∞

|ak | converges to the ratio test, since

Part (a) If L < 1, then the series

k=1

|an+1 | an+1

= L < 1.

lim = lim

n’∞ |an | an

n’∞

∞

Hence, the series ak converges absolutely.

k=1

∞

|ak | diverges by the ratio test if L > 1,

Part (b) As in Part (a), the series

k=1

since

|an+1 | an+1

= L > 1.

= lim

lim

n’∞ |an | an

n’∞

∞

|ak | converges by the root test, since

Part (c) If M < 1, then the series

k=1

1/n

lim |ak | = M < 1.

n’∞

∞

|ak | diverges by the root test as in

Part (d) If M > 1, then the series

k=1

Part (c).

∞ ∞

∞

1 1

1

and , L = M = 1, but diverges

Part (e) For the series

k2

k k

k=1 k=1

k=1

∞

1

and converges by the p-series test. Thus, L = 1 and M = 1 fail to

k2

k=1

determine convergence or divergence.

346 CHAPTER 8. INFINITE SERIES

This completes the proof of Theorem 8.5.3.

Exercises 8.3 Determine the region of convergence of the following series.

∞ ∞

(’1)n xn (’1)n (x + 2)n

71. 72.

3n n2

2n

n=1 n=1

∞ ∞

(’1)n (x ’ 1)n (’1)n n!(x ’ 1)n

73. 74.

5n

n!

n=1 n=1

∞ ∞

(x + 2)n

nn

75. (’2) x 76.

2n n2

n=0 n=1

∞ ∞

+ 1)n (’1)n (x ’ 3)n

n (x

77. 78.

(’1)

3n n3 n3/2

n=1 n=1

∞ ∞

(2x)n (’1)n xn

79. 80.

n! (2n)!

n=1 n=1

∞ ∞

(n + 1)!(x ’ 1)n (’1)n (2n)!xn

81. 82.

4n n!

n=1 n=1

∞ ∞

(’1)n n!(x ’ 1)n

2 n

83. n (x + 1) 84.

1 · 3 · · · 5 · · · (2n + 1)

n=1 n=1

∞ ∞

(’1)n (n!)2 (x ’ 1)n (’1)n 3n xn

85. 86.

3n (2n)! 23n

n=1 n=1

∞ ∞

(’1)n (x + 1)n ln(n + 1)2n (x + 1)n

87. 88.

(n + 1) ln(n + 1) n+2

n=1 n=1

∞ ∞

(’1)n (ln n)3n xn (’1)n 1 · 3 · 5 · · · (2n + 1) n

89. 90. x

4n n2 2 · 4 · 6 · · · (2n + 2)

n=1 n=1

8.6. POWER SERIES 347

8.6 Power Series

De¬nition 8.6.1 If a0 , a1 , a2 , . . . is a sequence of real numbers, then the

series ∞ ak xk is called a power series in x. A positive number r is called

k=1

the radius of convergence and the interval (’r, r) is called the interval of

convergence of the power series if the power series converges absolutely for

all x in (’r, r) and diverges for all x such that |x| > r. The end point x = r is

included in the interval of convergence if ∞ ak rk converges. The end point

k=1

x = ’r is included in the interval of convergence if the series ∞ (’1)k ak rk

k=1

converges. If the power series converges only for x = 0, then the radius of

convergence is de¬ned to be zero. If the power series converges absolutely

for all real x, then the radius of convergence is de¬ned to be ∞.

∞

cn xn converges for x = r = 0, then the

Theorem 8.6.1 If the series

n=1

∞

cn xn converges absolutely for all numbers x such that |x| < |r|.

series

n=0

∞

cn rn converges. Then, by the Divergence Test,

Proof. Suppose that

n=0

lim cn rn = 0.

n’∞

For = 1, there exists some natural number m such that for all n ≥ m,

|cn rn | < = 1.

Let

M = max{|cn rn | + 1 : 1 ¤ n ¤ m}.

Then, for each x such that |x| < |r|, we get |x/r| < 1 and

∞ ∞

x n

n n

|cn x | = |cn r | ·

r

n=0 n=0

∞

x n

¤ M

r

n=0

M

< ∞.

