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INTERNAL ASSESSMENT – I FOR ENGINEERING MATHEMATICS-II
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N-ENGINEERING MATHEMATICS – II
SOLUTIONS TO ASSIGNMENT – I FOR ENGINEERING MATHEMATICS – II
Soln: Given x2 + y2 – 4x + 6y + 8 = 0 ––––– (1) and x2 + y2 – 10x – 6y +14 = 0 ––––– (2) From ( 1 ) 2g1 = -4 2f1 = 6 c1 = 8 g1 = -2 f1= 3 centre is C1 = ( – g1 ,
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N-SCHEME-ENGINEERING MATHEMATICS-II MODEL EXAM QUESTION PAPER/2021
MODEL EXAM QUESTION PAPER 40022 ENGINEERING MATHEMATICS – II Time : 3.00 Hours Date: 22-05-2021 Max.Marks: 100 Note: 1. Answer all question in PART A. Each question carries one mark. 2. Answer any ten questions in PART B. Each question carries two marks. 3. Answer all question by selecting either A or B. Each question
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N – 2.2 – Product of two vectors – Exercise Problems with solutions
= 1(1) + 1(1) + 0 = 1 + 1 = 2 2(p) + 1(3) – 5 (-2) = 0 2p + 3 +10 = 0 2p + 13 = 0 Soln: = 2 ( 1 ) + 1 ( -4) + 3 ( – 6 ) = 2 – 4 – 18 = – 20
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N – 2.1 – Vector Introduction – Exercise Problems with Solutions
The given triangle is an isosceles triangle.
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2.1 – N – Vector Introduction – Exercise Problems
7. Show that the points whose position vectors 8. Prove that the points
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1.1 – N – Analytical Geometry – II Exercise Problems With Solutions
From the two given equations of the circles, we observe that the constant term alone differs ∴The given circles are concentric circles. The required equation of the circle is x2 + y2 + 4x + 6y – 12 = 0
1.1 – N – Analytical Geometry – II Exercise Problems With Solutions Read More »
1.1 – N – Analytical Geometry – II Exercise Problems
1. Find the equation of the circle with centre (1, -2) and radius 5 units. 2. Find the centre and radius of the circle x2 + y2 + 10x + 8y + 5 = 0 . 9. Find the equation of the circle passing through the point A (2, 3) and having its centre at C
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Diploma Continuous Assessment Test -2/ April – 2021
Course: First year Diploma course in Engineering & Technology Subject & Code : Engineering Mathematics – II (40022) Time : 2 Hours Date: 27-04-2021 Max. Marks: 50 PART – A Answer all questions ( 6×1=6 marks) Soln: = 1 ( 1 – 0 ) – 1 ( 0 – 1 ) + 0 ( 0
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DEFINITE INTEGRALS
Definition of Definite Integrals: Example 1: Example 2: Example 4: Example 5: Example 6: Example 7: Example 8: Example 9: Example 10: Example 11: Example 12: Example 13: du = cos x dx Example 14: Adding ( 1 ) & ( 2 )
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BERNOULLI’S FORMULA
If u and v are functions x, then Bernoulli’s form of integration by parts formula is Where u΄, u΄΄,u΄΄΄….. are successive differentiation of the function u and v, v1, v2, v3, …………. the successive integration of the function dv. Note: The function ‘u’ is differentiated up to constant. Example 1: Example 2: Example 3: Example
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4.1 INTEGRATION BY PARTS
Introduction: When the integrand is a product of two functions and the method of decomposition or substitution can not be applied, then the method of by parts is used. Integraiton by parts formula: The above formula is used by taking proper choice of ‘u’ and ‘dv’. ‘u’ should be chosen based on thefollowing order of
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3.3 STANDARD INTEGRALS
List of Formulae: Put u = 3x – 2 Exercise Problems
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3.2 METHODS OF INTEGRATION – INTEGRATION BY SUBSTITUTION
So far we have dealt with functions, either directly integrable using integration formula (or) integrable after decomposing the given functions into sums & differences. which cannot be decomposed into sums (or) differences of simple functions. In these cases, using proper substitution, we shall reduce the given form into standard form, which can be integrated using
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3.1 INTEGRATION – DECOMPOSITION METHOD (Text)
Sir Sardar Vallabhai Patel, called the Iron Man of India integrated several princely states together while forming our country Indian Nation after independence. Like that in Maths while finding area under a curve through integration, the area under the curve is divided into smaller rectangles and then integrating (i.e) summing of all the area of
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APPLICATION OF VECTOR DIFFERENTIATION
is a vector function, defined and differentiable at each point (x, y, z)in a certain region of space [i.e., A defines a vector field], then the divergence of (abbreviated as ‘Div ‘) is defined as, Basic properties of Divergence: If A, B are vector functions and ‘f’ is a scalar function, then Example: Soln: =
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VECTOR DIFFERENTIATION
Vector point function and Vector field: Let P be any point in a region ‘D’ of space. Let r be the position vector of P. If there exists a vector function F corresponding to each P, then such a function F is called a vector function and the region D is called a vector field.
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3.1 PRODUCT OF THREE AND FOUR VECTORS
Scalar triple Product Properties of Scalar triple Product: Example : Soln: = 1 ( 1 – 0 ) – 1 ( 0 – 1 ) + 0 ( 0 – 1 ) = 1 ( 1 ) -1 (- 1 ) + 0 = 1 + 1 = 2 Example : Soln: = 2 (
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2.3 APPLICATION OF SCALAR AND VECTOR PRODUCT
Application of Scalar Product Work done PART – B Example : Soln: = 3 ( 2 ) + 5 ( – 1 ) + 7 ( 1 ) = 6 – 5 + 7 Work done = 8 units Example : Soln: = 4 ( 2 ) + 9 ( – 7 ) + 4
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2.2 PRODUCT OF VECTORS
SCALAR PRODUCT Definition: Properties of Scalar Product: = 2(1) – 4 (6) + 8 (12) = 2 – 24 + 96 = 74 3(-6) – 1(m) + 5 (4) = 0 -18 – m + 20 = 0 -m + 2 = 0 = 1(0) – 1(4) + 2 (2) = 0 = 0(-10) +
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2.1 VECTOR – INTRODUCTION
Vectors constitute one of the several Mathematical systems which can be usefully employed to provide mathematical handling for certain types of problems in Geometry, Mechanics and other branches of Applied Mathematics. Vectors facilitate mathematical study of such physical quantities as possess Direction in addition to Magnitude. Velocity of a particle, for example, is one such
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1.2 CONICS
Conic: A conic is defined as the locus of a point which moves such that its distance from a fixed point is always ‘e’ times its distance from a fixed straight line. Focus: The fixed point is called the focus of the conic. Directrix: The fixed straight line is called the directrix of the conic.
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