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= 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 […]
N – 2.2 – Product of two vectors – Exercise Problems with solutions Read More »
The given triangle is an isosceles triangle.
N – 2.1 – Vector Introduction – Exercise Problems with Solutions Read More »
7. Show that the points whose position vectors 8. Prove that the points
2.1 – N – Vector Introduction – Exercise Problems Read More »
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. 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
1.1 – N – Analytical Geometry – II Exercise Problems Read More »
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
Diploma Continuous Assessment Test -2/ April – 2021 Read More »
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 )
DEFINITE INTEGRALS Read More »
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
BERNOULLI’S FORMULA Read More »
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
4.1 INTEGRATION BY PARTS Read More »
List of Formulae: Put u = 3x – 2 Exercise Problems
3.3 STANDARD INTEGRALS Read More »
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
3.2 METHODS OF INTEGRATION – INTEGRATION BY SUBSTITUTION Read More »
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
3.1 INTEGRATION – DECOMPOSITION METHOD (Text) Read More »
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: =
APPLICATION OF VECTOR DIFFERENTIATION Read More »
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.
VECTOR DIFFERENTIATION Read More »
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 (
3.1 PRODUCT OF THREE AND FOUR VECTORS Read More »
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
2.3 APPLICATION OF SCALAR AND VECTOR PRODUCT Read More »
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) +
2.2 PRODUCT OF VECTORS Read More »
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
2.1 VECTOR – INTRODUCTION Read More »
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.
EQUATION OF CIRCLE Definition: A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle. Equation of the circle
1.1 ANALYTICAL GEOMETRY II Read More »
UNIT – I ANALYTICAL GEOMETRY 1.1 ANALYTICAL GEOMETRY I Straight Line: When a variable point moves in accordance with a geometrical law, the point will trace some curve. This curve is known as the locus of the variable point. If a relation in x and y represent a curve then (i) The co-ordinates of every
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