=

1’ x

r

348 CHAPTER 8. INFINITE SERIES

∞

|cn xn | converges for x such that |x| <

By the comparison test the series

n=0

|r|. This completes the proof of Theorem 8.6.1.

∞

cn (x ’ a)n converges for some x ’ a =

Theorem 8.6.2 If the series

n=0

∞

cn (x ’ a)n converges absolutely for all x such that

r = 0, then the series

n=0

|x ’ a| < |r|.

∞

cn un converges for some u = r. Then

Proof. Let x’a = u. Suppose that

n=0

∞

cn un converges absolutely for all u such that

by Theorem 8.6.1, the series

n=0

∞

cn (x ’ a)n converges absolutely for all

|u| < |r|. It follows that the series

n=0

x such that |x ’ a| < |r|. This completes the proof of the theorem.

∞

cn xn be any power series. Then exactly one of the

Theorem 8.6.3 Let

n=0

following three cases is true.

(i) The series converges only for x = 0.

(ii) The series converges for all x.

(iii) There exists a number R such that the series converges for all x with

|x| < R and diverges for all x with |x| > R.

Proof. Suppose that cases (i) and (ii) are false. Then there exist two

∞ ∞

n

cn q n diverges.

nonzero numbers p and q such that cn p converges and

n=0 n=0

By Theorem 8.6.1, the series converges absolutely for all x such that |x| < |p|.

Let

∞

cn pn converges}.

A = {p :

n=0

8.6. POWER SERIES 349

The set A is bounded from above by q. Hence A has a least upper bound, say

R. Clearly |p| ¤ R < q and hence R is a positive real number. Furthermore,

∞

cn xn converges for all x such that |x| < R and diverges for all x such that

n=0

|x| > R. We de¬ne R to be 0 for case (i) and R to be ∞ for case (ii). This

completes the proof of Theorem 8.6.3.

∞

cn (x ’ a)n be any power series. Then exactly one

Theorem 8.6.4 Let

n=0

of the following three cases is true:

(i) The series converges only for x = a and the radius of convergence is 0.

(ii) The series converges for all x and the radius of convergence is ∞.

(iii) There exists a number R such that the series converges for all x such

that |x ’ a| < R and diverges for all x such that |x ’ a| > R.

∞

cn un . The

Let u = x ’ a and use Theorem 8.6.3 on the series

Proof.

n=0

details of the proof are left as an exercise.

∞

cn rn converges for |x| < R, then

Theorem 8.6.5 If R > 0 and the series

n=0

∞ ∞

ncn xn’1 , obtained by term-by-term di¬erentiation of c n xn ,

the series

n=1 n=0

converges absolutely for |x| < R.

Proof. For each x such that |x| < R, choose a number r such that |x| < r <

∞

cn xn converges, lim cn rn = 0 and hence {cn rn }∞ is bounded.

R. Then n=0

n’∞

n=0

There exists some M such that |cn rn | ¤ M for each natural number n. Then

∞ ∞

1x n’1

n’1 n

n|cn r | · ·

|ncn x |=

rr

n=1 n=1

∞

M x n’1

¤ .

n

r r

n=1

350 CHAPTER 8. INFINITE SERIES

∞

x

x n’1

converges by the ratio test, since < 1. It follows

The series n

r r

n=1

∞

ncn xn’1 converges absolutely for all x such that |x| < R. This

that

n=1

completes the proof of this theorem.

∞

cn (x ’ a)n converges for all x

Theorem 8.6.6 If R > 0 and the series

n=0

∞

cn (x’a)n may be di¬erentiated with

such that |x’a| < R, then the series

n=0

respect to x any number of times and each of the di¬erential series converges

for all x such that |x ’ a| < R.

∞

cn un converges for all u such that |u| < R. By

Proof. Let u = x’a. Then

n=0

∞

ncn un’1 converges for all u such that |u| < R.

Theorem 8.6.5, the series

n=1

This term-by-term di¬erentiation process may be repeated any number of

times without changing the radius of convergence. This completes the proof

of this theorem.

Theorem 8.6.7 Suppose that R > 0 and f (x) = ∞ cn xn and R is radius

n=0

∞ n

of convergence of the series n=0 cn x . Then f (x) is continuous for all x

such that |x| < R.

