ppendices - Meyers' Mathmeyersmath.com/wp-content/uploads/2014/01/11-End.pdfSection A1 Formulas from Precalculus Mathematics 563 . 6. Completing the Square If . a =1= 0, we can rewrite

A1

What you'll learn about

Algebra

Geometry

Trigonometry

. and why

These basic ideas and formulas from precalculus are useful in calculus.

562

ppendices

Fonnulas frOlTI Precalculus Mathematics

Algebra 1. Laws of Exponents

1, a-m = _l_If a '* 0, am 2. Zero Division by zero is not defined.

If a '* 0: Q= 0 aO 1, oa o a '

For any number a: a 0 0 a 0

3. Fractions

a e ad + be a e ae alb a d -a a a

-+-=, --,b+d bd b d bd' eld b e b b -b

(alb) + (c/d) (alb) + (c/d) (ad + be)fh= (elf) + (glh) (elf) + (glh) bdfh (eh + fg)bd

4. The Binomial Theorem

For any positive integer n,

(a + b)n n + n-1b + n(n - 1) n-2b2 a na 1 _2 a

n(n - l)(n 2)+ an- 3b3 + ... + nabn- 1 + bn.

1-2,3

For instance, (a + b)l = a + b,

(a + b)2 = a2 + 2ab + b2,

(a + b)3 = a3 + 3a 2b + 3ab2 + b3,

(a + b)4 a4 + 4a3b + 6a2b2 + 4ab3 + b4

S. Differences of Like Integer Powers, n > 1

bnan - (a b)(an- 1 + an- 2b + an- 3b2 + ... + abn- 2 + br.-I)

2 b2For instance, a (a - b)(a + b).

a 3 b3 = (a - b)(a2 + ab + b2 ), 4 b4a = (a - b)(a3 + a2b + ab2 + b3 ).

Section A1 Formulas from Precalculus Mathematics 563

6. Completing the Square If a =1= 0, we can rewrite the quadratic ax2 + bx + c in the form au 2 + C by a process called completing the square:

Factor a from the ax2 + bx + c = a(x2 + ;x) + c first two terms.

Add and subtract the square of half the coefficient of x.

2 b b )= a x 2 + -. x + -- + c( a 42

2 . b = au + C -=X 23

7. The Quadratic Formula By completing the square on the first two terms of the equation

ax2 + bx+ c and solving the resulting equation for x (details omitted), we obtain

-b x =

This equation is the quadratic formula.

The solutions of the equation 2x2 + 3x - 1 are

.....---'----- 2a

-3 Y(3)2 - 4(2)(-1) x = 2(2)

or

-3 + Vi7 -3

x and x=----

4 4

Geometry (A = area, B area of base, C = circumference, h = height, S lateral area or surface area, V = volume)

1. Triangle 2. Similar Triangles 3. Pythagorean Theorem

"'"-___...1._ __ 0

1

lh 1

I

a

A 1bh 2

564 Appendices

4. Parallelogram 5. Trapezoid 6. Circle

/ a

A bh I

A = 2(1 + b)h

7. Any Cylinder or Prism with Parallel Bases 8. Right Circular Cylinder

Th

1 V=Bh

9. Any Cone or Pyramid 10. Right Circular Cone 11. Sphere

Th

Trigonometry 1. Definitions of Fundamental Identities

y

Sine: sin e= 1. T esc e P(x, y)

1Cosine: cos (J = X T sec e

1.Tangent: tan (J x cot e

--+-----~--~--~--~x

Section A 1 Formulas from Precalculus Mathematics 565

2. Identities

sin (-0) = -sin e, cos (-e) = cos 0,

sin2 0 + cos2 e= 1, sec2 e= 1 + tan2 e, csc2 e 1 + cot2 e

sin 20 = 2 sin ecos e, cos 2e cos2 e sin2 e

1 + cos 2e cos 2e

cos2 e= 2 2

tan A + tanBsin (A + B) sin A cos B + cos A sin B tan (A + B) = ----

1 - tan A tan B sin (A B) = sin A cos B -cos A sin B

tanA-tanB cos (A + B) = cos A cos B - sin A sin B tan (A - B) = 1 + tan A tan B cos (A - B) = cos A cos B + sin A sin B

sin (A - ;) = -cosA, COS(A ;)=SinA

sin (A + ~) = cos A, cos (A + ;) -sin A

1 1

sin A sin B Zcos (A B) Zcos (A + B)

cos A cos B = Z1 cos (A B) + z1 cos (A + B) sin A cos B ~ sin (A - B) + ~ sin (A + B)

sin A + sin B = 2 sin t(A + B) cos t(A B)

sin A sin B = 2 cos t(A + B) sin t(A - B)

1 1 cos A + cos B 2 cos Z(A + B) cos Z(A - B)

cos A - cos B = -2 sin ~ + B) sin t(A - B)

3. Common Reference Triangles

2

'!!: 3

4. Angles and Sides of a Triangle

2 a2 + b2Law of cosines: e = - 2ab cos C

sin A sin B sin C B~ALaw of sines:

a b e b1 . BArea = tbe sin A Zaesm tab sin C

566 Appendices

A2


Mathematical Induction Principle

Other Starting Integers

... and why

Mathematical induction is an important method to provide proofs.

Mathematical Induction

Mathematical Induction Principle Many formulas, like

n(n + I)I + 2 + ... + n = ---'--2--'-'

can be shown to hold for every positive integer n by applying an axiom called the mathematical induction principle. A proof that uses this axiom is a proofby mathematical induction or a proofby induction.

The steps in proving a formula by induction are the following.

Step 1: Check that the formula holds for n = 1.

Step 2: Prove that if the formula holds for any positive integer n k, then it also holds for the next integer, n (k + I).

Once these steps are completed (the axiom says), we know that the formula holds for all positive integers n. By step I it holds for n = 1. By step 2 it holds for n 2, and therefore by step 2 also for n 3, and by step 2 again for n = 4, and so on. If the first domino falls, and the kth domino always knocks over the (k + Ost when it falls, all the dominoes fall.

From another point of view, suppose we have a sequence of statements SI' S2, ... , S/I' ... , one for each positive integer. Suppose we can show that assuming anyone of the statements to be true implies that the next statement in line is true. Suppose that we can also show that SI is true. Then we may conclude that the statements are true from SI on.

EXAMPLE 1 Sum of the First n Positive Integers

Show that for every positive integer n,

n(n + I)

1+2+"'+n= 2 .

SOLUTION

We accomplish the proof by carrying out the two steps.

Step 1: The formula holds for n = I because

= 1(1 + 1)

1 2 .

Step 2: If the formula holds for n = k, does it hold for n (k + l)? The answer is yes, and here's why: If

1+2+"'+k

then

1 + 2 + ... + k + (k + 1) = k(k; I) + (k + 1)

k2 + k + 2k + 2 2

(k + I)k + 1) + 1) 2

The last expression in this string of equalities is the expression n(n + 1)/2 for n (k + 1).

continued

Section A2 Mathematical Induction 567

The mathematical induction priniciple now guarantees the original formula for all positive integers n. All we have to do is carry out steps 1 and 2. The mathematical induction principle does the rest. Now try Exercise 1.

EXAMPLE 2 Sums of Powers of 1/2

Show that for all positive integers n,

1+1+"'+~=1 1 2n 2n

SOLUTION

We accomplish the proof by carrying out the two steps of mathematical induction.

Step 1: The formula holds for n 1 because

Step 2: If 1 1 1-+-+ ... + 1 - 2k '21 22

then

1 1 1+"'+-+--=1- +2k 2k+1

12 1 1 - 2k. 2 + 2k+l

=1- 2 +

= 1

Thus, the original formula holds for n = (k + 1) whenever it holds for n k. With these steps verified, the mathematical induction principle now guarantees the formula for every positive integer n. Now try Exercise 5.

Other Starting Integers Instead of starting at n = 1, some induction arguments start at another integer. The steps for such an argument are as follows.

Step 1: Check that the formula holds for n = n l (the first appropriate integer).

Step 2: Prove that if the formula holds for any integer n k 2:: n1, then it also holds for n = (k + 1).

Once these steps are completed, the mathematical induction principle guarantees the formula for all n nl'

EXAMPLE 3 Factorial Exceeding Exponential

Show that n! > 3n if n is large enough. continued

568 Appendices

Section A2 Exercises

SOLUTION

How large is large enough? We experiment:

It looks as if n! > 3" for n 7. To be sure, we apply mathematical induction. We take n 1 = 7 in step 1 and try for step 2.

Suppose k! > 3k for some k 7. Then

(k + 1)1 (k + l)(k!) > (k + 1)3k > 7 3k > 3k+l.

Thus, for k ~ 7, (k + I)! > 3k+l.

The mathematical induction principle now guarantees n! > 3n for all n ~ 7. Now try Exercise 7.

1. General Triangle Inequality Assuming that the triangle inequality I a + b I :::; Ia I + Ib I holds for any two numbers a and b, show that

IXI +x2 + ... +xnl:::; IXII + IX21 + ... + IX"i

for any n numbers.

2. Partial Sums of Geometric Series Show that if r '" 1, then I r n+1

I + r + r2 + .. + rJl I r

for every positive integer n.

3. Positive Integer Power Rule Use the Product Rule, d dv du ~(uv) = u~ + dx dx dx

and the fact that d ~(x)dx

to show that d_(x") nxn- 1 dx

for every positive integer n.

4. Products into Sums Suppose that a function lex) has the property that!(x1x2) = !(x l ) + !(X2) for any two positive numbers x 1 and x 2' Show that

!(x 1x2 .. xn ) !(xt) +!(x2) + + !(xn ) for the product of any n positive numbers XI' x2 , . , XII'

5. Show that 2 2+ ... +

3" 3"

for all positive integers n.

6. Show that n! > n3 if n is large enough. 7. Show that 2n > III if n is large enough. 8. Show that 2n ?o: 1/8 for n?o: - 3.

9. Sums of Squares Show that the sum of the squares of the first n positive integers is

n(n + +)(n + 1) 3

10. Sums of Cubes Show that the sum of the cubes of the first /l positive integers is (n(n + 0/2)2.

11. Rules for Finite Sums Show that the following finite sum rules hold for every positive integer n.

n "n

(a) 2: (ak + bk) = 2: ak + 2: bk k=l k=1 k=l

n rl It

(b) 2: (ak - bk) = 2: ak 2: bk k=1 k=1

(e) 2:/I ca. = c 2:" ak (Any number c) k=1 k=1

(d) 2:n

ak n c, if ak has the constant value c.

k=1

12. Absolute Values Show that Ix" I Ixln for every positive integer n and every real number x.

A3


Limit Definition

Finding Deltas for Given Epsilons

Proving Limit Theorems

... and why

This section provides practice using the formal definition of limit.

y

1L+ro

[(x)

L

~~____-+____~__~x~~~.x o c-o c c+o

Figure A3.1 A preliminary stage in the development of the definition of limit.

i 1 L + TO r----------,I--

L

L

The challenge:

Make - L I< e = 10

L+sl L .

