Calculus & analytic geometry B Sc. MATHEMATICS 2011 Admission onwards IV SEMESTER CORE COURSE UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION CALICUT UNIVERSITY.P.O., MALAPPURAM, KERALA, INDIA – 673 635 352 School of Distance Education UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION STUDY MATERIAL B Sc. MATHEMATICS 2011 Admission onwards IV Semester CORE COURSE CALCULUS & ANALYTIC GEOMETRY Prepared by: Sri.Nandakumar.M., Assistant Professor, Dept. of Mathematics, N.A.M. College, Kallikkandy. Layout & Settings Computer Section, SDE © Reserved Calculus and Analytic Geometry Page2 School of Distance Education CONTENTS Module I 1 Natural Logarithms 5 2 The Exponential Function 12 x 3 a and log x 18 a 4 Growth and Decay 24 5 L’Hopital’s Rule 29 6 Hyperbolic Functions 38 Module II 7 Sequences 45 8 Theorems for Calculating Limits of Sequences 50 9 Series 53 10 Alternating Series 67 Module III 11 Power series 76 12 Taylor and Maclaurin Series 82 13 Convergence of Taylor series- Error Estimate 88 Module IV 14 Conic Sections and Quadratic Equations 91 15 Classifying Conic Section by Eccentricity 95 16 Quadratic Equations and Rotations 98 17 Parametrization of Plane Curves 102 18 Calculus with Parametrized Curves 105 19 Polar Coordinates 110 20 Graphing in Polar Coordinates 115 21 Polar Equations for Conic Sections 118 22 Area of Polar Curves 124 23 Length of Polar Curves 127 24 Area of Surface of Revolution 129 Calculus and Analytic Geometry Page3 School of Distance Education Calculus and Analytic Geometry Page4 School of Distance Education MODULE I CHAPTER 1: NATURAL LOGARITHMS x1 The natural logarithm of a positive number x is the value of the integral  dt . It is 1 t written as lnx . i.e.,, x1 lnx   dt, x 0 …(1) 1 t Remarks 1. If x 1, then lnx is the area under the curve y 1/t from t 1 to t  x . 0  x 1 2. For , lnx gives the negative of the area under the curve from xto 1. 11 3. For x 1, ln1 dt0, as upper and lower limits equal. t 1 x  0 4. The natural logarithm function is not defined for . The Derivative of y = ln x Usingthe first part of the Fundamental Theorem of Calculus,for every positive value ofx, d d x1 1 lnx   dt  . dx dx 1 t x If u is a differentiable function of x whose values are positive, so that ln u is defined, then applying the Chain Rule dy dy du  dx du dx y  lnu to the function ( with u 0)gives d d du lnu lnu dx du dx d 1 du or simply lnu dx u dx Problem Evaluate 1n(x2 1). Solution d 1 d Using Eq.(1), with u  x2 1, ln(x2 1)  (x2 1) dx x2 1 dx 1 2x  2x x2 1 x2 1 Calculus and Analytic Geometry Page5 School of Distance Education Properties of Logarithms For any numbers a0andx0, 1. lnaxlnalnx (Product Rule) a 2. ln lnalnx (Quotient Rule) x 1 3. ln lnx (Reciprocal Rule) x 4. lnxn nlnx (Power Rule) Theorem lnax lnalnx. Proof We first note that lnaxand lnx have the same derivative. Using Corollary to the Mean Value Theorem, then, the functions must differ by a constant, which means that lnaxlnxC for some C. It remains only to show that C equalslna. Equation holds for all positive values of x, so it must in particular hold for x1. Hence, ln(a1)ln1C lna0C, since ln10 C lna Hence, substituting C lna, lnaxlnalnx a Theorem ln lnalnx. x Proof We use lnaxlnalnx Withareplaced by 1/x gives 1 1  ln lnxln x x x  =ln10, hence 1 ln lnx x 1 x replacedby gives x a  1 1 ln lna lnaln x  x x lnalnx Calculus and Analytic Geometry Page6 School of Distance Education Theorem lnxn nlnx (assuming n rational). Proof: For all positive value of x, d 1 d lnxn  (xn), using Eq. (1) with uxn dx xn dx 1  nxn1, here is where we need n to be rational. xn 1 d n  (nlnx) x dx Since ln xn and nlnx have the same derivative, by corollary to the Mean Value Theorem, lnxn nlnxC for some constant C. Taking x1, we obtain ln1nln1C or C 0.Hence the proof. The Graph and Range of ln x d 1 The derivative (lnx) is positive for x0, so lnx is an increasing function of x. The dt x second derivative, 1/x2, is negative , so the graph of lnx is concave down. We can estimate ln2 by numerical integration to be about 0.69 and, obtain 1 n ln2n nln2n  2 2 1 n and ln2n nln2n  . 