Proof. For each number c such that ’R < c < R, we have

∞

xn ’ c n

f (x) ’ f (c)

= cn

x’c x’c

n=0

∞

cn nan’1

= n

n=1

∞

n cn an’1

¤ n

n=1

8.6. POWER SERIES 351

for some an between c and x, for each natural number n, by the Mean Value

∞

n |cn an |n’1 converges. Hence,

Theorem. By Theorem 8.6.6, the series

n=1

∞

n cn an’1

lim |f (x) ’ f (c)| = lim |x ’ c| |c0 ’ c| + n

x’c x’c

n=1

∞

n cn an’1

=0· |c0 ’ c| + n

n=1

= 0.

Hence, f (x) is continuous at each number c such that ’R < c < R. This

completes the proof of this theorem.

∞

cn xn and R is the radius

Theorem 8.6.8 Suppose that R > 0, f (x) =

n=0

∞

cn xn . For each x such that |x| < R, we de¬ne

of convergence of the series

n=0

x

F (x) = f (t)dt.

0

Then, for each x such that |x| < R, we get

∞

xn+1

.

F (x) = cn

n+1

n=0

352 CHAPTER 8. INFINITE SERIES

Proof. Suppose that |x| < |r| < R. Then

m

m x x

xn+1

tn

f (t)dt ’

lim F (x) ’ = lim cn

cn

n+1 n’∞

m’∞ 0 0

n=0

n=0

m

x

cn tn

f (t) ’ dt

= lim

m’∞ 0 n=0

∞

x

cn tn dt

= lim

m’∞ 0 n=m+1

∞

x

|cn tn | dt

¤ lim

m’∞ 0 n=m+1

∞

x

|cn rn | dt

¤ lim

m’∞ 0 n=m+1

∞ x

n

¤ lim |cn r | 1 dt

m’∞ 0

n=m+1

= 0 · |x|

= 0,

∞

|cn rn | converges.

since

n=0

It follows that

∞

x x

cn tn dt

f (t)dt =

0 0 n=0

∞

xn+1

= cn .

n+1

n=0

This completes the proof of the this theorem.

∞

cn xn for all |x| < R, where R > 0

Theorem 8.6.9 Suppose that f (x) =

n=0

∞

cn xn . Then f (x) has continuous

is the radius of convergence of the series

n=0

8.6. POWER SERIES 353

derivatives of all orders for |x| < R that are obtained by successive term-by-

∞

c n xn .

term di¬erentiations of

n=0

Proof. For each |x| < R, we de¬ne

∞

ncn xn’1 .

g(x) =

n=1

∞

ncn xn’1 .

Then, by Theorem 8.6.5, R is the radius of convergence of the series

n=1

By Theorem 8.6.7, g(x) is continuous. Hence,

∞

x

cn xn = f (x).

c0 + g(x)dx = c0 +

0 n=1

By the fundamental theorem of calculus, f (x) = g(x). This completes the

proof of this theorem.

De¬nition 8.6.2 The radius of convergence of the power series

∞

ak (x ’ a)k

k=1

is

(a) zero, if the series converges only for x = a;

(b) r, if the series converges absolutely for all x such that |x ’ a| < r and

diverges for all x such that |x ’ a| > r.

(c) ∞, if the series converges absolutely for all real number x.

If the radius of convergence of the power series in (x ’ a) is r, 0 < r < ∞,

then the interval of convergence of the series is (a ’ r, a + r). The end points

x = a + r or x = a ’ r are included in the interval of convergence if the

corresponding series ∞ ak rk or ∞ (’1)k ak rk converges, respectively. If

k=1

k=1

r = ∞, then the interval of convergence is (’∞, ∞).

354 CHAPTER 8. INFINITE SERIES

Exercises 8.4 In problem 1“12, determine the Taylor series expansion for

each function f about the given value of a.

