GALGOTIAS COLLEGE OF ENGINEERING & TECHNOLOGY

1, Knowledge Park-II, Greater Noida

Engineering Mathematics-I

KAS-103T

Session 2020-2021 (Odd Semester)

Program Outcomes (POs) Program Outcome Engineering graduate will be able to:

PO1(K3) Engineering knowledge: Apply the knowledge of mathematics, science, engineering fundamentals, and an engineering specialization to the solution of complex engineering problems.

PO2(K4) Problem analysis: Identify, formulate, review research literature, and analyse complex engineering problems reaching substantiated conclusions using first principles of mathematics, natural sciences, and engineering sciences.

PO3(K6) Design/development of solutions: Design solutions for complex engineering problems and design system components or processes that meet the specific needs with appropriate considerations for the public health and safety, and the cultural, societal, and environmental considerations.

PO4(K5) Conduct investigations of complex problems: Use research-based knowledge and research methods including design of experiments, analysis and interpretation of data, and synthesis of the information to provide conclusions

PO5(K4) Modern tool usage: Create, select, and apply appropriate techniques, resources, and modern engineering and IT tools including prediction and modelling to complex engineering activities with an understanding of the limitations

PO6(K3) The engineer and society: Apply reasoning informed by the contextual knowledge to assess societal, health, safety, legal and cultural issues and the consequent relevant to the professional engineering practices

PO7(K2) Environment and sustainability: Understand the impact of the professional engineering solutions in societal and environmental contexts, and demonstrate the knowledge of, and need for sustainable development

PO8(K3) Ethics: Apply ethical principles and commit to professional ethics and responsibilities and norm of the engineering practices

PO9 Individual and team work: Function effectively as an individual, and as a member or leader in diverse teams, and in multidisciplinary settings

PO10 Communications: Communicate effectively on complex engineering activities with the engineering community and with society at large, such as, being able to comprehend and write effective reports and design documentation, make effective presentations, and give and receive clear instructions

PO11(K

3)

Project management and finance: Demonstrate knowledge and understanding of the engineering and management principles and apply these to one’s own work, as a member and leader in a team, to manage projects and in multidisciplinary environments.

PO12 Life-long learning: Recognize the need for, and have the preparation and ability to engage in independent and life learning in the broadest context of technological change.

Course Outcome: Engineering Mathematics-I (KAS103T) Bloom’s Knowledge Level ==> K1 - Remember, K2 - Understand, K3 - Apply, K4 - Analyse, K5 - Evaluate, K6 – Create

CO No. Course Outcome : Statement Bloom’s

Knowledge Level

Contents of Syllabus

CO-1 Remember the basics of matrices and apply the concept of rank for solving linear simultaneous equations K3

CO-2 Apply the concept of limit, continuity and differentiability in the study of Rolle’s, Lagrange’s, Cauchy mean value theorem and

Leibnitz theorem K3

CO-3 Apply partial differentiation for evaluating extreme values, expansion of function and Jacobian K3

CO-4 Apply the methods of multiple integral for finding area, volume, centre of mass and centre of gravity K3

CO-5 Apply the concept of vector for evaluating directional

derivatives, tangent and normal planes, line, surface and volume integrals

K3

CO-POs Mapping Matrices

You should be able to explain and justify the relationship between COs and POs.

1. slight (low) 2. moderate (medium) 3.substantial (high)

CO

PO1

PO2

PO3

PO4

PO5

PO6

PO7

PO8

PO9

PO10

PO11

PO12

CO-1 3 3 3 - - - - - - - - 3

CO-2 3 3 3 - - - - - - - - 3

CO-3 3 3 3 - - - - - - - - 3

CO-4 3 3 3 - - - - - - - - 3

CO-5 3 3 3 - - - - - - - - 3

Average 3 3 3 - - - - - - - - 3

Engineering Mathematics-I

Subject Code : KAS-103T

MODULE-1

MATRICES

Course outcome KAS-103T (CO-I)

Remember the basics of matrices and apply the concept of rank for solving linear simultaneous equations.

Lecture 1

Lecture 2

Lecture 3

Lecture 4

Lecture 5

Lecture 6

Lecture 7

Lecture 8

Lecture 9

Lecture 10

Problems (True and False)

MODULE - II

DIFFERENTIAL CALCULUS-I

Course outcome KAS-103T (CO-2)

Apply the concept of limit, continuity and differentiability in the study of Rolle’s, Lagrange’s, Cauchy mean value theorem and Leibnitz theorem

Lecture 1 Limits of a function

Let f be a function defined in a domain which we take to be an interval, say, I.

