a an a m n, a 0 1, a m a 2. Zero Division by zero is not defined. For any number a: a 0 0 a 0 a b, c d b a b c d

Transcription

1 Appendices A Wht ou ll lern bout Algebr Geometr Trigonometr... nd wh These bsic ides nd formuls from preclculus re useful in clculus. Formuls from Preclculus Mthemtics Algebr. Lws of Eponents m n mn, b m m b m, m n mn, mn n m If, m n mn,, m m. Zero Division b zero is not defined. If :,, For n number : 3. Frctions b c d bc, d bd b cd e f gh b c c b b, d d b d c, b b, c d b b cd b dfh d bcfh e f gh bdfh eh fgbd 4. The Binomil Theorem For n positive integer n, b n n n n b nn n b nn n 3 n3 b 3 nb n b n. For instnce, b b, b b b, b b 3b b 3, b b 6 b 4b 3 b Differences of Like Integer Powers, n n b n b n n b n3 b b n b n For instnce, b b b, 3 b 3 b b b, 4 b 4 b 3 b b b 3.

2 Section A Formuls from Preclculus Mthemtics Completing the Squre If, we cn rewrite the qudrtic b c in the form u C b process clled completing the squre: b c ( b ) c ( b b 4 ( ( ) c ) ( b 4) c 4 b Fctor from the first two terms. Add nd subtrct the squre of hlf the coefficient of. b b b Bring out the 4 4. b b 4) c b 4 Cll this This is ( b ). prt C. u b C u 7. The Qudrtic Formul B completing the squre on the first two terms of the eqution b c nd solving the resulting eqution for (detils omitted), we obtin b b 4c. This eqution is the qudrtic formul. The solutions of the eqution 3 re or 3 7 nd Geometr A re, B re of bse, C circumference, h height, S lterl re or surfce re, V volume. Tringle. Similr Tringles 3. Pthgoren Theorem h c c b c b b b A bh b c b c b c

3 564 Appendices 4. Prllelogrm 5. Trpezoid 6. Circle h b A bh h b A ( b) h r A r, C r 7. An Clinder or Prism with Prllel Bses 8. Right Circulr Clinder r h h h V Bh V r h, S rh 9. An Cone or Prmid. Right Circulr Cone. Sphere h h h s r r V Bh 3 V r h, S rs 3 V 4 3 pr 3, S 4pr Trigonometr. Definitions of Fundmentl Identities Sine: sin u r csc u P(, ) Cosine: Tngent: cos u r sec u tn u cot u O r

4 Section A Formuls from Preclculus Mthemtics 565. Identities sin u sin u, cos u cos u, sin u cos u, sec u tn u, csc u cot u sin u sin u cos u, cos u cos u sin u cos u c os u, sin u c os u sin A B sin A cos B cos A sin B sin A B sin A cos B cos A sin B cos A B cos A cos B sin A sin B cos A B cos A cos B sin A sin B sin ( A p ) cos A, cos sin ( A p ) cos A, cos tn A tn B tn A B tn A tn B tn A tn B tn A B tn A tn B ( A p ) sin A ( A p ) sin A sin A sin B cos A B cos A B cos A cos B cos A B cos A B sin A cos B sin A B sin A B sin A sin B sin A B cos A B sin A sin B cos A B sin A B cos A cos B cos A B cos A B 3. Common Reference Tringles cos A cos B sin A B sin A B Angles nd Sides of Tringle Lw of cosines: c b b cos C Lw of sines: sin A sin B sin C b c B c Are bc sin A c sin B b sin C C b A

5 566 Appendices A Wht ou ll lern bout Mthemticl Induction Principle Other Strting Integers... nd wh Mthemticl induction is n importnt method to provide proofs. Mthemticl Induction Mthemticl Induction Principle Mn formuls, like n nn, cn be shown to hold for ever positive integer n b ppling n iom clled the mthemticl induction principle. A proof tht uses this iom is proof b mthemticl induction or proof b induction. The steps in proving formul b induction re the following. Step : Check tht the formul holds for n. Step : Prove tht if the formul holds for n positive integer n k, then it lso holds for the net integer, n k. Once these steps re completed (the iom ss), we know tht the formul holds for ll positive integers n. B step it holds for n. B step it holds for n, nd therefore b step lso for n 3, nd b step gin for n 4, nd so on. If the first domino flls, nd the kth domino lws knocks over the k st when it flls, ll the dominoes fll. From nother point of view, suppose we hve sequence of sttements S, S,, S n,, one for ech positive integer. Suppose we cn show tht ssuming n one of the sttements to be true implies tht the net sttement in line is true. Suppose tht we cn lso show tht S is true. Then we m conclude tht the sttements re true from S on. EXAMPLE Sum of the First n Positive Integers Show tht for ever positive integer n, n nn. SOLUTION We ccomplish the proof b crring out the two steps. Step : The formul holds for n becuse. Step : If the formul holds for n k, does it hold for n k? The nswer is es, nd here s wh: If k kk, then k k kk k k k k k k k k. The lst epression in this string of equlities is the epression nn for n k. continued

