Chapter 1 Mathematics In Physics PDF

60 2 Mathematics In Physics E3 0.1 Introduction. Mathematics is the language of physics. It becomes very easier to describe, understand and apply the physical principles, if we have a good knowledge of mathematics. ID For example : Newton’s law of gravitation states that every body in this universe...

60 2 Mathematics In Physics E3 0.1 Introduction. Mathematics is the language of physics. It becomes very easier to describe, understand and apply the physical principles, if we have a good knowledge of mathematics. ID For example : Newton’s law of gravitation states that every body in this universe attracts every other body with a force which is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them. m 1m 2 U This law can be expressed by a single mathematical relationship F  r 2 or F  Gm 1 m 2 r2 D YG The techniques of mathematics such as algebra, trigonometry, calculus, graph and logarithm can be used to make predictions from the basic equation. If we are poor at grammar and vocabulary, it would be difficult for us to communicate our feelings, similarly for better understanding and expressing of physics the basic knowledge of mathematics is must. In this introductory chapter we will learn some fundamental mathematics. 0.2 Algebra. U (1) Quadratic equation : An equation of second degree is called a quadratic equation. Standard quadratic equation ax 2  bx  c  0 ST Here a is called the coefficient of x2, b is called the coefficient of x and c is a constant term, x is the variable whose value (roots of the equation) are to be determined Roots of the equation are : x   b  b 2  4 ac 2a This formula can be written as x  Coefficien t of x  (Coefficien t of x ) 2  4 (Coefficien t of x 2 )  (Constant term) 2(Coefficien t of x 2 ) Note :  If  and  be the roots of the quadratic equation then Mathematics In Physics 3 Sum of roots  +  = – b c and product of roots = a a Solve the equation 10 x 2  27 x  5  0 Solution : By comparing the given equation with standard equation a = 10, b = – 27, and c = 5  (27 )  (27 )2  4  10  5 27  23  b  b 2  4 ac   2a 2  10 20  x1  27  23 1 27  23 5 and x 2    20 5 20 2  Roots of the equation are 5 1 and. 2 5 E3 x 60 Problem 1. (1  x )n  1  nx  ID (2) Binomial theorem : If n is any number positive, negative or fraction and x is any real number, such that x < 1 i.e. x lies between – 1 and + 1 then according to binomial theorem n(n  1) 2 n(n  1)(n  2) 3 x  x ..... 2! 3! U Here 2 ! (Factorial 2) = 2  1, 3 ! (Factorial 3) = 3  2  1 and 4 ! (Factorial 4) = 4  3  2  1 Note :  If |x|