Engineering Mathematics

This subject page now includes a complete Module 1 study chapter for matrices and linear algebra in a textbook-style format aligned with common engineering mathematics syllabi.

Module 1: Matrices and Linear Algebra

Chapter Objective

Master matrix operations, solve linear systems, find eigenvalues and eigenvectors, apply the Cayley-Hamilton theorem, diagonalize matrices, and analyze quadratic forms.

• Exam weightage: 15 to 20 percent
• Target: 4 to 5 long questions and 3 to 4 short questions

1.1 Matrix Basics and Properties

Matrices are rectangular arrays of numbers arranged in rows and columns. They are used in solving linear equations, modeling transformations, circuit analysis, optimization, and numerical methods.

Important properties

• Two matrices are equal only if they have the same order and the same corresponding entries.
• Matrix addition is defined only for matrices of the same order.
• Matrix multiplication is defined when columns of the first matrix equal rows of the second matrix.
• Matrix multiplication is not commutative in general.
• (AB)^T = B^T A^T
• (AB)^-1 = B^-1 A^-1 whenever A and B are invertible.

Exam note

Always state the condition |A| != 0 before using the inverse formula.

1.2 Rank of a Matrix

The rank of a matrix is the maximum number of linearly independent rows or columns. It is denoted by rho(A) or r(A).

Methods to find rank

1. Minor method: largest non-zero minor order gives the rank. 2. Echelon form method: reduce to row echelon form and count non-zero rows. 3. Normal form method: reduce to PAQ = [I_r 0; 0 0], then rank is r.

Solved Example 1

Q: Find the rank of

A = [1 2 3
     2 4 6
     3 6 9]

Solution

• R2 -> R2 - 2R1 gives [0 0 0]
• R3 -> R3 - 3R1 gives [0 0 0]

Echelon form:

[1 2 3
 0 0 0
 0 0 0]

Number of non-zero rows = 1, therefore

rank(A) = 1

Quick reminders

• If determinant of a 3 x 3 matrix is non-zero, rank is 3.
• Elementary row operations do not change rank.
• In exams, echelon form is usually the fastest reliable method.

1.3 Consistency of Linear Systems AX = B

Let A be the coefficient matrix and [A | B] the augmented matrix.

Rouche-Capelli Theorem

Solved Example 2

Q: Solve

x + y + z = 6
2x + 3y + z = 11
3x + y + 2z = 10

Solution

Augmented matrix:

[1 1 1 |  6]
[2 3 1 | 11]
[3 1 2 | 10]

Apply:

• R2 -> R2 - 2R1
• R3 -> R3 - 3R1

[1  1  1 |  6]
[0  1 -1 | -1]
[0 -2 -1 | -8]

Now apply

• R3 -> R3 + 2R2

[1 1  1 |  6]
[0 1 -1 | -1]
[0 0 -3 | -10]

Hence,

rho(A) = 3, rho([A | B]) = 3, n = 3

So the system has a unique solution.

Back substitution:

z = 10/3
y - z = -1
y = 7/3
x + y + z = 6
x = 1/3

Hence,

x = 1/3, y = 7/3, z = 10/3

Since rho(A) = rho([A | B]) = 3, the system is consistent and has a unique solution.

1.4 Eigenvalues and Eigenvectors

If there exists a non-zero vector X such that

AX = lambda X

then lambda is an eigenvalue and X is the corresponding eigenvector.

Characteristic equation

|A - lambda I| = 0

Key properties

• Sum of eigenvalues = Tr(A)
• Product of eigenvalues = |A|
• For symmetric matrices, all eigenvalues are real.
• Eigenvectors corresponding to distinct eigenvalues are linearly independent.
• For real symmetric matrices, eigenvectors corresponding to distinct eigenvalues are orthogonal.

Solved Example 3

Q: Find eigenvalues and eigenvectors of

A = [ 2 -1  2
     -1  2 -1
      2 -1  2]

Solution

Characteristic equation:

|A - lambda I| =
|2-lambda  -1        2     |
|-1        2-lambda  -1    | = 0
|2         -1        2-lambda|

On expansion,

-(lambda - 1)^2 (lambda - 4) = 0

So eigenvalues are

lambda = 1, 1, 4

For lambda = 1:

(A - I)X = 0

This gives the relation

x1 - x2 + x3 = 0

Two independent eigenvectors may be taken as

[1 1 0]^T  and  [0 1 1]^T

For lambda = 4, a corresponding eigenvector is

[1 -1 1]^T

Exam tip

Always verify one eigenvector by substitution. Many errors in exams come from copying the wrong sign while solving (A - lambda I)X = 0.

1.5 Cayley-Hamilton Theorem

Statement

Every square matrix satisfies its own characteristic equation.

If

p(lambda) = 0

is the characteristic equation of A, then

p(A) = 0

Applications

• Verify the theorem
• Find A^-1
• Find higher powers like A^2, A^3, A^4
• Solve matrix equations

Solved Example 4

For

A = [1 2
     3 4]

verify Cayley-Hamilton theorem and find A^-1.

Solution

Characteristic equation:

|A - lambda I|
= (1 - lambda)(4 - lambda) - 6
= lambda^2 - 5lambda - 2 = 0

Replace lambda with A:

A^2 - 5A - 2I = 0

Now

A^2 = [1 2   [1 2   = [7 10
      3 4] *  3 4]     15 22]

Then

A^2 - 5A - 2I
= [7 10     [5 10     [2 0
   15 22] -   15 20] -   0 2]
= [0 0
   0 0]

So the theorem is verified.

From

A^2 - 5A = 2I
A(A - 5I) = 2I

we obtain

A^-1 = (A - 5I) / 2

Hence,

A^-1 = [-2    1
        3/2 -1/2]

1.6 Diagonalization and Quadratic Forms

Diagonalization

If A has n linearly independent eigenvectors, then there exists an invertible matrix P such that

P^-1 A P = D

where D is a diagonal matrix of eigenvalues.