6 Section A Mthemticl Induction 567 The mthemticl induction priniciple now gurntees the originl formul for ll positive integers n. All we hve to do is crr out steps nd. The mthemticl induction principle does the rest. Now tr Eercise. EXAMPLE Sums of Powers of Show tht for ll positive integers n, SOLUTION n n. We ccomplish the proof b crring out the two steps of mthemticl induction. Step : The formul holds for n becuse Step : If then. k k, k k k k k k k k k. Thus, the originl formul holds for n k whenever it holds for n k. With these steps verified, the mthemticl induction principle now gurntees the formul for ever positive integer n. Now tr Eercise 5. Other Strting Integers Insted of strting t n, some induction rguments strt t nother integer. The steps for such n rgument re s follows. Step : Check tht the formul holds for n n (the first pproprite integer). Step : Prove tht if the formul holds for n integer n k n, then it lso holds for n k. Once these steps re completed, the mthemticl induction principle gurntees the formul for ll n n. EXAMPLE 3 Fctoril Eceeding Eponentil Show tht n! 3 n if n is lrge enough. continued

7 568 Appendices SOLUTION How lrge is lrge enough? We eperiment: n n! n It looks s if n! 3 n for n 7. To be sure, we ppl mthemticl induction. We tke n 7 in step nd tr for step. Suppose k! 3 k for some k 7. Then k! k k! k 3 k 7 3 k 3 k. Thus, for k 7, k! 3 k k! 3 k. The mthemticl induction principle now gurntees n! 3 n for ll n 7. Now tr Eercise 7. Section A Eercises. Generl Tringle Inequlit Assuming tht the tringle inequlit b b holds for n two numbers nd b, show tht n n for n n numbers.. Prtil Sums of Geometric Series Show tht if r, then r r r n r n r for ever positive integer n. 3. Positive Integer Power Rule Use the Product Rule, d uv u d v v d u, d d d nd the fct tht d d to show tht d d n n n for ever positive integer n. 4. Products into Sums Suppose tht function f hs the propert tht f f f for n two positive numbers nd. Show tht f n f f f n for the product of n n positive numbers,,, n. 5. Show tht n 3 n for ll positive integers n. 6. Show tht n! n 3 if n is lrge enough. 7. Show tht n n if n is lrge enough. 8. Show tht n 8 for n Sums of Squres Show tht the sum of the squres of the first n positive integers is. Sums of Cubes Show tht the sum of the cubes of the first n positive integers is nn.. Rules for Finite Sums Show tht the following finite sum rules hold for ever positive integer n. n n n () k b k k k (b) n k (c) n k (d) n k k k b k n c k c n k n( n ) n 3 k k k n k b k b k k (An number c) k n c, if k hs the constnt vlue c.. Absolute Vlues Show tht n n for ever positive integer n nd ever rel number..

