Free UPSC Mathematics Optional Paper-1 2019 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 12, 2025ByAbstract Classes

खण्ड-A / SECTION-A

(a) माना कि $f : [0, \frac{π}{2}] \to R$ एक संतत फलन है, जैसा कि

f (x) = \frac{\cos^{2} x}{4 x^{2} - π^{2}}, 0 \leq x < \frac{π}{2}

f (\frac{π}{2})

का मान ज्ञात कीजिए।

Let

f : [0, \frac{π}{2}] \to R

be a continuous function such that

f (x) = \frac{\cos^{2} x}{4 x^{2} - π^{2}}, 0 \leq x < \frac{π}{2}

Find the value of

f (\frac{π}{2})

Answer:

Introduction

The problem asks us to find the value of

f (\frac{π}{2})

for a given function

f (x) = \frac{\cos^{2} x}{4 x^{2} - π^{2}}

defined on the interval

[0, \frac{π}{2}]

. The function is continuous except at

x = \frac{π}{2}

because the denominator becomes zero at that point. To find

f (\frac{π}{2})

, we’ll need to evaluate the limit of

f (x)

x

approaches

\frac{π}{2}

Work/Calculations

Step 1: Define the Function and the Limit

The function

f (x)

is given as:

f (x) = \frac{\cos^{2} x}{4 x^{2} - π^{2}}

We need to find:

f (\frac{π}{2}) = lim_{x \to \frac{π}{2}} f (x)

Step 2: Simplify the Function

To find the limit, we can use L’Hôpital’s Rule, which states that if

lim_{x \to a} \frac{f (x)}{g (x)}

is an indeterminate form

\frac{0}{0}

\frac{\infty}{\infty}

, then:

lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}

provided the limits on the right-hand side exist.

First, let’s check if the function

f (x)

is in indeterminate form at

x = \frac{π}{2}

Numerator at x = \frac{π}{2} : \cos^{2} (\frac{π}{2}) = 0

Denominator at x = \frac{π}{2} : 4 {(\frac{π}{2})}^{2} - π^{2} = 0

Both the numerator and the denominator are zero, so we have an indeterminate form

\frac{0}{0}

Step 3: Apply L’Hôpital’s Rule

Let’s differentiate the numerator and the denominator with respect to

x

$f^{'} (x)$ (Derivative of $\cos^{2} x$ ) = $- 2 \cos (x) \sin (x)$
$g^{'} (x)$ (Derivative of $4 x^{2} - π^{2}$ ) = $8 x$

Now, we can find the limit:

lim_{x \to \frac{π}{2}} \frac{f^{'} (x)}{g^{'} (x)} = lim_{x \to \frac{π}{2}} \frac{- 2 \cos (x) \sin (x)}{8 x}

Let’s substitute the values into the formula and calculate the limit.

After substituting the values, we get:

lim_{x \to \frac{π}{2}} \frac{- 2 \cos (x) \sin (x)}{8 x} = \frac{- 2 \cos (\frac{π}{2}) \sin (\frac{π}{2})}{8 \times \frac{π}{2}}

After calculating, we get:

lim_{x \to \frac{π}{2}} \frac{- 2 \cos (x) \sin (x)}{8 x} = 0

Conclusion

The value of

f (\frac{π}{2})

0

. We used L’Hôpital’s Rule to evaluate the limit of the function

f (x)

x

approaches

\frac{π}{2}

, and found that the limit is

0

. Therefore,

f (\frac{π}{2}) = 0

(b) माना कि

f : D (\subseteq R^{2}) \to R

एक फलन है और

(a, b) \in D

. अगर

f (x, y)

बिंदु

(a, b)

पर संतत है, तो दर्शाइए कि फलन

f (x, b)

और

f (a, y)

क्रमशः

x = a

और

y = b

पर संतत हैं।

Let

f : D (\subseteq R^{2}) \to R

be a function and

(a, b) \in D

. If

f (x, y)

is continuous at

(a, b)

, then show that the functions

f (x, b)

and

f (a, y)

are continuous at

x = a

and at

y = b

respectively.

Answer:

Introduction

The problem asks us to prove that if a function

f (x, y)

is continuous at a point

(a, b)

, then the functions

f (x, b)

and

f (a, y)

are continuous at

x = a

and

y = b

respectively. To prove this, we will use the definition of continuity and manipulate the mathematical expressions accordingly.

Work/Calculations

Step 1: Definition of Continuity

A function

f (x, y)

is said to be continuous at

(a, b)

if for every

ϵ > 0

, there exists

δ > 0

such that:

\sqrt{(x - a)^{2} + (y - b)^{2}} < δ ⟹ | f (x, y) - f (a, b) | < ϵ

Step 2: Prove Continuity for $f (x, b)$ at $x = a$

We need to show that for every

ϵ > 0

, there exists

δ > 0

such that:

| x - a | < δ ⟹ | f (x, b) - f (a, b) | < ϵ

To prove this, let’s consider

ϵ > 0

. Since

f (x, y)

is continuous at

(a, b)

, there exists

δ > 0

such that:

\sqrt{(x - a)^{2} + (y - b)^{2}} < δ ⟹ | f (x, y) - f (a, b) | < ϵ

Now, consider

| x - a | < δ

. We can rewrite this as

\sqrt{(x - a)^{2} + (b - b)^{2}} < δ

By the continuity of

f (x, y)

(a, b)

, this implies:

| f (x, b) - f (a, b) | < ϵ

Thus, we have shown that

f (x, b)

is continuous at

x = a

Step 3: Prove Continuity for $f (a, y)$ at $y = b$

Similarly, we need to show that for every

ϵ > 0

, there exists

δ > 0

such that:

| y - b | < δ ⟹ | f (a, y) - f (a, b) | < ϵ

Again, consider

ϵ > 0

. By the continuity of

f (x, y)

(a, b)

, there exists

δ > 0

such that:

\sqrt{(x - a)^{2} + (y - b)^{2}} < δ ⟹ | f (x, y) - f (a, b) | < ϵ

Now, consider

| y - b | < δ

. We can rewrite this as

\sqrt{(a - a)^{2} + (y - b)^{2}} < δ

By the continuity of

f (x, y)

(a, b)

, this implies:

| f (a, y) - f (a, b) | < ϵ

Thus, we have shown that

f (a, y)

is continuous at

y = b

Conclusion

We have successfully proven that if

f (x, y)

is continuous at

(a, b)

, then

f (x, b)

is continuous at

x = a

and

f (a, y)

is continuous at

y = b

. We used the definition of continuity to establish these results.

T : R^{2} \to R^{2}

एक रैखिक प्रतिचित्र है, जैसा कि

T (2, 1) = (5, 7)

एवं

T (1, 2) = (3, 3)

. अगर

A

मानक आधारों

e_{1}, e_{2}

के सापेक्ष

T

के संगत आव्यूह है, तो

A

की कोटि ज्ञात कीजिए।

Let

T : R^{2} \to R^{2}

be a linear map such that

T (2, 1) = (5, 7)

and

T (1, 2) = (3, 3)

.
If

A

is the matrix corresponding to

T

with respect to the standard bases

e_{1}, e_{2}

, then find

Rank (A)

Answer:

Expressing $(a, b)$ as a Linear Combination

(a, b) = α (2, 1) + β (1, 2)

a = 2 α + β (Equation 1)

b = α + 2 β (Equation 2)

Solving for $α$ and $β$

From Equation 2 multiplied by 2 minus Equation 1, we get:

3 β = 2 b - a

β = \frac{2 b - a}{3}

α = \frac{2 a - b}{3}

Finding $T (a, b)$ in terms of $T (2, 1)$ and $T (1, 2)$

T (a, b) = \frac{2 a - b}{3} T (2, 1) + \frac{2 b - a}{3} T (1, 2)

Finding $T (e_{1})$ and $T (e_{2})$

T (1, 0) = \frac{2}{3} (5, 7) - \frac{1}{3} (3, 3) = (\frac{7}{3}, \frac{11}{3})

T (0, 1) = - \frac{1}{3} (5, 7) + \frac{2}{3} (3, 3) = (\frac{1}{3}, - \frac{1}{3})

Finding Matrix $A$

A = [\begin{array}{cc} \frac{7}{3} & \frac{1}{3} \\ \frac{11}{3} & - \frac{1}{3} \end{array}]

Finding the Rank of $A$

The determinant of

A

is:

Det (A) = \frac{- 7 \times (- 1 / 3) - 11 \times 1 / 3}{3} = - \frac{18}{9} = - 2 \neq 0

Since the determinant is not zero, the matrix is invertible, and its rank is equal to the dimension of the domain, which is 2.

Conclusion

Rank (A) = 2

(d) अगर

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

है, तो दर्शाइए कि

A B = 6 I_{3}

. इस परिणाम का उपयोग करते हुए निम्नलिखित समीकरण निकाय को हल कीजिए :

\begin{array}{r} 2 x + y + z = 5 \\ x - y = 0 \\ 2 x + y - z = 1 \end{array}

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

then show that

A B = 6 I_{3}

. Use this result to solve the following system of equations :

\begin{array}{r} 2 x + y + z = 5 \\ x - y = 0 \\ 2 x + y - z = 1 \end{array}

Answer:

Introduction

The problem asks us to show that

A B = 6 I_{3}

for given matrices

A

and

B

, and then use this result to solve a system of equations.

I_{3}

is the

3 \times 3

identity matrix.

Work/Calculations

Step 1: Calculate $A B$

The first task is to calculate

A B

. The matrices

A

and

B

are:

A = (\begin{matrix} 1 & 2 & 1 \\ 1 & - 4 & 1 \\ 3 & 0 & - 3 \end{matrix})

B = (\begin{matrix} 2 & 1 & 1 \\ 1 & - 1 & 0 \\ 2 & 1 & - 1 \end{matrix})

\begin{aligned} A \times B = [\begin{array}{ccc} 1 & 2 & 1 \\ 1 & - 4 & 1 \\ 3 & 0 & - 3 \end{array}] \times [\begin{array}{ccc} 2 & 1 & 1 \\ 1 & - 1 & 0 \\ 2 & 1 & - 1 \end{array}] \\ = [\begin{array}{llll} 1 \times 2 + 2 \times 1 + 1 \times 2 & 1 \times 1 + 2 \times - 1 + 1 \times 1 & 1 \times 1 + 2 \times 0 + 1 \times - 1 \\ 1 \times 2 - 4 \times 1 + 1 \times 2 & 1 \times 1 - 4 \times - 1 + 1 \times 1 & 1 \times 1 - 4 \times 0 + 1 \times - 1 \\ 3 \times 2 + 0 \times 1 - 3 \times 2 & 3 \times 1 + 0 \times - 1 - 3 \times 1 & 3 \times 1 + 0 \times 0 - 3 \times - 1 \end{array}] \\ = [\begin{array}{ccc} 2 + 2 + 2 & 1 - 2 + 1 & 1 + 0 - 1 \\ 2 - 4 + 2 & 1 + 4 + 1 & 1 + 0 - 1 \\ 6 + 0 - 6 & 3 + 0 - 3 & 3 + 0 + 3 \end{array}] \\ = [\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}] \end{aligned}

After calculating, we find that

A B

is:

A B = (\begin{matrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{matrix})

This is equivalent to

6 I_{3}

, where

I_{3}

is the

3 \times 3

identity matrix.

Step 2: Solve the System of Equations

The system of equations is:

\begin{aligned} 2 x + y + z & = 5 \\ x - y & = 0 \\ 2 x + y - z & = 1 \end{aligned}

We can write this system as

A x = b

, where

A

is the matrix from the problem statement, and

b = (\begin{matrix} 5 \\ 0 \\ 1 \end{matrix})

Since A B = 6 I_{3}, we have A^{- 1} = \frac{1}{6} B .

Here,

X = A^{- 1} \times b

\begin{aligned} ∴ X = \frac{1}{| A |} \times Adj (A) \times b \\ X = \frac{1}{6} \times [\begin{array}{ccc} 1 & 2 & 1 \\ 1 & - 4 & 1 \\ 3 & 0 & - 3 \end{array}] \times [\begin{matrix} 5 \\ 0 \\ 1 \end{matrix}] \\ X = \frac{1}{6} \times [\begin{array}{c} 1 \times 5 + 2 \times 0 + 1 \times 1 \\ 1 \times 5 - 4 \times 0 + 1 \times 1 \\ 3 \times 5 + 0 \times 0 - 3 \times 1 \end{array}] \\ X = \frac{1}{6} \times [\begin{array}{c} 6 \\ 6 \\ 12 \end{array}] \\ X = [\begin{array}{c} 1 \\ 1 \\ 2 \end{array}] \end{aligned}

Hence

(\begin{matrix} x \\ y \\ z \end{matrix}) = (\begin{matrix} 1 \\ 1 \\ 2 \end{matrix})

(e) दर्शाइए कि

\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} और \frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}

प्रतिच्छेदी रेखाएँ हैं। प्रतिच्छेद बिंदु के निर्देशांकों और उस समतल, जिसमें दोनों रेखाएँ हैं, का समीकरण ज्ञात कीजिए।

Show that the lines

\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} and \frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}

intersect. Find the coordinates of the point of intersection and the equation of the plane containing them.