. [tx)

L-e

x -O~------c-_~o~~cr---c-+~o--+X

Section A3 Using the Limit Definition 569

Using the Limit Definition

Limit Definition . We begin by setting the stage for the definition of limit Recall that the limit off ofx as x approaches c equals L (limX--7J(x) L) means that the values f(x) of the function f approach or equal L as the values of x approach (but do not equal) c. Suppose we are watching the values of a functionf(x) as x approaches c (without taking on the value of c itself). Certainly we want to be able to say thatf(x) stays within one-tenth of a unit of L as soon as x stays within some distance i5 of c (Figure A3.l). But that in itself is not enough, because as x continues on its course toward c, what is to preventf(x) from jittering about within the interval from L - 1/10 to L + 1/10 without tending toward L? .

We can insist that f(x) stay within 1/100 or 1/1000 or 1/100,000 of L. Each time, we find a new i5-interval about c so that keeping x within that interval keepsf(x) within E = 1/100 or 1/1000 or 1/100,000 of L. And each time the possibility exists that c jitters away from L at the last minute.

Figure A3.2 illustrates the problem. You can think of this as a quarrel between a skeptic and a scholar. The skeptic presents E-challenges to prove that the limit does not exist or, more precisely, that there is room for doubt, and the scholar answers every challenge with a a-interval around c. y

y

Figure A3.2 The first of a possibly Figure A3.3 The relation of the 8 and endless sequence of challenges. in the definition of limit.

How do we stop this seemingly endless sequence of challenges and responses? By proving that for every E-distance that the challenger can produce, we can find, calculate, or conjure a matching a-distance that keeps x "close enough" to c to keep f{x) within that distance of L (Figure A3.3).

The following definition that we made in Section 2.1 provides a mathematical way to say that the closer x gets to c, the closer f(x) must get to L.

DEFINITION Limit

Let c and L be real numbers. The function f has limit L as x approaches c if, given any positive number E, there is a positive number asuch that for all x

0< Ix cl < (} ==> If(x) - LI < E. We write

limj(x) L. x-,>c

570 Appendices

y y = 5x - 3

2+e:

Figure A3.4 Iff(x) = 5x - 3, then 0< Ix - 11 < d5 guarantees that If(x) - 21 < E. (Example 1)

\.

\ Intersection \ X = .49751244 Y = 2.01

[0.48, 0.52J by [1.98:2.02]

(a}

\

\"

Intersection X =.50251256 Y = 1.99

[0.48, 0.52J by [1.98, 2.02J

(b)

Figure A3.5 We can see from the two graphs that if 0.498 < x < 0.502, then 1.99 1(5x 3) 2. In Example 1, we confirmxthis result using the definition of limit.

I EXAMPLE 1 Using the Definition of Limit

Show that iim ->1(5x - 3) = 2.x

SOLUr'ON

Set c 1, f(x) 5x - 3, and L = 2 in the definition of limit. For any given 10 > 0 we have to find a suitable 0 > 0 so that if x * 1 and x is within distance 8 of c = 1, that is. if

0< Ix - 11 < 8,

then f(x) is within distance 10 of L = 2, that is,

If(x) 21 < E.

We find 0 by working backward from the -inequality:

1(5x-3) 21 15x 511(5x - 3) = 2. Now try Exercise 5.

The value of 0 = 10/5 is not the only value that will make 0 < Ix - 11 < 0 imply If(x) 21 = 15x - 51 < in Example 1. Any smaller positive 8 will do as well. The definition does not ask for a "best" positive 0, just one that will work.

We can use graphs to find a 0 for a specific as in Example 2.

EXAMPLE 2 Finding a (j Graphically

For the limit limHO.5(l/x) = 2, find a {j that works for = 0.01. That is, find a

8 > 0 such that for all x

0< Ix - 0.51 < 0 =:;> If(x) - 21 < 0.01.

SOLUTION

Heref(x) = l/x, c = 0.5, and L 2. Figure A3.5 shows the graphs of! and the two horizontal lines

y = L - = 2 O.oI 1.99 and y = L + = 2 + 0.01 = 2.01. Figure A3.5a shows that the graph off intersects the horizontal line y = 2.01 at about (0.49751244,2.01), and Figure A3.5b shows that the graph of f intersects the horizontal line y = 1.99 at about (0.50251256, 1.99). It follows that

0< Ix 0.51 < 0.002 =:;> If(x) - 21 < 0.01.

Thus, {j = 0.002 works. Now try Exercise 3.

http:0.49751244,2.01http:1.98:2.02

A3.6 The function in Example 3.


EXAMPLE 3 Finding 0 Algebraically

Prove that limx-->d(x) 4 if

f(x) = {x2, x ~2

1, x - 2.

SOLUTION

Our task is to show that given E > 0 there exists a 0 > 0 such that for all x

O

572 Appendices

THEOREM 1 Properties of Limits

If L, M, c, and k are real numbers and

limJ(x) L and lim g(x) = M, then X-7C X-7C

1. Sum Rule: lim (j(x) + g(x = L + M X-7C

2. Difference Rule: lim (j(x) - g(x = L - M ~\-tc

3. Product Rule: lim (j(x) g(x = L M X""""c

4. Constant Multiple Rule: lim k J(x) = k L x-tc

S. Quotient Rule: lim J(x) = , M* 0 .Hc g(x) M

6. Power Rule: If rand s are integers, s *0, then lim (j(x)Yls U ls X-7C

provided Lrls is a real number.

Proof of the limit Sum Rule We need to show that for any E > 0, there is a 0> 0 such that for all x in the common domain D ofJand g,

O


Proof of the Limit Product Rule We show that for any E > 0, there is a 8> 0 such that for all x in the common domain D of f and g,

0< Ix cl < 8 =? If(x)g(x) - LMI < E. Write f(x) and g(x) asf(x) = L + (f(x) - L), g(x) M + (g(x) - M).

MUltiply these expressions together and subtract LM:

f(x) g(x) - LM = (L + (f(x) L(M + (g(x) M - LM

= LM + L(g(x) - M) + M(f(x) - L) + (f(x) L)(g(x) M) - LM (1)

= L(g(x) - M) + M(f(x) L) + (f(x) - L)(g(x) M)

Since f and g have limits Land Mas X-7C, there exist positive numbers 8 t, 82, 83, and 84 such that for all x in D

0< Ix - ci c x->c g(x) M M

by the Limit Product Rule.

Let E > 0 be given. To show that limHC (11g(x 11M, we need to show that there exists a 8 > 0 such that for all x

< Ix cl < 8 =? 1 _~IM 0, there exists a positive number 8) such that for all x

IMI0< Ix cl < 8 t =? Ig(x) - MI < 2' (3)

continued'

574 Appendices

For any numbers A and B it can be shown that

IAI-IBI slA BI and IBI-IAI IA - BI,

from which it follows thatl!AI IBllslA BI, With A g(x) and B=M, this becomes

Ilg(x)1 -IMII s Ig(x) MI,

which can be combined with the inequality on the right in (3) to get, in turn,

Ilg(x)1 - IMII < TIMI

_ IMI < Ig(x) I - IMI < IMI

2 2

IMI 3

T

,d


These inequalities combine with the inequality g(x) S f{x) S hex) to give

L - e < g(x) sf(x) s hex) < L + e,

L-e 0 such that for all x the inequality c - 0 < x < c implies

L - E < g(x) < L + E and L - e < hex) < L + E.

We conclude as before that for all x, c - 0 < x < c implies If(x) LI < fE. Therefore, limx->c-f(x) = L.

Proof for Two-sided Limits If limHc g(x) = limx.... hex) = L, then g(x) andc hex) both approach L as x~c+ and x~c-; so lim.HC+f(x) L and lim.Hc~f(x) L. Hence limx-'lcf(x) exists and equals L. IiI!!i

Section A3 Exercises

til Exercises I and 2, sketch the interval (a, b) on the x-axis with the point c inside. Then find a value of 0 > 0 such that for all x,

0< Ix - cl < 0 ==- a < x < b.

1. a 4/9, b 4/7, c 1/2

2. a = 2.7591, b = 3.2391, c 3

In Exercises 3 and 4, use the graph to find a 0 > 0 such that for all 10< Ix cl < 0 ==- If(x) - LI < E.

3. Y fix) = 2rx+1 4. y c 3

L 4

8 0.2

Y 2rx+1 4.2

4

3.81-----

~--I----~~--~~x

2.61 3 3.41

NOT TO SCALE

Exercises 5-8 give a function f(x) and numbers [, c, and E. Find an open interval about c on which the inequality I!(x) [I < E holds. Then give a value for 0 > 0 such that for all x satisfying 0< Ix - cl < fi the inequality If(x) [I < E holds. Use algebra to find your answers.

5. f(x) = 2x - 2, [ -6, c -2, E = 0.02

6. f(x) Vx+t, [' = 1, c 0, 0.1E 7. fex) =~, [= 3, c = 10, E = 1

8. f(x) = Xl, [= 4. c -2, E 0.5

Exercises 9-12 give a function f(x), a point c, and a positive number E. (a) Find L limHJ(x). Then (b) find a number fi > 0 such that for all x

0< Ix cl < fi ==- If(x) - [I < E.

9.fex)=X2+6xt5 c -5,'E 0.05 x+

10.f(x) = {~x-:~', : ~ i: c = 1, E = 0.5 11. f(x) sin x, c 1, E = 0.01

x12.f(x) = c=-l, E=O.l

4

576 Appendices

In Exercises 13 and 14, nse the definition of limit to prove the limit statement.

Ix2, x'* 113. lim f(x) = 1 if f(x)

x->I 2, x = 1 . I I

14. hm -, =-3 X-4V'3 x

15. Relating to Limits Given E > 0, (a) find an interval 1= (5,5 + 8),8> 0, such that if x lies in I, then

v7=5 < E. (b) What limit is being verified?

16. Relating to Limits Given E > 0, (a) find an interval 1= (4 - 8, 4), 8 > 0, such that if x lies in I, then

~ < E. (b) What limit is being verified?

17. Prove the Constant Multiple Rule for limits.

18. Prove the Difference Rule for limits.

19. Generalized Limit Sum Rule Suppose that functions fl (x), f2(x), andfJ(x) have limits L I, L 2 , and L 3, respectively, as X-7C- Show that their sum has limit LI + L2 + Use mathematical induction (Appendix A2) to generalize this result to the sum of any finite number of functions.

20. Generalized Limit Product Rule Use mathematical induction and the Limit Product Rule in Theorem 1 to show that if functionsfl (x),f2(x), ... ,fn(x) have limits L 1, ,Ln. respectively. as X-7C, then

21. Positive Integer Power Rule Use the fact that Iimx->cx == C and the result of Exercise 20 to show that lim x .... c x" e" for any integer n > 1. .