8 Section A3 Using the Limit Definition 569 A3 Wht ou ll lern bout Limit Definition Finding Delts for Given Epsilons Proving Limit Theorems... nd wh This section provides prctice using the forml definition of limit. L L L f() f() lies in here c for ll in here c c c Figure A3. A preliminr stge in the development of the definition of limit. Using the Limit Definition Limit Definition We begin b setting the stge for the definition of limit. Recll tht the limit of f of s pproches c equls L lim c f L mens tht the vlues f of the function f pproch or equl L s the vlues of pproch (but do not equl) c. Suppose we re wtching the vlues of function f s pproches c (without tking on the vlue of c itself). Certinl we wnt to be ble to s tht f sts within one-tenth of unit of L s soon s sts within some distnce d of c (Figure A3.). But tht in itself is not enough, becuse s continues on its course towrd c, wht is to prevent f from jittering bout within the intervl from L to L without tending towrd L? We cn insist tht f st within or or, of L. Ech time, we find new d-intervl bout c so tht keeping within tht intervl keeps f within e or or, of L. And ech time the possibilit eists tht c jitters w from L t the lst minute. Figure A3. illustrtes the problem. You cn think of this s qurrel between skeptic nd scholr. The skeptic presents e-chllenges to prove tht the limit does not eist or, more precisel, tht there is room for doubt, nd the scholr nswers ever chllenge with d-intervl round c. L L L c f() The chllenge: Mke f() L Figure A3. The first of possibl endless sequence of chllenges. L L L f() f() lies in here c for ll c in here c Figure A3.3 The reltion of the d nd e in the definition of limit. How do we stop this seemingl endless sequence of chllenges nd responses? B proving tht for ever e-distnce tht the chllenger cn produce, we cn find, clculte, or conjure mtching d-distnce tht keeps close enough to c to keep f within tht distnce of L (Figure A3.3). The following definition tht we mde in Section. provides mthemticl w to s tht the closer gets to c, the closer f must get to L. c DEFINITION Limit Let c nd L be rel numbers. The function f hs limit L s pproches c if, given n positive number e, there is positive number d such tht for ll c d f () L e. We write lim f () L. c

9 57 Appendices Finding Delts for Given Epsilons From our work in Chpter we know tht lim 5 3. In Emple, we confirm this result using the definition of limit. 5 3 EXAMPLE Show tht lim 5 3. Using the Definition of Limit SOLUTION Set c, f 5 3, nd L in the definition of limit. For n given e we hve to find suitble d so tht if nd is within distnce d of c, tht is, if d, 5 5 then f is within distnce e of L, tht is, f () e. We find d b working bckwrd from the e-inequlit: e 5 e 3 e5 Figure A3.4 If f 5 3, then e5 gurntees tht f e. (Emple ) Thus we cn tke d e5 (Figure A3.4). If d e5, then e5 e. This proves tht lim 5 3. Now tr Eercise 5. The vlue of d e5 is not the onl vlue tht will mke d impl f 5 5 e in Emple. An smller positive d will do s well. The definition does not sk for best positive d, just one tht will work. We cn use grphs to find d for specific e s in Emple. Intersection X = Y =. [.48,.5] b [.98,.] () EXAMPLE Finding Grphicll For the limit lim.5, find d tht works for e.. Tht is, find d such tht for ll SOLUTION.5 d f.. Here f, c.5, nd L. Figure A3.5 shows the grphs of f nd the two horizontl lines Intersection X =.5556 Y =.99 [.48,.5] b [.98,.] (b) Figure A3.5 We cn see from the two grphs tht if.498.5, then.99 f.. (Emple ) L e..99 nd L e... Figure A3.5 shows tht the grph of f intersects the horizontl line. t bout ,., nd Figure A3.5b shows tht the grph of f intersects the horizontl line.99 t bout.5556,.99. It follows tht.5. f.. Thus, d. works. Now tr Eercise 3.

10 Section A3 Using the Limit Definition 57 EXAMPLE 3 Prove tht lim f 4 if Finding Algebricll, f {, (, 4) (, ) 4 Figure A3.6 The function in Emple 3. SOLUTION Our tsk is to show tht given e there eists d such tht for ll d f 4 e. Step : Solve the inequlit f 4 e to find n open intervl bout c on which the inequlit holds for ll c. For c, we hve f, nd the inequlit to solve is 4 e: 4 e e 4 e 4 e 4 e 4 e 4 e Assume e 4. 4 e 4 e An open intervl bout tht solves the inequlit The inequlit f 4 e holds for ll in the open intervl 4, e 4 e (Figure A3.6). Step : Find vlue of d tht plces the centered intervl d, d inside the open intervl 4, e 4. e Tke d to be the distnce from c to the nerer endpoint of 4, e 4. e In other words, tke d min { 4, e 4 e }, the minimum (the smller) of the two numbers 4 e nd 4 e. If d hs this or n smller positive vlue, the inequlit d will utomticll plce between 4 e nd 4 e to mke For ll, f 4 e. d f 4 e. This completes the proof. Now tr Eercise 3. Wh ws it ll right to ssume e 4 in Emple 3? Becuse, in finding d such tht, for ll, d implied f 4 e 4, we found d tht would work for n lrger e s well. Finll, notice the freedom we gined in letting d min { 4, e 4 e }. We did not hve to spend time deciding which, if either, number ws the smller of the two. We just let d represent the smller nd went on to finish the rgument. Proving Limit Theorems We use the limit definition to prove prts, 3, nd 5 of Theorem (Properties of Limits) from Section..