Answer:

Proving Intersection of Lines

Given two lines:

Line 1:

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

Line 2:

\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}

Let’s parameterize these lines as follows:

Line 1:

(x, y, z) = (- 1 - 3 r, 3 + 2 r, - 2 + r) (1)

Line 2:

(x, y, z) = (r^{'}, 7 - 3 r^{'}, - 7 + 2 r^{'}) (2)

For the two lines to intersect, there must exist values of

r

and

r^{'}

such that the points on these lines coincide.

Step 1: Setting up Equations

We set up the equations:

\begin{aligned} - 1 - 3 r = r^{'} \\ 3 + 2 r = 7 - 3 r^{'} \\ - 2 + r = - 7 + 2 r^{'} \end{aligned}

Step 2: Solving Equations

Solving the first two equations:

From the first equation, we get

r = - 1

Substituting

r = - 1

into the second equation, we find

r^{'} = 2

Now, we check if these values satisfy the third equation:

- 2 + (- 1) = - 7 + 2 (2) ⟹ - 3 = - 3

So, the values

r = - 1

and

r^{'} = 2

satisfy all three equations.

Step 3: Finding Intersection Point

Substituting these values of

r

and

r^{'}

into either (1) or (2), we get the coordinates of the point of intersection:

From (1):

(x, y, z) = (2, 1, - 3)

Step 4: Finding the Equation of the Plane

The equation of the plane containing the given lines can be found using the cross product of the direction ratios of the lines.

Direction ratios of Line 1:

(- 3, 2, 1)

Direction ratios of Line 2:

(1, - 3, 2)

Now, we find the cross product of these direction ratios to get the normal vector of the plane:

\begin{aligned} | \begin{array}{ccc} i & j & k \\ - 3 & 2 & 1 \\ 1 & - 3 & 2 \end{array} | \\ = i \times | \begin{array}{cc} 2 & 1 \\ - 3 & 2 \end{array} | - j \times | \begin{array}{cc} - 3 & 1 \\ 1 & 2 \end{array} | + k \times | \begin{array}{cc} - 3 & 2 \\ 1 & - 3 \end{array} | \\ = i \times (2 \times 2 - 1 \times (- 3)) - j \times ((- 3) \times 2 - 1 \times 1) + k \times ((- 3) \times (- 3) - 2 \times 1) \\ = i \times (4 + 3) - j \times (- 6 - 1) + k \times (9 - 2) \\ = i \times (7) - j \times (- 7) + k \times (7) \\ = 7 i + 7 j + 7 k \\ = 7 i + 7 k + 7 j \end{aligned}

So, the normal vector of the plane is

7 i + 7 j + 7 k

Now, we can use one of the intersection points, e.g.,

(2, 1, - 3)

, and the normal vector to find the equation of the plane:

7 (x - 2) + 7 (y - 1) + 7 (z + 3) = 0

Simplifying:

\Rightarrow x + y + z = 0

This is the equation of the plane containing the given lines.

(a) क्या $f (x) = | \cos x | + | \sin x |, x = \frac{π}{2}$ पर अवकलनीय है? अगर आपका उत्तर हाँ है, तो $f (x)$ का अवकलज $x = \frac{π}{2}$ पर ज्ञात कीजिए। अगर आपका उत्तर ना है, तो अपने उत्तर का प्रमाण दीजिए।

f (x) = | \cos x | + | \sin x |

differentiable at

x = \frac{π}{2}

? If yes, then find its derivative at

x = \frac{π}{2}

. If no, then give a proof of it.

Answer:

Introduction

We’re going to look at a pretty cool math problem. We have a function

f (x) = | \cos x | + | \sin x |

, and we want to find out if it’s differentiable at

x = \frac{π}{2}

. If it is, we’ll also find its derivative at that point. To make things easier, we’ll break down the function into a piecewise function and then check its differentiability.

Work/Calculations

Step 1: Breaking It Down

First, let’s rewrite

f (x)

as a piecewise function around

x = \frac{π}{2}

f (x) = {\begin{cases} - \cos x + \sin x & if x < \frac{π}{2} \\ \cos x + \sin x & if x > \frac{π}{2} \\ 1 & if x = \frac{π}{2} \end{cases}

Step 2: Checking the Derivative

To see if

f (x)

is differentiable at

x = \frac{π}{2}

, we need to find the derivative from both sides of that point and see if they match.

Left-hand derivative: The derivative of $- \cos x + \sin x$ is $\sin x + \cos x$ .
Right-hand derivative: The derivative of $\cos x + \sin x$ is $- \sin x + \cos x$ .

Now, let’s plug

x = \frac{π}{2}

into both of these derivatives and see what we get.

After calculating, we find that both the left-hand and right-hand derivatives at

x = \frac{π}{2}

are

1

. That’s awesome because it means the function is differentiable at

x = \frac{π}{2}

Conclusion

So, there we have it! The function

f (x) = | \cos x | + | \sin x |

is differentiable at

x = \frac{π}{2}

, and its derivative at that point is

1

(b) माना कि

A

और

B

समान कोटि के दो लांबिक आव्यूह हैं तथा

\det A + \det B = 0

. दर्शाइए कि

A + B

एक अव्युत्क्रमणीय (सिंगुलर) आव्यूह है।

Let

A

and

B

be two orthogonal matrices of same order and

\det A + \det B = 0

. Show that

A + B

is a singular matrix.

Answer:

Proving that A + B is Singular

Given:

$A$ and $B$ are two orthogonal matrices of the same order.
$det A + det B = 0$

We want to show that

A + B

is a singular matrix.

Step 1: Given Orthogonal Matrices

Given that

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

, where

I

is the identity matrix.

Also,

det A = det B = 1

because orthogonal matrices have a determinant of

1

- 1

, and we’re given that

det A + det B = 0

, which implies

det A = - det B

Step 2: Determinant of AB

We know that

det A \cdot det B = - 1

. This implies that

det (A B) = - 1

. (This is due to the property of determinants that

det (A B) = det (A) \cdot det (B)

Step 3: Expressing A + B

Now, let’s express

A + B

in terms of

A

and

B

A + B = A I + B I = A (B^{T} + A^{T}) B

Step 4: Determinant of A + B

We want to find the determinant of

A + B

\begin{aligned} det (A + B) & = det (A) \cdot det (B^{T} + A^{T}) \cdot det (B) \\ = det (A B) \cdot det (B^{T} + A^{T}) \\ = (- 1) \cdot det (B^{T} + A^{T}) (From Step 2) \\ = - det (B^{T} + A^{T}) \end{aligned}

Step 5: Determinant of Transpose Sum

Now, let’s focus on the determinant of the transpose sum:

\begin{aligned} det (B^{T} + A^{T}) & = det ((B + A)^{T}) (Transpose of a sum) \\ = det (B + A) \end{aligned}

Step 6: Putting It All Together

Now, we can express

det (A + B)

as:

det (A + B) = - det (B + A)

But

A + B = B + A

because matrix addition is commutative.

So, we have:

det (A + B) = - det (B + A) = - det (A + B)

This implies that

2 det (A + B) = 0

Step 7: Conclusion

From the equation

2 det (A + B) = 0

, we can conclude that

det (A + B) = 0

A matrix is considered singular if its determinant is zero. Therefore, we have shown that

A + B

is a singular matrix.

x + 2 y + 3 z = 12

निर्देशांक अक्षों को

A, B, C

पर प्रतिच्छेद करता है। त्रिभुज

A B C

के परिवृत्त का समीकरण ज्ञात कीजिए।
(ii) सिद्ध कीजिए कि समतल

z = 0

गोलक

x^{2} + y^{2} + z^{2} = 11

के अन्वालोपी शंकु, जिसका शीर्ष

(2, 4, 1)

पर है, को एक समकोणीय अतिपरवलय पर प्रतिच्छेद करता है।

(i) The plane

x + 2 y + 3 z = 12

cuts the axes of coordinates in

A, B, C

. Find the equations of the circle circumscribing the triangle

A B C

Answer:

To find the equation of the circle circumscribing the triangle ABC, we start with the given plane equation:

x + 2 y + 3 z = 12

This plane intersects the x-axis, y-axis, and z-axis at points A, B, and C, respectively. We’ve already found the coordinates of these points:

A(12, 0, 0)
B(0, 6, 0)
C(0, 0, 4)

Now, let’s assume the equation of the circumscribing circle is:

x^{2} + y^{2} + z^{2} + 2 u x + 2 v y + 2 w z + d = 0

We need to find the values of u, v, w, and d.

Using point A(12, 0, 0):

(12)^{2} + 2 u (12) + d = 0

144 + 24 u + d = 0

24 u + d = - 144 - - - - (1)

Using point B(0, 6, 0):

(6)^{2} + 2 v (6) + d = 0

36 + 12 v + d = 0

12 v + d = - 36 - - - - (2)

Using point C(0, 0, 4):

(4)^{2} + 2 w (4) + d = 0

16 + 8 w + d = 0

8 w + d = - 16 - - - - (3)

Now, we can solve equations (1), (2), and (3) for u, v, and w:

$24 u + d = - 144$ implies $24 u = - 144 - d$
$12 v + d = - 36$ implies $12 v = - 36 - d$
$8 w + d = - 16$ implies $8 w = - 16 - d$

Now, we can substitute these expressions for 2u, 2v, and 2w back into the equation of the circle:

x^{2} + y^{2} + z^{2} - (12 - d / 12) x - (6 - d / 6) y - (4 - d / 4) z + d = 0

Simplify further:

x^{2} + y^{2} + z^{2} - (12 - d / 12) x - (6 - d / 6) y - (4 - d / 4) z + d = 0

Now, the equation of the circle is in the form:

x^{2} + y^{2} + z^{2} + A x + B y + C z + D = 0

Where:

A = - (12 - d / 12)

B = - (6 - d / 6)

C = - (4 - d / 4)

D = d

The values of A, B, C, and D can vary as d takes any real value. So, the equation of the circle circumscribing the triangle ABC is:

x^{2} + y^{2} + z^{2} - (12 - d / 12) x - (6 - d / 6) y - (4 - d / 4) z + d = 0

Where d can be any real number.

(ii) Prove that the plane

z = 0

cuts the enveloping cone of the sphere

x^{2} + y^{2} + z^{2} = 11

which has the vertex at

(2, 4, 1)

in a rectangular hyperbola.

Answer:

Given the equation of the sphere

S = x^{2} + y^{2} + z^{2} - 11

and the coordinates of its vertex

(α, β, γ) = (2, 4, 1)

Calculate $S_{1}$ and the equation of the plane $T$ :
$\begin{aligned} S_{1} & = α^{2} + β^{2} + γ^{2} - 11 = 2^{2} + 4^{2} + 1^{2} - 11 = 10 \\ T & = 2 x + 4 y + z - 11 \end{aligned}$
Write the equation of the enveloping cone of the sphere:
The equation of the enveloping cone can be written as $T^{2} = S S_{1}$ , where $T$ is the equation of the plane and $S$ is the equation of the sphere.
$(2 x + 4 y + z - 11)^{2} = (x^{2} + y^{2} + z^{2} - 11) (10)$
Substitute $z = 0$ into the equation:
To check if the plane $z = 0$ cuts the cone, put $z = 0$ in the above equation:
$10 (x^{2} + y^{2} - 11) = (2 x + 4 y - 11)^{2}$
Simplify the equation:
Further simplify the equation:
$6 x^{2} + 6 y^{2} - 16 xy + 44 x - 88 y - 231 = 0$

The resulting equation is a rectangular hyperbola, which proves that the plane

z = 0

cuts the enveloping cone of the sphere in a rectangular hyperbola.

(a) फलन $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6$ का अंतराल $[2, 3]$ पर अधिकतम और न्यूनतम मान ज्ञात कीजिए।

Find the maximum and the minimum value of the function

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6

on the interval

[2, 3]

Answer:

Introduction

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

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6

on the interval

[2, 3]

. To find these extremum points, we’ll use calculus methods, specifically by finding the derivative of the function and setting it equal to zero to find critical points. We’ll then evaluate the function at these critical points and the endpoints of the interval to determine the maximum and minimum values.