22. Limits of Polynomials Use the fact that Iim ->c k k for anyxnumber k together with the results of Exercises 19 and 21 to show that limx->J(x) = fee) for any polynomial function

23. Limits of Rational Functions Use Theorem 1 and the result of Exercise 22 to show that if f(x) and g(x) are polynomial functions and gee) '* 0, then

lim f(x) = f(c) x .... c g(x) g(e)'

24. Composites of Continuous Functions A3.7 gives the diagram for a proof that the composite of two continuous functions is continuous. Reconstruct the proof from the diagram. The statement to be proved is this: If f is continuous at x = C and g is continuous at fCc), then g 0 f is continuous at c.

Assume that c is an interior point of the domain of f and that fCc) is an interior point of the domain of g. This will make the limits involved two-sided. (The argument for the cases that involve one-sided limits are similar.)

go!

Figure A3.7 The continuity of composites holds for any finite number of functions. The only requirement is that each function be continuous where it is applied. Here,fis to be continuous at c and g atf(e).

A4


Error in the Approximation M=df

The Proof

.,. and why

This section helps explain approximating change in function values.

y y = I(x)

Secant slope

layi

---I I I I

~~~-L~~__~________-+x a

Figure A4.1 The graph of y as a function of x. The derivative of y with respect to x at x = a is lim (.1.y/.1.x).ax.....o

Section A4 Proof of the Chain Rule 577

Proof of the Chain Rule

Error in the Approximation At""" dt . Letf(x) be differentiable at x = a and suppose that Ax is an increment of x. We know that the differential j'(a)Ax is an approximation for the change Af = (f(a + Ax) f(a)) infas x changes from a to (a + AX). How well does dfapproximate Af?

We measure the approximation error by subtracting dffrom Af:

Approximation error = Af df = Af - j'(a)Ax

f(a + Ax) f(a) j'(a)Ax ::.f + ..lx) - f(a) Af

- (f(a + Ax) - f(a) _ f'()) A - A- Ax a x - E' X

Cail this

As AX-70, the difference quotient (f(a + Ax) - f(a))/Ax approachesj'(a) (remember the definition ofj'(a)), so the quantity in parentheses becomes a very small number (which is why we called it E). In fact, E-70 as AX-70. When Ax is small, the approximation error EAx is smaller stilL

Af j'(a) Ax + EAx (I)

change esiimaied

cnange

The Proof We want to show if f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x. More precisely, if g is differentiable at a and f is differentiable at g(a), then the composite is differentiable at a and

:z Ix=a = j'(g(a)) g'(a). Let Ax be an increment in x and let Au and Ay be the corresponding increments in u andy. As you can see in FigureA4.l,

dyl r dx x=a = ~~ Ax

so our goal is to show that the limit is j'(g(a)) g'(a).

By Equation 1,

where El -70 as AX-70. Similarly, since f is differentiable at g(a),

where E2-70 as AU-70. Notice also that AU-70 as AX-70. Combining the equations for Au and Ay gives Ay (f'(g(a)) + E2)(g'(a) + E1)Ax, so

~~ = j'(g(a))g'(a) + E2 g'(a) + j'(g(a))EI + E2E!. Since EJ and EZ go to zero as Ax goes to zero, three of the four terms on the right vanish in the limit, leaving

lim A = j'(g(a))g'(a).Ilx.....O uX

This concludes the proof.

578 Appendices

A5~:


Circles

Interiors and Exteriors of Circles

Parabolas

Ellipses

Axes of an Ellipse

Hyperbolas

Asymptotes and Drawing

Reflective Properties

Other Applications

... and why

This section provides some basic information about equations of conic sections and how to sketch their graphs.

Circle: plane perpendicular to cone axis

Conic Sections

Overview

Conic sections are the paths traveled by planets, satellites, and other bodies (even electrons) whose motions are driven by inverse-square forces. Once we know that the path of a moving body is a conic section, we immediately have infonnation about the body's velocity and the force that drives it. In this appendix, we study the connections between conic sections and quadratic equations and classify conic sections by eccentricity (Pluta's orbit is highly eccentric while Earth's is nearly circular).

AS.1 Conic Sections and Quadratic Equations Circles

The Greeks of Plato's time defined conic sections as the curves formed by cutting through a double cone with a plane (Figure A5. 1). Today, we define conic sections with the distance function in the coordinate plane.

Ellipse: plane oblique Parabola: plane parallel to cone axis to side of cone

(a)

Hyperbola: plane cuts both halves of cone

Point plane through cone vertex only

Single line: plane tangent to cone

(b)

Pair of intersecting line!

Figure A5.1 The standard conic sections (a) are the curves in which a plane cuts a double cone. Hyperbolas come in two parts, called branches. The point and lines obtained by passing the plane through the cone's vertex (b) are degenerate conic sections.

Section A5 Conic Sections 579

DEFINITION Circle

A circle is the set of points in a plane whose distance from a given fixed point in the plane is constant. The fixed point is the center of the circle; the constant distance is the radius.

Ifa> 0, the equation x2 + y2 a 2 represents all the points (x, y) in the plane whose distance from the origin is

Y(X-0)2+(y-0)2 Yx2 +y2 W=a.

These are the points of the circle of radius a centered at the origin. If we shift the circle to place its center at the point (h, k), its equation becomes (x - h)2 + (y - k)2 = a2

Circle of Radius a Centered at (h, k)

(x h)2 + (y k)2 = a 2

Exterior: (x /z)2 + (y - k)2 > a2

2 a

Figure A5.2 The interior and exterior of circle (x - h)2 + (y - k)2 = a2

EXAMPLE 1 Finding Center and Radius

Find the center and radius of the circle

x2 + y2 + 4x - 6y - 3 O.

SOLUTION

We convert the equation to standard form by completing the squares in x and y:

Start with the x 2 + y2 + 4x - 6y - 3 0 given equation.

Gather terms. Move (x2 + 4x) + (y2 6y) 3 ihe constant to the

right-hand side.

of half

(X2 + 4x + (~n + k -6y + ( ~6n (4)2 ( 6)2=3+ 2 +2 novi perfect squares.

(Xl + 4x + 4) + (y2 - 6y + 9) 3 + 4 + 9

Write each quadratic as i}(X+2)2+(y 3)2=16 squared linear expression.

With the equation now in standard form, we read off the center's coordinates and the radius: (h, k) (-2, 3) and a = 4. Now try Exercise 3.

Interiors and Exteriors of Circles

The points that lie inside the circle (x - h)2 + (y k)2 = a 2 are the points less than a units from (h, k). They satisfy the inequality

ex - h)2 + (y - k)2 < a2 They make up the region we call the interior of the circle (Figure AS.2).

580 Appendices

The circle's exterior consists of the points that lie more than a units from (h, k). These points satisfy the inequality

I EXAMPLE 2 Interpreting Inequalities

Inequality

I x2 + I

Interior of the unit circle Unit circle plus its interior Exlerior of the unit circle Unit circle its exterior

Now try Exercise 5.

Parabolas

DEFINITION Parabola

A set that consists of all the points in a plane equidistant from a given fixed point and a given fixed line in the plane is a parabola. The fixed point is the focus of the parabola. The fixed line is the directrix.

y

If the focus F lies on the directrix L, the parabola is the line through F perpendicular to L. We consider this to be a degenerate case and assume henceforth that F does not lie on L.

A parabola has its simplest equation when its focus and directrix straddle one of the Focus F(O,p) coordinate axes. For example, suppose that the focus lies at the point F(O, p) on the posi

p tive y-axis and that the directrix is the line y = -p (Figure AS.3). In the notation of the -=-".j-"""----'---...... x figure, a point P(x, y) lies on the parabola if and only if PF = PQ. From the distance

The vertex lies ~

halfway between I' formula,

directrix and focus.

-----------+--~~L------L Directrix: y = -I' Q(x, -I') PF

2 PQ=Figure AS.3 The parabola x 4py.

When we equate these expressions, square, and simplify, we get

y or = 4py. Standard form (I)

These equations reveal the parabola's symmetry about the y-axis. We call the y-axis the Directrix: y = I' axis of the parabola (short for "axis of symmetry").

The point where a parabola crosses its axis is the vertex. The vertex of the parabola Vertex at origin '-. 4py lies at the origin (Figure AS.3). The positive number p is the parabola's focal

--------~~~~~--------~x length.

If the parabola opens downward, with its focus at (0, -p) and its directrix the line y P, '-.. Equations 1 becomeFocus (0, -I')

41')' y= or -4py

(FigureA5A). We obtain similar equations for parabolas opening to the right orto the left Figure AS.4 The parabola x 2 = -4py. (FigureA5.S and TableA5.1).


y y

Directrix Directrix x -p x= p

Vertex

---f----'+----il-------+x

Vertex

Figure AS.S (a)The parabola y2 = 4px. (b) The parabolay2 -4px.

liMpt!\.,. Standard-form equations for parabolas with vertices at the origin (p> 0)

Equation Focus Directrix Axis 4py (O,p) y =-p y-axis Up

x 2 = -4py (0, -p) y= p y-axis Down y2 = 4px (p,O) x= -p x-axis To the right

= -4px (-p, 0) x= p x-axis To the left

EXAMPLE 3 Finding Focus and Directrix

Find the focus and directrix of the parabola y2 = lOx.

SOLUTION

We find the value of p in the standard equation y2 = 4px:

4p 10, so

Then we find the focus and directrix for this value ofp:

- \ Focus: (p,O) = (%,0) Directrix: x = -p or

Now try Exercise 7.

Ellipses

DEFINITION EllipseVertex Vertex

An ellipse is the set of points ina plane whose distances from two fixed points in the .plane have a constant sum. The fixed points are the foci of the ellipse. The line through the foci is the focal axis. The point on the axis halfway between the foci is the center. The points where the focal axis and ellipse cross are the vertices (Figure AS.6). Figure AS.6 Points on the focal axis of

an ellipse.

x

S82 Appendices

Figure AS.7 How to draw an ellipse.

y

b

Focus ~--~~----~------~--~~X

FI(-c,O)

Figure AS.8 The ellipse defined by the equation PF1 + PFz = 2a is the graph of the equation (x2fa2) + (y2/b2) = 1.

Vertex Vertex (-4,0) (4, 0)

Focus --~--~----~~-----.--~~X

Center (..ft, 0)

(0,-3)

Figure AS.9 Major axis horizontal. (Example 4)

The quickest way to construct an ellipse uses the definition. Put a loop of string around two tacks F\ and F2, pull the string taut with a pencil point P, and move the pencil around to trace a closed curve (Figure AS.7). The curve is an ellipse because the sum PF[ + PF