1. f (x) = e’2x , a = 0 2. f (x) = cos(3x), a = 0

4. f (x) = (1 + x)’2 , a = 0

3. f (x) = ln(x), a = 1

5. f (x) = (1 + x)’3/2 , a = 0 6. f (x) = ex , a = 2

π π

7. f (x) = sin x, a = 8. f (x) = cos x, a =

6 4

π

10. f (x) = x1/3 , a = 8

9. f (x) = sin x, a =

3

1 1

11. f (x) = sin x ’ 12. f (x) = cos x ’

,a = 0 ,a = 0

2 2

n

(x ’ a)k

(k)

In problems 13-20, determine .

f (a)

k!

k=0

2

14. f (x) = x2 e’x , a = 0, n = 3

13. f (x) = ex , a = 0, n = 3

1

15. f (x) = , a = 0, n = 2 16. f (x) = arctan x, a = 0, n = 3

1 ’ x2

17. f (x) = e2x cos 3x, a = 0, n = 4 18. f (x) = arcsin x, a = 0, n = 3

20. f (x) = (1 + x)1/2 , a = 0, n = 5

19. f (x) = tan x, a = 0, n = 3

8.7 Taylor Polynomials and Series

Theorem 8.7.1 (Taylor™s Theorem) Suppose that f, f , · · · , f (n+1) are all

continuous for all x such that |x ’ a| < R. Then there exists some c between

a and x such that

f (x) = Pn (x) + Rn (x)

where

n

(x ’ a)n+1

(x ’ a)k (n+1)

(k)

Pn (x) = , Rn (x) = f (c) .

f (a)

k! (n + 1)!

k=0

8.7. TAYLOR POLYNOMIALS AND SERIES 355

The polynomial Pn (x) is called the nth degree Taylor polynomial approxima-

tion of f . The term Rn (x) is called the Lagrange form of the remainder.

Proof. We de¬ne a function g of a variable z such that

f (z)(x ’ z) f (z)(x ’ z)2

g(z) = [f (x) ’ f (z)] ’ ’ ’ ···

1! 2!

f (n) (z)(x ’ z)n (x ’ z)n+1

’ ’ Rn (x) .

(x ’ a)n+1

n!

Then

n

f (k) (a)

(x ’ a)k + Rn (x)

g(a) = f (x) ’ = 0,

k!

k=0

and

g(x) = f (x) ’ f (x) = 0.

By the Mean Value Theorem for derivatives there exists some c between a

and x such that g (c) = 0. But

f (z)(x ’ z)2

’ ···

g (z) = ’f (z) ’ [’f (z) + f (z)(x ’ z)] ’ ’f (z)(x ’ z) +

2!

f n (z)(x ’ z)n’1 f (n+1) (z)(x ’ z)n (n + 1)(x ’ z)n

’’ + + Rn (x)

(x ’ a)n+1

n! n!

(x ’ z)n (n + 1)(x ’ z)n

(n+1)

= ’f (z) + Rn (x)

(x ’ a)n+1

n!

(x ’ c)n (n + 1)(x ’ c)n

(n+1)

g (c) = 0 = ’f (c) + Rn (x) .

(x ’ a)n+1

n!

Therefore,

(x ’ a)n+1 f (n+1) (c) (x ’ a)n+1

(n+1)

·

Rn (x) = =f (c)

n+1 n! (n + 1)!

as required. This completes the proof of this theorem.

Theorem 8.7.2 (Binomial Series) If m is a real number and |x| < 1, then

∞

m(m ’ 1) · · · (m ’ k + 1) k

m

(1 + x) = 1 + x

k!

k=1

m(m ’ 1) 2 m(m ’ 1)(m ’ 2) 3

x + ··· .

= 1 + mx + x+

2! 3!

356 CHAPTER 8. INFINITE SERIES

This series is called the binomial series. If we use the notation

m(m ’ 1) · · · (m ’ k + 1)

m

=

k k!

m

then is called the binomial coe¬cient and

k

∞

mk

(1 + x)m = 1 + x.

k

k=1

If m is a natural number, then we get the binomial expansion

m

mk

m

(1 + x) = 1 + x.

k

k=1

Proof. Let f (x) = (1 + x)m . Then for all natural numbers n,

f (x) = m(1 + x)m’1 , f (x) = m(m ’ 1)(1 + x)m’2 , · · · ,

f (n) (x) = m(m ’ 1) · · · (m ’ n + 1)(1 + x)m’n .