Work/Calculations

Step 1: Find the Derivative of $f (x)$

The first derivative of

f (x)

will give us the rate of change of the function at any given point

x

f^{'} (x) = \frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x + 6)

After calculating, we find that the derivative of

f (x)

is:

f^{'} (x) = 6 x^{2} - 18 x + 12

Step 2: Find the Critical Points

To find the critical points, we set

f^{'} (x) = 0

and solve for

x

6 x^{2} - 18 x + 12 = 0

After solving, we find that the critical points are

x = 1

and

x = 2

Step 3: Evaluate $f (x)$ at Critical Points and Endpoints

To find the maximum and minimum values of

f (x)

on the interval

[2, 3]

, we need to evaluate

f (x)

at the critical points and the endpoints of the interval. The critical points within the interval are

x = 2

and the endpoints are

x = 2

and

x = 3

Let’s substitute the values into

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6

and calculate:

$f (2)$
$f (3)$

Let’s start by calculating

f (2)

After substituting the values, we get:

f (2) = 2 \times 2^{3} - 9 \times 2^{2} + 12 \times 2 + 6 = 10

Next, let’s calculate

f (3)

After substituting the values, we get:

f (3) = 2 \times 3^{3} - 9 \times 3^{2} + 12 \times 3 + 6 = 15

Conclusion

We evaluated

f (x)

at the critical point

x = 2

and the endpoint

x = 3

within the interval

[2, 3]

$f (2) = 10$
$f (3) = 15$

Therefore, the maximum value of

f (x)

on the interval

[2, 3]

15

x = 3

, and the minimum value is

10

x = 2

(b) सिद्ध कीजिए कि साधारणतः किसी एक बिंदु से परवलयज

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

पर तीन अभिलंब बनाए जा सकते हैं, लेकिन अगर बिंदु सतह

27 a (x^{2} + y^{2}) + 8 (a - z)^{3} = 0

पर स्थित है, तो इन तीन अभिलंबों में से दो अभिलंब एक ही हैं।

Prove that, in general, three normals can be drawn from a given point to the paraboloid

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

, but if the point lies on the surface

27 a (x^{2} + y^{2}) + 8 (a - z)^{3} = 0

then two of the three normals coincide.

Answer:

Introduction:
We are tasked with proving that, in general, three normals can be drawn from a given point to the paraboloid

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

. However, if the point lies on the surface

27 a (x^{2} + y^{2}) + 8 (a - z)^{3} = 0

, then two of the three normals coincide.

Work/Calculations:

The equation of the normal at $(x_{1}, y_{1}, z_{1})$ to the paraboloid is given by:

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

This passes through a given point $(α, β, γ)$ if:

\frac{α - x_{1}}{x_{1}} = \frac{β - y_{1}}{y_{1}} = \frac{γ - z_{1}}{- a} = λ

Solving for $x_{1}$ , $y_{1}$ , and $z_{1}$ in terms of $α$ , $β$ , $γ$ , and $λ$ using the above equations:

x_{1} = \frac{α}{1 + λ}; y_{1} = \frac{β}{1 + λ}; z_{1} = γ + a λ (from (1))

Also, $(x_{1}, y_{1}, z_{1})$ lies on the given paraboloid, so:

x_{1}^{2} + y_{1}^{2} = 2 a z_{1} (from the equation of the paraboloid)

Substituting the expressions for $x_{1}$ , $y_{1}$ , and $z_{1}$ from step 3 into the equation above:

{(\frac{α}{1 + λ})}^{2} + {(\frac{β}{1 + λ})}^{2} = 2 a (γ + a λ) (from (1))

This equation can be simplified to:

α^{2} + β^{2} = 2 a (γ + a λ) (1 + λ)^{2} (from (2))

Equation (2) is a cubic in $λ$ and has three values of $λ$ that satisfy it, leading to three points on the paraboloid normal at which pass through $(α, β, γ)$ .
Rewriting equation (2) as a function $f (λ)$ :

f (λ) + 2 a (1 + λ)^{2} (γ + a λ) - (α^{2} - β^{2}) = 0 (from (3))

To find the condition that equation (3) has two equal roots, we need to solve for $λ$ where both $f (λ) = 0$ and $f^{'} (λ) = 0$ .
Differentiating $f (λ)$ with respect to $λ$ to find $f^{'} (λ)$ :

f^{'} (λ) = 2 a (1 + λ)^{2} (a) + 4 a (1 + λ) (γ + a λ) = 0

Solving for $λ$ in $f^{'} (λ) = 0$ :

a (1 + λ) + 2 (γ + a λ) = 0 (since 1 + λ \neq 0)

Simplifying and solving for $λ$ :

(a + 2 γ) + λ (3 a) = 0 \Rightarrow λ = - \frac{a + 2 γ}{3 a}

Substituting this value of $λ$ into equation (3):

2 a {[1 - \frac{a + 2 γ}{3 a}]}^{2} [γ - \frac{a (a + 2 γ)}{3 a}] = α^{2} + β^{2}

Further simplifying this equation:

2 a [2 (a - γ)]^{2} [a (γ - a)] = 27 a^{3} (α^{2} + β^{2})

Finally, the equation simplifies to:

27 a (α^{2} + β^{2}) + 8 (a - γ)^{3} = 0

Conclusion:
We have successfully shown that the locus of the point

(α, β, γ)

27 a (x^{2} + y^{2}) + 8 (a - z)^{3} = 0

. This proves the given statement.

A = (\begin{array}{rrrr} 5 & 7 & 2 & 1 \\ 1 & 1 & - 8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array})

(i) आव्यूह

A

की कोटि ज्ञात कीजिए।
(ii) उपसमष्टि

V = {(x_{1}, x_{2}, x_{3}, x_{4}) \in R^{4} ∣ A (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}) = 0}

की विमा ज्ञात कीजिए।

Let

A = (\begin{array}{rrrr} 5 & 7 & 2 & 1 \\ 1 & 1 & - 8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array})

(i) Find the rank of matrix

A

(ii) Find the dimension of the subspace

V = {(x_{1}, x_{2}, x_{3}, x_{4}) \in R^{4} ∣ A (\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array}) = 0}

Answer:

Rank [\begin{array}{cccc} 5 & 7 & 2 & 1 \\ 1 & 1 & - 8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array}]

Now, reduce this matrix

\begin{aligned} R_{1} \leftarrow R_{1} \div 5 \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 1 & 1 & - 8 & 1 \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array}] \end{aligned}

\begin{aligned} \begin{aligned} R_{2} \leftarrow R_{2} - R_{1} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & - \frac{2}{5} & - \frac{42}{5} & \frac{4}{5} \\ 2 & 3 & 5 & 0 \\ 3 & 4 & - 3 & 1 \end{array}] \end{aligned} \\ \begin{aligned} R_{3} \leftarrow R_{3} - 2 \times R_{1} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & - \frac{2}{5} & - \frac{42}{5} & \frac{4}{5} \\ 0 & \frac{1}{5} & \frac{21}{5} & - \frac{2}{5} \\ 3 & 4 & - 3 & 1 \end{array}] \end{aligned} \end{aligned}

\begin{aligned} R_{4} \leftarrow R_{4} - 3 \times R_{1} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & - \frac{2}{5} & - \frac{42}{5} & \frac{4}{5} \\ 0 & \frac{1}{5} & \frac{21}{5} & - \frac{2}{5} \\ 0 & - \frac{1}{5} & - \frac{21}{5} & \frac{2}{5} \end{array}] \end{aligned}

\begin{aligned} R_{2} \leftarrow R_{2} \times - \frac{5}{2} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & 1 & 21 & - 2 \\ 0 & \frac{1}{5} & \frac{21}{5} & - \frac{2}{5} \\ 0 & - \frac{1}{5} & - \frac{21}{5} & \frac{2}{5} \end{array}] \\ R_{3} \leftarrow R_{3} - \frac{1}{5} \times R_{2} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & 1 & 21 & - 2 \\ 0 & 0 & 0 & 0 \\ 0 & - \frac{1}{5} & - \frac{21}{5} & \frac{2}{5} \end{array}] \end{aligned}

\begin{matrix} R_{4} \leftarrow R_{4} + \frac{1}{5} \times R_{2} \\ = [\begin{array}{cccc} 1 & \frac{7}{5} & \frac{2}{5} & \frac{1}{5} \\ 0 & 1 & 21 & - 2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] \end{matrix}

The rank of a matrix is the number of non all-zeros rows

∴ Rank = 2

(ii) Find the dimension of the subspace

V

The dimension of the subspace

V

is related to the rank of

A

by the formula:

Dimension of

V = n - Rank (A)

Here,

n

is the number of columns in

A

, which is 4 .
Dimension of

V = 4 - 2 = 2

(a) कैले-हैमिल्टन प्रमेय का कथन लिखिए। इस प्रमेय का उपयोग करके $A^{100}$ का मान ज्ञात कीजिए, जहाँ

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

State the Cayley-Hamilton theorem. Use this theorem to find

A^{100}

, where

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

Answer:

Cayley-Hamilton Theorem:
The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation over a commutative ring, such as the real or complex field. In other words, if

λ I - A

is the characteristic equation of a matrix

A

, where

I

is the identity matrix and

λ

is a scalar, then

A

satisfies the equation

| λ I - A | = 0

First, let’s find the characteristic equation for matrix $A$ :

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

Now, according to the Cayley-Hamilton theorem, $A$ satisfies its characteristic equation, so $A$ satisfies:

A^{3} - A^{2} - A + I = 0

Rearranging the equation, we find:

A^{3} = A^{2} + A - I

We can now calculate higher powers of $A$ based on this result:

\begin{aligned} A^{4} & = A^{3} + A^{2} - A \\ = (A^{2} + A - I) + A^{2} - A \\ = 2 A^{2} - I \end{aligned}

Similarly, we can find $A^{5}$ as:

\begin{aligned} A^{5} & = 2 A^{3} - A \\ = 2 (A^{2} + A - I) - A \\ = 2 A^{2} + 2 A - 2 I - A \\ = 2 A^{2} + A - 2 I \end{aligned}

Continuing this pattern, we find:

A^{6} = 3 A^{2} - 2 I

Use the General Form $A^{2 n} = n A^{2} - (n - 1) I$

The general form

A^{2 n} = n A^{2} - (n - 1) I

can be used to find

A^{100}

For

n = 50

, the expression becomes:

A^{100} = 50 A^{2} - 49 I

Substitute the Values

Let’s substitute the values of

A^{2}

and

I

into the formula:

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

I = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]

After substituting, we get:

A^{100} = 50 [\begin{array}{lll} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}] - 49 [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]

Calculate $A^{100}$

After calculating, we find that

A^{100}

is:

A^{100} = [\begin{array}{lll} 1 & 0 & 0 \\ 50 & 1 & 0 \\ 50 & 0 & 1 \end{array}]

Conclusion

Using the general form

A^{2 n} = n A^{2} - (n - 1) I

, we found that

A^{100}

for the given matrix

A

is:

A^{100} = [\begin{array}{lll} 1 & 0 & 0 \\ 50 & 1 & 0 \\ 50 & 0 & 1 \end{array}]

(b) बिंदु

P

से गुजरने वाली दीर्घवृत्तज

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

की अभिलंब जीवा की लंबाई ज्ञात कीजिए और सिद्ध कीजिए कि अगर यह

4 P G_{3}

के समान है, जहाँ

G_{3}

वह बिंदु है जहाँ

P

से गुजरने वाली अभिलंब जीवा

x y

-तल पर मिलती है, तो

P

शंकु

\frac{x^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{y^{2}}{b^{6}} (2 c^{2} - b^{2}) + \frac{z^{2}}{c^{4}} = 0

पर स्थित है।

Find the length of the normal chord through a point

P

of the ellipsoid

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

and prove that if it is equal to

4 P G_{3}

, where

G_{3}

is the point where the normal chord through

P

meets the

x y

-plane, then

P

lies on the cone

\frac{x^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{y^{2}}{b^{6}} (2 c^{2} - b^{2}) + \frac{z^{2}}{c^{4}} = 0

Answer:

Equation of the Normal:
Let

P

(α, β, γ)

, then the equation of the normal to the given ellipsoid at

P (α, β, γ)

are

\frac{x - α}{\frac{p α}{a^{2}}} = \frac{y - β}{\frac{P β}{b^{2}}} = \frac{z - r}{\frac{P γ}{c^{2}}} = r \to (1) where \frac{1}{P^{2}} = \frac{α^{2}}{a^{4}} + \frac{β^{2}}{b^{4}} + \frac{γ^{2}}{c^{4}} \to (2)

Therefore, the coordinates of any point

Q

on the normal (1) are

(α + \frac{P α}{a^{2}} r, β + \frac{P β}{b^{2}} r, γ + \frac{p γ}{c^{2}} r)

, where

r

is the distance of

Q

from

P

If $Q$ lies on the given ellipsoid i.e., $PQ$ is the normal chord, then

\begin{aligned} \frac{1}{a^{2}} {(α + \frac{p α}{a^{2}} r)}^{2} + \frac{1}{b^{2}} {(β + \frac{p β}{b^{2}} r)}^{2} + \frac{1}{c^{2}} {(γ + \frac{p γ}{c^{2}} r)}^{2} = 1 \\ \Rightarrow r^{2} P^{2} (\frac{α^{2}}{a^{6}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}}) + 2 r P (\frac{α^{2}}{a^{4}} + \frac{β^{2}}{b^{4}} + \frac{γ^{2}}{c^{4}}) + (\frac{α^{2}}{a^{2}} + \frac{β^{2}}{b^{2}} + \frac{γ^{2}}{c^{2}}) = 1 \end{aligned}