2,

being the length of the loop minus the distance between the tacks, remains constant. The ellipse's foci lie at F[ and F2

If the foci are F[(-e, 0) and F2(e, 0) (Figure AS.8) and PF1 + PF2 is denoted by 2a, then the coordinates of a point P on the ellipse satisfy the equation

Vex + e)2 + y2 + Vex - eF + y2 = 2a. To simplify this equation, we move the second radical to the right-hand side, square, iso. late the remaining radical, and square again, obtaining

(2)

Since PF! + PF2 is greater than the length Fl Fz (triangle inequality for triangle PF!F1 ), the number 2a is greater than 2e. Accordingly, a > e and the number a2 - e2 in Equation 2 is positive.

The steps leading to Equation 2 can be reversed to show that every point P whose coordinates satisfy an equation of this form with 0 < c < a also satisfies the equation PF1 + 2a. A point therefore lies on the ellipse if and only if its coordinates satisfy Equation 2.

If (3)b=

then a2 b 2 and Equation 2 takes the form

x2 y2 +--;;'2=1. (4)

Equation 4 reveals that this ellipse is symmetric with respect to the origin and both coordinate axes. It lies inside the rectangle bounded by the lines x = :!:a and y = :!:h. It crosses the axes at the points 0) and (0, :!:b). The tangents at these points are perpendicular to the axes because

Obtained from Eq. 4 by implicit differentiation

is zero if x 0 and infinite if y O.

Axes of an Ellipse

The major axis of the ellipse in Equation 4 is the line segment of length 2a joining the points (:!:a, 0). The minor axis is the line segment of length 2b joining the points (0, :!:b). The number a itself is the semimajor axis, the number b the semiminor axis. The number e found from Equation 3 as

e=

is the center- to-focus distance of the ellipse.

EXAMPLE 4 Major Axis Horizontal

The ellipse 2x (5)-+

16 9

(Figure AS.9) has

Semimajor axis: a

Semiminor axis: b V9 3 continued

4

y

x2 y29" + 16 = 1 (0,4) Vertex

;:::: (0,-Y7)

(3,0)(-3,0) ~~--------~-------+~X

Center

Focus (0, --Y7)

Vertex (0, -4)

Figure A5.10 Major axis vertical. (Example 5)

e

r 1

I


Center-to-focus distance: c = Vi6="9 = V?

Foci: (:tc,O) = (:tV?, 0)

Vertices: (:ta, 0) = (:t4, 0)

Center: (0,0). Now try Exercise 13.

EXAMPLE 5 Major Axis Vertical

The ellipse

(6)

obtained by interchanging x and y in Equation 5, has its major axis vertical instead of horizontal (FigureA5.1O). With a 2 still equal to 16 and P equal to 9, we have

Semimajor axis: a = VT6 = 4 Semiminor axis: b = V9 = 3 Center-to-focus distance: c = Vi6="9 = V? Foci: (0, :tc) = (0, :tV?)

Vertices: (0, :ta) = (0, :t4)

Center: (0,0). Now try Exercise 17.

There is never any cause for confusion in analyzing equations like (5) and (6). We simply find the intercepts on the coordinate axes; then we know which way the major axis runs because it is the longer of the two axes. The center always lies at the origin and the foci lie on the major axis.

Standard-Form Equations for Ellipses Centered at the Origin

Foci on the x-axis:

Va 2 b 2Center-to-focus distance: c = Foci: (:tc,O)

Vertices: (:ta,O)

Foci on the y-axis:

b 2Center-to-focus distance: c = Va 2 Foci: (0, :tc)

Vertices: (0, :ta)

In each case, a is the semimajor axis and b is the semiminor axis.

Hyperbolas

DEFINITION Hyperbola

A hyperbola is the set of points in a plane whose distances from two fixed points in the plane have a constant difference. The two fixed points are the foci of the hyperbola.

ed

http:FigureA5.1O

584 Appendices

/

Figure AS.11 Hyperbolas have two branches. For points on the right-hand branch of the hyperbola shown here, PF1 PF2 2a. For points on the left-hand branch, PF2 PF1 = 2a.

Figure A5.12 Points on the focal axis of a hyperbola.

If the foci are FI(-c, 0) and F2 (e, 0) (Figure AS.ll) and the constant difference is 2a, then a point (x, y) lies on the hyperbola if and only if

V(x + c)2 + y2 ::t2a. (7) To simplify this equation, we move the second radical to the right-hand side, square, isolate the remaining radical, and square again, obtaining

x2 y2

-2+ 2 2 1. (8)a a - e

e2So far, this looks just like the equation for an ellipse. But now a2 - is negative because 2a, being the difference of two sides of triangle PF1F2 is less than 2e, the third side.

The algebraic steps leading to Equation 8 can be reversed to show that every point P whose coordinates satisfy an equation of this form with 0 < a < c also satisfies Equation 7. A point therefore lies on the hyperbola if and only if its coordinates satisfy Equation 8.

2If we let b denote the positive square root of c a2,

(9)

then a2 - and Equation 8 takes the more compact form

x2 y2

al - b2 = 1. (10)

The differences between Equation 10 and the equation for an ellipse (Equation 4) are the minus sign and the new relation

2 2 + b2c = a From Eq. 9

Like the ellipse, the hyperbola is symmetric with respect to the origin and coordinate axes. It crosses the x-axis at the points (::ta, 0). The tangents at these points are vertical because

implicit I1ITT,prPllT1;lTlCm

is infinite when y O. The hyperbola has no y-intercepts; in fact, no part of the curve lies between the lines x -a and x a.

DEFINITION Parts of a Hyperbola

The line through the foci of a hyperbola is the focal axis. The point on the axis halfway between the foci is the hyperbola's center. The points where the focal axis and hyperbola cross are the vertices (Figure AS.12).

Asymptotes and Drawing

The hyperbola

(11)


has two asymptotes, the lines b

y = ::-x. a

The asymptotes give the guidance we need to draw hyperbolas quickly. (See the drawing lesson.) The fastest way to find the equations of the asymptotes is to replace the I III Equation II by 0 and solve the new equation for y:

b y = -x.

a

:-;"/perbO!3 asymptotes

Standard-Form Equations for Hyperbolas Centered at the Origin

Foci on the x-axis:

Center-to-focus distance: c=Va 2 +P Foci: (c,O) Vertices: (a,O)

Asymptotes: x2 y2-;;'2-17=0 ,or y =

b-x

a

Foci on the y-axis:

Center-to-focus distance: c = Va 2 + b 2 Foci: (0, c) Vertices: (0, a)

y2 x2 aAsymptotes: -;;'2-17=0 or Y = +-x-b

Notice the difference in the asymptote equations (bla in the first, alb in the second).

x2 y2 DRAWING LESSON How to Graph the Hyperbola - b2 = 1a 2 1. Mark the points (a, 0) and (0, b) with line segments and complete the rec

tangle they determine.

2. Sketch the asymptotes by extending the rectangle'S diagonals.

3. Use the rectangle and asymptotes to guide your drawing.

y y

y

b

__-....:8+-+-f'8:....--;..X ---=+-+---t=''---;''X

-b

586 Appendices

y

i5 x2

~---&-+---:*:--:t-

Figure A5.16 An elliptical mirror (shown here in profile) reflects light from one focus to the other.

Parabola

Primary mirror

Figure AS.17 Schematic drawing of a reflecting telescope.


This property is used by flashlight, headlight, and spotlight reflectors and by microwave broadcast antennas to direct radiation from point sources into narrow beams. Conversely, electromagnetic waves arriving parallel to a parabolic reflector's axis are directed toward the reflector's focus. This property is used to intensify signals picked up by radio telescopes and television satellite dishes, to focus arriving light in telescopes, and to concen

. trate sunlight in solar heaters. If an ellipse is revolved about its major axis to generate a surface (a surface called an

ellipsoid) and the interior is silvered to produce a mirror, light from one focus will be reflected to the other focus (Figure As.I6). Ellipsoids reflect sound the same way, and this property is used to construct whispering galleries, rooms in which a person standing at one focus can hear a whisper from the other focus. Statuary Hall in the U.S. Capitol building is a whispering gallery. Ellipsoids also appear in instruments used to study aircraft noise in wind tunnels (sound at one focus can be received at the other focus with relatively little interference from other sources).

Light directed toward one focus of a hyperbolic mirror is reflected toward the other focus. This property of hyperbolas is combined with the reflective properties of parabolas and ellipses in designing modern telescopes. In Figure As.I7 starlight reflects off a primary parabolic mirror toward the mirror's focus Fp. It is then reflected by a small hyperbolic mirror, whose focus is Ff{ = Fp, toward the second focus of the hyperbola, FE = Ff{. Since this focus is shared by an ellipse, the light is reflected by the elliptical mirror to the ellipse's second focus to be seen by an observer.

As past experience with NASA's Hubble space telescope shows, the mirrors have to be nearly perfect to focus properly. The aberration that caused the malfunction in Hubble's primary mirror (now corrected with additional mirrors) amounted to about half a wavelength of visible light, no more than 1/50 the width of a human hair.

Other Applications

Water pipes are sometimes designed with elliptical cross sections to allow for expansion when the water freezes. The triggering mechanisms in some lasers are elliptical, and stones on a beach become more and more elliptical as they are ground down by waves. There are also applications of ellipses to fossil formation. The ellipsolith, once thought to be a separate species, is now known to be an elliptically deformed nautilus.

Hyperbolic paths arise in Einstein's theory of relativity and form the basis for the (unrelated) LORAN radio navigation system. (LORAN is short for "long range navigation.") Hyperbolas also form the basis for a new system the Burlington Northern Railroad developed for using synchronized electronic signals from satellites to track freight trains. Computers aboard Burlington Northern locomotives in Minnesota can track trains to within one mile per hour of their speed and to within feet of their actual location.

588 Appendices

Section A5.l Exercises

In Exercises 1 and 2, find an equation for the circle with center e(h, k) and radius a. Sketch the circle in the xy-plane. Label the circle's center and x- and y-intercepts (if any) with their coordinate pairs.

1. e(O, 2), a = 2 2. e(-I, 5), a VlO

In Exercises 3 and 4, find the center and radius of the circle. Then sketch the circle.

3. x 2 + y2 + 4x - 4y + 4 =0 4. x 2 + y2 - 4x + 4y 0

In Exercises 5 and 6, describe the regions defined by the inequalities and pairs of inequalities.

5. (x - 1)2 + y2 : 1, x 2 + y2 < 4