Quadratic form

Q(x, y, z) = X^T A X

where A is symmetric.

Nature test

Solved Example 5

Reduce

Q = 3x^2 + 3y^2 + 3z^2 - 2xy - 2yz - 2zx

to canonical form.

Solution

The associated matrix is

A = [ 3 -1 -1
     -1  3 -1
     -1 -1  3]

Its characteristic equation is

-(lambda - 2)^2 (lambda - 5) = 0

So eigenvalues are

2, 2, 5

Hence the canonical form is

2y1^2 + 2y2^2 + 5y3^2

All eigenvalues are positive, so the quadratic form is positive definite.

Exercise Sets

Basic

1. Find rank of [[1,2,3],[2,3,4],[3,4,5]]. 2. Solve 2x + y - z = 3, x + 2y + z = 4, x + y + z = 3. 3. Find eigenvalues of [[4,1],[2,3]]. 4. Verify A * adj(A) = |A| I for a 2 x 2 matrix.

Intermediate

1. Use Cayley-Hamilton theorem to find A^-1 for [[2,1,1],[1,2,1],[1,1,2]]. 2. Diagonalize [[6,-2,2],[-2,3,-1],[2,-1,3]]. 3. Reduce x^2 + y^2 + z^2 - 2xy - 2yz + 2zx to canonical form and state nature. 4. Check consistency of x + 2y + 3z = 1, 2x + 3y + 4z = 2, 3x + 4y + 5z = 3.

Advanced

1. If A^3 - 6A^2 + 11A - 6I = 0, find A^-1 and A^4. 2. For a symmetric matrix with eigenvalues 2, -1, 0, find index and signature. 3. Prove eigenvectors of distinct eigenvalues are linearly independent. 4. Solve AX = B where A is singular using a generalized inverse approach.

Quick Reference Sheet

• rho(A) = number of non-zero rows in echelon form
• AX = B is consistent if and only if rho(A) = rho([A | B])
• sum of eigenvalues = Tr(A)
• product of eigenvalues = |A|
• Cayley-Hamilton: P(A) = 0, useful for A^-1 and higher powers
• A matrix is diagonalizable if it has n independent eigenvectors
• Q(X) = X^T A X, and its nature is decided by the signs of eigenvalues
• A^-1 exists if and only if |A| != 0 and rho(A) = n

Exam Focus and PYQ Pattern

Pro tip

Many university papers repeat standard matrix patterns. Practice 3 x 3 symmetric matrices and inverse by Cayley-Hamilton thoroughly.

Study Strategy and Common Mistakes

Revision Order

1. Learn determinant, adjoint, inverse, and rank. 2. Practice consistency questions using augmented matrices. 3. Solve at least five eigenvalue problems, especially symmetric matrices. 4. Practice one Cayley-Hamilton inverse problem daily. 5. Finish with diagonalization and quadratic form nature tests.

Module Outcome

After finishing this module, you should be able to:

• test whether a matrix is invertible,
• compute rank quickly using echelon form,
• decide whether a linear system is consistent,
• find eigenvalues and eigenvectors with confidence,
• use Cayley-Hamilton to simplify matrix powers,
• diagonalize standard matrices, and
• classify quadratic forms correctly.

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Module 2: Differential Calculus (Single Variable)

Chapter Objective

Master higher-order differentiation, mean value theorems, series expansions, curvature concepts, and curve tracing techniques.

• Exam weightage: 10 to 15 percent
• Target: 3 to 4 long questions and 2 to 3 short questions

2.1 Successive Differentiation and Leibniz Theorem

Successive differentiation means finding second, third, and higher-order derivatives of a function. It is especially useful in series expansion, differential equations, and repeated derivative questions.

Solved Example 1

Q: Find y_n if

y = x^2 e^(2x)

Solution

Take

u = e^(2x), v = x^2

Then

u_k = 2^k e^(2x)
v_0 = x^2, v_1 = 2x, v_2 = 2, v_k = 0 for k >= 3

By Leibniz theorem,

y_n = nC0 u_n v_0 + nC1 u_(n-1) v_1 + nC2 u_(n-2) v_2

So

y_n = 2^n e^(2x) x^2 + n . 2^(n-1) e^(2x) . 2x + [n(n-1)/2] . 2^(n-2) e^(2x) . 2

Therefore,

y_n = 2^(n-2) e^(2x) [4x^2 + 4nx + n(n-1)]

Basic Example

Q: Find y_n if

y = log(ax + b)

Solution

Step 1: Differentiate successively

y_1 = d/dx [log(ax+b)] = a / (ax+b)
y_2 = d/dx [a(ax+b)^(-1)] = -a^2 (ax+b)^(-2)
y_3 = d/dx [-a^2(ax+b)^(-2)] = 2a^3 (ax+b)^(-3)
y_4 = -6a^4 (ax+b)^(-4)

Step 2: Identify the pattern

• Coefficients: 1, -1, 2, -6, ... = (-1)^(n-1) (n-1)!
• Power of a: a^n
• Denominator: (ax+b)^n

Final answer:

y_n = (-1)^(n-1) (n-1)! a^n / (ax+b)^n

2.2 Mean Value Theorems

Mean value theorems connect continuity and differentiability with the existence of a special point inside an interval.

Solved Example 2

Q: Verify LMVT for

f(x) = x^3 - 2x^2 - x + 3 on [0, 2]

Solution

First,

f(0) = 3
f(2) = 8 - 8 - 2 + 3 = 1

Then

[f(2)-f(0)] / (2-0) = (1-3)/2 = -1

Now

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

Set f'(c) = -1:

3c^2 - 4c - 1 = -1
3c^2 - 4c = 0
c(3c - 4) = 0

So

c = 0 or c = 4/3

Since 4/3 lies in (0,2), LMVT is verified.

Basic Example

Q: Verify Rolle's theorem for

f(x) = x(x-1)(x-2) on [0,2]

Solution

Step 1: Check conditions

• Polynomial function, so it is continuous on [0,2]
• Polynomial function, so it is differentiable on (0,2)
• f(0) = 0 and f(2) = 2(1)(0) = 0

Hence all conditions of Rolle's theorem are satisfied.