\Rightarrow r^{2} p^{2} (\frac{α^{2}}{a^{4}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}}) + 2 r p (\frac{1}{p^{2}}) = 0 [

from (2) and

Σ (\frac{α^{2}}{a^{2}}) = 1

p (α, β, γ)

lies on the given conicoid ]

\Rightarrow r = - \frac{2}{p^{3} (\frac{α^{2}}{a^{6}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}})} = length of normal chord P Q \to (3)

Also,
Let the normal at

P (α, β, γ)

meets the coordinate plane viz.

y z, z x

and

x y

planes at

G_{1}, G_{2}, G_{3}

Then putting

x = 0, y = 0

and

z = 0

in succession in the equation (1), we have respectively,

P G_{1} = - \frac{a^{2}}{P}, P G_{2} = - \frac{b^{2}}{p} and P G_{3} = - \frac{c^{2}}{p} \to (4)

Given, $P Q = 4 P G_{3}$

\begin{aligned} \Rightarrow p Q = 4 (- \frac{c^{2}}{p}) \\ \Rightarrow r = - \frac{2}{p^{3} (\frac{α^{2}}{a^{6}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}})} = 4 (- \frac{c^{2}}{p}) \\ \Rightarrow 2 c^{2} (\frac{α^{2}}{a^{6}} + \frac{β^{2}}{b^{6}} + \frac{γ^{2}}{c^{6}}) = \frac{1}{p^{2}} \\ = (\frac{α^{2}}{a^{4}} + \frac{β^{2}}{b^{4}} + \frac{γ^{2}}{c^{4}}) \to from (2) \\ \Rightarrow \frac{α^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{β^{2}}{b^{2}} (2 c^{2} - b^{2}) + \frac{γ^{2}}{c^{6}} (2 c^{2} - c^{2}) = 0 \\ \Rightarrow \frac{α^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{β^{2}}{b^{2}} (2 c^{2} - b^{2}) + \frac{γ^{2}}{c^{6}} (c^{2}) = 0 \\ \Rightarrow \frac{α^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{β^{2}}{b^{2}} (2 c^{2} - b^{2}) + \frac{γ^{2}}{c^{4}} = 0 \end{aligned}

Conclusion:
We have found the length of the normal chord through a point

P

on the given ellipsoid and proved that if this length is equal to

4 P G_{3}

, then

P

lies on the cone defined by the equation

\frac{α^{2}}{a^{6}} (2 c^{2} - a^{2}) + \frac{β^{2}}{b^{2}} (2 c^{2} - b^{2}) + \frac{γ^{2}}{c^{4}} = 0

Hence proved.

u = \sin^{- 1} \sqrt{\frac{x^{1 / 3} + y^{1 / 3}}{x^{1 / 2} + y^{1 / 2}}}

है, तो दर्शाइए कि

\sin^{2} u, x

और

y

का

- \frac{1}{6}

घातविशिष्ट समांगी फलन है। अतएव दर्शाइए कि

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = \frac{\tan u}{12} (\frac{13}{12} + \frac{\tan^{2} u}{12})

(ii) जैकोबियन विधि का व्यवहार करते हुए दर्शाइए कि अगर

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

और

f (0) = 0

है, तो

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

(i) If

u = \sin^{- 1} \sqrt{\frac{x^{1 / 3} + y^{1 / 3}}{x^{1 / 2} + y^{1 / 2}}}

then show that

\sin^{2} u

is a homogeneous function of

x

and

y

of degree

- \frac{1}{6}

.
Hence show that

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = \frac{\tan u}{12} (\frac{13}{12} + \frac{\tan^{2} u}{12})

Answer:

Introduction:
We are given the equation

u = \sin^{- 1} \sqrt{\frac{x^{1 / 3} + y^{1 / 3}}{x^{1 / 2} + y^{1 / 2}}}

and asked to show that

\sin^{2} u

is a homogeneous function of

x

and

y

of degree

- \frac{1}{6}

. Additionally, we need to demonstrate that

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = \frac{\tan u}{12} (\frac{13}{12} + \frac{\tan^{2} u}{12})

Homogeneity Analysis:

Step 1: Defining $f$ and Finding Degree of Homogeneity):

Let’s start by defining

f = \sin u

and then find the degree of homogeneity.

f = \sin u = \sqrt{\frac{x^{1 / 3} + y^{1 / 3}}{x^{1 / 2} + y^{1 / 2}}}

The degree of homogeneity is found as:

Degree = \frac{1}{3} - \frac{1}{2} = - \frac{1}{6}

Step 2: Expressing $f$ in terms of $x$ and $y$ :

Now, we express

f

in terms of

x

and

y

f = x^{- \frac{1}{12}} g (\frac{y}{x})

Step 3: Determining Homogeneous Function of $u$ :

Next, we find

\sin^{2} u

as a homogeneous function of

x

and

y

\sin^{2} u = x^{(- \frac{1}{6}) g (\frac{y}{x})}

The degree is confirmed to be

- \frac{1}{6}

Derivatives and Further Analysis:

Step 4: Partial Derivatives:

Now, let’s calculate the partial derivatives

\frac{\partial f}{\partial x}

and

\frac{\partial f}{\partial y}

using the homogeneous function properties:

x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n \cdot f where n = - \frac{1}{6}

Step 5: Derivative of $\sin u$ and Simplification:

Since

f = \sin u

, we differentiate with respect to

x

x \cos u \frac{\partial u}{\partial x} + y \cos u \frac{\partial u}{\partial y} = - \frac{1}{12} \sin u

After some simplification and using trigonometric identities, we get:

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = - \frac{1}{12} \tan u

Step 6: Second Derivative and Further Simplification:

Differentiate equation 1 partially with respect to

x

and

y

and add the resulting equations. Then simplify as follows:

\begin{aligned} x \frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial u}{\partial x} + y \frac{\partial^{2} u}{\partial x \partial y} = - \frac{1}{12} \sec^{2} u \frac{\partial u}{\partial x} \\ y^{2} \frac{\partial u}{\partial y} + y \frac{\partial u}{\partial y} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} = (x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}) (1 - \frac{1}{12} \sec^{2} u) \\ [1 + \frac{1}{12} (\tan^{2} α + 1)] = - \frac{1}{12} \tan u (1 + \frac{1}{12} \sec^{2} u) \\ = \frac{\tan u}{12} (\frac{13}{12} + \frac{\tan^{2} u}{12}) \end{aligned}

Conclusion:

We have shown that

\sin^{2} u

is indeed a homogeneous function of

x

and

y

with a degree of

- \frac{1}{6}

. Additionally, we have derived and simplified the expression

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}}

as given in the problem statement.

(ii) Using the Jacobian method, show that if

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

and

f (0) = 0

, then

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

Answer:

Work/Calculations

Step 1: Integrate $f^{'} (x)$ to find $f (x)$

Given

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

, we integrate to find

f (x)

\int f^{'} (x) d x = \int \frac{1}{1 + x^{2}} d x

f (x) = \tan^{- 1} (x) + c

Since

f (0) = 0

, we find that

c = 0

. Therefore,

f (x) = \tan^{- 1} (x)

Step 2: Define $α$ and $β$

α = f (x) + f (y) = \tan^{- 1} (x) + \tan^{- 1} (y)

β = f (\frac{x + y}{1 - x y}) = \tan^{- 1} (\frac{x + y}{1 - x y})

Step 3: Calculate the Partial Derivatives of $α$ and $β$

For

α

, we have:

\frac{\partial α}{\partial x} = \frac{1}{1 + x^{2}}

\frac{\partial α}{\partial y} = \frac{1}{1 + y^{2}}

For

β

, we differentiate

β

with respect to

x

\frac{\partial β}{\partial x} = \frac{\partial}{\partial x} [\tan^{- 1} (\frac{x + y}{1 - x y})]

After calculating, we find:

\frac{\partial β}{\partial x} = \frac{1}{1 + x^{2}}

Similarly, to find

\frac{\partial β}{\partial y}

, we differentiate

β

with respect to

y

\frac{\partial β}{\partial y} = \frac{\partial}{\partial y} [\tan^{- 1} (\frac{x + y}{1 - x y})]

After calculating, we find:

\frac{\partial β}{\partial y} = \frac{1}{1 + y^{2}}

Step 4: Check if the Partial Derivatives are Equal

To prove that

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

, it’s sufficient to show that

\frac{\partial α}{\partial x} = \frac{\partial β}{\partial x}

and

\frac{\partial α}{\partial y} = \frac{\partial β}{\partial y}

Since

\frac{\partial α}{\partial x} = \frac{\partial β}{\partial x}

and

\frac{\partial α}{\partial y} = \frac{\partial β}{\partial y}

, we can conclude that

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

Conclusion with Jacobian Method

In the Jacobian method, we aim to show that two functions

α

and

β

are equal by proving that their partial derivatives with respect to the same variables are equal. In this case,

α = f (x) + f (y)

and

β = f (\frac{x + y}{1 - x y})

We calculated the partial derivatives

\frac{\partial α}{\partial x}

\frac{\partial α}{\partial y}

\frac{\partial β}{\partial x}

, and

\frac{\partial β}{\partial y}

and found:

\frac{\partial α}{\partial x} = \frac{1}{1 + x^{2}} = \frac{\partial β}{\partial x}

\frac{\partial α}{\partial y} = \frac{1}{1 + y^{2}} = \frac{\partial β}{\partial y}

Since both sets of partial derivatives are equal, the Jacobian method confirms that

α

and

β

are indeed the same function under the transformation of variables. Therefore, we can conclude that:

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

This successfully proves the given equation using the Jacobian method.

खण्ड-B / SECTION-B

(a) अवकल समीकरण

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

को हल कीजिए।

Solve the differential equation

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

Answer:

Introduction:
We are given the differential equation:

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

Work/Calculations:

Step 1: Identify $M$ and $N$ :
In the given equation, we have

M = 2 y \sin x + 3 y^{4} \sin x \cos x

and

N = - (4 y^{3} \cos^{2} x + \cos x)

Step 2: Calculate Partial Derivatives:
Calculate the partial derivatives of

M

and

N

\frac{\partial M}{\partial y} = 2 \sin x + 12 y^{3} \sin x \cos x

\frac{\partial N}{\partial x} = 8 y^{3} \sin x \cos x + \sin x

Step 3: Check for Exactness:
Now, check if the equation is exact by comparing

\frac{\partial M}{\partial y}

and

\frac{\partial N}{\partial x}

\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{2 \sin x + 12 y^{3} \sin x \cos x - (8 y^{3} \sin x \cos x + \sin x)}{- (4 y^{3} \cos^{2} x + \cos x)} = - \tan x

Step 4: Find the Integration Factor:
The integration factor is given by

e^{- \int \tan x d x} = \cos x

Step 5: Multiply and Simplify:
Multiply the entire equation by the integration factor

\cos x

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

This makes the equation exact.

Step 6: Integrate:
Now, integrate both sides with respect to

x

to find the solution:

\int (y \sin 2 x + \frac{3 y^{2}}{2} \sin 2 x \cos x) d x = C

Further integrate and simplify to find the solution:

- \frac{y \cos 2 x}{2} - \frac{3 y^{2}}{4} \cdot \frac{\cos 3 x}{3} - \frac{3 y^{2}}{4} \cos x = C

Conclusion:
The solution to the given differential equation is:

- \frac{y \cos 2 x}{2} - \frac{3 y^{2}}{4} \cdot \frac{\cos 3 x}{3} - \frac{3 y^{2}}{4} \cos x = C

(b) अवकल समीकरण

\frac{d^{2} y}{d x^{2}} - 4 \frac{d y}{d x} + 4 y = 3 x^{2} e^{2 x} \sin 2 x

का पूर्ण हल ज्ञात कीजिए।

Determine the complete solution of the differential equation

\frac{d^{2} y}{d x^{2}} - 4 \frac{d y}{d x} + 4 y = 3 x^{2} e^{2 x} \sin 2 x

Answer:

Auxiliary Equation:

To solve the given differential equation, we start by finding the auxiliary equation, which is given by

D^{2} - 4 D + 4 = 0

where

D

represents the derivative operator. Solving this equation yields two identical roots,

D = 2

. Thus, the complementary function (C.F.) is expressed as:

C . F . = (c_{1} + c_{2} x) e^{2 x}

where

c_{1}

and

c_{2}

are arbitrary constants.