Match the parabolas in Exercises 7-10 with the following equations:

x2 = 2y, x 2 = -6y, y2 = 8x, y2 -4x. Then find the parabola's focus and directrix.

7. 8.y y

/ --t----+x -----t--~"x

9. y 10. y A

11\ ) x llL.

Match each conic section in Exercises 11-14 with one of these equations:

x 2 Xl 1, + 1,4+ 9 2

2 L - x 2 = 1 L4 ' 4 9

Then find the conic section's foci and vertices. If the conic section is a hyperbola, find its asymptotes as welL

11. y 12. y

----+-~--~--~x---~+-+------+x

13. y 14. y

----+----->-x-+------+----~x

Exercises 15 and 16 give equations of parabolas. Find each parabola's focus and directrix. Then sketch the parabola. Include the focus and directrix in your sketch.

15. y2 = 12x 16. Y 4x2

Exercises 17 and 18 give equations for ellipses. Put each equation in standard form. Then sketch the ellipse. Include the foci in your sketch.

17. 16x2 + 25y2 = 400 18. 3x2 + 2y2 := 6

Exercises 19 and 20 give information about the foci and vertices of ellipses centered at the origin of the xy-plane. In each case, find the ellipse's standard-form equation from the given information.

19. Foci: (V2,0) 20. Foci: (0, 4)

Vertices: (2, 0) Vertices: (0, 5)

" Exercises 21 and 22 give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch.

21. x2 - y2 = 1 22. 8y2 - 2X2 = 16

Exercises 23 and 24 give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the xy-plane. In each case, find the hyperbola's standard-form equation from the information given.

23. Foci: (0. 24. Vertices: (3,0)

4

Asymptotes: y x Asymptotes: y = 3x

25. The parabola 8x is shifted down 2 units and right

1 unit to generate the parabola (y + 2)2 = 8(x - O.

(a) Find the new parabola's vertex, focus, and directrix. (b) Plot the new vertex, focus, and directrix, and sketch in the parabola.

26. The ellipse xV16 + y2/9 = 1 is shifted 4 units to the right and 3 units up to generate the ellipse

(x 4)2 (y - 3)2

16 + 9 := L

(a) Find the foci, vertices, and center of the new ellipse.

(b) Plot the new foci, vertices, and center, and sketch in the new ellipse.

27. The hyperbola x 2/16 - y2/9 = 1 is shifted 2 units to the right to generate the hyperbola

(x - 2)2 y2 16 -9= I.

(a) Find the center, foci, vertices, and as:ymptotes of the new hyperbola. (b) Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola.

Exercises 28-31 give equations for conic sections and tell how many units up or down and to the right or left each is to be shifted. Find an equation for the new conic section and find the new vertices, foci, directrices, center, and asymptotes, as appropriate.

28. y2 = 4x, left 2, down 3

Xl y2 29. 6 +9= I, left 2, down I

x2 y2 30. 4 - "5 = I, right 2, up 2

31. y2 -x2 = 1, left I, up I

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of each conic section in Exercises 32-36.

32. x 2 + 4x + y2 = 12

33. 2x2 + 2y2 - 28x + 12y + 114 0

34. x 2 + 2x + 4y - 3 = 0

35. x 2 + Sy2 + 4x = 1

36. x 2 - y2 - 2x + 4y = 4

37. Archimedes' Formula for the Volume of a Parabolic Solid The region enclosed by the parabola y (4h/b 2)x2 and the line y = h is revolved about the to generate the solid shown here. Show that the volume of the solid is 3/2 the volume of the corresponding cone.

4h 2 2X

b

x

Section AS Conic Sections 589

38. Comparing Volumes If lines are drawn parallel to the coordinate axes through a point P on the parabola y2 = kx, k > 0, the parabola partitions the rectangular region bounded by these lines and the coordinate axes into two smaller regions, A and B.

y

y2= kx

p

-0-+----------------. x

(a) If the two smaller regions are revolved about the

show that they generate solids whose volumes have the

ratio 4: 1.

(b) What is the ratio of the volumes of the solids gellenlted by revolving the regions about the x-axis?

39. Perpendicular Tangents Show that the tangents to the curve y2 = 4px from any point on the line x -p are perpendicular.

40. Maximizing Area Find the dimensions of the rectangle of largest area that can be inscribed in the x 2 + 4y2 = 4 with its sides parallel to the coordinate axes. What is the area of the rectangle?

41. Volume Find the volume of the solid generated by revolving the region enclosed by the ellipse 9x 2 + 36 about the (a) x-axis, (b) y-axis.

42. Volume The "triangular" region in the first quadrant bounded by the x-axis, the line x 4, and the hyperbola 9x2 4y2 = 36 is revolved about the x-axis to generate a solid. Find the volume of the solid.

43. Volume The region bounded on the left by the y-axis, on the 2right by the hyperbola x y2 I, and above and below by the

lines y = ::t3 is revolved about the y-axis to generate a solid. Find the volume of the solid.

Extending the Ideas 44. Suspension Bridge Cables The suspension bridge cable

shown here supports a uniform load of w pounds per horizontal foot. It can be shown that if H is the horizontal tension of the cable at the origin, then the curve of the cable satisfies the equation

Show that the cable hangs in a parabola by solving this differential equation subject to the initial condition that y 0 when x = O.

o x

590 Appendices

45. Ripple Tank Circular waves were made by touching the surface of a ripple tank, first at a point A and shortly thereafter at a nearby point B. As the waves expanded, their points of intersection appeared to trace a hyperbola. Did they really do that? To find out, we can model the waves with circles in the plane centered at nearby points labeled A and B as in the accompanying figure.

At time t, the point P is rA(t) units from A and rn(t) units from B. Since the radii of the circles increase at a constant rate, the rate at which the waves are traveling is

E!t1. = drs dt dt .

Conclude from this equation that rA rH has a constant value, so that P must lie on a hyperbola with foci at A and B.

46. How the Astronomer Kepler Used String to Draw Parabolas Kepler's method for drawing a parabola (with more modem tools) requires a string the length of a T square and a table whose edge can serve as the parabola's directrix. Pin one end of the string to the point where you want the focus to be and the other end to the upper end of the T square. Then, holding the string taut against the T square with a pencil, slide the T square along the table's edge. As the T square moves, the pencil will trace a parabola. Why?

B Directrix

Figure A5.18 The ellipse changes from a circle to a line segment as c increases

AS.2 Classifying Conic Sections by Eccentricity What you'll learn about Ellipses and Orbits Ellipses and Orbits We now associate with each conic section a number called the eccentricity. The eccentric Hyperbolas ity tells whether the conic is a circle, ellipse, parabola, or hyperbola, and, in the case of

Focus-Directrix Equation ellipses and hyperbolas, describes the conic's proportions. We begin with the ellipse. Although the center-to-focus distance c does not appear in the equation ... and why

This section provides basic 1, (a> b)

information about eccentricity

of conic sections. for an ellipse, we can still determine c from the equation

c=

If we fix a and vary c over the interval 0 :::;; c a, the resulting ellipses will vary in shape (Figure AS.18). They are circles if c 0 (so that a = b) and flatten as c increases. If c = a, the foci and vertices overlap and the ellipse degenerates into a line segment.

We use the ratio of c to a to describe the various shapes the ellipse can take. We call this ratio the ellipse's eccentricity.

DEFINITION Eccentricity of Ellipse

The eccentricity of the ellipse x2/a 2 + y2/b2 = 1 (a> b) is

Va 2 2 e =. = ba a

ISM'!!...,J Eccentricities of planetary orbits

Mercury 0.21 Venus 0.Q1

Earth O.oz Mars 0.09

Jupiter 0.05

Saturn 0.06

Uranus 0.05

Neptune om Pluto 0.25

j comet

IEdmund Halley (1656-1742; pronouncedI "haw-ley"), British biologist, geologist, J sea captain, pirate, spy, Antarctic voy-I ager, astronomer, adviser on fortifica; tions, company founder and director, 1and the author of the first actuarial j mortality tables, was also the mathe1matician who pushed and harried ! Newton into writing his PrinCipia.1Despite these accomplishments, Halley . is known today chiefly as the man who ! calculated the orbit of the great comet lj. of 1682: "wherefore if according to what

we have already said [the comet] should j return again about the year 1758, can-I did posterity will not refuse to acknowl~I' edge that this was first discovered by ~ an Englishman." Indeed, candid posterity Idid not refuse-ever since the comet's

return in 1758, it has been known as

Halley's comet.

Last seen rounding the sun during the winter and spring of 1985-1986, the comet is due to return in the year .2062. The comet has made about 2000 ,cycles so far with about the same number to go before the sun erodes it away.


The planets in the solar system revolve around the sun in elliptical orbits with the sun at one focus. Most of the orbits are nearly circular, as can be seen from the eccentricities in TableA5.2. Pluto has a fairly eccentric orbit, with e 0.25, as does Mercury, with e 0.21. Other members of the solar system have orbits that are even more eccentric. Icarus, an asteroid about I mile wide that revolves around the sun every 409 Earth days, has an orbital

. eccentricil:'J of 0.83 (Figure A5 .19).

Mars

Earth

Icarus

Figure A5.19 The orbit of the asteroid Icarus is highly eccentric. Earth's orbit is so nearly circular that its foci lie inside the sun.

EXAMPLE 1 Finding the Eccentricity of an Orbit

The orbit of Halley's comet is an ellipse 36.18 astronomical units long by 9.12 astro

nomical units wide. (One astronomical unit [AU] is 149,597,870 km, the semimajor

axis of Earth's orbit.) Its eccentricity is

Y(36.18/2)2 - (9.12/2)2 e

a (1/2)(36.18)

Y(18.09)2 - (4.56)2

18.09

=0.97.

Now try Exercise 1.

EXAMPLE 2 Locating Vertices

Locate the vertices of an ellipse of eccentricity 0.8 whose foci lie at the points (0, 7).

SOLUTION

Since e cia, the vertices are the points (0, a) where

c 7

a = --; = Qg = 8.75,

or (0, 8.75). Now try Exercise 5.

http:1/2)(36.18

592 Appendices

Whereas a parabola has one focus and one directrix, each ellipse has two foci and two directrices. These are the lines perpendicular to the major axis at distances ale from the center. The parabola has the property that

PF = lPD (l)

for any point P on it, where F is the focus and D is the point nearest P on the directrix. For an ellipse, it can be shown that the equations that replace (1) are