2 2 Hence, it follows that limlnx and limlnx. x x The domain of ln xis the set of positive real numbers; the range is the entire real line. Logarithmic Differentiation The derivatives of positive functions given by formulas that involve products, quotients, and powers can often be found more quickly if we take the natural logarithm of both sides before differentiating. This enables us to use the properties of natural logarithm to simplify the formulas before differentiating. The process, called logarithmic differentiation, is illustrated in the coming examples. dy Problem Find where y (sinx)cosx 0 dx Solution Given y (sinx)cosx. Taking logarithms on both sides, we obtain ln y cosxlnsinx Calculus and Analytic Geometry Page7 School of Distance Education Now differentiating both sides with respect to x, we obtain d d d d lny (cosxlnsinx) (cosx)lnsinxcosx (lnsinx) dx dx dx dx 1 d sinxlnsinxcosx (sinx). sinx dx 1 dy i.e., cotxcosxsinxlnsinx y dx dy   ycotxcosxsinxlnsinx dx dy i.e., (sinx)cosxcotxcosxsinxlnsinx. dx dy x2 x1 Problem Find , where y  . dx x2 x1 1/2 Solution Given yx2x1  2  x x1 Taking logarithms on both sides, we get lny = ½ [ln(x2 + x + 1)  ln(x2  x+ 1)]. Now differentiating both sides with respect to x,we obtain 1 dy 1 1 d   1 1 d      x2x1  x2x1. y dx 2 x2x1 dx 2 x2x1dx 1 1 1 1        2x1  2x1. 2 x2x1 2 x2x1 dy  2x1 2x1   y     dx 2 x2x1 2 x2x1 dy 1x2 or  . dx x2 x11/2x2 x13/2 The Integral (1/u)du Ifu is a nonzero differentiable function, 1  duln u C. u Proof Whenu is a positive differentiable function, Eq. (1) leads to the integral formula 1  dulnuC , u Ifu is negative, then u is positive and Calculus and Analytic Geometry Page8 School of Distance Education 1 1  du d(u) u (u) ln(u)C We can combine the above equations into a single formula by noticing that in each case the expression on the right is ln u C. Proof. lnuln u because u 0; ln(u)ln u because u0. Hence whether u is positive or negative, the integral of (1/u)du is ln u C. This completes the proof. We recall that un1 undu C, n1. n1 The case of n1is given in Eq. (9).Hence, un1 C, n1  n1 undu  1n|u|, n1  Integration Using Logarithms Integrals of a certain form lead to logarithms. That is, f (x)  dx ln f (x) C f (x) whenever f(x) is a differentiable function that maintains a constant sign on the domain given for it. 2 2x Problem Evaluate dx. 0 x2 5 Answer 2 2x 1du 1  dx ,lettingu x2 5, du2xdx, u(0)5, u(2)1 ln u 0 x2 5 5 u 5 ln 1ln 5 ln1ln5ln5. /2 4cos Problem Evaluate  d. /232sin Calculus and Analytic Geometry Page9 School of Distance Education Solution /2 4cos 52  d  du,takingu32sin ,du2cosd, /232sin 1 u u( /2)1, u( /2)5. 5 2ln u  1 2ln 5 2ln1 2ln5. The Integrals of tan x and cot x Problem Evaluate tanxdx and cotxdx Answer sinx du (i)tanxdx dx , taking ucosx, dusinxdx. cosx u du  ln u C, using Eq.(9) u 1 ln cosx C ln C, by Reciprocal Rule cosx ln secx C. cosxdx du (ii)cotxdx  , taking u sinx, du cosxdx sinx u ln u C ln sinx C ln cscx C. In general, we have tanuduln cosu C ln secu C cotuduln sinu C ln cscu C /6 Problem Evaluate  tan2xdx. 0 Answer /6 /3 du  tan2xdx tanu , taking u2x, dxdu/2, u(0)0, 0 0 2 u( /6) /3 1 /3   tanudu 2 0 1 /3  ln secu  2 0 1 1  (ln2ln1) ln2. 2 2 Calculus and Analytic Geometry Page10