11 57 Appendices THEOREM Properties of Limits If L, M, c, nd k re rel numbers nd lim f L nd lim g M, c c then. Sum Rule: lim c f g L M. Difference Rule: lim c f g L M 3. Product Rule: lim c f g L M 4. Constnt Multiple Rule: lim c k f k L f L 5. Quotient Rule: lim, M c g M 6. Power Rule: If r nd s re integers, s, then lim c f rs L rs provided L rs is rel number. Proof of the Limit Sum Rule We need to show tht for n e, there is d such tht for ll in the common domin D of f nd g, c d f g L M e. Regrouping terms, we get f g L M f L g M f L g M. b b Here we hve pplied the tringle inequlit, which sttes tht for ll rel numbers nd b, b b. Since lim c f L, there eists number d such tht for ll in D c d f L e. Similrl, since lim c g M, there eists number d such tht for ll in D c d g M e. Let d min {d, d }, the smller of d nd d. If c d then nd c d, so f L e, c d, so g M e. Therefore, f g L M e e e. This shows tht lim c f g L M.

12 Section A3 Using the Limit Definition 573 Proof of the Limit Product Rule We show tht for n e, there is d such tht for ll in the common domin D of f nd g, c d f g LM e. Write f nd g s f L f L, g M g M. Multipl these epressions together nd subtrct LM: f g LM L f LM g M LM LM Lg M M f L f Lg M LM () Lg M M f L f Lg M Since f nd g hve limits L nd M s c, there eist positive numbers d, d, d 3, nd d 4 such tht for ll in D c d f L e3 c d g M e3 e c d 3 f L () 3 M e c d 4 g M. 3 L If we tke d to be the smllest of the numbers d through d 4, the inequlities on the righthnd side of () will hold simultneousl for c d. Then, ppling the tringle inequlit to Eqution, we hve for ll in D, c d implies f g LM Lg M M f L f Lg M Lg M M f L f Lg M e e 3 3 e 3 e e. Vlues from () 3 This completes the proof of the Limit Product Rule. Proof of the Limit Quotient Rule We show tht lim c g M. We cn then conclude tht f lim lim c g c ( f g ) lim f lim L L M c c g M b the Limit Product Rule. Let e be given. To show tht lim c g M, we need to show tht there eists d such tht for ll c d g M e. Since M, there eists positive number d such tht for ll c d g M M. (3) continued

13 574 Appendices For n numbers A nd B it cn be shown tht A B A B nd B A A B, from which it follows tht A B A B. With A g nd B M, this becomes g M g M, which cn be combined with the inequlit on the right in (3) to get, in turn, g M M M g M M M g 3 M Therefore, c d implies tht g 3. Multipl b Mg. (4) g M g M M M g g M g M g M g. Inequlit (4) (5) M M Since M e, there eists number d such tht for ll in D c d M g e M. (6) If we tke d to be the smller of d nd d, the conclusions in (5) nd (6) both hold for ll such tht c d. Combining these conclusions gives c d g M e. This completes the proof of the Limit Quotient Rule. The lst proof we give is of the Sndwich Theorem (Theorem 4) of Section.. THEOREM 4 The Sndwich Theorem If g f h for ll c in some intervl bout c, nd lim g lim h L, c c then lim f L. c Proof for Right-hnd Limits Suppose tht lim c g lim c h L. Then for n e there eists d such tht for ll the inequlit c c d implies L e g L e nd L e h L e. continued