Step 2: Find c such that f'(c) = 0

f(x) = x^3 - 3x^2 + 2x
f'(x) = 3x^2 - 6x + 2

Set

3c^2 - 6c + 2 = 0

Then

c = [6 +- sqrt(36-24)] / 6
  = [6 +- sqrt(12)] / 6
  = 1 +- 1/sqrt(3)

So

c_1 approx 0.423, c_2 approx 1.577

Both values lie in (0,2), therefore Rolle's theorem is verified.

Intermediate Example

Q: Verify LMVT for

f(x) = sqrt(x) on [1,4]

Solution

Step 1: Check conditions

• sqrt(x) is continuous on [1,4]
• f'(x) = 1 / (2sqrt(x)) exists for all x in (1,4)

Step 2: Apply LMVT

[f(4)-f(1)] / (4-1) = (2-1)/3 = 1/3

Now

f'(c) = 1 / (2sqrt(c))

Set

1 / (2sqrt(c)) = 1/3

Then

2sqrt(c) = 3
sqrt(c) = 3/2
c = 9/4

Since 9/4 lies in (1,4), LMVT is verified at c = 9/4.

2.3 Taylor and Maclaurin Series

Series expansion helps approximate functions near a point and is used in numerical methods, limits, and engineering models.

Standard expansions

• e^x = 1 + x + x^2/2! + x^3/3! + ...
• sin x = x - x^3/3! + x^5/5! - ...
• cos x = 1 - x^2/2! + x^4/4! - ...
• log(1+x) = x - x^2/2 + x^3/3 - x^4/4 + ... for |x| < 1
• (1+x)^n = 1 + nx + n(n-1)x^2/2! + ...

Solved Example 3

Q: Expand

f(x) = e^x sin x

up to x^4 using Maclaurin series.

Solution

f(0) = 0

Differentiate:

f'(x) = e^x (sin x + cos x)      => f'(0) = 1
f''(x) = 2 e^x cos x             => f''(0) = 2
f'''(x) = 2 e^x (cos x - sin x)  => f'''(0) = 2
f^(4)(x) = -4 e^x sin x          => f^(4)(0) = 0

Using Maclaurin expansion,

f(x) = f(0) + x f'(0) + x^2/2! f''(0) + x^3/3! f'''(0) + x^4/4! f^(4)(0) + ...

Hence,

f(x) = x + x^2 + x^3/3

up to the required order.

Basic Example

Q: Expand tan^(-1)x up to x^5 using Maclaurin series.

Solution

Use series integration:

d/dx [tan^(-1)x] = 1 / (1+x^2)
                 = 1 - x^2 + x^4 - x^6 + ...

Integrate term by term:

tan^(-1)x = integral(1 - x^2 + x^4 - ...) dx
          = x - x^3/3 + x^5/5 - x^7/7 + ...

Final answer up to x^5:

tan^(-1)x = x - x^3/3 + x^5/5 + O(x^7)

Intermediate Example

Q: Expand e^(2x) cos x up to x^3 using Maclaurin series.

Solution

Use standard expansions:

e^(2x) = 1 + 2x + (2x)^2/2! + (2x)^3/3! + ...
       = 1 + 2x + 2x^2 + (4/3)x^3 + ...

cos x = 1 - x^2/2 + ...

Multiply up to the required order:

(1 + 2x + 2x^2 + 4x^3/3)(1 - x^2/2)
= 1 + 2x + 2x^2 + 4x^3/3 - x^2/2 - x^3 + ...
= 1 + 2x + (3/2)x^2 + (1/3)x^3

Final answer:

e^(2x) cos x approx 1 + 2x + (3/2)x^2 + (1/3)x^3

2.4 Curvature and Radius of Curvature

Curvature measures how sharply a curve bends. The radius of curvature tells how large the osculating circle is at a point.

Solved Example 4

Q: Find the radius of curvature for

y = x^3

at (1,1).

Solution

y' = 3x^2   => y'(1) = 3
y'' = 6x    => y''(1) = 6

Using the Cartesian formula,

rho = [1 + 3^2]^(3/2) / 6
    = 10^(3/2) / 6
    = 10 sqrt(10) / 6
    = 5 sqrt(10) / 3

Therefore,

rho = 5 sqrt(10) / 3 approx 5.27 units

Basic Example

Q: Find the radius of curvature for

y = sin x

at x = pi/2.

Solution

y' = cos x      => y'(pi/2) = 0
y'' = -sin x    => y''(pi/2) = -1

Using the Cartesian formula,

rho = [1 + (y')^2]^(3/2) / |y''|
    = [1 + 0^2]^(3/2) / 1
    = 1

Final answer:

rho = 1 unit

2.5 Curve Tracing Essentials

Curve tracing means analyzing the shape and behavior of a curve before sketching it.

Standard order for tracing a curve

1. Check symmetry. 2. Check whether the curve passes through the origin. 3. Find intercepts on the axes. 4. Look for asymptotes. 5. Identify the region in which the curve exists. 6. Find critical points and tangents. 7. Sketch the final plot.

Common checks

• Symmetry about the x-axis
• Symmetry about the y-axis
• Symmetry about the origin
• Existence of loops or cusps
• Asymptotic behavior for large values

Intermediate Example

Q: Trace the curve

y^2 = x^2(1-x)

Solution

At x = 2/3:

y^2 = (4/9)(1/3) = 4/27
y = +- 2/(3sqrt(3))

So the curve has local extreme points at

(2/3, +- 2/(3sqrt(3)))

Sketch guide

• Start from the left side where the branches extend outward.
• The curve crosses the origin with tangents y=x and y=-x.
• It bulges upward and downward symmetrically.
• It reaches maximum and minimum height at x=2/3.
• It closes at (1,0).

The overall shape is like a teardrop pointing right, with symmetry about the x-axis.

