_{1} d x + \int (N_{1} \sim x) d y = 0 \\ \int (4 x^{- 5} y^{3} + 3 x^{- 4} y^{2}) d x + \int 0 d y = 0 \end{aligned}

\begin{aligned} \frac{4 x^{- 4}}{- 4} y^{3} - \frac{3 x^{- 3}}{3} y^{3} + 0 = C \\ \frac{- y^{3}}{x^{4}} - \frac{y^{3}}{x^{3}} = C \\ \frac{y^{3}}{x^{3}} + \frac{y^{3}}{x^{4}} = - C \end{aligned}

\frac{y^{3}}{x^{3}} + \frac{y^{3}}{x^{4}} = k

where

k = - c

(a) मान लीजिये कि $\vec{v} = v_{1} \vec{i} + v_{2} \vec{j} + v_{3} \vec{k}$ है। दर्शाइये कि $curl (curl \vec{v}) = grad (div \vec{v}) - \nabla^{2} \vec{v}$ .

Let

\vec{v} = v_{1} \vec{i} + v_{2} \vec{j} + v_{3} \vec{k}

. Show that curl(curl

\vec{v}) = grad (div \vec{v}) - \nabla^{2} \vec{v}

Answer:

Introduction:
We are given a vector field

\vec{v} = v_{1} \hat{i} + v_{2} \hat{j} + v_{3} \hat{k}

, and we need to show that

\nabla \times (\nabla \times \vec{v}) = grad (div \vec{v}) - \nabla^{2} \vec{v}

Work/Calculations:

First, we find the curl of

\vec{v}

using the determinant form:

\begin{aligned} \nabla \times \vec{v} & = | \begin{array}{c} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ v_{1} & v_{2} & v_{3} \end{array} | \end{aligned}

Expanding the determinant, we get:

\begin{aligned} \nabla \times \vec{v} & = (\frac{\partial v_{3}}{\partial y} - \frac{\partial v_{2}}{\partial z}) \hat{i} + (\frac{\partial v_{1}}{\partial z} - \frac{\partial v_{3}}{\partial x}) \hat{j} + (\frac{\partial v_{2}}{\partial x} - \frac{\partial v_{1}}{\partial y}) \hat{k} \end{aligned}

Now, we compute

\nabla \times (\nabla \times \vec{v})

using the same determinant form:

\begin{aligned} \nabla \times (\nabla \times \vec{v}) & = | \begin{array}{c} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \frac{\partial v_{3}}{\partial y} - \frac{\partial v_{2}}{\partial z} & \frac{\partial v_{1}}{\partial z} - \frac{\partial v_{3}}{\partial x} & \frac{\partial v_{2}}{\partial x} - \frac{\partial v_{1}}{\partial y} \end{array} | \end{aligned}

Expanding this determinant and simplifying, we get:

\begin{aligned} \nabla \times (\nabla \times \vec{v}) & = \sum [{\frac{\partial}{\partial y} (\frac{\partial v_{2}}{\partial x} - \frac{\partial v_{1}}{\partial y}) - \frac{\partial}{\partial z} (\frac{\partial v_{1}}{\partial z} - \frac{\partial v_{3}}{\partial x})} \hat{i}] \end{aligned}

Continuing the simplification:

\begin{aligned} = \sum [{(\frac{\partial^{2} V_{2}}{\partial y \partial x} + \frac{\partial^{2} V_{3}}{\partial z \partial x} - (\frac{\partial^{2} V_{1}}{\partial y^{2}} + \frac{\partial^{2} v_{1}}{\partial z^{2}})} i]] & = \sum [{\frac{\partial}{\partial x} (\frac{\partial v_{2}}{\partial y} + \frac{\partial v_{3}}{\partial z}) - (\frac{\partial^{2} v_{1}}{\partial y^{2}} + \frac{\partial^{2} v_{1}}{\partial z^{2}})} \hat{i}] \end{aligned}

Now, further simplifying:

\begin{aligned} = \sum [{\frac{\partial}{\partial x} (\frac{\partial v_{2}}{\partial y} + \frac{\partial v_{3}}{\partial z}) - (\frac{\partial^{2} v_{1}}{\partial y^{2}} + \frac{\partial^{2} v_{1}}{\partial z^{2}})} \hat{i}] \\ = \sum [{\frac{\partial}{\partial x} (\nabla \cdot \vec{v}) - \nabla^{2} v_{1}} \hat{i}] \end{aligned}

Using the sum notation, we get:

\nabla \times (\nabla \times \vec{v}) = \sum [{\frac{\partial}{\partial x} (\nabla \cdot \vec{v}) - \nabla^{2} v_{1}} \hat{i}]

Now, let’s write this in a more concise form:

\nabla \times (\nabla \times \vec{v}) = \nabla (\nabla \cdot \vec{v}) - \nabla^{2} \vec{v}

Conclusion:
Therefore, we have shown that

\nabla \times (\nabla \times \vec{v}) = \nabla (\nabla \cdot \vec{v}) - \nabla^{2} \vec{v}

, as required.