We shall study the concept of limit of f at a point ‘a’ in I. We say lim → 푓(푥) is the expected

value of f at x = a given the values of f near to the left of a. This value is called the left hand limit

of f at a.

We say lim → 푓(푥) is the expected value of f at x = a given the values of f near to the right of

a. This value is called the right hand limit of, f at a.

If the right and left hand limits coincide, we call the common value as the limit of f at x = a and

denote it by lim → 푓(푥).

Some properties of limits Let f and g be two functions such that (Algebra of limits)

1. The limit of a sum (difference) is equal to the sum (difference) of the limits.

2. The limit of product is equal to the product of the limits.

3. The limit of a quotient is equal to the quotient of the limits provided that the limit of the

denominator is not equal to zero.

Problems:

Q1.Evaluate lim → . Ans 3/2.

Q2. Evaluate lim →( ) / ( ) /

. Ans 2/3.

Q3. Evaluate lim → (1 + 푥) . Ans e.

Q4. Evaluate (i) lim → , (ii).lim → . Ans (i) 1, (ii) 0.

Q5. Evaluate lim → sin . Ans does not exist

Q6. Let 푓(푥) = 푥, 푖푓푥푖푠푟푎푡푖표푛푎푙−푥, 푖푓푥푖푠푖푟푟푎푡푖표푛푎푙. Show that the limit exists only when 푎 = 0.

Continuity:

Definition 1 Suppose f is a real function on a subset of the real numbers and let c be a point in the

domain of f. Then f is continuous at c if lim → 푓(푥) = 푓(푐). otherwise the function is

discontinuous at 푥 = 푐.

More elaborately, if the left hand limit, right hand limit and the value of the function at x = c

exist and equal to each other, then f is said to be continuous at x = c. Recall that if the right hand

and left hand limits at x = c coincide, then we say that the common value is the limit of the

function at x = c.

Hence we may also rephrase the definition of continuity as follows: a function is continuous at x

= c if the function is defined at x = c and if the value of the function at x = c equals the limit of

the function at x = c. If f is not continuous at c, we say f is discontinuous at c and c is called a

point of discontinuity of f.

Algebra of continuous functions

Theorem 1: Suppose f and g be two real functions continuous at a real number c. Then

(1) f + g is continuous at x = c.

(2) f – g is continuous at x = c.

(3) f . g is continuous at x = c. (4) f /g is continuous at x = c, (provided g (c) ≠ 0).

Problems:

Q1. Discuss the continuity of the function f given by f(x) = | x | at x = 0.

Q2. Discuss the continuity of the function f defined by f (x) = 1/x , x ≠ 0.

Q3. Test the following function for continuity 푓(푥) = 푥 sin , 푥 ≠ 0, 푓(0) = 0,푎푡푥 = 0.

Q4. Examine the function defined below for continuity at 푥 = 푎, 푓(푥) = cosec, 푥 ≠푎, 푓(푥) = 0, 푎푡푥 = 푎.

Q5. Examine the function defined below for continuity 푥 = 0, 푓(푥) = ,푥 ≠ 0,푓(0) = 1.

Lecture 2 Differentiability:

Definitions

Derivative at a point:

Let I, denote the open interval]푎, 푏[ in R and let 푥 ∈ 퐼. The a function 푓: 퐼 → 푅 is said to be differentiable at 푥 iff

lim →( ) ( ) or equivalently lim →

( ) ( ) exists finitly and this limit, if it exists

finitly, is called the differential coefficient or derivative of f with respect to 푥 = 푥 . It is denoted by 푓 (푥 )표푟퐷푓(푥 ).

Progressive and regressive derivative:

The progressive derivative of f at 푥 = 푥 is given by

lim →( ) ( ) , ℎ > 0. it is also called right hand differential coefficient and is denoted

by 푅푓 (푥 )표푟푏푦푓 (푥 + 0).

The regressive derivative of f at 푥 = 푥 is given by

lim →( ) ( ) , ℎ > 0. it is also called left hand differential coefficient and is denoted by

퐿푓 (푥 )표푟푏푦푓 (푥 − 0).

It is obvious that that 풇is differentiable at 풙 = 풙ퟎ, iff 푹풇 (풙ퟎ) and 푳풇 (풙ퟎ) both exists and are equal.

Differentiability in an interval:

Open interval ]풂,풃[ : A function 푓: ]푎,푏[ → 푅 is said to be differentiable in ]푎,푏[ iff it is differentiable at every point of ]푎, 푏[.