Module 2 Quick Reference Sheet

• Leibniz theorem: (uv)_n = sum nCr u_(n-r) v_r
• MVT conditions: continuous on [a,b] and differentiable on (a,b)
• Maclaurin series: f(0) + x f'(0) + x^2/2! f''(0) + ...
• rho in Cartesian form: (1+y'^2)^(3/2) / |y''|
• rho in parametric form: (x'^2+y'^2)^(3/2) / |x' y'' - y' x''|
• Curve tracing order: symmetry, origin, intercepts, asymptotes, region, critical points, plot
• Memorize expansions of e^x, sin x, cos x, log(1+x), and (1+x)^n
• A^-1 style shortcut is not used here; focus on standard derivative and expansion patterns
• tan^(-1)x = x - x^3/3 + x^5/5 + ...
• rho = (1+y'^2)^(3/2) / |y''| after substituting the point carefully

Module 2 Practice Add-On

Basic

1. Find y_n if y = log(ax+b). 2. Verify Rolle's theorem for f(x)=x(x-1)(x-2) on [0,2]. 3. Expand tan^(-1)x up to x^5. 4. Find the radius of curvature for y = sin x at x = pi/2.

Intermediate

1. Use Leibniz theorem to find y_n for y = e^x x^2. 2. Verify LMVT for f(x)=sqrt(x) on [1,4]. 3. Expand e^(2x) cos x up to x^3. 4. Trace y^2 = x^2(1-x).

Module 2 Exam Focus

Module 2 Study Strategy

Module 2 Key Takeaways

Module 2 Outcome

After finishing this module, you should be able to:

• compute higher-order derivatives quickly,
• apply Leibniz theorem for product functions,
• verify Rolle's theorem and LMVT,
• expand functions using Taylor and Maclaurin series,
• find radius of curvature in standard forms, and
• analyze and sketch basic curves systematically.

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Module 3: Partial Differentiation and Applications

Chapter Objective

Master partial derivatives, chain rule, Euler's theorem, Jacobians, unconstrained and constrained optimization, and Taylor's expansion for multivariable functions.

• Exam weightage: 15 percent
• Target: 3 to 4 long questions and 2 to 3 short questions

3.1 Partial Derivatives and Higher Order Derivatives

For a function z = f(x,y), partial differentiation means differentiating with respect to one variable while treating the other as constant.

Solved Example 1

Q: If

z = x^3 y^2 + e^(xy)

find

partial^2 z / partial x partial y

Solution

First differentiate with respect to x:

partial z / partial x = 3x^2 y^2 + y e^(xy)

Now differentiate with respect to y:

partial^2 z / partial x partial y
= partial/partial y [3x^2 y^2 + y e^(xy)]
= 6x^2 y + [1 . e^(xy) + y . x e^(xy)]
= 6x^2 y + e^(xy)(1 + xy)

3.2 Total Differential and Chain Rule

When z depends on more than one variable, small changes in each variable contribute to the overall change in z.

Solved Example 2

Q: If z = f(u,v), where

u = x + y, v = x - y

prove that

partial^2 z / partial x^2 - partial^2 z / partial y^2 = 4 partial^2 z / partial u partial v

Solution

First-order partials:

partial z / partial x = f_u . 1 + f_v . 1 = f_u + f_v
partial z / partial y = f_u . 1 + f_v . (-1) = f_u - f_v

Second-order partials:

partial^2 z / partial x^2
= partial/partial x (f_u + f_v)
= (f_uu + f_uv) + (f_vu + f_vv)
= f_uu + 2f_uv + f_vv

partial^2 z / partial y^2
= partial/partial y (f_u - f_v)
= (f_uu - f_uv) - (f_vu - f_vv)
= f_uu - 2f_uv + f_vv

Subtracting,

partial^2 z / partial x^2 - partial^2 z / partial y^2 = 4f_uv

Hence,

partial^2 z / partial x^2 - partial^2 z / partial y^2 = 4 partial^2 z / partial u partial v

3.3 Euler's Theorem on Homogeneous Functions

A function f(x,y,z) is homogeneous of degree n if

f(tx, ty, tz) = t^n f(x,y,z)

Euler's theorem states:

x partial f/partial x + y partial f/partial y + z partial f/partial z = n f

Important corollaries

• If u is homogeneous of degree n, then

x partial u/partial x + y partial u/partial y = nu

• For second-order derivatives:

x^2 u_xx + 2xy u_xy + y^2 u_yy = n(n-1)u

• If v = log u and u is homogeneous of degree n, then

x partial v/partial x + y partial v/partial y = n

Solved Example 3

Q: Verify Euler's theorem for

u = x^3 + y^3 + z^3 - 3xyz

Solution

Each term has degree 3, so the function is homogeneous of degree 3.

partial u / partial x = 3x^2 - 3yz
partial u / partial y = 3y^2 - 3xz
partial u / partial z = 3z^2 - 3xy

Now,

x(partial u / partial x) + y(partial u / partial y) + z(partial u / partial z)
= x(3x^2 - 3yz) + y(3y^2 - 3xz) + z(3z^2 - 3xy)
= 3x^3 - 3xyz + 3y^3 - 3xyz + 3z^3 - 3xyz
= 3(x^3 + y^3 + z^3 - 3xyz)
= 3u

Hence Euler's theorem is verified.

3.4 Jacobians and Their Properties

The Jacobian is the determinant formed by partial derivatives of transformed variables.

J = partial(u,v) / partial(x,y)

For two variables,

partial(u,v) / partial(x,y)
= | partial u/partial x   partial u/partial y |
  | partial v/partial x   partial v/partial y |

Key properties

• partial(u,v)/partial(x,y) . partial(x,y)/partial(u,v) = 1
• Chain rule:

partial(u,v)/partial(x,y) = partial(u,v)/partial(r,s) . partial(r,s)/partial(x,y)

• If u and v are functionally dependent, then the Jacobian is zero.