(2)

Here, e is the eccentricity, P is any point on the ellipse, Fj and are the foci, and Dr and D2 are the points on the directrices nearest P (Figure A5.20).

yDirectril( I Directril( 2

a a x x e e b

Figure A5.20 The foci and directrices of the ellipse x 2/a 2 + y2/b 2 = 1. Directrix 1 corresponds to focus F j , and directrix 2 to focus F2.

In each equation in (2) the directrix and focus must correspond; that if we use the distance from P to FI , we must also use the distance from P to the directrix at the same end of the ellipse. The directrix x = corresponds to Fj (-c, 0), and the directrix x ale corresponds to F2(c, 0).

Hyperbolas

The eccentricity of a hyperbola is also e cia, only in this case c equals instead of In contrast to the eccentricity of an ellipse, the eccentricity of a hyperbola is always greater than 1.

DEFINITION Eccentricity of Hyperbola

The eccentricity of the hyperbola x 21a2 y21b2 1 is

c e=-=---

a a

In both ellipse and hyperbola, the eccentricity is the ratio of the distance between the foci to the distance between the vertices (because cia 2e12a).

y Directrix 1 Directrix 2

ax= -e a

x = e +--+--tr:-----:;f/P(x, y)

_----4!>"-H-----:+-+-+---8-----,------'>-x

Figure A5.21 The foci and directrices of the hyperbola xZ/a2 - yZ/b 2 = 1. No matter where P lies on the hyperbola, PFt = e PDt, and PFz = e PD2


In an ellipse, the foci are closer together than the vertices and the ratio is less than 1. In a hyperbola, the foci are farther apart than the vertices and the ratio is greater than 1.

EXAMPLE 3 Finding Eccentricity

Find the eccentricity of the hyperbola 9xz - 16yZ = 144.

SOLUTION

We divide both sides of the hyperbola's equation by 144 to put it in standard form,

obtaining

or

With a Z = 16 and b Z= 9, we find that c = v'a Z+ b2 = v'l6+9 = 5,

so

c 5

e=-=

a 4' Now try Exercise 15.

As with the ellipse, it can be shown that the lines x = "a/e act as directrices for the hyperbola and that

(3)

Here P is any point on the hyperbola, F[ and Fz are the foci, and D) and D2 are the points nearest P on the directrices (Figure A5.21).

Focus-Directrix Equation

To complete the picture, we define the eccentricity of a parabola to be e = 1. Equations 1-3 then have the common form PF = e PD.

DEFINITION Eccentricity of Parabola

The eccentricity of a parabola is e = 1.

The focus-directrix equation PF = e PD unites the parabola, ellipse, and hyperbola in the following way. Suppose that the distance PF of a point P from a fixed point F (the focus) is a constant multiple of its distance from a fixed line (the directrix). That is, suppose

PF=ePD, (4)

where e is the constant of proportionality. Then the path traced by P is

(a) a parabola if e = 1,

(b) an ellipse of eccentricity e if e < 1, and (c) a hyperbola of eccentricity e if e > 1.

Equation 4 may not look like much to get excited about. There are no coordinates in it and when we try to translate it into coordinate fonn it translates in different ways, depending on the size of e. At least, that is what happens in Cartesian coordinates. However, in polar coordinates, the equation PF = e PD translates into a single equation regardless of the value of e, an equation so simple that it has been the equation of choice of astronomers and space scientists for nearly 300 years.

594 Appendices

y

x=l

Figure A5.22 The hyperbola in Example 4.

Section A5.2 Exercises

Given the focus and corresponding directrix of a hyperbola centered at the origin and with foci on the x-axis, we can use the dimensions shown in Figure AS.21 to find e. Knowing e, we can derive a Cartesian equation for the hyperbola from the equation PF = e PD, as in the next example. We can find equations for ellipses centered at the origin and with foci on the x-axis in a similar way, using the dimensions shown in Figure AS.20.

EXAMPLE 4 Using Focus and Directrix

Find a Cartesian equation for the hyperbola centered at the origin that has a focus at (3, 0) and the line x 1 as the corresponding directrix.

SOLUTION

We first use the dimensions shown in Figure AS.21 to find the hyperbola's eccentricity. The focus is

(c,O) (3,0), so c = 3.

The directrix is the line

x a

1, so a = e. e

When combined with the equation e cia that defines eccentricity, these results give

e c 3

so e2 = 3 and e \13. a e

Knowing e, we can now derive the equation we want from the equation PF e PD. In the notation of Figure AS.22, we have

PF ePD Eq. 4

V3lx -II e

x 2 - 6x + 9 + y2 3(x2 - 2x + 1)

2x2 y2 6

2x 6 = 1.3

Now try Exercise 19.

In Exercises 1-4, find the eccentricity, foci, and directrices of the ellipse.

1. 16x2 + 400 2. 2x2 + y2 2 3. 3x2 + 2y2 6 4. 6x2 + 9y2 54

Exercises 5-8 give the foci or vertices and the eccentricities of ellipses centered at the origin of the xy-plane. In each case, find the ellipse's standard-form equation.

5. Foci: (0,

Eccentricity: 0.5

6. Foci: (:t8,0)

Eccentricity: 0.2

7. Vertices: (:t 10,0)

Eccentricity: 0.24

8. Vertices: (0, :t70)

Eccentricity: 0.1


Exercises 9 and 10 give foci and corresponding directrices of ellipses centered at the origin of the xy-plane. In each case, use the dimensions in Figure AS.20 to find the eccentricity of the ellipse. Then find the ellipse's standard-form equation.

9. Focus: (v5,0)

". 9

DIrectnx: x = v5

10. Focus: (-4,0)

Directrix: x = -16

11. Draw an ellipse of eccentricity 4/S. Explain your procedure.

12. Draw the orbit of Pluto (eccentricity 0.2S) to scale. Explain your procedure.

13. The endpoints of the major and minor axes of an ellipse are (1, I), (3,4), (1,7), and (-1,4). Sketch the ellipse, give its equation in standard form, and find its foci, eccentricity, and directrices.

14. Find an equation for the ellipse of eccentricity 2/3 that has the line x = 9 as a directrix and the point (4, 0) as the corresponding focus.

In Exercises IS-18, find the eccentricity, foci, and directrices of the hyperbola.

15. 9x2 - 16y2 = 144

x 216. y2 - = 8

17. 8x2 - 2y2 = 16 18. Sy2 - 2x2 = 16

Exercises 19 and 20 give the eccentricities and the vertices or foci of hyperbolas centered at the origin of the .1.)'-plane. In each case, find the hyperbola's standard-form equation.

19. Eccentricity: 3

Vertices: (0, 2: 1)

20. Eccentricity: 3

Foci: (2:3,0)

Exercises 21 and 22 give foci and corresponding directrices of hyperbolas centered at the origin of the xy-plane. In each case, find the hyperbola's eccentricity. Then find the hyperbola's standard-form equation.

21. Focus: (4, 0)

Directrix: x = 2

22. Focus: (-2,0)

Directrix: x 2

23. A hyperbola of eccentricity 3/2 has one focus at (I, -3), The corresponding directrix is the line y 2. Find an equation for the hyperbola.

Explorations 24. The Effect of Eccentricity on a Hyperbola's Shape

What happens to the graph of a hyperbola as its eccentricity increases? To find out, rewrite the equation x 2/a 2 - y2/b2 = I in terms of a and e instead of a and h. Graph the hyperbola for various values of e and describe what you find.

25. Determining Constants What values of the constants a, b, and c make the ellipse

4x2 + y2 + ax + by + c = 0 lie tangent to the x-axis at the origin and pass through the point (-1, 2)? What is the eccentricity of the ellipse?

Extending the Ideas 26. The Reflective Property of Ellipses An ellipse is revolved

about its major axis to generate an ellipsoid. The inner surface of the ellipsoid is silvered to make a mirror. Show that a my of light emanating from one focus will be reflected to the other focus. Sound waves also follow such paths, and this property is used in constructing "whispering galleries." (Hint: Place the ellipse in standard position in the xy-plane and show that the lines from a point P on the ellipse to the two foci make congruent angles with the tangent to the ellipse at P.)

27. The Reflective Property of Hyperbolas Show that a ray of light directed toward one focus of a hyperbolic mirror, as in the accompanying figure, is reflected toward the other focus. (Hint: Show that the tangent to the hyperbola at P bisects the angle made by segments PF1 and PF2.)

y

28. A Confocal Ellipse and Hyperbola Show that an ellipse and a hyperbola that have the same foci A and B, as in the accompanying figure, cross at right angles at their points of intersection. [Hint: A ray of light from focus A that met the hyperbola at P would be reflected from the hyperbola as if it came directly from B (Exercise 27). The same ray would be reflected off the ellipse to pass through B (Exercise 26).J

596 Appendices


Quadratic Curves

Cross Product Term

Rotating Axes to Eliminate Bxy

Possible Graphs of Quadratic Equations

Discriminant Test

Technology Application

... and why

This section provides basic information about general quadratic equations in two variables and their graphs.

y

----------+-~~~---------+x

Figure A5.23 The focal axis of the hyperbola 2xy 9 makes an angle of 'Tf/4 radians with the positive x-axis.

y

"

Figure AS.24 A counterclockwise rotation through angle a about the origin.

A5.3 Quadratic Equations and Rotations Quadratic Curves

In this section, we examine one of the most amazing results in analytic geometry, which is that the Cartesian graph of any equation

AX2 + Bxy + Cy2 + Dx + Ey + F 0, (I)

in which A, B, and C are not all zero, is nearly always a conic section. The exceptions are the cases in which there is no graph at all or the graph consists of two parallel lines. It is conventional to call all graphs of Equation 1, curved or not, quadratic curves.

Cross Product Term

You may have noticed that the term Bxy did not appear in the equations for the conic sections in Section AS.I. This happened because the axes of the conic sections ran parallel to (in fact, coincided with) the coordinate axes.

To see what happens when the parallelism is absent, let us write an equation for a hyperbola with a 3 and foci at F1(-3, -3) and F2(3, 3) (Figure AS.23). The equation IPF1 - PF2 1 2a becomes IPF1 - PF2 1 2(3) = 6 and

:t6.

When we transpose one radical, square, solve for the remaining radical and square again, the equation reduces to

2xy = 9, (2)

a case of Equation I in which the cross product term is present. The asymptotes of the hyperbola in Equation 2 are the x- and y-axes, and the focal axis makes an angle of'Tf/4 radians with the positive x-axis. As in this example, the cross product term is present in Equation 1 only when the axes of the conic are tilted.

