SREENIVASA INSTITUTE of TECHNOLOGY and MANAGEMENT STUDIES (autonomous)

(ENGINEERING MATHEMATICS-III)

Course Material

II- B.TECH / I - SEMESTER regulation: r18

Course Code: 18SAH211

Compiled by

Department OF MATHEMATICS

Unit-I: Numerical Integration

Source:https://www.intmath.com/integration/integration-intro.php

Numerical Integration:

Simpson’s 1/3- Rule

Note: While applying the Simpson’s 1/3 rule, the number of sub-intervals (n) should be taken as multiple of 2.

Simpson’s 3/8- Rule

Note: While applying the Simpson’s 3/8 rule, the number of sub-intervals (n) should be taken as multiple of 3.

Numerical solution of ordinary differential equations

Taylor’s Series Method

Picard’s Method

Euler’s Method

Runge-Kutta Formula

UNIT-II Multiple Integrals

1. Double Integration

Evaluation of Double Integration

Triple Integration

UNIT-III Partial Differential Equations

Partial differential equations are those equations which contain partial differential coefficients, independent variables and dependent variables. The independent variables are denoted by x and y and dependent variable by z. the partial differential coefficients are denoted as follows

The order of the partial differential equation is the same as that of the order of the highest differential coefficient in it.

UNIT-IV Vector Differentiation

Elementary Vector Analysis Definition (Scalar and vector): Scalar is a quantity that has magnitude but not direction. For instance mass, volume, distance

Vector is a directed quantity, one with both magnitude and direction. For instance acceleration, velocity, force

Basic Vector System

Magnitude of vectors:

Let P = (x, y, z). Vector OPP is defined by

OP  p  x i + y j + z k

 []x, y, z with magnitude (length)

OP  p  x2 + y 2 + z 2

Calculation of Vectors

Vector Equation

Two vectors are equal if and only if the corresponding components are equals

Let a  a1i + a2 j + a3 k and b  b1i + b2 j + b3 k. Then a  b  a  b , a  b , a  b 1 1 2 2 3 3 Addition and Subtraction of Vectors

a b  (a1 b1)i +(a2 b2 ) j +(a3 b3)k

Multiplication of Vectors by Scalars

If  is a scalar, then b  (b )i + (b ) j + (b )k 1 2 3

Example 1

Given p  5 i + j - 3 k and q  4 i - 3 j + 2 k . Find

a) p + q b) p - q cp) Magnitude of vector d) 2 q - 10 p

Vector Products

If aaiajak1 + 2 + 3 and bbibjbk  1 + 2 + 3 , ~ ~ ~ ~ ~ ~ ~~ 1) Scalar Product (Dot product)

a b  a1 b 1 + a 2 b 2 + a 3 b 3 ~~ or a.b | a || b | cos,  is the angle between a and b ~ ~ ~ ~

2) Vector Product (Cross product)

i j k ~~~

a b a1 a 2 a 3 ~~ b1 b 2 b 3

ab23 - abi 32 - ab 13 - abj 31 + ab 12 - abk 21  ~~ ~

Differentiation of Two Vectors

Del Operator Or Nabla (Symbol )

Operator  is called vector differential operator, defined as

        i + j + k . x ~ y ~ z ~  

Grad (Gradient of Scalar Functions)

If  x,y,z is a scalar function of three variables and f is differentiable, the gradient of f is defined as

   grad     i + j + k . ~ ~ x y ~ z

* is a scalar function

*  is a vector function

Example.2

If   x2 yz3 + xy2 z 2 , determine grad  at P  (1,3,2).

Solution Given   x2 yz3 + xy2 z 2 , hence   2xyz3 + y 2 z 2 x   x2 z3 + 2xyz2 y   3x2 yz2 + 2xy2 z z Therefore,      i + j + k x ~ y ~ z ~  (2xyz3 + y 2 z 2 ) i + (x2 z3 + 2xyz2 ) j ~ ~ + (3x2 yz2 + 2xy2 z) k . ~ At P  (1,3,2), we have   (2(1)(3)(2)3 + (3)2 (2)2 ) i + ((1)2 (2)3 + 2(1)(3)(2)2 ) j ~ ~ + (3(1)2 (3)(2)2 + 2(1)(3)2 (2))k . ~  84i + 32 j + 72k . ~ ~ ~ Example.2 If   x3 yz+ xy2 z3, determine grad  at point P  (1,2,3). Given   x3 yz + xy2 z3 , then    x    y    z  Grad      At P  (1,2,3),  126i +111 j +110k . ~ ~ ~ Grad Properties

If A and B are two scalars, then

1) (A+ B)  A+ B 2) (AB)  A(B) + B(A)

Divergence of a Vector

If A  ax i + ay j + az k, the divergence of A is ~ ~ ~ ~ ~ defined as div A  . A ~ ~        i + j + k .(ax i + ay j + az k)  x ~ y ~ z ~  ~ ~ ~ a a a  div A  . A  x + y + z . ~ ~ x y z

Example.1

If A  x2 y i - xyz j + yz2 k, ~ ~ ~ ~ determine div A at point (1,2,3). ~ Answer a a a div A  . A  x + y + z ~ ~ x y z  2xy- xz + 2yz. At point (1,2,3), div A  2(1)(2) - (1)(3) + 2(2)(3) ~ 13. Example.2

If A  x3 y 2 i + xy2 z j - yz3 k, ~ ~ ~ ~ determine div A at point (3,2,1). ~ a a a Answer div A  . A  x + y + z ~ ~ x y z   At point (3,2,1), div A   ~  114. Remarks

A is a vector function, but div A is a scalar function. ~ ~ If div A  0, vector A is called solenoidvector. ~ ~ Curl of a Vector

If A  ax i + ay j + az k, the curl of A is defined by ~ ~ ~ ~ ~ curl A   A ~ ~        i + j + k (ax i + ay j + az k)  x ~ y ~ z ~  ~ ~ ~ i j k ~ ~ ~     curl A   A  . ~ ~ x y z

ax ay az

Example 1

If A  (y 4 - x2 z 2 ) i + (x2 + y 2 ) j - x2 yzk, ~ ~ ~ ~ determine curl A at (1,3,-2). ~

Solution i j k ~ ~ ~    curl A   A  ~ ~ x y z y 4 - x2 z 2 x2 + y 2 - x2 yz

  2  2 2    (-x yz) - (x + y ) i  y z  ~     -  (-x2 yz) - (y4 - x2 z 2 ) j  x z  ~

  2 2  4 2 2  +  (x + y ) - (y - x z )k  x y  ~  -x2 z i - (-2xyz+ 2x2 z) j + (2x - 4y3 ) k . ~ ~ ~