Solved Example 4

Q: If

u = x(1-y), v = xy

show that

partial(u,v) / partial(x,y) = x

Solution

partial u / partial x = 1-y
partial u / partial y = -x
partial v / partial x = y
partial v / partial y = x

So

J = |1-y   -x|
    | y     x|

J = (1-y)x - (-x)(y)
  = x - xy + xy
  = x

Hence proved.

3.5 Maxima, Minima, and Lagrange Multipliers

Unconstrained optimization

For f(x,y), critical points are obtained by solving

f_x = 0, f_y = 0

Then compute

D = f_xx f_yy - (f_xy)^2

Constrained optimization

To maximize or minimize f(x,y) subject to g(x,y)=c, form

F(x,y,lambda) = f(x,y) - lambda[g(x,y)-c]

and solve

F_x = 0, F_y = 0, F_lambda = 0

Solved Example 5

Q: Find the maximum of

f(x,y) = xy

subject to

x + y = 10

Solution

Constraint:

g(x,y) = x + y - 10 = 0

Lagrangian:

F = xy - lambda(x + y - 10)

Now,

F_x = y - lambda = 0   => y = lambda
F_y = x - lambda = 0   => x = lambda

So

x = y

Using the constraint:

x + x = 10
2x = 10
x = 5, y = 5

Hence,

f(5,5) = 25

So the maximum value is 25.

3.6 Taylor's Theorem for Two Variables and Error Approximation

For a function f(x,y), the Taylor expansion about (a,b) is

f(x+h, y+k)
= f + h f_x + k f_y + 1/2! (h^2 f_xx + 2hk f_xy + k^2 f_yy) + ...

The approximate error formula is

Delta f approx df = f_x Delta x + f_y Delta y

Solved Example 6

Q: Expand

f(x,y) = e^x cos y

around (0,0) up to second degree.

Solution

At (0,0):

f(0,0) = 1
f_x = e^x cos y      => f_x(0,0) = 1
f_y = -e^x sin y     => f_y(0,0) = 0
f_xx = e^x cos y     => f_xx(0,0) = 1
f_xy = -e^x sin y    => f_xy(0,0) = 0
f_yy = -e^x cos y    => f_yy(0,0) = -1

Using Taylor's expansion:

f(x,y)
= 1 + x(1) + y(0) + 1/2 [x^2(1) + 2xy(0) + y^2(-1)]

Hence,

f(x,y) approx 1 + x + (x^2 - y^2)/2

Module 3 Quick Reference Sheet

• p = partial z/partial x, q = partial z/partial y
• r = partial^2 z/partial x^2, s = partial^2 z/partial x partial y, t = partial^2 z/partial y^2
• Mixed partials are equal if continuous
• Euler theorem: x f_x + y f_y + z f_z = nf
• partial(u,v)/partial(x,y) . partial(x,y)/partial(u,v) = 1
• Max/Min test: D = rt - s^2
• Lagrange multipliers: grad f = lambda grad g
• Taylor two-variable expansion:

f + h f_x + k f_y + 1/2(h^2 f_xx + 2hk f_xy + k^2 f_yy)

• Error approximation:

Delta f approx f_x Delta x + f_y Delta y

Module 3 Exam Focus

Module 3 Study Strategy

Module 3 Key Takeaways

Module 3 Outcome

After finishing this module, you should be able to:

• compute first and second order partial derivatives,
• apply the total differential and chain rule correctly,
• verify Euler's theorem for homogeneous functions,
• evaluate Jacobians and use their properties,
• solve unconstrained and constrained optimization problems, and
• expand multivariable functions up to second degree.

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Module 4: Multiple Integrals and Applications

Chapter Objective

Master double and triple integrals, change of order and variables, Beta and Gamma functions, and apply them to area, volume, mass, and center of mass calculations.

• Exam weightage: 15 percent
• Target: 3 to 4 long questions and 2 to 3 short questions

4.1 Double Integrals in Cartesian and Polar Form

Double integrals are used to compute area, mass, centroid, and volume under surfaces.

Solved Example 1

Q: Evaluate

double integral over R of (x+y) dA

where R is bounded by x=0, y=0, and x+y=1.

Solution

The triangular region is:

0 <= x <= 1
0 <= y <= 1-x

So

integral from 0 to 1 [ integral from 0 to 1-x (x+y) dy ] dx

Integrate with respect to y:

= integral from 0 to 1 [xy + y^2/2] from 0 to 1-x dx

= integral from 0 to 1 [x(1-x) + (1-x)^2/2] dx

= integral from 0 to 1 [x - x^2 + (1 - 2x + x^2)/2] dx

= integral from 0 to 1 [1/2 - x^2/2] dx

= [x/2 - x^3/6] from 0 to 1
= 1/2 - 1/6
= 1/3

Hence,

double integral over R of (x+y) dA = 1/3

4.2 Change of Order of Integration

Some double integrals become easier only after reversing the order of integration.

Standard steps

1. Draw the region from the given limits. 2. Identify the bounding curves. 3. Rewrite the region in the opposite order. 4. Rebuild the integral carefully with new limits.

Solved Example 2

Q: Change the order of integration:

integral from 0 to 1 [ integral from x^2 to x f(x,y) dy ] dx

Solution

Given:

• x ranges from 0 to 1
• y ranges from x^2 to x

The boundary curves are:

• y = x^2
• y = x

Their points of intersection are (0,0) and (1,1).

Now take y as the outer variable.

For a fixed y:

• left boundary is x = y
• right boundary is x = sqrt(y)

Also,

0 <= y <= 1

So the new integral is

integral from 0 to 1 [ integral from y to sqrt(y) f(x,y) dx ] dy

4.3 Triple Integrals and Coordinate Systems

Triple integrals are used to compute volume, mass, and moments in three dimensions.