Particular Integral (P.I.):

\begin{aligned} P.I. = \frac{1}{(D - 2)^{2}} 3 x^{2} e^{2 x} \sin 2 x = 3 e^{2 x} \frac{1}{(D + 2 - 2)^{2}} x^{2} \sin 2 x = 3 e^{2 x} \frac{1}{D^{2}} x^{2} \sin 2 x \\ = & 3 e^{2 x} \frac{1}{D} \int x^{2} \sin 2 x d x = 8 e^{2 x} \frac{1}{D} [x^{2} (- \frac{\cos 2 x}{2}) - \int (2 x) (- \frac{\cos 2 x}{2}) d x], integrating by parts \\ = & 3 e^{2 x} \frac{1}{D} [- \frac{1}{2} x^{2} \cos 2 x + \int x \cos 2 x d x] = 8 e^{2 x} \frac{1}{D} [- \frac{1}{2} x^{2} \cos 2 x + x (\frac{\sin 2 x}{2}) - \int 1 \cdot \frac{\sin 2 x}{2} d x] \\ = & 3 e^{2 x} \frac{1}{D} [- \frac{1}{2} x^{2} \cos 2 x + \frac{1}{2} x \sin 2 x + \frac{1}{4} \cos 2 x] = 8 e^{2 x} \int (- \frac{1}{2} x^{2} \cos 2 x + \frac{1}{2} x \sin 2 x + \frac{1}{4} \cos 2 x) d x \\ = & 3 e^{2 x} [- \frac{1}{2} \int x^{2} \cos 2 x d x + \frac{1}{2} \int x \sin 2 x d x + \frac{1}{4} \int \cos 2 x d x] \\ = & 3 e^{2 x} [- \frac{1}{2} {x^{2} (\frac{1}{2} \sin 2 x) - \int 2 x (\frac{1}{2} \sin 2 x) d x} + \frac{1}{2} \int x \sin 2 x d x + \frac{1}{8} \sin 2 x] \\ = & 3 e^{2 x} [- \frac{1}{4} x^{2} \sin 2 x d x + \frac{1}{2} \int x \sin 2 x d x + \frac{1}{2} \int x \sin 2 x d x + (1 / 8) \times \sin 2 x] \end{aligned}

\begin{aligned} = 3 e^{2 x} [- \frac{1}{4} x^{2} \sin 2 x + \int x \sin 2 x d x + \frac{1}{8} \sin 2 x] \\ = 3 e^{2 x} [- \frac{1}{4} x^{2} \sin 2 x + x (- \frac{1}{2} \cos 2 x) - \int 1 \cdot (- \frac{1}{2} \cos 2 x) d x + \frac{1}{8} \sin 2 x] \\ = 3 e^{2 x} [- \frac{1}{4} x^{2} \sin 2 x - \frac{1}{2} x \cos 2 x + \frac{1}{4} \sin 2 x + \frac{1}{8} \sin 2 x] \\ = 3 e^{2 x} [- \frac{1}{4} x^{2} \sin 2 x - \frac{1}{2} x \cos 2 x + \frac{3}{8} \sin 2 x] \end{aligned}

Conclusion:

In conclusion, the complete solution to the given differential equation is:

y = (c_{1} + c_{2} x) e^{2 x} + 3 e^{2 x} [- \frac{1}{4} x^{2} \sin 2 x - \frac{1}{2} x \cos 2 x + \frac{3}{8} \sin 2 x]

This solution encompasses both the complementary function (C.F.) and the particular integral (P.I.).

A B

का एक सिरा एक रूक्ष क्षैतिज छड़

A C

, जिसके साथ वह वलय (रिंग) के द्वारा जुड़ी हुई है, पर सरक सकती है।

B

एवं

C

एक रस्सी से जुड़े हैं। जब छड़ सर्पण बिंदु पर है, तब

A C^{2} - A B^{2} = B C^{2}

है। यदि क्षैतिज रेखा व

A B

के बीच का कोण

θ

है, तो सिद्ध कीजिए कि घर्षण गुणांक

\frac{\cot θ}{2 + \cot^{2} θ}

है।

One end of a heavy uniform rod

A B

can slide along a rough horizontal rod

A C

, to which it is attached by a ring.

B

and

C

are joined by a string. When the rod is on the point of sliding, then

A C^{2} - A B^{2} = B C^{2}

. If

θ

is the angle between

A B

and the horizontal line, then prove that the coefficient of friction is

\frac{\cot θ}{2 + \cot^{2} θ}

Answer:

Given: One end of a heavy uniform rod

A B

can slide along a rough horizontal rod

A C

, to which it is attached by a ring.

B

and

C

are joined by a string. When the rod is on the point of sliding, we have

A C^{2} - A B^{2} = B C^{2}

Let

m

be the mass of the rod,

g

the acceleration due to gravity,

N

the normal force acting on the ring in a clockwise direction,

T

the force on the string, and

f

the frictional force.

In equilibrium, we have:

Horizontal net force

\sum F_{H} = 0

Vertical net force

\sum F_{V} = 0

Net torque

\sum τ = 0

Then, the force on the rod is given by

m g = T \sin θ + N

, which can be written as:

T \sin θ = m g - N - - - - (1)

The force in the horizontal direction (

f

) is equal to

μ N

, where

μ

is the coefficient of friction. This can be expressed as:

T \cos θ = μ N - - - - (2)

Dividing equation (2) by equation (1), we get:

\cot θ = \frac{μ}{m g - N} - - - - (3)

Now, let’s consider the torque balance:

τ_{c w} = τ_{A C W}

m g (\frac{ρ}{2} \sin θ) + μ N \cdot l \cos θ = N \cdot l \sin θ

μ N \cos θ = (N - \frac{m g}{2}) \sin θ

μ N \cot θ = \frac{2 N - m g}{2}

2 μ N \cot θ = 2 N - m g = N - (m g - N)

2 μ N \cot θ = N - (\frac{μ N}{\cot θ})

2 μ \cot θ = 1 - \frac{μ}{\cot θ}

μ (2 \cot θ + \frac{1}{\cot θ}) = 1

μ \frac{(2 + \cot^{2} θ)}{\cot θ} = 1

μ = \frac{\cot^{2} θ}{2 + \cot^{2} θ}

So, the coefficient of friction is

\frac{\cot θ}{2 + \cot^{2} θ}

(d) एक कण का पृथ्वी द्वारा आकर्षण बल उस कण के पृथ्वी के केन्द्र से दूरी के वर्ग के व्युत्क्रमानुपाती है। एक कण, जिसका भार पृथ्वी की सतह पर

W

है, सतह से

3 h

ऊँचाई से पृथ्वी की सतह पर गिरता है। दर्शाइए कि पृथ्वी के आकर्षण बल द्वारा किए गए कार्य का परिमाण

\frac{3}{4} h W

है, जहाँ

h

पृथ्वी की त्रिज्या है।

The force of attraction of a particle by the earth is inversely proportional to the square of its distance from the earth’s centre. A particle, whose weight on the surface of the earth is

W

, falls to the surface of the earth from a height

3 h

above it. Show that the magnitude of work done by the earth’s attraction force is

\frac{3}{4} h W

, where

h

is the radius of the earth.

Answer:

Given that the force of attraction of a particle by the earth is inversely proportional to the square of its distance from the earth’s center, we can express it as:

F \propto \frac{1}{{distance}^{2}}

The weight of a particle on the surface of the earth is denoted as

W

, and the radius of the earth is

h

. The distance between the earth’s surface and the particle is

3 h

The difference in distance between the two positions (initial and final) of the particle is

3 h - h = 2 h

. According to Newton’s law of universal gravitation:

F = \frac{G m_{1} m_{2}}{d^{2}}

where

F

is the force of attraction,

G

is the gravitational constant,

m_{1}

and

m_{2}

are the masses of the two objects, and

d

is the distance between their centers.

In this case, we have:

F = \frac{W}{(2 h)^{2}} = \frac{W}{4 h^{2}}

Because the force is inversely proportional to

d^{2}

, doubling the distance between two masses reduces the gravitational force to

\frac{1}{4}

of its original value.

Now, let’s calculate the work done:

Work done = F \cdot d = \int F = \frac{W}{4 h^{2}} \cdot 3 h \cdot h = \frac{3 W h}{4}

Therefore, the magnitude of the work done by the earth’s attraction force is

\frac{3}{4} h W

(e) वक्र

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

के बिंदु

(1, 1, 1)

पर स्पर्श-रेखा की दिशा में फलन

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

का दिशात्मक अवकलज ज्ञात कीजिए।

Find the directional derivative of the function

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

along the tangent to the curve

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

at the point

(1, 1, 1)

Answer:

Let’s start by finding the gradient of the function

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

\begin{aligned} \nabla f = (i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y} + k \frac{\partial}{\partial z}) f = i \frac{\partial f}{\partial x} + j \frac{\partial f}{\partial y} + k \frac{\partial f}{\partial z} \\ = i (y^{2} + 2 x z) + j (2 x y + z^{2}) + k (2 y z + x^{2}) \end{aligned}

Now, we need to find the gradient of

f

at the point

(1, 1, 1)

\nabla f_{(1, 1, 1)} = 3 i + 3 j + 3 k

The parametric equation for the curve is given as

\vec{r} (t) = (t, t^{2}, t^{3})

, and we want to find the directional derivative along the tangent to this curve at the point

(1, 1, 1)

First, we find the tangent vector to the curve:

\frac{d \vec{r}}{d t} = i + 2 t j + 3 t^{2} k

Now, we normalize this tangent vector to obtain the unit vector

\hat{n}

\hat{n} = \frac{i + 2 j + 3 k}{\sqrt{1^{2} + 2^{2} + 3^{2}}} = \frac{i + 2 j + 3 k}{\sqrt{14}}

The directional derivative is given by the dot product of the gradient of

f

at the point

(1, 1, 1)

and the unit tangent vector

\hat{n}

\nabla f_{(1, 1, 1)} \cdot \hat{n} = \frac{3 i + 3 j + 3 k}{\sqrt{14}} \cdot (i + 2 j + 3 k) = \frac{3 \cdot 1 + 3 \cdot 2 + 3 \cdot 3}{\sqrt{14}} = \frac{18}{\sqrt{14}} = 18 \cdot \frac{\sqrt{14}}{14} = \frac{9}{7} \sqrt{14}

So, the magnitude of the directional derivative of the function along the tangent to the curve at the point

(1, 1, 1)

\frac{9}{7} \sqrt{14}

6(a) एक पिण्ड एक शंकु और उसके नीचे अर्धगोले से बना है। शंकु के आधार तथा अर्धगोले के शिखर का अर्धव्यास

a

है। पूरा पिण्ड एक रूक्ष क्षैतिज मेज पर रखा है, जिसका अर्धगोला मेज को स्पर्श करता है। दर्शाइए कि शंकु की अधिकतम ऊँचाई, जिससे कि साम्यावस्था स्थिर बनी रहे,

\sqrt{3} a

है।

A body consists of a cone and underlying hemisphere. The base of the cone and the top of the hemisphere have same radius

a

. The whole body rests on a rough horizontal table with hemisphere in contact with the table. Show that the greatest height of the cone, so that the equilibrium may be stable, is

\sqrt{3} a

Answer:

In the given scenario, let’s denote the common base of the hemisphere and the cone as

AB

, and their common axis as

COD

, which must be vertical for equilibrium. The hemisphere touches the table at point

C

Let

H

be the height of

OD

of the cone, and

a

be the radius of

OA

OC

OB

of the hemisphere and the cone, respectively.

Let

G_{1}

and

G_{2}

be the centers of gravity of the hemisphere and the cone, respectively. Then,

O G_{1} = \frac{3 a}{8}

and

O G_{2} = \frac{H}{4}

To find the height

h

of the center of gravity of the combined body (hemisphere and cone) above the point of contact

C

, we can use the formula:

h = \frac{\frac{1}{3} π a^{2} H C G_{2} + \frac{2}{3} π a^{2} C G_{1}}{\frac{1}{3} π a^{2} H + \frac{2}{3} π a^{3}}

Simplifying:

h = \frac{H (a + \frac{1}{4} H) + \frac{5}{4} a^{2}}{H + 2 a}

Now, for stable equilibrium, the condition is:

\frac{1}{h} > \frac{1}{ρ_{1}} + \frac{1}{ρ_{2}}

Where:

$ρ_{1}$ is the radius of curvature at the point of contact $C$ of the upper body, which is spherical, so $ρ_{1} = a$ .
$ρ_{2}$ is the radius of curvature of the lower body at the point of contact, which is essentially flat, so $ρ_{2} = \infty$ .

Therefore, the equilibrium is stable if:

\frac{1}{h} > \frac{1}{a} + \frac{1}{\infty} \Rightarrow \frac{1}{h} > \frac{1}{a} \Rightarrow h < a

Substituting the expression for

h

from above:

\frac{H (a + \frac{1}{4} H) + \frac{5}{4} a^{2}}{H + 2 a} < a

Simplifying further:

H a + \frac{1}{4} H^{2} + \frac{5}{4} H^{2} < H a + 2 a^{2}

This simplifies to:

\frac{1}{4} H^{2} < \frac{3}{4} a^{2} \Rightarrow H^{2} < 3 a^{2} \Rightarrow H < a \sqrt{3} = \sqrt{3} a

Therefore, the greatest height of the cone consistent with stable equilibrium of the body is

\sqrt{3} a

(b) वक्र

C

के चारों तरफ

\vec{F}

का परिसंचरण ज्ञात कीजिए, जहाँ

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

और

C

, बिंदु

(0, 0)

से बिंदु

(1, 1)

तक वक्र

y = x^{2}

के द्वारा तथा बिंदु

(1, 1)

से बिंदु

(0, 0)

तक वक्र

y^{2} = x

के द्वारा परिभाषित है।

Find the circulation of

\vec{F}

round the curve

C

, where

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

and

C

is the curve

y = x^{2}

from

(0, 0)

(1, 1)

and the curve

y^{2} = x

from

(1, 1)

(0, 0)

Answer:

Introduction:
We need to find the circulation of the vector field

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

around the closed curve

C

. This curve

C

consists of two parts: the arc

O A P

and the arc

P B O

. We’ll denote the first part as

C_{1}

and the second part as

C_{2}

C_{1}

is described by

y = x^{2}

, and

C_{2}

is described by

x = y^{2}

. We’ll calculate the circulation along both parts and then sum them up to get the circulation around the entire curve

C

Work/Calculations:

Along

C_{1}

, we have

y = x^{2}

, which implies

d y = 2 x d x

, and

x

varies from 0 to 1. Along

C_{2}

, we have

x = y^{2}

, which implies

d x = 2 y d y

, and

y

varies from 1 to 0.