Thus, f (n) (0) = m(m ’ 1) · · · (m ’ n + 1), and

∞

m(m ’ 1)(m ’ 2) · · · (m ’ n + 1) n

x

f (x) =

n!

n=0

∞

mn

= x

n

n=0

m m

= m(m ’ 1) · · · (m ’ n + 1) is called the nth

where = 1 and

0 n

binomial coe¬cient. By the ratio test we get

m(m ’ 1) · · · (m ’ n)xn+1 n!

·

lim

m(m ’ 1) · · · (m ’ n + 1)xn

(n + 1)!

n’∞

m’n

= |x| lim

n+1

n’∞

m

’1

= |x| lim n 1

n’∞ 1 +

n

= |x|,

8.7. TAYLOR POLYNOMIALS AND SERIES 357

and, hence, the series converges for |x| < 1.

This completes the proof of the theorem.

Theorem 8.7.3 The following power series expansions of functions are valid.

∞ ∞

1. (1 ’ x)’1 = 1 + xk and (1 + x)’1 = 1 + (’1)k xk , |x| < 1.

k=1 k=1

∞

∞ k

xk kx

’x

2. ex = 1 + , |x| < ∞.

, e =1+ (’1)

k! k!

k=1

k=1

∞

x2k+1

k

, |x| < ∞.

3. sin x = (’1)

(2k + 1)!

k=0

∞

x2k k

, |x| < ∞.

4. cos x = (’1)

(2k)!

k=0

∞

x2k’1

, |x| < ∞.

5. sinh x =

(2k + 1)!

k=0

∞

x2k

, |x| < ∞.

6. cosh x =

(2k)!

k=0

∞

xk+1 k

, ’1 < x ¤ 1.

7. ln(1 + x) = (’1)

k+1

k=0

∞

x2k+1

1 1+x

, ’1 < x < 1.

8. ln =

1’x

2 2k + 1

k=0

∞

x2k+1 k

, ’1 ¤ x ¤ 1.

9. arctan x = (’1)

2k + 1

k=0

∞ 2k+1

’1/2 kx

, |x| ¤ 1.

10. arcsin x = (’1)

2k + 1

k

k=0

358 CHAPTER 8. INFINITE SERIES

Proof.

Part 1. By the geometric series expansion, for all |x| < 1, we have

∞ ∞

1 1

1

xk (’1)k xk .

=1+ = =1+

and

1’x 1 ’ (’x)

1+x

k=1 k=1

Part 2. If f (x) = ex , then f (n) (x) = ex and f (n) (0) = 1 for each n =

0, 1, 2, · · · . Thus

∞

xn

x

.

e=

n!

n=0

By the ratio test the series converges for all x.

xn+1 n! 1

· n = |x| lim = 0.

lim

n’∞ (n + 1)! x n’∞ n + 1

Part 3. Let f (x) = sin x. Then f (x) = cos x, f (x) = ’ sin x, f (3) (x) =

’ cos x and f (4) (x) = sin x. It follows that, for each n = 0, 1, 2, 3, · · · , we

have

f (4n) (0) = 0, f (4n+1) (0) = 1, f (4n+2) (0) = 0 and f (4n+3) (0) = ’1.

Hence,

x3 x5

’ ···

sin x = x ’ +

3! 5!

∞

x2n+1

n

= .

(’1)

(2n + 1)!

n=0

By the ratio test, the series converges for all |x| < ∞:

x2n+3 (2n + 1)!

n+1

lim (’1)

(2n + 3)! x2n+1

n’∞

1

= x2 lim

n’∞ (2n + 3)(2n + 2)

= 0.

Part 4. By term-by-term di¬erentiation we get

∞

x2n

n

, |x| < ∞.

cos x = (sin x) = (’1)

(2n)!

n=0

8.7. TAYLOR POLYNOMIALS AND SERIES 359

Part 5. For all |x| < ∞, we get

1x

(e ’ e’x )

sinh x =

2

∞ ∞

xn n

1 nx

’

= (’1)

2 n=0 n! n=0 n!

∞

x2n+1

= .