Solved Example 3

Q: Evaluate

triple integral over V of z dV

where V is the hemisphere

x^2 + y^2 + z^2 <= a^2, z >= 0

Solution

Use spherical coordinates:

0 <= rho <= a
0 <= phi <= pi/2
0 <= theta <= 2pi

Also,

z = rho cos phi
dV = rho^2 sin phi drho dphi dtheta

Therefore,

integral from 0 to 2pi
integral from 0 to pi/2
integral from 0 to a
(rho cos phi)(rho^2 sin phi) drho dphi dtheta

= [integral from 0 to 2pi dtheta]
   [integral from 0 to pi/2 sin phi cos phi dphi]
   [integral from 0 to a rho^3 drho]

= 2pi . [sin^2 phi / 2] from 0 to pi/2 . [rho^4/4] from 0 to a

= 2pi . (1/2) . (a^4/4)
= pi a^4 / 4

Hence,

triple integral over V of z dV = pi a^4 / 4

4.4 Change of Variables and Jacobian

When a region or integrand becomes simpler under substitution, use a new coordinate system together with the Jacobian.

double integral over R of f(x,y) dx dy
= double integral over S of f(x(u,v), y(u,v)) |J| du dv

where

J = partial(x,y) / partial(u,v)

Solved Example 4

Q: Evaluate

double integral over R of e^[(y-x)/(y+x)] dA

where R is the triangle bounded by x=0, y=0, and x+y=1.

Solution

Take

u = y-x
v = y+x

Then

x = (v-u)/2
y = (v+u)/2

Jacobian:

J = | partial x/partial u   partial x/partial v |
    | partial y/partial u   partial y/partial v |

  = | -1/2   1/2 |
    |  1/2   1/2 |

J = (-1/2)(1/2) - (1/2)(1/2) = -1/2

So

|J| = 1/2

The triangle transforms to

0 <= v <= 1
-v <= u <= v

Hence,

integral from 0 to 1
integral from -v to v
e^(u/v) (1/2) du dv

= (1/2) integral from 0 to 1 [v e^(u/v)] from -v to v dv

= (1/2) integral from 0 to 1 v(e - e^(-1)) dv

= (e - 1/e)/2 . [v^2/2] from 0 to 1
= (e - 1/e)/4
= (e^2 - 1)/(4e)

4.5 Beta and Gamma Functions

These special functions help evaluate integrals that are difficult by elementary methods.

Solved Example 5

Q: Evaluate

integral from 0 to infinity x^4 e^(-2x) dx

Solution

Let

2x = t

Then

x = t/2
dx = dt/2

So

integral from 0 to infinity x^4 e^(-2x) dx
= integral from 0 to infinity (t/2)^4 e^(-t) (dt/2)

= 1/32 integral from 0 to infinity t^4 e^(-t) dt

= 1/32 Gamma(5)
= 1/32 . 4!
= 24/32
= 3/4

Hence,

integral from 0 to infinity x^4 e^(-2x) dx = 3/4

4.6 Volume, Mass, and Standard Applications

Double and triple integrals are central in geometric and physical applications.

Standard formulas

• Area:

A = double integral over R 1 dA

• Volume under z = f(x,y):

V = double integral over R f(x,y) dA

• Mass:

M = triple integral rho dV

• Center of mass:

(1/M) triple integral r rho dV

Worked volume example

Evaluate

V = double integral over x^2+y^2<=4 of (4 - x^2 - y^2) dA

Use polar coordinates:

V = integral from 0 to 2pi integral from 0 to 2 (4-r^2) r dr dtheta

= 2pi integral from 0 to 2 (4r - r^3) dr

= 2pi [2r^2 - r^4/4] from 0 to 2

= 2pi (8 - 4)
= 8pi

Hence,

V = 8pi

Exercise Sets

Basic

1. Evaluate integral from 0 to 2 integral from 0 to x^2 x y dy dx. 2. Change order of integral from 0 to 1 integral from y to sqrt(2-y^2) f(x,y) dx dy. 3. Evaluate integral from 0 to infinity e^(-3x) x^2 dx using Gamma function. 4. Find the area bounded by y=x and y=x^2 using a double integral.

Intermediate

1. Evaluate double integral over first quadrant of e^(-(x^2+y^2)) dA using polar form. 2. Use u=x+y, v=x-y to evaluate a transformed double integral over a triangle. 3. Find the volume of the solid bounded by z=xy, x+y=1, x=0, y=0, z=0. 4. Prove B(1/2, 1/2) = pi using the Beta-Gamma relation.

Advanced

1. Evaluate triple integral over sphere of (x^2+y^2+z^2) dV. 2. Find the center of mass of a hemisphere with uniform density. 3. Show that integral from 0 to 1 x^m (log x)^n dx = (-1)^n n! / (m+1)^(n+1). 4. Evaluate double integral over annulus of (x^2+y^2)^(-3/2) dA.

Module 4 Quick Reference Sheet

• Polar area element: dA = r dr dtheta
• Cylindrical volume element: dV = r dr dtheta dz
• Spherical volume element: dV = rho^2 sin phi drho dphi dtheta
• Change order method: sketch region, swap limits, recheck boundaries
• Use |J| while changing variables
• Gamma(n+1)=nGamma(n)
• Gamma(1/2)=sqrt(pi)
• B(m,n)=Gamma(m)Gamma(n)/Gamma(m+n)
• Volume under surface: double integral f(x,y) dA
• Standard trick: x^2+y^2 suggests polar coordinates, x+y and x-y suggest substitution

Module 4 Exam Focus

Module 4 Study Strategy

Module 4 Key Takeaways

Module 4 Outcome

After finishing this module, you should be able to:

• evaluate double and triple integrals in multiple coordinate systems,
• reverse the order of integration correctly,
• use Jacobians for variable transformation,
• apply Beta and Gamma functions to non-elementary integrals, and
• solve area, volume, mass, and center of mass problems.

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Module 5: Ordinary Differential Equations (ODEs)

Chapter Objective

Master solution techniques for first-order and higher-order linear ODEs, special equations such as Cauchy-Euler equations, variation of parameters, and real-world engineering applications.