Rotating Axes to Eliminate Bxy

To eliminate the xy-term from the equation of a conic, we rotate the coordinate axes to eliminate the "tilt" in the axes of the conic. The equations for the rotations we use are derived in the following way. In the notation of Figure AS.24, which shows a counterclockwise rotation about the origin through an angle a,

x OM = OP cos (8 + a) OP cos 8 cos a OP sin 8 sin a (3)

y = MP OP sin (8 + a) OP cos 8 sin a + OP sin 8 cos a.

Since

OPcos 8 = OM' x' and

OP sin 8 = M'P = y', the equations in (3) reduce to the following.

Equations for Rotating Coordinate Axes

x = x' cos a - y' sin a (4)

y x' sin a + y' cos a

Figure A5.25 The hyperbola in Example 1 (x' and y' are the new coordinates) .


EXAMPLE 1 Changing an Equation

The x- and y-axes are rotated through an angle of 7T/4 radians about the origin. Find an equation for the hyperbola 2xy = 9 in the new coordinates.

SOLUTION

Since cos 7T/4 = sin 7T/4 = 1IV2, we substitute

x' - y' x' + y'

y=-x= V2 ' V2 from Equations 4 into the equation 2xy = 9, obtaining

See Figure A5.25. Now try Exercise 35.

If we apply Equations 4 to the quadratic Equation 1, we obtain a new quadratic equation

A'X'2 + B'x'y' + C'y'2 + D'x' + E'y' + F' = O. (5)

The new and old coefficients are related by the equations

A' = A cos2 a + B cos a sin a + C sin2 a B' = B cos 2a + (C - A) sin 2a C' = A sin2 a - B sin a cos a + C cos2 a (6) D' = D cos a + E sin a E' = -D sin a + E cos a F'=F.

Equations 6 show, among other things, that if we start with an equation for a curve in which the cross product term is present (B =1= 0), we can find a rotation angle a that produces an equation in which no cross product term appears (B' = 0). To find a, we set B' = 0 in the second equation in (6) and solve the resulting equation,

B cos 2a + (C - A) sin 2a = 0,

for a. In practice, this means determining a from one of the two equations.

A-C B cot 2a = -B-- or tan 2a = -A--' (7)-C

EXAMPLE 2 Eliminating a Cross Product Term

The coordinate axes are to be rotated through an angle a to produce an equation for the curve

2x2 + V3xy + y2 - 10 = 0

that has no cross product term. Find a suitable a and the couesponding new equation. Identify the curve.

continued

598 Appendices

2

2a I

Figure A5.26 This triangle identifies 2a = coc l (1/\13) as 7T/3. (Example 2)

Figure A5.27 The conic section in Example 2.

SOLUTION

The equation 2x2 + \/3xy + y2 10 = 0 has A 2, B = \/3, and C = 1. We substitute these values into Equation 7 to find 0::

A-C 2 1 1 cot 20: = -B-

\/3 \/3 From the right triangle in Figure A5.26, we see that one appropriate choice of angle is 20: 11'/3, so we take 0: = 11'/6. Substituting 0: = 11'/6, A = 2, B \/3, C = 1,

D E = 0, and F -10 into Equations 6 gives

A' ~, B' 0, C'=t, D'=E'=O, F' -10. Equation S then gives

,0S +L= 14 20 .

The curve is an ellipse with foci on the new y'-axis (Figure AS.27).


Possible Graphs of Quadratic Equations

We now return to the graph of the general quadratic equation. Since axes can always be rotated to eliminate the cross product term, there is no loss of

generality in assuming that this has been done and that the equation has the form

AX2 + Cy2 + Dx + Ey + F = O. (8)

Equation 8 represents

(a) a circle if A = C '* 0 (special cases: the graph is a point or there is no graph at all); (b) a parabola if Equation 8 is quadratic in one variable and linear in the other;

(c) an ellipse if A and C are both positive or both negative (special cases: circles, a single point, or no graph at all);

(d) a hyperbola if A and C have opposite signs (special case: a pair of intersecting lines);

(e) a straight line if A and C are zero and at least one of D and E is different from zero;

(f) one or two straight lines if the left-hand side of Equation 8 can be factored into the product of two linear factors.

See Table A5.3 (on page 600) for examples.

Discriminant Test

We do not need to eliminate the xy-term from the equation

AX2 + Bxy + Cy2 + Dx + Ey + F = 0 (9)

to tell what kind of conic section the equation represents. If this is the only information we want, we can apply the following test instead.

As we have seen, if B '* 0, then rotating the coordinate axes through an angle 0: that satisfies the equation

A-C cot 20: (10)

B

will change Equation 9 into an equivalent form

A'x'l + C'y'2 + D'x' + E'y' + F' = 0 (11)

without a cross product term.


Now, the graph of Equation 11 is a (real or degenerate)

(a) parabola if A' or C' = 0; that is, if A'C' = 0;

(b) ellipse if A' and C' have the same sign; that is, if A'C' > 0; (c) hyperbola if A' and C' have opposite signs; that is, if A'C' < O.

It can also be verified from Equations 6 that for any rotation of axes,

B2 - 4AC = B'2 - 4A'C'. (12)

This means that the quantity B2 - 4A C is not changed by a rotation. But when we rotate through the angle a given by Equation 10, B' becomes zero, so

. B2 - 4AC = -4A'C'.

Since the curve is a parabola if A'C' = 0, an ellipse if A'C' > 0, and a hyperbola if A'C' < 0, the curve must be a parabola if B2 - 4AC = 0, an ellipse if B2 - 4AC < 0, and a hyperbola if B2 - 4AC > O. The number B2 - 4AC is called the discriminant of Equation 9.

Discriminant Test

With the understanding that occasional degenerate cases may arise, the quadratic curve AX2 + Bxy + Cy2 + Dx + Ey + F = 0 is

(a) a parabola if B2 - 4AC = 0,

(b) an ellipse if B2 - 4AC < 0, (c) a hyperbola if B2 - 4AC > O.

EXAMPLE 3 Applying the Discriminant Test

(a) 3x2 - 6xy + 3y2 + 2x - 7 = 0 represents a parabola because

B2 - 4AC = (-6)2 - 433 = 36 - 36 = O.

(b) x 2 + xy + y2 - I = 0 represents an ellipse because

B2 - 4AC = (1)2 - 4 I I = -3 < O.

(c) xy - y2 - 5y + I = 0 represents a hyperbola because

B2 - 4AC = (1)2 - 4(0)(-1) = 1 > O.


Technology Application

How Some Calculators Use Rotations to Evaluate Sines and Cosines Some calculators use rotations to calculate sines and cosines of arbitrary angles. The procedure goes something like this: The calculator has, stored,

1. ten angles or so, say

= sin- 1 (10-1), = sin- 1 (10-2 ), .. , alD = sin- I (10- 10 ),a l a 2 and

2. twenty numbers, the sines and cosines of the angles ai' a 2 , ... , alD'

To calculate the sine and cosine of an arbitrary angle e, we enter e(in radians) into the calculator. The calculator substracts or adds mUltiples of 271" to eto replace eby the angle between 0 and 271" that has the same sine and cosine as e(we continue to call the angle e).

600 Appendices

(cosO, sinO) The calculator then "writes" 0 as a sum of multiples of a t (as many as possible without overshooting) plus multiples of al (again, as many as possible), and so on, working its way to a 10' This gives

The calculator then rotates the point (1, 0) through mt copies of at (through aI' m t times in succession), plus m 2 copies of a2' and so on, finishing off with mtO copies of a lO (Figure AS.28). The coordinates of the final position of (1,0) on the unit circle are the val

(1,0) x ues the calculator gives for (cos 0, sin 0).o

Figure A5.28 To calculate the sine and cosine of an angle ebetween 0 and 21T, the calculator rotates the point (I, 0) to an appropriate location on the unit circle and displays the resulting coordinates.

1ff1"M"I' Examples of quadratic curves AX2 + Bxy + Cy2 + Dx + Ey + F 0

A B C D E F Equation Remarks

Circle -4 x 2 + y2 = 4 A = C;F

2


Section A5.3 Exercises

Use the discriminant B2 - 4AC to decide whether the equations in Exercises 1-16 represent parabolas, ellipses, or hyperbolas.

1. x - 3xy + y" -- x = 0 2. 3x2 - ISxy + 27y2 - 5x + 7y = -4 3. 3x2 - txy + Vl7y2 = 1 4. 2x2 - Vl5xy + 2y2 + X + Y = 0 5. x 2 + 2xy + y2 + 2x - y + 2 = 0 6. 2x2 - y2 + 4xy - 2x + 3y = 6 7. x 2 + 4xy + 4y2 - 3x = 6 S. x2 + y2 + 3x - 2y = 10 9. xy + y2 - 3x = 5

10. 3x2 + 6xy + 3y2 - 4x + 5y = 12 11. 3x2 - 5xy + 2y2 - 7x - 14y = -I 12. 2x2 - 4.9xy + 3y2 - 4x = 7 13. x 2 - 3xy + 3y2 + 6y = 7 14. 25x2 + 21xy + 4y2 - 350x = 0 15. 6x2 + 3xy + 2y2 + 17y + 2 = 0 16. 3x2 + 12xy + 12y2 + 435x - 9y + 72 = 0

In Exercises 17-26, rotate the coordinate axes to change the given equation into an equation that has no cross product (xy) term. Then identify the graph of the equation. (The new equations will vary with the size and direction of the rotation you use.)

17. xy = 2 IS. x2 + xy -+ y2 = I 19. 3x2 + 2V3xy + y2 - Sx + SV3y = 0 20. x 2 - V3xy + 2y2 = I

2 21. x 2xy + y2 = 2 22. 3x2 - 2V3xy + y2 = I 23. Yzx2 + 2Yzxy + Yzy2 - Sx + Sy = 0 24. xy - y - x + I = 0 25. 3x2 + 2xy + 3y2 = 19 26. 3x2 + 4V3xy - y2 = 7

27. Find the sine and cosine of an angle through which the coordinate axes can be rotated to eliminate the cross product term from the equation

14x2 + 16xy + 2y2 - lOx + 26,370y - 17 = O. Do not carry out the rotation.

2S. Find the sine and cosine of an angle through which the coordinate axes can be rotated to eliminate the cross product term from the equation

4x2 - 4xy + y2 - SVsx - 16Vsy = O. Do not carry out the rotation.

The conic sections in Exercises 17-26 were chosen to have rotation angles that were "nice" in the sense that once we knew cot 2a or tan 2a we could idemify 2a and find sin a and cos a from familiar triangles. The conic sections encountered in practice may not have such nice rotation angles, and we may have to use a calculator to determine a from the value of cot 2a or tan 2a.

In Exercises 29-34, use a calculator to find an angle a through which the coordinate axes can be rotated to change the given equation into a quadratic equation that has no cross product term. Then find sin a and cos a to two decimal places and use Equations 6 to find the coefficients of the new equation to the nearest decimal place. In each case, say whether the conic section is an ellipse, hyperbola, or parabola.

29. x 2 - xy + 3y2 + X - y - 3 = 0 30. 2x2 + xy - 3y2 + 3x - 7 = 0 31. x 2 - 4xy + 4y2 - 5 = 0 32. 2x2 - 12xy + lSy2 - 49 = 0 33. 3x2 + 5xy + 2y2 - Sy - I = 0 34. 2x2 + txy + 9)'2 + 20x - S6 = 0

35. The Hyperbola xy = a The hyperbola xy = I is one of many hyperbolas of the form xy = a that appear in science and mathematics.

(a) Rotate the coordinate axes through an angle of 45 to change the equation xy = I into an equation with no xy-term. What is the new equation?

(b) Do the same for the equation xy = a.