(2n + 1)!

n=0

Part 6. By di¬erentiating term-by-term, we get

∞

x2n

, l |x| < ∞.

cosh x = (sinh x) =

(2n)!

n=0

Part 7. For each |x| < 1, by performing term by integration, we get

x

1

ln(1 + x) = dx

1+x

0

∞

∞

(’1)n xn dx

=

0 n=0

∞

xn+1 n

= (’1) .

n+1

n=0

Part 8. By Part 7, for all |x| < 1, we get

1 1+x 1

[ln(1 + x) ’ ln(1 ’ x)]

ln =

1’x

2 2

∞ ∞

xn+1 n+1

1 n (’x)

n

’

= (’1) (’1)

2 n + 1 n=0 n+1

n=0

∞

(’1)n

1

(1 ’ (’1)n+1 )xn+1

=

2 n+1

n=0

∞

x2k+1

= .

2k + 1

k=0

1 1+x

Recall that arctanh x = ln .

1’x

2

360 CHAPTER 8. INFINITE SERIES

Part 9. For each |x| ¤ 1, we perform term-by-term integration to get

x

1

arctan x = dx

1 + x2

0

∞

x

(’1)k x2k dx

=

0 k=0

∞

x2k+1 k

= (’1) .

(2k + 1)

k=0

Part 10. By performing term-by-term integration of the binomial series, we

get

x

1

√ dx

arcsin x =

1 ’ x2

0

x

(1 ’ x2 )’1/2 dx

=

0

∞

x

’1/2

(’x2 )k dx

=

k

0 k=0

∞

x2k+1

’1/2 k

= (’1) .

k (2k + 1)

k=0

This series converges for all |x| ¤ 1.

This completes the proof of this theorem.

8.8 Applications

Chapter 9

Analytic Geometry and Polar

Coordinates

A double right-circular cone is obtained by rotating a line about a ¬xed axis

such that the line intersects the axis and makes the same angle with the

axis. The intersection point of the line and the axis is called a vertex. A

conic section is the intersection of a plane and the double cone. Some of

the important conic sections are the following: parabola, circle, ellipse and a

hyperbola.

9.1 Parabola

De¬nition 9.1.1 A parabola is the set of all points in the plane that are

equidistant from a given point, called the focus, and a given line called the

directrix. A line that passes through the focus and is perpendicular to the

directrix is called the axis of the parabola. The intersection of the axis with

the parabola is called the vertex.

Theorem 9.1.1 Suppose that v(h, k) is the vertex and the line x = h ’ p is

the directrix of a parabola. Then the focus is F (h + p, k) and the axis is the

horizontal line with equation y = k. The equation of the parabola is

(y ’ k)2 = 4p(x ’ h).

361

362CHAPTER 9. ANALYTIC GEOMETRY AND POLAR COORDINATES

Theorem 9.1.2 Suppose that v(h, k) is the vertex and the line y = k ’ p is

the directrix of a parabola. Then the focus is F (h, k + p) and the axis is the

vertical line with equation x = h. The equation of the parabola is

(x ’ h)2 = 4p(y ’ k).

9.2 Ellipse

De¬nition 9.2.1 An ellipse is the locus of all points, the sum of whose

distances from two ¬xed points, called foci, is a ¬xed positive constant that

is greater than the distance between the foci. The midpoint of the line

segment joining the two foci is called the center. The line segment through

the foci and with end points on the ellipse is called the major axis. The

line segment, through the center, that has end points on the ellipse and is

perpendicular to the major axis is called the minor axis. The intersections

of the major and minor axes with the ellipse are called the vertices.

Theorem 9.2.1 Let an ellipse have center at (h, k), foci at (h ± c, k), ends

of the major axis at (h ± a, k) and ends of the minor axis at (h, k ± b), where

a > 0, b > 0, c > 0 and a2 = b2 + c2 . Then the equation of the ellipse is

(x ’ h)2 (y ’ k)2

+ = 1.

a2 b2

The length of the major axis is 2a and the length of the minor axis is 2b.

Theorem 9.2.2 Let an ellipse have center at (h, k), foci at (h, k ± c), ends

of the major axis at (h, k ± a), and the ends of the minor axis at (h ± b, k),

where a > 0, b > 0, c > 0 and a2 = b2 + c2 . Then the equation of the ellipse

is

(y ’ k)2 (x ’ h)2

+ = 1.

a2 b2

The length of the major axis is 2a and the length of the minor axis is 2b.

Remark 24 If c = 0, then a = b, foci coincide with the center and the

ellipse reduces to a circle.