• Exam weightage: 20 to 25 percent
• Target: 4 to 5 long questions and 2 short questions

5.1 First Order ODEs: Types and Solutions

First-order differential equations appear in growth-decay models, circuits, heat transfer, and many engineering systems.

Solved Example 1

Q: Solve

dy/dx + y tan x = sec x

Solution

This is a linear equation with

P = tan x, Q = sec x

Integrating factor:

IF = e^(integral tan x dx)
   = e^(ln|sec x|)
   = sec x

Multiply the equation by sec x:

y sec x = integral sec x . sec x dx + C

y sec x = integral sec^2 x dx + C

y sec x = tan x + C

Hence,

y = cos x (tan x + C)
  = sin x + C cos x

5.2 Higher Order Linear ODEs with Constant Coefficients

A higher-order linear ODE with constant coefficients can be written as

a y'' + b y' + c y = f(x)

or in operator form,

F(D)y = f(x)

The complete solution is

y = CF + PI

Complementary Function

Solve the auxiliary equation

a m^2 + b m + c = 0

Particular Integral shortcuts

Solved Example 2

Q: Solve

y'' - 4y' + 4y = e^(2x)

Solution

Auxiliary equation:

m^2 - 4m + 4 = 0

(m-2)^2 = 0

So the root is repeated:

m = 2, 2

Hence,

CF = (c1 + c2 x)e^(2x)

For the particular integral,

PI = 1/(D-2)^2 e^(2x)

This is a case of failure because F(2)=0, so multiply by x. For repeated failure, the final result becomes

PI = (x^2/2)e^(2x)

Therefore,

y = (c1 + c2 x + x^2/2)e^(2x)

5.3 Variation of Parameters

Variation of parameters is used when the RHS is not suitable for standard PI shortcuts.

For the equation

y'' + P(x)y' + Q(x)y = R(x)

if y1 and y2 are linearly independent solutions of the corresponding homogeneous equation, then

y_p = -y1 integral(y2 R / W) dx + y2 integral(y1 R / W) dx

where

W = y1 y2' - y2 y1'

is the Wronskian.

Solved Example 3

Q: Solve

y'' + y = sec x

Solution

Auxiliary equation:

m^2 + 1 = 0

So,

m = +- i

Hence,

CF = c1 cos x + c2 sin x

Take

y1 = cos x, y2 = sin x, R = sec x

Wronskian:

W = y1 y2' - y2 y1'
  = cos x . cos x - sin x(-sin x)
  = cos^2 x + sin^2 x
  = 1

Now,

y_p = -cos x integral(sin x sec x) dx + sin x integral(cos x sec x) dx

    = -cos x integral(tan x) dx + sin x integral(1) dx

    = -cos x ln|sec x| + x sin x

Therefore,

y = c1 cos x + c2 sin x + x sin x - cos x ln|sec x|

5.4 Cauchy-Euler (Equidimensional) Equation

A Cauchy-Euler equation has the form

x^2 y'' + a x y' + by = f(x)

Use the substitution

x = e^z

Then

x y' = D y
x^2 y'' = D(D-1)y

which converts the equation into one with constant coefficients.

Solved Example 4

Q: Solve

x^2 y'' + x y' - y = x^2

Solution

Under the substitution, the equation becomes

[D(D-1) + D - 1]y = e^(2z)

(D^2 - 1)y = e^(2z)

Auxiliary equation:

m^2 - 1 = 0

So,

m = +- 1

Hence,

CF = c1 e^z + c2 e^(-z)
   = c1 x + c2 / x

Particular integral:

PI = 1/(D^2 - 1) e^(2z)
   = e^(2z)/(4-1)
   = e^(2z)/3
   = x^2/3

Therefore,

y = c1 x + c2/x + x^2/3

5.5 Engineering Applications of ODEs

Ordinary differential equations model many engineering systems.

Solved Example 5

Q: An RL circuit has L = 2 H, R = 10 ohm, E = 20 V. Find i(t) if i(0)=0.

Solution

The governing equation is

2 di/dt + 10 i = 20

Divide by 2:

di/dt + 5i = 10

This is a first-order linear equation.

Integrating factor:

IF = e^(integral 5 dt) = e^(5t)

Multiply through:

i e^(5t) = integral 10 e^(5t) dt + C

i e^(5t) = 2 e^(5t) + C

i(t) = 2 + C e^(-5t)

Use i(0)=0:

0 = 2 + C

So,

C = -2

Hence,

i(t) = 2(1 - e^(-5t)) A

Exercise Sets

Basic

1. Solve dy/dx = (x^2 + y^2)/(xy). 2. Solve y'' - 5y' + 6y = 0. 3. Solve dy/dx + 2y/x = x^2. 4. Find the particular integral of y'' + 4y = cos 2x.

Intermediate

1. Solve (x^2 + 1) dy/dx + 2xy = 4x^2. 2. Solve y'' - 2y' + y = x e^x. 3. Use variation of parameters to solve y'' + 4y = tan 2x. 4. Solve x^2 y'' - 2x y' + 2y = x^3.

Advanced

1. Solve y''' - 3y'' + 3y' - y = e^x. 2. Solve x^2 y'' + 3x y' + y = 1/(1-x)^2. 3. A 10 kg mass stretches a spring 0.5 m. If displaced 0.2 m upward and released, find the equation of motion. 4. Solve dy/dx = (x + y - 1)/(x - y + 1) using translation.

Module 5 Quick Reference Sheet

• Linear first-order ODE:

IF = e^(integral P dx)

• Exact equation condition:

partial M/partial y = partial N/partial x

• CF roots:

distinct, repeated, and complex cases must be recognized quickly

• PI shortcuts:

F(D)e^(ax) -> e^(ax)/F(a)

• VOP formula:

y_p = -y1 integral(y2 R/W) dx + y2 integral(y1 R/W) dx

• Wronskian:

W = y1 y2' - y2 y1'

• Cauchy-Euler substitution:

x = e^z

• RL and RC circuits reduce to first-order linear equations

Module 5 Exam Focus

Module 5 Study Strategy

Module 5 Key Takeaways

Module 5 Outcome

After finishing this module, you should be able to:

• solve standard first-order differential equations,
• build CF and PI for higher-order linear equations,
• use variation of parameters when required,
• solve Cauchy-Euler equations systematically, and
• interpret ODEs in engineering applications such as circuits and vibration models.