36. Writing to Learn Can anything be said about the graph of the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 if

. AC < O? Give reasons for your answer.

37. Writing to Learn Does any nondegenerate conic section Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 have all of the following properties?

(a) It is symmetric with respect to the origin.

(b) It passes through the point (1, 0).

(c) It is tangent to the line y = 1 at the point (-2. I).

Give reasons for your answer.

3S. When A = C Show that rotating the axes through an angle of 7T/4 radians will eliminate the xy-term from Equation 1 whenever A = C.

39. Identifying a Conic Section

(a) What kind of conic section is the curve xy + 2x - y = O? (b) Solve the equation xy + 2x - y = 0 for y and sketch the curve as the graph of a rational function of x.

(c) Find equations for the lines parallel to the line y = -2x that are normal to the curve. Add the lines to your sketch.

602 Appendices

40. Sign of AC Prove or find counterexamples to the following statements about the graph of

AX2 + Bxy + Cy2 + Dx + Ey + F O.

(a) If AC > 0, the graph is an ellipse. (b) If AC > 0, the graph is a hyperbola. (c) If AC < 0, the graph is a hyperbola.

Explorations 41. 90 0 Rotations What effect does a 900 rotation about the

origin have on the equations of the following conic sections? Give the new equation in each case.

(a) The ellipse x 2/a 2 + y2/b z =., I (a > b) (b) The hyperbola x2faz y2/b2 =., 1

a 2(c) The circlex2 + y2 =., (d) The line y mx

(e) The line y mx + b

42. 180 0 Rotations What effect does a 1800 rotation about the origin have on the equations of the following conic sections? Give the new equation in each case.

(a) The ellipse x2/a 2 + y2/P = 1 (a> b) (b) The hyperbola x2/a2 - y2/b2

a2(c) The circle x 2 + y2 =., (d) The line y = mx {e} The line y = mx + b

Extending the Ideas 43. Degenerate Conic Section

(a) Decide whether the equation

x 2 + 4xy + 4y2 + 6x + 12y + 9 0

represents an ellipse, a parabola, or a hyperbola.

(b) Show that the graph of the equation in part (a) is the line 2y -x - 3.

44. Degenerate Conic Section

(a) Decide whether the conic section with equation

9x2 + 6xy + y2 12x - 4y + 4 0

represents a parabola, an ellipse, or a hyperbola.

(b) Show that the graph of the equation in part (a) is the line y -3x + 2.

45. A Nice Area Formula for Ellipses When B2 4AC is negative, the equation

Ax2 + Bxy + Cy2 = I represents an ellipse. If the ellipse's semi-axes are a and b, its area is Tfab (a standard formula). Show that the area is also 2Tf/V4AC - B2. (Hint: Rotate the coordinate axes to eliminate thexy-term and apply Equation 12 to the new equation.)

46. Other Rotation Invariants We describe the fact that B'2 - 4A'C' equals B2 4AC after a rotation about the origin by saying that the discriminant of a quadratic equation is an invariant of the equation. Use Equations 6 to show that the numbers (a) A + C and (b) D2 + E2 are also invariants, in the sense that

A' + C' A. + C and D'2 + E'2 = D2 + E2. We can use these equalities to check against numerical errors when we rotate axes. They can also be helpful in shortening the work required to find values for the new coefficients.

A6


Background

Definitions

Identities

Derivatives and Integrals

Inverse Hyperbolic Functions

Identities for sech-1 x, csch-1 X, coth-1 X

Derivatives of Inverse Hyperbolic Functions; Associated Integrals

. ,. and why

This section provides basic information about hyperbolic functions and their derivatives and integrals.

Section A6 Hyperbolic Functions 603

Hyperbolic Functions

Background .suspension cables like th~se of the Golden Gate Bridge, which support a constant load per horizontal foot, hang in parabolas (Section AS.I, Exercise 44). Cables like power line cables, which hang freely, hang in curves called hyperbolic cosine curves.

Besides describing the shapes of hanging cables, hyperbolic functions describe the motions of waves in elastic solids, the temperature distributions in metal cooling fins, and the motions of falling bodies that encounter air resistance proportional to the square of the velocity. If a hanging cable were turned upside down (without changing shape) to form an arch, the internal forces, then reversed, would once again be in eqUilibrium, making the inverted hyperbolic cosine curve the ideal shape for a self-standing arch. The center line of the Gateway Arch to the West in St. Louis follows a hyperbolic cosine curve.

Definitions The hyperbolic cosine and sine functions are defined by the first two equations in Table A6.1 . The table also defines the hyperbolic tangent, cotangent, secant, and cosecant. As we will see, the hyperbolic functions bear a number of similarities to trigonometric functions after which they are named.

li'MQt411 The six basic hyperbolic functions

Hyperbolic cosine of x: cosh x = eX + e-X

2

Pronouncing "cosh" and "sinh" Hyperbolic sine of x:

"Cosh" is often pronounced "kosh," rhyming with "gosh" or "gauche." "Sinh" is pronounced as if spelled "cinch" or "shine."

Hyperbolic tangent:

Hyperbolic cotangent:

h tan x =

cothx

sinh x cosh x

cosh x sinh x

Hyperbolic secant: sech x = 1 h

cos x 2

Hyperbolic cosecant: cschx sinh x

2

1i!11d4itJ Identities for hv'oelrbllilic functions

sinh 2x = 2 sinh x cosh x cosh 2x = cosh2 X + sinh2 x

cosh 2x + 1h2cos x = 2

. h? cosh 2x sm -x= 2

cosh2 x sinh2 x

tanh 2 x sech2 x

coth2 x = 1 + csch2 X

See Figure A6.1 for graphs.

Identities Hyperbolic functions satisfy the identities in Table A6.2. Except for differences in sign, these are identities we already know for trigonometric functions.

Derivatives and Integrals The six hyperbolic functions, being rational combinations of the differentiable functions eX and e-X, have derivatives at every point at which they are defined (Table A6.3 on the following page). Again, there are similarities with trigonometric functions. The derivative formulas in Table A6.3 lead to the integral formulas seen there.

604 Appendices

ISnrE'I!t+. Derivatives and companion integrals

:x (sinh u) =

! (cosh u) = d

dx (tanhu) =

d dx (coth u) =

d--(sech u) = dx

u) =

cosh u ~:

sinh u ~:

du sech2 u dx

? du -csch- u dx

du-sech u tanh u

dx

du -csch u coth u dx

Isinh u du cosh It + C Icosh II du sinh u + C Isech2 Ii dlt tanh II + C Icsch2 U dlt = -coth It + C Isech It tanh It dlt -sech It + C Icsch It coth It dlt -csch It + C

(a) The hyperbolic sine and its component exponentials.

y

2

y y

y = coth x = lItanh x.

y

(b) The hyperbolic cosine and its component exponentials.

coshx

(c) The graphs of y tanh x and

Y cschx

(e) The graphs of y sinhx and y = csch x = lfsinh x.

(d) The graphs of y = cosh x and y sech x = IIcosh x.

Figure A6.1 The graphs of the six hyperbolic functions.

---


y

y=sinhx y=x

)

(a)

y t

Y = cosh x. 8- f x;;;:: 0

/ y=x

7 :

6

5

/4

3

2

x 0

0)

y

y sech- 1x (x sech y. y;;;:: 0) y x

3

o

Figure A6.2 The graphs of the inverse hyperbolic sine, cosine, and secant of x. Notice the symmetries about the line y=x.

EXAMPLE 1 Finding a Derivative

~(tanh Vi+f2) = sech2 Vi+f2 ~(Vi+f2)& &

~sech2Vi+f2 1 + (2


EXAMPLE 2 Integrating a Hyperbolic Cotangent

du

Jcosh 5x dx = I u = sinh 5x,

Jcoth 5x dx du 5 cash 5x dxsinh 5x 5 u

= +In Iu I + C = +In Isinh 5xI + C Now try Exercise 41.

EXAMPLE 3 Using an Identity to Integrate

Evaluate ilsinh2 x dx.

SOLUTION

Solve Numerically To five decimal places,

NINT(sinh x)2, x, 0, 1) 0.40672.

Confirm Analytically

l ilsinh2 x dx i cosh ~ - 1 dx Table A6.2

11 f 1 I sinh 2x sinh 2 12J (cosh 2x l)dx = 2 2 x] 2 = 0.406724o 0


Inverse Hyperbolic Functions We use the inverses of the six basic hyperbolic functions in integration. Since d(sinhx)ldx cosh x > 0, the hyperbolic sine is an increasing function of x. We denote its inverse by

y = sinh-! x.

For every value of x in the interval -00 < x < 00, the value of y = sinh-1 x is the number whose hyperbolic sine is x (Figure A6.2a).

The function y cosh x is not one-to-one, as we can see from the graph in Figure A6.1. But the restricted function y cosh x, x ::::: 0, is one-to-one and therefore has an inverse, denoted by

y = cosh-1 x.

For every value of x::::: 1, y = cosh-1 X is the number in the interval 0 ~ y < 00 whose hyperbolic cosine is x (FigureA6.2b).

Like y = cosh x, the function y = sech x l/cosh x fails to be one-to-one, but its restriction to nonnegative values of x does have an inverse, denoted by

y = sech-1 x.

For every value of x in the interval (0, 1], y = sech- I x is the nonnegative number whose hyperbolic secant is x (Figure A6.2c).

http:FigureA6.2b

606 Appendices

The hyperbolic tangent, cotangent, and cosecant are one-to-one on their domains and therefore have inverses, denoted by

Y = tanh-1 x, Y = coth-1 x, Y = csch- l X (Figure A6.3).

y y y.. x = cothy x = cschy

y = x = tanhy

y = coth-Ix y = csch-Ixtanh-Ix

------::1-----'::;" x-~--.~-~--_l+-~O--~----~X \0 (b) (c)(a)

Figure A6.3 The graphs of the inverse hyperbolic tangent, cotangent, and cosecant ofx.

~@!I"'W""". Viewing Inverses

Let Xl (t) t, YI(t)=t,

x 2(t) t, Y2(t) l/cosh t,

X3(t) = h(t), Y3(t) x 2(t).

1. Graph the parametric equations simultaneously in a square viewing window that contains x : 6, : Y : 4. Set tMin = 0, tMax 6, and t-step 0.05. Explain what you see. Explain the domain of each function.

2. Let x 4(t) = t, Y4(t) = cosh-l (lit). Graph and compare (x]' Y3) and (Y4' X4)'

Predict what you should see, and explain what you do see.

Ii'MMft' Identities for inverse hyperbolic functions

sech- I x cosh-J ~ X

csch-1 x sinh-1 ~ x

1coth- l X = tanh- 1 x

Identities for sech-1 x, csch-1 X, coth-1 X We use the identities in Table A6.4 to calculate the values of sech-1 x, csch-l x, and coth-1x on calculators that give only COSh-I x, sinh-1 x, and tanh-I x.

Derivatives of Inverse Hyperbolic Functions; Associated Integrals The chief use of inverse hyperbolic functions lies 'in integrations that reverse the derivative formulas in Table A6.5.


ItnDtF'.t5JOl Derivatives of inverse hyperbolic functions

d(sinh- l u) du dx dx

d(cosh- I u) du u>!

dx dx

d(tanh- l u) du lui < 1dx

d(coth- I u) du lui> 1dx dx'

d(sech- I u)

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ppendices - Meyers' Mathmeyersmath.com/wp-content/uploads/2014/01/11-End.pdfSection A1 Formulas from Precalculus Mathematics 563 . 6. Completing the Square If . a =1= 0, we can rewrite