9.3. HYPERBOLA 363

9.3 Hyperbola

De¬nition 9.3.1 A hyperbola is the locus of all points, the di¬erence of

whose distances from two ¬xed points, called foci, is a ¬xed positive constant

that is less than the distance between the foci. The mid point of the line

segment joining the two foci is called the center. The line segment, through

the foci, and with end points on the hyperbola is called the major axis. The

end points of the major axis are called the vertices.

Theorem 9.3.1 Let a hyperbola have center at (h, k), foci at (h ± c, k),

√

vertices at (h ± a, k), where 0 < a < c, b = c2 ’ a2 , then the equation of

the hyperbola is

(x ’ h)2 (y ’ k)2

’ = 1.

a2 b2

Theorem 9.3.2 Let a hyperbola have center at (h, k), foci at (h, k ± c),

√

vertices at (h, k ± a), where 0 < a < c, b = c2 ’ a2 , then the equation of

the hyperbola is

(y ’ k)2 (x ’ h)2

’ = 1.

a2 b2

9.4 Second-Degree Equations

De¬nition 9.4.1 The transformations

x = x cos θ ’ y sin θ

y = x sin θ + y cos θ

and

x = x cos θ + y sin θ

y = ’x sin θ + y cos θ

are called rotations. The point P (x, y) has coordinates (x , y ) in an x y -

coordinate system obtained by rotating the xy-coordinate system by an angle

θ.

Theorem 9.4.1 Consider the equation ax2 + bxy + cy 2 + dx + ey + f =

0, b = 0. Let cot 2θ = (a ’ c)/b and x y -coordinate system be obtained

364CHAPTER 9. ANALYTIC GEOMETRY AND POLAR COORDINATES

through rotating the xy-coordinate system through the angle θ. Then the

given second degree equation

ax2 + bxy + cy 2 + dx + ey + f = 0

becomes

ax2+cy2+dx+ey+f =0

where

= a cos2 θ + b cos θ sin θ + c sin2 θ

a

= a sin2 θ ’ b sin θ cos θ + c cos2 θ

c

d = d cos θ + e sin θ

= ’d sin θ + e cos θ

e

f =f

Furthermore, the given second degree equation represents

(i) an ellipse, a circle, a point or no graph if b2 ’ 4ac < 0;

(ii) a hyperbolic or a pair of intersecting lines if b2 ’ 4ac > 0;

(iii) a parabola, a line, a pair of parallel lines, or else no graph if b2 ’4ac = 0.

9.5 Polar Coordinates

De¬nition 9.5.1 Each point P (x, y) in the xy-coordinate plane is assigned

the polar coordinates (r, θ) that satisfy the following relations:

x2 + y 2 = r2 , y = r cos θ, y = r sin θ.

The origin is called the pole and the positive x-axis is called the polar axis.

The number r is called the radial coordinate and the angle θ is called the

angular coordinates. The polar coordinates of a point are not unique as the

rectangular coordinates are. In particular,

(r, θ) ≡ (r, θ + 2nπ) ≡ (’r, θ + (2m + 1)π)

where n and m are any integers. There does exist a unique polar represen-

tation (r, θ) if r ≥ 0 and 0 ¤ θ < 2π.

9.6. GRAPHS IN POLAR COORDINATES 365

9.6 Graphs in Polar Coordinates

Theorem 9.6.1 A curve in polar coordinates is symmetric about the

(a) x-axis if (r, θ) and (r, ’θ) both lie on the curve;

(b) y-axis if (r, θ) and (r, π ’ θ) both lie on the curve;

(c) origin if (r, θ), (r, θ + π) and (’r, θ) all lie on the curve.

Theorem 9.6.2 Let e be a positive number. Let a ¬xed point F be called the

focus and a ¬xed line, not passing through the focus, be called a directrix. If

P is a point in the plane, let P F stand for the distance between P and the

focus F and let P D stand for the distance between P and the directrix. Then

the locus of all points P such that P F = eP D is a conic section representing

(a) an ellipse if 0 < e < 1;

(b) a parabola if e = 1;

(c) a hyperbola if e > 1;

The number e is called the eccentricity of the conic.