Closed interval [풂,풃]: A function 푓: [풂,풃] → 푅 is said to be differentiable in [푎,푏] iff 푅푓 (푥 ) and 퐿푓 (푥 )exists and it is differentiable at every point of ]푎,푏[.

Problems:

Q1. Continuity is necessary and sufficient condition for the existence of a finite derivative.

Q2. Prove that the function 푓(푥) = |푥| is continuous at 푥 = 0, but not differentiable at 푥 = 0, where |푥| means the numerical or absolute value of x. Also draw the graph.

Q3. Show that the function푓(푥) = |푥| + |푥 − 1| is not differentiable at 푥 = 0,푥 = 1.

Q4. Prove that the function 푓(푥) = 푥 tan 푓표푟푥 ≠ 0,푓(0) = 0. is continuous at 푥 = 0, but not differentiable at 푥 = 0.

Lecture 3 Rolle’s Theorem:

Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). If

f(a) = f(b), then there is at least one point c in (a, b) where f '(c) = 0.

(The tangent to a graph of f where the derivative vanishes is parallel to x-axis, and so is the line

joining the two "end" points (a, f(a)) and (b, f(b)) on the graph. The line that joins to points on a

curve -- a function graph in our context -- is often referred to as a secant. Thus Rolle's theorem

claims the existence of a point at which the tangent to the graph is parallel to the secant, provided

the latter is horizontal.)

Problems:

Q1. Discuss the applicability of Rolle’s theorem for 푓(푥) = 2 + (푥 − 1) / in the interval [0, 2].

Q2. Discuss the applicability of Rolle’s theorem for 푓(푥) = |푥| in the interval [−1, 1].

Q3. Verify the Rolle’s theorem for the following functions

(i) 푓(푥) = 2푥 + 푥 − 4푥 − 2, (ii) 푓(푥) = sin 푥 푖푛[0,휋]

Lagrange’s Mean Value Theorem:

Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). Then

there is at least one point c in (a, b) where

f '(c) = (f(b) - f(a)) / (b - a).

(The Mean Value Theorem claims the existence of a point at which the tangent is parallel to the

secant joining (a, f(a))and (b, f(b)). Rolle's theorem is clearly a particular case of the MVT in

which f satisfies an additional condition, f(a) = f(b).)

Problems:

Q1. If 푓(푥) = (푥 − 1)(푥 − 2)(푥 − 3) and 푎 = 0,푏 = 4,find ‘c’ using Mean value theorem.

Q2. . If 푓(푥) = 푥(푥 − 1)(푥 − 2) and 푎 = 0,푏 = ,find ‘c’ using Mean value theorem.

Q3. Using Lagrange’s Mean value theorem prove that 1 + 푥 < 푒 < 1 + 푥푒 ∀푥 > 0.

Cauchy’s Mean Value Theorem:

Cauchy's mean-value theorem is a generalization of the usual mean-value theorem. It states that

if and are continuous on the closed interval , if , and if both functions

are differentiable on the open interval , then there exists at least one with such

that

Problems:

Q1. Verify the Cauchy’s Mean theorem for the functions 푥 and 푥 in the interval [1,2].

Q2. In Cauchy's mean-value theorem, we write 푓(푥) = 푒 and 푔(푥) = 푒 ,show that ‘c’ is the

arithmetic mean between a and b.

Lecture 4

Lecture 5

Lecture 6

Lecture 7 EVOLUTE

Corresponding to each point on a curve we can find the curvature of the curve at that point.

Drawing the normal at these points, we can find Centre of Curvature corresponding to each of

these points. Since the curvature varies from point to point, centers of curvature also differ. The

totality of all such centres of curvature of a given curve will define another curve and this curve

is called the evolute of the curve.

“The Locus of centers of curvature of a given curve is called the evolute of that curve.”

The locus of the centre of curvature C of a variable point P on a curve is called the evolute of the

curve. The curve itself is called involute of the evolute.

Here, for different points on the curve, we get different centre of curvatures. The locus of all

these centers of curvature is called as Evolute.

The external curve which satisfies all these centers of curvature is called as Evolute.

Here Evolute is nothing but an curve equation. To find Evolute, the following models exist.

1. If an equation of the curve is given and If we are asked to show / prove L.H.S=R.H.S,

Then do as follows.

First find Centre of Curvature 퐶(푋,푌), where 푋 = 푥 − [ ]

푌 = 푦 + [ ]

And then consider L.H.S: In that directly substitute 푋 in place of푥 and 푌in place of .

