Calculus & analytic geometry

130 Pages·2013·1.3 MB·English

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
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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
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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
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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
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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
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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
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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
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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
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