In particular an equation of the form

ek

r=

1 ± e cos θ

represents a conic with eccentricity e, a focus at the pole (origin), and a

directrix perpendicular to the polar axis and k units to the right of the pole,

in the case of + sign, and k units to the left of the pole, in the case of ’

sign.

Also, an equation of the form

ek

r=

1 ± e sin θ

represents a conic with eccentricity e, a focus at the pole, and a directrix

parallel to the polar axis and k units above the pole, in the case of + sign,

and k units below the pole, in the case of ’ sign.

366CHAPTER 9. ANALYTIC GEOMETRY AND POLAR COORDINATES

9.7 Areas in Polar Coordinates

Theorem 9.7.1 Let r = f (θ) be a curve in polar coordinates such that f is

continuous and nonnegative for all ± ¤ θ ¤ β where ± ¤ β ¤ 2π + ±. Then

the area A bounded by the curves r = f (θ), θ = ± and θ = β is given by

β β

12 1

(f (θ))2 dθ.

A= r dθ =

2 2

± ±

Theorem 9.7.2 Let r = f (θ) be a curve in polar coordinates such that f

and f are continuous for ± ¤ θ ¤ β, and there is no overlapping, the arc

length L of the curve from θ = ± to θ = β is given by

β

(f (θ))2 + (f (θ))2 dθ

L=

±

2

β

dr

r2 +

= dθ

dθ

±

9.8 Parametric Equations

De¬nition 9.8.1 A parametrized curve C in the xy-plane has the form

C = {(x, y) : x = f (t), y = g(t), t ∈ I}

for some interval I, ¬nite or in¬nite.

The functions f and g are called the coordinate functions and the variable

t is called the parameter.

Theorem 9.8.1 Suppose that x = f (t), y = g(t) are the parametric equa-

tions of a curve C. If f (t) and g (t) both exist and f (t) = 0, then

dy g (t)

= .

dx f (t)

Also, if f (t) and g (t) exist, then

d2 y f (g)g (t) ’ g (t)f (t)

= .

dx2 (f (t))2

At a point P0 (f (t0 ), g(t0 )), the equation of

9.8. PARAMETRIC EQUATIONS 367

(a) the tangent line is

g (t0 )

y ’ g(t0 ) = (x ’ f (t0 ))

f (t0 )

(b) the normal line is

f (t0 )

(x ’ f (t0 ))

y ’ g(t0 ) = ’

g (t0 )

provided g (t0 ) = 0 and f (t0 ) = 0.

Theorem 9.8.2 Let C = {(x, y) : x = f (t), y = g(t), a ¤ t ¤ b} where f (t)

and g (t) are continuous on [a, b]. Then the arc length L of C is given by

b

[(f (t))2 + (g (t))2 ]1/2 dt

L=

a

2 1/2

2

b

dy

dx

+

= dt.

dt dt

a

Theorem 9.8.3 Let C = {(x, y) : x = f (t), y = g(t), a ¤ t ¤ b}, where f (t)

and g (t) are continuous on [a, b].

(a) If C lies in the upper half plane or the lower half plane and there is no

overlapping, then the surface area generated by revolving C around the

x-axis is given by

b

2πg(t) (f (t))2 + (g (t))2 dt.

a

(b) If 0 ¤ f (t) on [a, b], (or f (t) ¤ 0 on [a, b]) and there is no overlapping,

then the surface area generated by revolving C around the y-axis is

b

2πf (t) (f (t))2 + (g (t))2 dt.

a

368CHAPTER 9. ANALYTIC GEOMETRY AND POLAR COORDINATES

De¬nition 9.8.2 Let C = {(x(t), y(t)) : a ¤ t ¤ b} for some interval I.

Suppose that x (t), y (t), x (t) and y (t) are continuous on I.

(a) The arc length s(t) is de¬ned by

t

[(x (t))2 + (y (t))2 ]1/2 dt.

s(t) =

a

(b) The angle of inclination, φ, of the tangent line to the curve C is de¬ned

by

y (t) dy

φ(t) = arctan = arctan .

x (t) dx

(c) The curvature κ(t), read kappa of t, is de¬ned by

|x (t)y (t) ’ y (t)x (t)|

dφ

= .

[(x (t))2 + (y (t))2 ]3/2

ds

(d) The radius of curvature, R, is de¬ned by

1

R(t) = .

κ(t)