Similarly for R.H.S. and then show that L.H.S=R.H.S

2. If a curve is given and if we are asked to find the evolute of the given curve, then do as

follows:

First find Centre of curvature퐶(푋,푌) and then re-write as 푥 in terms of푋 and푦 in terms

of . and then substitute in the given curve, which gives us the required evolute.

3. If a curve is given, which is in parametric form, then first find Centre of curvature, which

will be in terms of parameter. Then using these values of 푋 and 푌 eliminate the

parameter, which gives us evolute.

Problems:

Q1. Find the coordinates of centre of curvature at any point of the parabola 푦 = 4푎푥and

also show its evolute is given by27푎푦 = 4(푥 − 2푎) .

Q2. Determine the parametric equations for the evolute of the curve 푥 = , 푦 = .

Ans: 푋 = − 푡 − 푡 ,푌 = 푡 + 푡 .

Q3. Find the evolute of the hyperbola − = 1. Deduce the evolute of a rectangular

hyperbola. Ans: (푎푋) − (푏푌) = (푎 + 푏 ) , for rectangular hyperbola(푋) − (푌) =

(2푎) .

Q4. Find the evolute of the following curves (i) 푦 = 푥 ,(ii) 푦 = 푒 .

Ans: (푖)푋(푥) = − 푥 ,푌(푥) = 푥 − , (푖푖)푋(푥) = 푥 − 1 − 푒 ,푌(푥) = 2푒 +

푒 .

Lecture 8

Envelopes A curve which touches each member of a given family of curves is called envelope of

that family.

Procedure to find envelope for the given family of curves:

Case 1: Envelope of one parameter Let us consider 푦 = 푓(푥)being the given family of

curves.

Step 1: Differentiate w.r.t to the parameter partially, and find the value of the parameter

Step 2: By Substituting the value of parameter in the given family of curves, we get

required envelope.

Special Case: If the given equation of curve is quadratic in terms of parameter, then

envelope is given by 푑푖푠푐푟푖푚푖푛푎푛푡 = 0.

Case 2: Envelope of two parameter

Let us consider to be the given family of curves, and a relation connecting these two

parameters

Step 1: Obtain one parameter in terms of other parameter from the given relation

Step 2: Substitute in the given equation of curve, so that the problem of two parameter

converts to problem of one parameter.

Step 3: Use one parameter technique to obtain envelope for the given family of curve

Problems:

Q1. Find the envelope of the family of straight line 푦 = 푚푥 + √푎 푚 + 푏 ,푚 is the

parameter. Ans: + = 1.

Q2. Find the envelope of the one parameter family of curves 푦 = 푚푥 + 푎푚 , where m is

the parameter and a, p are constant. Ans: 푎푝 푦 = −푥 . 푝 + (−푥) .

Q3. Show that the family of straight lines 2푦 − 4푥 + 훼 = 0 has no envelope, where 훼

being the parameter.

Q4. Determine the envelope of 푥푠푖푛푡 − 푦 cos 푡 = 푎푡,where t is the parameter.

Q5. Find the envelope of family of straight line + = 1, where 푎, 푏 are two parameters

which are connected by the relation 푎 + 푏 = 푐. Ans: √푥 + 푦 = √푐.

Q6. Determine the envelop of two parameter family of parabolas + = 1, where

푎, 푏 are two parameters which are connected by the relation 푎 + 푏 = 푐, where c is a given

constant.

Lecture 9

Lecture 10

Differential

CalCulus-II Course Outcome KAS-103T (CO-3)

Apply partial differentiation for evaluating extreme values, expansion of function and Jacobian

Lecture 1

Lecture 2

Lecture 3

Lecture 4

Lecture 5

Lecture 6

Lecture 7

Lecture 8

Lecture 9

Lecture 10

Module-IV

MULTIPLE INTEGRALS

Course outcome KAS-103T (CO-IV)

Apply the methods of multiple integral for finding area, volume, centre of mass and centre of gravity.

Lecture 1

Lecture 2

Lecture 3

Lecture 4

Lecture 5

Lecture 6

Lecture 7

Module-V

VECTOR CALCULAS

Course outcome KAS-103T (CO-V)

Apply the concept of vector for evaluating directional derivatives, tangent and normal planes, line, surface and volume integrals

Lecture 1

Lecture 2

Lecture 3

Lecture 4

Lecture 5

Lecture 6

Lecture 7

Lecture 8