(b) स्टोकस प्रमेय का इस्तेमाल करते हुये रेखा समाकल

\int_{C} - y^{3} d x + x^{3} d y + z^{3} d z

का मान निकालिये। यहाँ सिलिन्डर

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

और समतल

x + y + z = 1

का प्रतिच्छेद

C

है।

C

पर अभिकिन्यास

x y

-समतल में वामावर्त गति के संगत है।

Evaluate the line integral

\int_{C} - y^{3} d x + x^{3} d y + z^{3} d z

using Stokes’ theorem. Here

C

is the intersection of the cylinder

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

and the plane

x + y + z = 1

. The orientation on

C

corresponds to counterclockwise motion in the

x y

-plane.

Answer:

Introduction:
We are given the line integral

\int_{C} (- y^{3} d x + x^{3} d y + z^{3} d z)

, where

C

is the intersection of the cylinder

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

and the plane

x + y + z = 1

. We will evaluate this integral using Stokes’ theorem.

Work/Calculations:

Let’s start by finding the unit normal vector

\hat{n}

for the plane

x + y + z = 1

. The orientation on

C

corresponds to counterclockwise motion in the

x y

-plane. Therefore,

\hat{n} = \frac{i + j + k}{\sqrt{1^{2} + 1^{2} + 1^{2}}} = \frac{1}{\sqrt{3}} i + \frac{1}{\sqrt{3}} j + \frac{1}{\sqrt{3}} k

We will apply Stokes’ theorem with

P = - y^{3}

Q = x^{3}

, and

R = z^{3}

. Calculate

\nabla \times F

\begin{aligned} \nabla \times F & = | \begin{array}{c} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ - y^{3} & x^{3} & z^{3} \end{array} | \\ = i [0] - j [0] + k [3 x^{2} + 3 y^{2}] \\ = 3 (x^{2} + y^{2}) k \end{aligned}

Now, we will use Stokes’ theorem:

∮_{C} (- y^{3} d x + x^{3} d y + z^{3} d z) = \iint_{S} (\nabla \times F) \cdot d S

Here,

S

is the surface bounded by

C

The projection of the surface

S

onto the

x y

-plane is the circle

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

with a radius of 1. Representing the equation of the plane in the form

z = 1 - x - y

and using the formula for the surface integral:

\begin{aligned} \iint_{S} d S & = \iint_{R} \sqrt{1 + {(\frac{\partial z}{\partial x})}^{2} + {(\frac{\partial z}{\partial y})}^{2}} d x d y \\ = \iint_{R} \sqrt{1 + (- 1)^{2} + (- 1)^{2}} d x d y \\ = \sqrt{3} \iint_{R} d x d y \\ = \sqrt{3} π (1^{2}) \\ = \sqrt{3} π \end{aligned}

Now, we can calculate the line integral using Stokes’ theorem:

\begin{aligned} ∮_{C} (- y^{3} d x + x^{3} d y + z^{3} d z) & = \iint_{S} (\nabla \times F) \cdot d S \\ = \sqrt{3} (\sqrt{3} π) \\ = 3 π \end{aligned}

Conclusion:
The line integral

\int_{C} (- y^{3} d x + x^{3} d y + z^{3} d z)

over the curve

C

is equal to

3 π

when evaluated using Stokes’ theorem.

\vec{F} = x y^{2} \vec{i} + (y + x) \vec{j}

है। ग्रीन के प्रमेय का इस्तेमाल करते हुये प्रथम चतुर्थाश में वक्रों

y = x^{2}

और

y = x

द्वारा परिबद्ध क्षेत्र पर

(\nabla \times \vec{F}) \cdot \vec{k}

का समाकलन कीजिये।

Let

\vec{F} = x y^{2} \vec{i} + (y + x) \vec{j}

. Integrate

(\nabla \times \vec{F}) \cdot \vec{k}

over the region in the first quadrant bounded by the curves

y = x^{2}

and

y = x

using Green’s theorem.