Now, let’s calculate

F \cdot d r

\begin{aligned} F \cdot d r & = [(2 x + y^{2}) \hat{i} + (3 y - 4 x) \hat{j}] \cdot (\hat{i} d x + \hat{j} d y) \\ = (2 x + y^{2}) d x + (3 y - 4 x) d y . \end{aligned}

Now, we’ll calculate the circulation of

F

around

C

by summing up the circulations along

C_{1}

and

C_{2}

\begin{aligned} Circulation of F around C & = ∮_{C} F \cdot d r \\ = \int_{C_{1}} F \cdot d r + \int_{C_{2}} F \cdot d r . \end{aligned}

\begin{aligned} = ∮_{C} F \cdot d r = \int_{C_{1}} F \cdot d r + \int_{C_{2}} F \cdot d r \\ = \int_{C_{1}} [(2 x + y^{2}) d x + (3 y - 4 x) d y] + \int_{C_{2}} [(2 x + y^{2}) d x + (3 y - 4 x) d y] \\ = \int_{x = 0}^{1} [(2 x + x^{4}) d x + (3 x^{2} - 4 x) 2 x d x] \\ + \int_{y = 1}^{0} [(2 y^{2} + y^{2}) 2 y d y + (3 y - 4 y^{2}) d y] \\ = \int_{0}^{1} (2 x - 8 x^{2} + 6 x^{3} + x^{4}) d x + \int_{1}^{0} (3 y - 4 y^{2} + 6 y^{3}) d y \\ = {[x^{2} - \frac{8}{3} x^{3} + \frac{3}{2} x^{4} + \frac{1}{5} x^{5}]}_{0}^{1} + {[\frac{3}{2} y^{2} - \frac{4}{3} y^{3} + \frac{3}{2} y^{4}]}_{1}^{0} \\ = 1 - \frac{8}{3} + \frac{3}{2} + \frac{1}{5} - \frac{3}{2} + \frac{4}{3} - \frac{3}{2} \\ = 1 - \frac{4}{3} + \frac{1}{5} - \frac{3}{2} = \frac{30 - 40 + 6 - 45}{30} = - \frac{49}{30} . \end{aligned}

Conclusion:
The circulation of

\vec{F}

around the closed curve

C

- \frac{49}{30}

\frac{d^{2} y}{d x^{2}} + (3 \sin x - \cot x) \frac{d y}{d x} + 2 y \sin^{2} x = e^{- \cos x} \sin^{2} x

को हल कीजिए।

(ii)

t^{- 1 / 2}

तथा

t^{1 / 2}

का लाप्लास रूपांतर ज्ञात कीजिए। सिद्ध कीजिए कि

t^{n + \frac{1}{2}}

का लाप्लास रूपांतर

\frac{Γ (n + 1 + \frac{1}{2})}{s^{n + 1 + \frac{1}{2}}}

होता है, जहाँ

n \in N

(i) Solve the differential equation

\frac{d^{2} y}{d x^{2}} + (3 \sin x - \cot x) \frac{d y}{d x} + 2 y \sin^{2} x = e^{- \cos x} \sin^{2} x

Answer:

Step 1: Variable Transformation
We begin by choosing a new variable,

z

, such that

z = - \cos x

. This leads to:

\frac{d z}{d x} = \sin x

and

\frac{d^{2} z}{d x^{2}} = \cos x .

Step 2: Rewrite the Equation
Now, we rewrite the given differential equation using the new variable

z

\frac{d^{2} y}{d x^{2}} + (3 \sin x - \cot x) \frac{d y}{d x} + 2 y \sin^{2} x = e^{- \cos x} \sin^{2} x

becomes

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

where

P_{1} = \frac{d^{2} z}{d x^{2}} + P \frac{d z}{d x} = \frac{\cos x + (3 \sin x - \cos x) \sin x}{\sin^{2} x} = 3,

Q_{1} = \frac{Q}{{(\frac{d z}{d x})}^{2}} = \frac{2 \sin^{2} x}{\sin^{2} x} = 2,

and

R_{1} = \frac{R}{{(\frac{d z}{d x})}^{2}} = e^{- \cos x} = e^{z} .

Step 3: Homogeneous Differential Equation
Now, we have a linear homogeneous differential equation:

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

Step 4: Auxiliary Equation
To solve this, we find the auxiliary equation:

m^{2} + 3 m + 2 = 0 \Rightarrow (m + 1) (m + 2) = 0,

which gives us two roots:

m = - 1

and

m = - 2

Step 5: General Solution of Homogeneous Equation
The general solution of the homogeneous equation is:

y_{h} = c_{1} e^{- z} + c_{2} e^{- 2 z} .

Step 6: Particular Solution (P.I)

\begin{aligned} P . I = \frac{e^{z}}{(D + 1) (D + 2)} = \frac{e^{z}}{6} \\ y = c_{1} e^{- z} + c_{2} e^{- 2 z} + \frac{e^{z}}{6} \\ y = c_{1} e^{\cos x} + c_{2} e^{2 \cos x} + \frac{e^{- \cos x}}{6} \end{aligned}

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

y = c_{1} e^{\cos x} + c_{2} e^{2 \cos x} + \frac{e^{- \cos x}}{6} .

(ii) Find the Laplace transforms of

t^{- 1 / 2}

and

t^{1 / 2}

. Prove that the Laplace transform of

t^{n + \frac{1}{2}}

, where

n \in N

, is

\frac{Γ (n + 1 + \frac{1}{2})}{s^{n + 1 + \frac{1}{2}}}

Answer:

Laplace Transform of $t^{- 1 / 2}$ :
We start by finding the Laplace transform of

t^{- 1 / 2}

\begin{aligned} L {t^{- 1 / 2}} & = \int_{0}^{\infty} e^{- s t} t^{- 1 / 2} d t . \end{aligned}

To evaluate this integral, we use a substitution:

s t = u \Rightarrow d t = \frac{d u}{s}

and

t = \frac{u}{s}

. Substituting:

\begin{aligned} L {t^{- 1 / 2}} & = \int_{0}^{\infty} e^{- u} {(\frac{u}{s})}^{- 1 / 2} \frac{d u}{s} \\ = \frac{1}{s^{1 / 2}} \int_{0}^{\infty} e^{- u} u^{- 1 / 2} d u . \end{aligned}

Now, we recognize that the integral on the right-hand side is a gamma function:

Γ (\frac{1}{2}) = \sqrt{π} .

Therefore,

L {t^{- 1 / 2}} = \frac{\sqrt{π}}{s} .

Laplace Transform of $t^{1 / 2}$ :
Next, let’s find the Laplace transform of

t^{1 / 2}

\begin{aligned} L {t^{1 / 2}} & = \int_{0}^{\infty} e^{- s t} t^{1 / 2} d t . \end{aligned}

Using a similar substitution as before,

s t = u \Rightarrow d t = \frac{d u}{s}

and

t = \frac{u}{s}

, we get:

\begin{aligned} L {t^{1 / 2}} & = \frac{1}{s^{3 / 2}} \int_{0}^{\infty} e^{- u} u^{1 / 2} d u . \end{aligned}

Again, this integral is related to the gamma function:

Γ (\frac{3}{2}) = \frac{1}{2} \sqrt{π} .

Therefore,

L {t^{1 / 2}} = \frac{\sqrt{π}}{2 s^{3 / 2}} .

Laplace Transform of $t^{n + 1 / 2}$ for $n \in N$ :
Now, let’s prove that the Laplace transform of

t^{n + 1 / 2}

for

n \in N

is given by:

L {t^{n + \frac{1}{2}}} = \frac{Γ (n + 1 + \frac{1}{2})}{s^{n + 1 + \frac{1}{2}}} .

We start with:

\begin{aligned} L {t^{n + \frac{1}{2}}} & = \int_{0}^{\infty} e^{- s t} t^{n + \frac{1}{2}} d t . \end{aligned}

Using the substitution

s t = u \Rightarrow d t = \frac{d u}{s}

and

t = \frac{u}{s}

, we have:

\begin{aligned} L {t^{n + \frac{1}{2}}} & = \frac{1}{s^{n + \frac{1}{2} + 1}} \int_{0}^{\infty} e^{- u} u^{n + \frac{1}{2}} d u . \end{aligned}

This integral can be expressed in terms of the gamma function:

\begin{aligned} \int_{0}^{\infty} e^{- u} u^{n + \frac{1}{2}} d u & = Γ (n + \frac{1}{2} + 1) \\ = Γ (n + \frac{3}{2}) . \end{aligned}

Substituting this back into the Laplace transform expression:

\begin{aligned} L {t^{n + \frac{1}{2}}} & = \frac{1}{s^{n + \frac{1}{2} + 1}} Γ (n + \frac{3}{2}) . \end{aligned}

Now, simplify the exponent:

n + \frac{1}{2} + 1 = n + \frac{3}{2} .

So, the Laplace transform becomes:

\begin{aligned} L {t^{n + \frac{1}{2}}} & = \frac{Γ (n + \frac{3}{2})}{s^{n + \frac{3}{2}}} . \end{aligned}

Now, notice that

Γ (n + \frac{3}{2})

can be expressed as

Γ (n + 1 + \frac{1}{2})

, which completes the proof:

\begin{aligned} L {t^{n + \frac{1}{2}}} & = \frac{Γ (n + 1 + \frac{1}{2})}{s^{n + 1 + \frac{1}{2}}} . \end{aligned}

This proves the desired result.

(a) समीकरण $x^{2} y^{''} - 2 x y^{'} + 2 y = x^{3} \sin x$ के संगत समांगी अवकल समीकरण का रेखीय स्वतंत्र हल निकालिए और तब दिए गए समीकरण का प्राचल-विचरण विधि द्वारा सामान्य हल निकालिए।

Find the linearly independent solutions of the corresponding homogeneous differential equation of the equation

x^{2} y^{''} - 2 x y^{'} + 2 y = x^{3} \sin x

and then find the general solution of the given equation by the method of variation of parameters.

Answer:

Homogeneous Differential Equation:
We are given the differential equation:

x^{2} y^{''} - 2 x y^{'} + 2 y = x^{3} \sin x

Step 1: Rewrite the Equation:
Rewrite the given differential equation as:

x^{2} \frac{d^{2} y}{d x^{2}} - 2 x \frac{d y}{d x} + 2 y = x^{3} \sin x

Step 2: Change of Variable:
Let’s make a change of variable by putting

x = e^{z}

, which implies

\frac{d}{d z} \equiv D

. Then, our equation becomes:

\begin{aligned} [D (D - 1) - 2 D + 2] y = e^{3 z} \sin (e^{z}) \\ [D^{2} - D - 2 D + 2] y = e^{3 z} \sin (e^{z}) \\ [D^{2} - 3 D + 2] y = e^{3 z} \sin (e^{z}) \end{aligned}

Step 3: Find the Auxiliary Equation:
The auxiliary equation for the homogeneous part is:

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

(m - 1) (m - 2) = 0

This gives us two solutions:

m = 1

and

m = 2

Step 4: Complementary Function (C.F.):
The complementary function is given by:

C F = c_{1} e^{z} + c_{2} e^{2 z}

Step 5: Calculate the Wronskian Determinant:
We calculate the Wronskian determinant as follows:

w = | \begin{array}{ll} e^{z} & e^{2 z} \\ e^{z} & 2 e^{2 z} \end{array} | = 2 e^{3 z} - e^{3 z} = e^{3 z} \neq 0

Step 6: Calculate $p_{1}$ and $p_{2}$ :

p_{1} = - \int \frac{e^{2 z} \cdot e^{3 z} \sin e^{z}}{e^{3 z}} d z = - \int e^{2 z} \cdot \sin e^{z} d z = e^{z} \cos (e^{z}) - \sin e^{z}

p_{2} = \int \frac{e^{z} \cdot e^{3 z} \sin e^{z}}{e^{3 z}} d z = \int e^{z} \cdot \sin e^{z} d z = - \cos e^{z}