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Module 6: Infinite Series and Convergence (with Fourier Basics)

Chapter Objective

Master convergence and divergence tests for infinite series, power series and radius of convergence, and the basics of Fourier series.

• Exam weightage: 10 to 15 percent
• Target: 2 to 3 long questions and 2 short questions

6.1 Convergence Tests for Positive Term Series

A positive term series has the form

sum a_n, where a_n > 0

The main task is to decide whether the series converges or diverges.

Solved Example 1

Q: Test the convergence of

sum (n!)/(n^n)

Solution

Let

a_n = n!/n^n

Use the ratio test:

a_(n+1)/a_n
= [(n+1)!/(n+1)^(n+1)] . [n^n/n!]

= (n+1)n^n / (n+1)^(n+1)
= n^n / (n+1)^n
= 1 / (1 + 1/n)^n

As n -> infinity,

L = 1/e

Since

1/e < 1

the series converges.

6.2 Alternating and Conditional Convergence

Alternating series change sign from term to term.

Leibniz Test

An alternating series

sum (-1)^n a_n

converges if:

1. a_n is decreasing 2. lim a_n = 0

Solved Example 2

Q: Test

sum (-1)^n / sqrt(n)

Solution

Let

a_n = 1/sqrt(n)

Then:

• a_n is decreasing
• lim 1/sqrt(n) = 0

So by Leibniz test, the series converges.

Now check absolute convergence:

sum |(-1)^n / sqrt(n)| = sum 1/sqrt(n)

This is a p-series with

p = 1/2 < 1

so it diverges.

Hence the series is conditionally convergent.

6.3 Power Series and Radius of Convergence

A power series has the form

sum a_n (x-c)^n

where c is the center.

Radius of convergence

The radius R is given by:

R = lim |a_n / a_(n+1)|

when that limit exists.

The interval of convergence is

(c-R, c+R)

and endpoints must always be checked separately.

Solved Example 3

Q: Find the radius and interval of convergence of

sum (x-3)^n / (n 2^n)

Solution

Here

a_n = 1/(n 2^n)

Use the ratio formula:

R = lim |a_n / a_(n+1)|

= lim [(n+1)2^(n+1)] / [n2^n]
= lim 2(1 + 1/n)
= 2

So the radius is

R = 2

The center is c = 3, so the open interval is

(1,5)

Check endpoints:

At x = 1:

sum (-2)^n / (n 2^n) = sum (-1)^n / n

This is the alternating harmonic series, so it converges.

At x = 5:

sum 2^n / (n 2^n) = sum 1/n

This is the harmonic series, so it diverges.

Hence the interval of convergence is

[1,5)

6.4 Fourier Series Basics

For a function of period 2L, the Fourier series is

f(x) = a_0/2 + sum [a_n cos(n pi x / L) + b_n sin(n pi x / L)]

Coefficients

Half-range rules

• Even function: cosine series only, so b_n = 0
• Odd function: sine series only, so a_n = 0

Solved Example 4

Q: Find the Fourier series of

f(x) = x on (-pi, pi)

Solution

The function f(x)=x is odd.

Therefore,

a_0 = 0
a_n = 0

Now,

b_n = (1/pi) integral from -pi to pi x sin(nx) dx

Since the product is even,

b_n = (2/pi) integral from 0 to pi x sin(nx) dx

Using integration by parts:

u = x, dv = sin(nx) dx
du = dx, v = -cos(nx)/n

Then

b_n = (2/pi)[ -x cos(nx)/n ] from 0 to pi + (2/pi) integral from 0 to pi cos(nx)/n dx

The second term becomes zero, so

b_n = (2/pi)[ -pi cos(npi)/n ]

Since cos(npi) = (-1)^n,

b_n = 2(-1)^(n+1)/n

Hence,

f(x) = sum from n=1 to infinity [2(-1)^(n+1)/n] sin(nx)

Exercise Sets

Basic

1. Test convergence of sum n^2 / 3^n. 2. Test sum (-1)^n / n^3. 3. Find the radius of convergence for sum (2x)^n / n!. 4. Check whether sum 1/(n ln n) converges or diverges.

Intermediate

1. Use Raabe's test for a factorial-type series. 2. Find the interval of convergence for sum (x+2)^n / sqrt(n+1). 3. Expand f(x)=|x| on (-pi,pi) as a Fourier series. 4. Test absolute and conditional convergence of sum (-1)^n (n+1)/(n^2+1).

Advanced

1. Prove that absolute convergence implies convergence of sum a_n sin(n theta). 2. Find the sum of sum (-1)^(n+1)/n^2 using Fourier series. 3. Determine convergence of sum n! x^n using the root test. 4. Find the half-range sine series of f(x)=x(pi-x) on (0,pi).

Module 6 Quick Reference Sheet

• Ratio test: L = lim |a_(n+1)/a_n|
• Root test: L = lim |a_n|^(1/n)
• Raabe test: R = lim n(|a_n/a_(n+1)| - 1)
• Leibniz test: alternating series converges if a_n decreases to 0
• Radius:

R = lim |a_n/a_(n+1)|

• Interval:

(c-R, c+R) plus endpoint testing

• Fourier coefficients:

a_0, a_n, b_n

• Even functions give cosine series, odd functions give sine series
• p-series:

sum 1/n^p converges if p > 1

• Geometric series:

sum ar^n converges if |r| < 1

Module 6 Exam Focus

Module 6 Study Strategy

Module 6 Key Takeaways

Module 6 Outcome

After finishing this module, you should be able to:

• decide whether a series converges or diverges using standard tests,
• classify alternating series as absolute or conditional,
• find radius and interval of convergence of power series, and
• compute Fourier series for simple functions using symmetry.