Answer:

Introduction

The problem asks us to integrate

(\nabla \times \vec{F}) \cdot \vec{k}

over the region in the first quadrant bounded by the curves

y = x^{2}

and

y = x

using Green’s theorem.
Green’s theorem states that for a vector field

\vec{F} = M \vec{i} + N \vec{j}

∮_{C} (M d x + N d y) = \iint_{R} (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) d A

Here,

C

is the positively oriented boundary curve of the region

R

.
Given

\vec{F} = x y^{2} \vec{i} + (y + x) \vec{j}

, we have

M = x y^{2}

and

N = y + x

Work/Calculations

Step 1: Compute $(\nabla \times \vec{F}) \cdot \vec{k}$

The curl of

\vec{F}

is given by:

\nabla \times \vec{F} = (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) \vec{k}

For

M = x y^{2}

and

N = y + x

, we have:

\frac{\partial N}{\partial x} = 1, \frac{\partial M}{\partial y} = 2 x y

Therefore,

(\nabla \times \vec{F}) \cdot \vec{k} = 1 - 2 x y

The curl of

\vec{F}

is given by:

(\nabla \times \vec{F}) \cdot \vec{k} = \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 1 - 2 x y

Step 2: Evaluate the Double Integral Using Green’s Theorem

The region

R

is bounded by

y = x^{2}

and

y = x

. These curves intersect at

x = 0

and

x = 1

Using Green’s theorem, we have:

\iint_{R} (1 - 2 x y) d A

To set up the integral, we note that

x

ranges from 0 to 1, and

y

ranges from

x^{2}

x

\int_{0}^{1} \int_{x^{2}}^{x} (1 - 2 x y) d y d x

After calculating, we get:

\int_{0}^{1} [y - x y^{2}]_{x^{2}}^{x} d x

Let’s substitute the values:

\int_{0}^{1} (x - x^{3} - x^{2} + x^{5}) d x

After calculating , the result is

\frac{1}{12}

Conclusion

The integral of

(\nabla \times \vec{F}) \cdot \vec{k}

over the region in the first quadrant bounded by the curves

y = x^{2}

and

y = x

\frac{1}{12}

(d)

f (y)

, जिसके लिये समीकरण

(2 x e^{y} + 3 y^{2}) d y + (3 x^{2} + f (y)) d x = 0

सथातथ्य है, ज्ञात कीजिये और हाल निकालिये।

Find

f (y)

such that

(2 x e^{y} + 3 y^{2}) d y + (3 x^{2} + f (y)) d x = 0

is exact and hence solve.

Answer:

Introduction:
We are given the differential equation

(2 x e^{y} + 3 y^{2}) d y + (3 x^{2} + f (y)) d x = 0

, and our objective is to determine the function

f (y)

such that this equation is exact. Once we find

f (y)

, we can proceed to solve the equation.

Work/Calculations:

To ensure that the equation is exact, we need to satisfy the condition $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ , where $M$ and $N$ are the coefficients of $d x$ and $d y$ terms, respectively.
In the given equation, we have $M = 3 x^{2} + f (y)$ and $N = 2 x e^{y} + 3 y^{2}$ .
Compute the partial derivatives $\frac{\partial M}{\partial y}$ and $\frac{\partial N}{\partial x}$ :

\frac{\partial M}{\partial y} = \frac{d}{d y} (3 x^{2} + f (y)) = f^{'} (y)

\frac{\partial N}{\partial x} = \frac{d}{d x} (2 x e^{y} + 3 y^{2}) = 2 e^{y}

Equate these expressions since the equation is exact:

f^{'} (y) = 2 e^{y}

Solve this first-order differential equation for $f (y)$ :

\begin{aligned} f^{'} (y) & = 2 e^{y} \\ \frac{d f}{d y} & = 2 e^{y} \\ d f & = 2 e^{y} d y \\ \int d f & = \int 2 e^{y} d y \\ f (y) & = 2 \int e^{y} d y \\ f (y) & = 2 e^{y} + C \end{aligned}

Here,

C

is the constant of integration.

We have found

f (y) = 2 e^{y} + C

such that the given differential equation is exact.

Now, we can rewrite the equation as:

(3 x^{2} + 2 e^{y}) d x + (2 x e^{y} + 3 y^{2}) d y = 0

To solve this exact equation, we can use the method of exact differentials. We recognize that the equation can be derived from a potential function, so we can write:

(3 x^{2} + 2 e^{y}) d x + (2 x e^{y} + 3 y^{2}) d y = 0

General solution:

\int M d x + \int (N \sim x) d y = 0

\begin{aligned} \int (3 x^{2} + 2 e^{y}) d x + \int 3 y^{2} d y = 0 \\ x^{3} + 2 x e^{y} + y^{3} = C \end{aligned}

Now, we find

F (x, y)

and add the constant of integration

C

F (x, y) = x^{3} + 2 x e^{y} + y^{3} + C

So, the solution to the differential equation is:

x^{3} + 2 x e^{y} + y^{3} + C = 0

where

C

is the constant of integration.