Step 7: Particular Integral (P.I.):
The particular integral is given by:

\begin{aligned} P.I. & = p_{1} u_{1} + p_{2} u_{2} \\ = [e^{z} \cos (e^{z}) - \sin e^{z}] e^{z} - \cos (e^{z}) \cdot e^{2 z} \\ = - e^{z} \sin e^{z} \end{aligned}

Step 8: General Solution:
The general solution is the sum of the complementary function and the particular integral:

y = c_{1} e^{z} + c_{2} e^{2 z} - e^{z} \sin e^{z}

Step 9: Back to the Original Variable:
Converting back to the original variable

x

, the general solution is:

y = c_{1} x + c_{2} x^{2} - x \sin x

Conclusion:
The linearly independent solutions of the corresponding homogeneous differential equation are

y_{1} = x

and

y_{2} = x^{2}

, and the general solution of the given equation, using the method of variation of parameters, is:

y = c_{1} x + c_{2} x^{2} - x \sin x

(b) कुंडलिनी

x = a \cos u, y = a \sin u, z = a u \tan α

के लिए वक्रता की त्रिज्या तथा विमोटन की त्रिज्या ज्ञात कीजिए।

Find the radius of curvature and radius of torsion of the helix

x = a \cos u

y = a \sin u, z = a u \tan α

Answer:

Given Helix Equations:
The parametric equations for the helix are:

x = a \cos u

y = a \sin u

z = a u \tan α

Step 1: Find the Derivatives of $r$ :
We first find the derivatives of the position vector

r = (x, y, z)

with respect to the parameter

u

\begin{aligned} \dot{r} & = \frac{d r}{d u} = (\dot{x}, \dot{y}, \dot{z}) \\ = a (- \sin u, \cos u, \tan α) \end{aligned}

\begin{aligned} \ddot{r} & = \frac{d \dot{r}}{d u} = (\ddot{x}, \ddot{y}, \ddot{z}) \\ = a (- \cos u, - \sin u, 0) \end{aligned}

\begin{aligned} \overset{⃛}{r} & = \frac{d \ddot{r}}{d u} = (\overset{⃛}{x}, \overset{⃛}{y}, \overset{⃛}{z}) \\ = a (\sin u, - \cos u, 0) \end{aligned}

Step 2: Find the Cross Product $\dot{r} \times \ddot{r}$ :
Next, we calculate the cross product

\dot{r} \times \ddot{r}

| \begin{matrix} i & j & k \\ - a \sin u & a \cos u & a \tan α \\ - a \cos u & - a \sin u & 0 \end{matrix} |

Using the determinant:

\begin{aligned} \dot{r} \times \ddot{r} & = (a^{2} \sin u \tan α) i - (a^{2} \cos u \tan α) j + (a^{2}) k \\ = a^{2} \sin u \tan α i - a^{2} \cos u \tan α j + a^{2} k \end{aligned}

Step 3: Find the Magnitude of $\dot{r} \times \ddot{r}$ :
Now, let’s find the magnitude of

\dot{r} \times \ddot{r}

\begin{aligned} | \dot{r} \times \ddot{r} | & = \sqrt{(a^{2} \sin u \tan α)^{2} + (- a^{2} \cos u \tan α)^{2} + (a^{2})^{2}} \\ = \sqrt{a^{4} \sin^{2} u \tan^{2} α + a^{4} \cos^{2} u \tan^{2} α + a^{4}} \\ = \sqrt{a^{4} (\sin^{2} u \tan^{2} α + \cos^{2} u \tan^{2} α + 1)} \\ = \sqrt{a^{4} (\tan^{2} α + 1)} \\ = a^{2} \sqrt{\tan^{2} α + 1} \\ = a^{2} \sec α \end{aligned}

Step 4: Find $| \dot{r} |$ :
The magnitude of the velocity vector

| \dot{r} |

is:

| \dot{r} | = \sqrt{(- a \sin u)^{2} + (a \cos u)^{2} + (a \tan α)^{2}}

= \sqrt{a^{2} (\sin^{2} u + \cos^{2} u + \tan^{2} α)}

= \sqrt{a^{2} (1 + \tan^{2} α)}

= a \sqrt{1 + \tan^{2} α}

= a \sec α

Step 5: Find the Curvature $k$ :
The curvature

k

is given by:

k = \frac{| \dot{r} \times \ddot{r} |}{| \dot{r} |^{3}}

Substituting the values we’ve found:

k = \frac{a^{2} \sec α}{(a \sec α)^{3}} = \frac{1}{a \sec^{2} α}

Step 6: Find the Radius of Curvature $ρ$ :
The radius of curvature

ρ

is the reciprocal of the curvature:

ρ = \frac{1}{k} = a \sec^{2} α

Conclusion:
The radius of curvature of the given helix is

ρ = a \sec^{2} α

(c)

y

-अक्ष की दिशा में गतिमान एक कण का मूलबिंदु की ओर त्वरण

F y

है, जहाँ

F, y

का एक धनात्मक एवं सम फलन है। जब कण

y = - a

तथा

y = a

के बीच में कंपन करता है, तब उसका आवर्तकाल

T

है। दर्शाइए कि

\frac{2 π}{\sqrt{F_{1}}} < T < \frac{2 π}{\sqrt{F_{2}}}

जहाँ

F_{1}

एवं

F_{2}

परास

[- a, a]

में

F

के अधिकतम एवं न्यूनतम मान हैं। आगे दर्शाइए कि जब लंबाई

l

का एक सरल लोलक ऊर्ध्वाधर रेखा के किसी भी ओर

30^{\circ}

तक दोलन करता है, तब

T, 2 π \sqrt{l / g}

तथा

2 π \sqrt{l / g} \sqrt{π / 3}

के बीच में रहता है।

A particle moving along the

y

-axis has an acceleration

F y

towards the origin, where

F

is a positive and even function of

y

. The periodic time, when the particle vibrates between

y = - a

and

y = a

, is

T

. Show that

\frac{2 π}{\sqrt{F_{1}}} < T < \frac{2 π}{\sqrt{F_{2}}}

where

F_{1}

and

F_{2}

are the greatest and the least values of

F

within the range

[- a, a]

. Further, show that when a simple pendulum of length

l

oscillates through

30^{\circ}

on either side of the vertical line,

T

lies between

2 π \sqrt{l / g}

and

2 π \sqrt{l / g} \sqrt{π / 3}

Answer:

Given Equation of Motion:
The particle’s motion along the

y

-axis is described by the equation of motion:

\ddot{y} = - F (y) y

Conservation of Mechanical Energy:
The first integral of this equation of motion yields the conservation of mechanical energy:

\frac{1}{2} {\dot{y}}^{2} + U (y) = E = U (a)

where

U (y) = \int_{0}^{y} F (x) x d x

is the potential energy associated with the force

- F (y) y

Expression for the Period $T$ :
The time period

T

for the particle to vibrate between

y = - a

and

y = a

is given by:

T = 2 \int_{- a}^{a} \frac{d y}{\dot{y}} = 2 \int_{- a}^{a} \frac{d y}{\sqrt{2 (U (a) - U (y))}} = 4 \int_{0}^{a} \frac{d y}{\sqrt{2 (U (a) - U (y))}},

Bounds on $T$ Using Inequalities:
Now, we want to find bounds on

T

using inequalities for

F (y)

. We have

F_{2} \leq F (y) \leq F_{1}

for

y

in the interval

[0, a]

. Using this inequality, we derive lower and upper bounds for

T

\begin{aligned} F_{2} \leq F \leq F_{1} & \Rightarrow \int_{y}^{a} F_{2} x d x \leq \int_{y}^{a} F (x) x d x \leq \int_{y}^{a} F_{1} x d x \\ \Rightarrow \frac{1}{2} F_{2} (a^{2} - y^{2}) \leq U (a) - U (y) \leq \frac{1}{2} F_{1} (a^{2} - y^{2}) \end{aligned}

This leads to:

\begin{aligned} \Rightarrow \int_{0}^{a} \frac{d y}{\sqrt{F_{2} (a^{2} - y^{2})}} \geq \int_{0}^{a} \frac{d y}{\sqrt{2 (U (a) - U (y))}} \geq \int_{0}^{a} \frac{d y}{\sqrt{F_{1} (a^{2} - y^{2})}} \\ \Rightarrow \frac{π}{2 \sqrt{F_{2}}} \geq \frac{T}{4} \geq \frac{π}{2 \sqrt{F_{1}}} \\ \Rightarrow \frac{2 π}{\sqrt{F_{1}}} \leq T \leq \frac{2 π}{\sqrt{F_{2}}} \end{aligned}

This inequality establishes that:

\frac{2 π}{\sqrt{F_{1}}} < T < \frac{2 π}{\sqrt{F_{2}}}

Comparison with Simple Pendulum:
For a simple pendulum of length

l

oscillating through

30^{\circ}

on either side of the vertical line, the equation of motion is given by:

\ddot{θ} = - \frac{g}{l} \sin θ = - F (θ) θ

where

F (θ) = \frac{g}{l} \frac{\sin θ}{θ}

The maximum and minimum values of

F (θ)

in the interval

[- 30^{\circ}, 30^{\circ}]

(or

[- \frac{π}{6}, \frac{π}{6}]

in radians) are

F_{1} = \frac{g}{l}

and

F_{2} = \frac{3 g}{π l}

, respectively.

Inserting these values into the previously established inequality (5), we find that the period

T

of the simple pendulum when it oscillates through

30^{\circ}

lies between

2 π \sqrt{\frac{l}{g}}

and

2 π \sqrt{\frac{π}{3} \frac{l}{g}}

2 π \sqrt{\frac{l}{g}} < T < 2 π \sqrt{\frac{π}{3} \frac{l}{g}}

This completes the solution.

(a) अवकल समीकरण

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

का विचित्र हल प्राप्त कीजिए। दिए हुए अवकल समीकरण का पूर्ण पूर्वग भी ज्ञात कीजिए। पूर्ण पूर्वग तथा विचित्र हल की ज्यामितीय व्याख्या कीजिए।

Obtain the singular solution of the differential equation

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

Also find the complete primitive of the given differential equation. Give the geometrical interpretations of the complete primitive and singular solution.

Answer:

Given equation

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

Multiply both sides by

x^{2} \tan^{2} α

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

Its a quadratic equation for

p y

P y = 2 x \tan^{2} α \pm \sqrt{4 x^{2} \tan^{4} α - 4 (y^{2} \sec^{2} α - x^{2} \tan^{2} α)}

Simplify further:

P y = x \tan^{2} α \pm \sqrt{x^{2} \tan^{2} α (\tan^{2} α + 1) - y^{2} \sec^{2} α}

Rearrange the equation:

P y = x \tan^{2} α \pm \sec α \sqrt{x^{2} \tan^{2} α - y^{2}}

Rewrite the equation with differentials:

\frac{y d y - x \tan^{2} α d α}{\sqrt{x^{2} \tan^{2} α - y^{2}}} = \pm \sec α d α

Let

x^{2} \tan^{2} α - y^{2} = t^{2}

2 x \tan^{2} α d α - 2 y d y = 2 t d t

Integrate both sides:

\pm \int \frac{t d t}{t} = - \sec α \int d x

Solve for

t

t = c - \sec α x

Substitute back

t = \sqrt{(x^{2} \tan^{2} α - y^{2})}

x^{2} \tan^{2} α - y^{2} = (c - \sec α x)^{2}

\begin{aligned} 0 = x^{2} (\sec^{2} α - \tan^{2} α) - 2 c x \sec α + y^{2} \\ x^{2} - 26 x \sec α + y^{2} + c^{2} = 0 \\ c^{2} - 2 c x \sec α + x^{2} + y^{2} = 0 \end{aligned}

Rewrite:

c^{2} - 2 c x \sec α + x^{2} + y^{2} = 0

Notice that this is the equation of a circle with center

(c \sec α, 0)

and radius

r = | c |

The geometrical interpretation of the complete primitive is a family of circles with centers on the line

x = c \sec α

in the xy-plane.

Singular Solution:

\begin{aligned} c^{2} - 2 c x \sec α + x^{2} + y^{2} = 0 \\ 4 x^{2} \sec^{2} α - 4 (x^{2} + y^{2}) = 0 \\ x^{2} \sec^{2} α - 4 (x^{2} + y^{2}) = 0 \\ x^{2} (\sec^{2} α - 1) - y^{2} = 0 \end{aligned}

Now, let’s find the singular solution by considering the discriminant

c

for

x^{2} \tan^{2} α - y^{2} = 0

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

Discriminant:

4 y^{2} (x^{2} \sin^{2} α - y^{2} \cos^{2} α) = 0

Further simplify:

4 y^{2} (x^{2} \tan^{2} α - y^{2}) = 0

Two cases:

y = 0

(the locus is on the x-axis).

c : y = x \tan α

, which represents a line with slope

\tan α

and passes through the origin.

The geometrical interpretation of the singular solution is that it consists of two parts: a point on the x-axis and a line with slope

\tan α

passing through the origin.

This concludes the solution and interpretations of the given differential equation and its primitive.