This concludes the solution to the given differential equation with

f (y) = 2 e^{y} + C

Free UPSC Mathematics Optional Paper-1 2018 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Introduction

Work/Calculations

Step 1: Dimensions of C C CCC

Step 2: Rank of C C CCC

Step 3: Determinant of C C CCC

Conclusion

Introduction

Work/Calculations

Step 1: Setting up the Equations for e 1 e 1 e_(1)e_1e1

Step 2: Solving for c 1 c 1 c_(1)c_1c1 and c 2 c 2 c_(2)c_2c2

Step 3: Setting up the Equations for e 2 e 2 e_(2)e_2e2

Step 4: Solving for d 1 d 1 d_(1)d_1d1 and d 2 d 2 d_(2)d_2d2

Conclusion

Introduction

Work/Calculations

Step 1: Find the Direction Vector of the Line

Step 2: Find the Normal Vector of the Plane

Step 3: Find the Projection Vector

Step 4: Find the Equation of the Projected Line

Conclusion

Introduction

Work/Calculations

Definition of Similar Matrices

Definition of Eigenvalues

Step 1: Assume A A AAA has an Eigenvalue λ λ lambda\lambdaλ

Step 2: Multiply Both Sides by P − 1 P − 1 P^(-1)P^{-1}P−1

Step 3: Use the Similarity Relation B = P − 1 A P B = P − 1 A P B=P^(-1)APB = P^{-1} A PB=P−1AP

Step 4: Show that λ λ lambda\lambdaλ is an Eigenvalue of B B BBB

Step 5: Reverse the Argument

Conclusion

Introduction

Work/Calculations

Step 1: Parametric Equation of the Parabola

Step 2: Distance Formula

Step 3: Minimize the Distance

Conclusion

Introduction

Work/Calculations

Step 1: Solve for y y yyy in terms of x x xxx

Step 2: Volume of Solid of Revolution

Step 3: Set up the Integral

Step 4: Evaluate the Integral

Conclusion

Introduction

Work/Calculations

Step 1: Form the Augmented Matrix

Step 2: Row Reduce the Matrix

Step 3: Determine the Rank

Step 4: Analyze the Solution

Conclusion

Introduction

Work/Calculations

Step 1: Partial Derivative with respect to x x xxx at ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1)

Step 2: Partial Derivative with respect to y y yyy at ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1)

Conclusion

Introduction

Work/Calculations

Step 1: Generators of the Surface

Step 2: Substituting the Point ( 1 , 1 , 1 ) ( 1 , 1 , 1 ) (1,1,1)(1,1,1)(1,1,1)

Step 3: Generators with Specific λ λ lambda\lambdaλ and μ μ mu\muμ

Step 4: Direction Ratios and Equations of the Lines

Conclusion

Introduction

Work/Calculations

Step 1: General Equation of a Sphere

Step 2: Substitute the Points into the Equation

Step 3: Solve for the Center and Radius

Step 4: Equation of the Sphere

Conclusion

Introduction

Work/Calculations

Step 1: Identify the Function and Interval

Step 2: Find the Critical Points

Step 3: Evaluate the Function at the Endpoints

Step 4: Identify the Maximum and Minimum Values

Conclusion

Introduction

Work/Calculations

Step 1: Identify the Region of Integration

Step 1: Dimensions of $C$

Step 2: Rank of $C$

Step 3: Determinant of $C$

Step 1: Setting up the Equations for $e_{1}$

Step 2: Solving for $c_{1}$ and $c_{2}$

Step 3: Setting up the Equations for $e_{2}$

Step 4: Solving for $d_{1}$ and $d_{2}$

Step 1: Assume $A$ has an Eigenvalue $λ$

Step 2: Multiply Both Sides by $P^{- 1}$

Step 3: Use the Similarity Relation $B = P^{- 1} A P$

Step 4: Show that $λ$ is an Eigenvalue of $B$

Step 1: Solve for $y$ in terms of $x$

Step 1: Partial Derivative with respect to $x$ at $(0, 1)$

Step 2: Partial Derivative with respect to $y$ at $(0, 1)$

Step 2: Substituting the Point $(1, 1, 1)$

Step 3: Generators with Specific $λ$ and $μ$

Step 4: Correcting the Formula for $θ$

Step 1: Rewrite the Differential Equation in Terms of $p$

Step 2: Differentiate $x$ with Respect to $y$

Step 3: Solve for $μ^{2}$

Step 4: Solve for the Periodic Time $T$

Step 2: Calculate Wronskian $W$

Step 2: Find the Complementary Function ( $y_{c}$ )

Step 3: Find the Particular Integral ( $y_{p}$ )

Step 4: Write the General Solution ( $y$ )

Step 1: First Derivative ${\vec{r}}^{'} (u)$

Step 2: Magnitude of ${\vec{r}}^{'} (u)$

Step 3: Second Derivative ${\vec{r}}^{″} (u)$

Step 4: Third Derivative ${\vec{r}}^{‴} (u)$

Step 5: Torsion $T$

Step 6: Compute $| {\vec{r}}^{'} \times {\vec{r}}^{″} |$

Step 1: Complementary Function ( $y_{c}$ )