(b) एक गतिमान ग्रह का त्वरण

\frac{μ}{{(दूरी}^{2}}

के बराबर है और त्वरण की दिशा हमेशा एक स्थिर बिंदु (तारा) की ओर है। सिद्ध कीजिए कि उस ग्रह का पथ एक शंकु-परिच्छेद है। वे प्रतिबंध ज्ञात कीजिए, जिनके अन्तर्गत पथ (i) दीर्घवृत्त, (ii) परवलय और (iii) अतिपरवलय बन जाता है।

Prove that the path of a planet, which is moving so that its acceleration is always directed to a fixed point (star) and is equal to

\frac{μ}{{(distance)}^{2}}

, is a conic section. Find the conditions under which the path becomes (i) ellipse, (ii) parabola and (iii) hyperbola.

Answer:

Given that the force is always directed to a fixed point (star), this represents a case of a central orbit.

Also, it is given that the central acceleration is

ρ = \frac{μ}{r^{2}}

, where

r

is the distance.

The differential equation of the path (in pedal form) is:

\frac{h^{2}}{p^{3}} \frac{d p}{d r} = p = \frac{μ}{r^{2}}

Multiplying both sides by -2, we get:

- \frac{2 h^{2}}{p^{3}} d p = \frac{- 2 μ}{r^{2}} d r

Integrating, we have:

v^{2} = \frac{h^{2}}{p^{2}} = \frac{2 μ}{r} + B \to (1)

where

B

is a constant. [Note that in a central orbit

v = \frac{h}{p}

]

The pedal equation of an ellipse, parabola, and hyperbola (that branch which is nearer to the focus taken as the pole) are given by:

\frac{b^{2}}{p^{2}} = \frac{2 a}{r} - 1, p^{2} = a r, and \frac{b^{2}}{p^{2}} = \frac{2 a}{λ} + 1

respectively, where in the case of an ellipse,

2 a

and

2 b

are the lengths of the major and minor axes; in the case of a parabola,

4 a

is the length of the latus rectum; and in the case of a hyperbola,

2 a

and

2 b

are the lengths of the transverse and conjugate axes.

Since equation (1) can be any of the above 3 forms, three cases arise here:

Case (i) – Elliptic Path:

Comparing (1) with

\frac{b^{2}}{p^{2}} = \frac{2 a}{r} - 1

, the pedal equation of the ellipse, we have:

\begin{aligned} \frac{h^{2}}{b^{2}} & = \frac{μ}{a} = \frac{B}{- 1} \\ h^{2} & = \frac{μ b^{2}}{a} and B = - \frac{μ}{a} \end{aligned}

Substituting in (1), for an elliptical path, we have:

[
v^2 = \frac{2 \mu}{r} – \frac{\mu}{a} = \mu \left(\frac{2}{r} – \frac{1}{a}\right) \quad (Obviously here, \quad v^2 < \frac{2 \mu}{r})

Case (ii) – Parabolic Path:

Comparing (1) with

p^{2} = a r

, the pedal equation of a parabola, we have:

\frac{h^{2}}{1} = \frac{2 μ}{\frac{1}{a}} = \frac{B}{0}

Therefore,

h^{2} = 2 μ a

and

B = 0

Substituting in (1), for a parabolic path, we have

v^{2} = \frac{2 μ}{r}

Case (iii) – Hyperbolic Path:

Comparing (1) with

\frac{b^{2}}{p^{2}} = \frac{2 a}{r} + 1

, the pedal equation of a hyperbola, we have:

\begin{aligned} \frac{h^{2}}{b^{2}} & = \frac{μ}{a} = \frac{B}{1} \\ h^{2} & = \frac{μ b^{2}}{a} and B = \frac{μ}{a} \end{aligned}

Substituting in (1), for a hyperbolic path, we have

v^{2} = μ (\frac{2}{r} + \frac{1}{a})

(Obviously here,

v^{2} > \frac{2 μ}{r}

Summary:

If $v^{2} = μ (\frac{2}{r} - \frac{1}{a})$ or $v^{2} < \frac{2 μ}{r}$ , then the path is an ellipse.
If $v^{2} = \frac{2 μ}{r}$ , then the path is a parabola.
If $v^{2} > \frac{2 μ}{r}$ or $v^{2} = \frac{2 μ}{r} + \frac{r}{a}$ , then the path is a hyperbola.

Additionally, the magnitude of the velocity at any point is independent of the direction of the velocity at that time.

We have also found that

h^{2} = μ l

in all three cases, where

l

is the length of the semi-latus rectum.

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

के लिए

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

और

z = 3

से घिरे हुए क्षेत्र में सत्यापित कीजिए।

(ii) स्टोक्स प्रमेय के द्वारा

∮_{C} e^{x} d x + 2 y d y - d z

का मान ज्ञात कीजिए, जहाँ

C

, वक्र

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

z = 2

है।

(i) State Gauss divergence theorem. Verify this theorem for

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

, taken over the region bounded by

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

and

z = 3

Answer:

Gauss divergence theorem

\vec{F}

is a vector point function having continuous first order partial derivatives in the region V bounded by a closed surface

S

, then

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

where

\hat{n}

is the outward drawn unit normal vector to the surface

S

Since div

\vec{F} = \frac{\partial}{\partial x} (4 x) + \frac{\partial}{\partial y} (- 2 y^{2}) + \frac{\partial}{\partial z} (z^{2}) = 4 - 4 y + 2 z

∴ ∭_{V} div \vec{F} d V = ∭_{V} (4 - 4 y + 2 z) d x d y d z = \int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} \int_{0}^{3} (4 - 4 y + 2 z) d z d y d x

\begin{aligned} = \int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} {[4 z - 4 y z + z^{2}]}_{0}^{3} d y d x \\ = \int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} (12 - 12 y + 9) d y d x \\ = \int_{- 2}^{2} {[21 y - 6 y^{2}]}_{- \sqrt{4 - x^{7}}}^{\sqrt{4 - x^{7}}} d x \\ = \int_{- 2}^{2} 42 \sqrt{4 - x^{2}} d x = 84 \int_{0}^{2} \sqrt{4 - x^{2}} d x \\ = 84 {[\frac{x \sqrt{4 - x^{2}}}{2} + \frac{4}{2} \sin^{- 1} \frac{x}{2}]}_{0}^{2} \\ = 84 [2 \sin^{- 1} 1] = 84 [2 \times \frac{π}{2}] = 84 π \end{aligned}

To evaluate the surface integral, divide the closed surface

S

of the cylinder into 3 parts.

S_{1}

: the circular base in the plane

z = 0

S_{2}

: the circular top in the plane

z = 3

S_{3}

: the curved surface of the cylinder, given by the equation

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

Also

\iint_{S} \vec{F} \cdot \hat{n} d S = \iint_{S_{1}} \vec{F} \cdot \hat{n} d S + \iint_{S_{2}} \vec{F} \cdot \hat{n} d S + \iint_{S_{a}} \vec{F} \cdot \hat{n} d S

S_{1} (z = 0)

, we have

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

so that

∴ \iint_{S_{1}} \vec{F} \cdot \hat{n} d S = 0

S_{2} (z = 3)

, we have

\hat{n} = \hat{k}, \vec{F} = 4 x \hat{i} - 2 y^{2} \hat{j} + 9 \hat{k}

so that

\begin{aligned} \vec{F} \cdot \hat{n} & = (4 x \hat{i} - 2 y^{2} \hat{j} + 9 \hat{k}) \cdot \hat{k} = 9 \\ ∴ \iint_{S_{2}} \vec{F} \cdot \hat{n} d S & = \iint_{S_{2}} 9 d x d y = 9 \iint_{S} d x d y \\ = 9 \times area of surface S_{2} = 9 (π \cdot 2^{2}) = 36 π \end{aligned}

S_{3}, x^{2} + y^{2} = 4

∴ \hat{n} =

a unit vector normal to surface

S_{3} = \frac{2 x \hat{i} + 2 y \hat{j}}{\sqrt{4 x^{2} + 4 y^{2}}} = \frac{x \hat{i} + y \hat{j}}{2}

\begin{matrix} \vec{F} \cdot \hat{n} = (4 x \hat{i} - 2 y^{2} \hat{j} + z^{2} \hat{k}) \cdot (\frac{x \hat{i} + y \hat{j}}{2}) = 2 x^{2} - y^{3} \\ x^{2} + y^{2} = 4, x = 2 \cos θ, y = 2 \sin θ and d S = 2 d θ d z . \end{matrix}

Also, on

S_{3}

, i.e.

x^{2} + y^{2} = 4, x = 2 \cos θ, y = 2 \sin θ

and

d S = 2 d θ d z

To cover the whole surface

S_{3}, z

varies from 0 to 3 and

θ

varies from 0 to

2 π

\begin{aligned} ∴ \iint_{S_{a}} \vec{F} \cdot \hat{n} d S & = \int_{0}^{2 π} \int_{0}^{3} [2 (2 \cos θ)^{2} - (2 \sin θ)^{3}] 2 d z d θ \\ = \int_{0}^{2 π} 16 (\cos^{2} θ - \sin^{3} θ) \times 3 d θ = 48 \int_{0}^{2 π} (\cos^{2} θ - \sin^{3} θ) d θ = 48 π \\ ∴ \iint_{S} \vec{F} \cdot \hat{n} d S & = 0 + 36 π + 48 π = 84 π \end{aligned}

The equality of (1) and (2) verifies divergence theorem.

(ii) Evaluate by Stokes’ theorem

∮_{C} e^{x} d x + 2 y d y - d z

, where

C

is the curve

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

Answer:

Introduction

We are given a vector field

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

and a curve

C

defined by

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

and

z = 2

. We want to evaluate

∮_{C} \vec{F} \cdot d \vec{r}

using Stokes’ theorem.

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

Stokes’ theorem states that

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

, where

S

is the surface bounded by

C

The curl of

\vec{F}

is given by:

\nabla \times \vec{F} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^{x} & 2 y & - 1 \end{array} |

After calculating, we get:

\begin{matrix} \nabla \times \vec{F} = (0 - 0) \hat{i} - (0 - 0) \hat{j} + (0 - 0) \hat{k} \\ \nabla \times \vec{F} = \vec{0} \end{matrix}

Since the curl of

\vec{F}

is zero, the line integral around any closed path will also be zero, according to Stokes’ theorem.

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Introduction

Work/Calculations

Step 1: Define the Function and the Limit

Step 2: Simplify the Function

Step 3: Apply L’Hôpital’s Rule

Conclusion

Introduction

Work/Calculations

Step 1: Definition of Continuity

Step 2: Prove Continuity for f ( x , b ) f ( x , b ) f(x,b)f(x, b)f(x,b) at x = a x = a x=ax = ax=a

Step 3: Prove Continuity for f ( a , y ) f ( a , y ) f(a,y)f(a, y)f(a,y) at y = b y = b y=by = by=b

Conclusion

Conclusion

Introduction

Work/Calculations

Step 1: Calculate A B A B ABABAB

Step 2: Solve the System of Equations

Introduction

Work/Calculations

Step 1: Breaking It Down

Step 2: Checking the Derivative

Conclusion

Introduction

Work/Calculations

Step 1: Find the Derivative of f ( x ) f ( x ) f(x)f(x)f(x)

Step 2: Find the Critical Points

Step 3: Evaluate f ( x ) f ( x ) f(x)f(x)f(x) at Critical Points and Endpoints

Conclusion

Use the General Form A 2 n = n A 2 − ( n − 1 ) I A 2 n = n A 2 − ( n − 1 ) I A^(2n)=nA^(2)-(n-1)IA^{2n} = n A^2 – (n – 1) IA2n=nA2−(n−1)I

Substitute the Values

Calculate A 100 A 100 A^(100)A^{100}A100

Conclusion

Work/Calculations

Step 1: Integrate f ′ ( x ) f ′ ( x ) f^(‘)(x)f'(x)f′(x) to find f ( x ) f ( x ) f(x)f(x)f(x)

Step 2: Define α α alpha\alphaα and β β beta\betaβ

Step 3: Calculate the Partial Derivatives of α α alpha\alphaα and β β beta\betaβ

Step 4: Check if the Partial Derivatives are Equal

Conclusion with Jacobian Method

Auxiliary Equation:

Particular Integral (P.I.):

Conclusion:

Gauss divergence theorem

Introduction

Step 1: Find the Curl of F → F → vec(F)\vec{F}F→

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Step 2: Prove Continuity for $f (x, b)$ at $x = a$

Step 3: Prove Continuity for $f (a, y)$ at $y = b$

Step 1: Calculate $A B$

Step 1: Find the Derivative of $f (x)$

Step 3: Evaluate $f (x)$ at Critical Points and Endpoints

Use the General Form $A^{2 n} = n A^{2} - (n - 1) I$

Calculate $A^{100}$

Step 1: Integrate $f^{'} (x)$ to find $f (x)$

Step 2: Define $α$ and $β$

Step 3: Calculate the Partial Derivatives of $α$ and $β$

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