14 Section A3 Using the Limit Definition 575 These inequlities combine with the inequlit g f h to give L e g f h L e, L e f L e, e f L e. Thus, for ll, the inequlit c c d implies f L e. Therefore, lim c f L. Proof for Left-hnd Limits Suppose tht lim c g lim c h L. Then for n e there eists d such tht for ll the inequlit c d c implies L e g L e nd L e h L e. We conclude s before tht for ll, c d c implies f L e. Therefore, lim c f L. Proof for Two-sided Limits If lim c g lim c h L, then g nd h both pproch L s c nd c ; so lim c f L nd lim c f L. Hence lim c f eists nd equls L. Section A3 Eercises In Eercises nd, sketch the intervl, b on the -is with the point c inside. Then find vlue of d such tht for ll, c d b.. 49, b 47, c..759, b 3.39, c 3 In Eercises 3 nd 4, use the grph to find d such tht for ll c d f L e. 3. f() + 4. c 3 L NOT TO SCALE f() c L Eercises 5 8 give function f nd numbers L, c, nd e. Find n open intervl bout c on which the inequlit f L e holds. Then give vlue for d such tht for ll stisfing c d the inequlit f L e holds. Use lgebr to find our nswers. 5. f, L 6, c, e. 6. f, L, c, e. 7. f 9, L 3, c, e 8. f, L 4, c, e.5 Eercises 9 give function f, point c, nd positive number e. () Find L lim c f. Then (b) find number d such tht for ll c d f L e. 9. f 6 5, c 5, e ,, f { 6 4,, c, e.5. f sin, c, e.. f, c, e. 4

15 576 Appendices In Eercises 3 nd 4, use the definition of limit to prove the limit sttement. 3., lim f if f {, 4. lim Relting to Limits Given e, () find n intervl I 5, 5 d, d, such tht if lies in I, then 5 e. (b) Wht limit is being verified? 6. Relting to Limits Given e, () find n intervl I 4 d, 4, d, such tht if lies in I, then 4 e. (b) Wht limit is being verified? 7. Prove the Constnt Multiple Rule for limits. 8. Prove the Difference Rule for limits. 9. Generlized Limit Sum Rule Suppose tht functions f, f, nd f 3 hve limits L, L, nd L 3, respectivel, s c. Show tht their sum hs limit L L L 3. Use mthemticl induction (Appendi A) to generlize this result to the sum of n finite number of functions.. Generlized Limit Product Rule Use mthemticl induction nd the Limit Product Rule in Theorem to show tht if functions f, f,,f n hve limits L, L,,L n, respectivel, s c, then lim c f f f n L L L n.. Positive Integer Power Rule Use the fct tht lim c c nd the result of Eercise to show tht lim c n c n for n integer n.. Limits of Polnomils Use the fct tht lim c k k for n number k together with the results of Eercises 9 nd to show tht lim c f f c for n polnomil function f n n n n. 3. Limits of Rtionl Functions Use Theorem nd the result of Eercise to show tht if f nd g re polnomil functions nd gc, then lim f f c. c g g c 4. Composites of Continuous Functions Figure A3.7 gives the digrm for proof tht the composite of two continuous functions is continuous. Reconstruct the proof from the digrm. The sttement to be proved is this: If f is continuous t c nd g is continuous t f c, then g f is continuous t c. Assume tht c is n interior point of the domin of f nd tht f c is n interior point of the domin of g. This will mke the limits involved two-sided. (The rgument for the cses tht involve one-sided limits re similr.) g f f g f f g g c f (c) g( f (c)) Figure A3.7 The continuit of composites holds for n finite number of functions. The onl requirement is tht ech function be continuous where it is pplied. Here, f is to be continuous t c nd g t f c.

16 Section A4 Proof of the Chin Rule 577 A4 Wht ou ll lern bout Error in the Approimtion f df The Proof... nd wh This section helps eplin pproimting chnge in function vlues. Proof of the Chin Rule Error in the Approimtion f df Let f be differentible t nd suppose tht is n increment of. We know tht the differentil df f is n pproimtion for the chnge f f f in f s chnges from to. How well does df pproimte f? We mesure the pproimtion error b subtrcting df from f: Approimtion error f df f f f f f f f f( ) f() ( f f f ) e f() Secnt slope Cll this prt e. As, the difference quotient f f pproches f (remember the definition of f ), so the quntit in prentheses becomes ver smll number (which is wh we clled it e). In fct, e s. When is smll, the pproimtion error e is smller still. f f e () The Proof true estimted error chnge chnge We wnt to show if f u is differentible function of u nd u g is differentible function of, then f g is differentible function of. More precisel, if g is differentible t nd f is differentible t g, then the composite is differentible t nd d d f g g. Let be n increment in nd let u nd be the corresponding increments in u nd. As ou cn see in Figure A4., d d lim, Figure A4. The grph of s function of. The derivtive of with respect to t is lim. so our gol is to show tht the limit is f g g. B Eqution, u g e g e, where e s. Similrl, since f is differentible t g, f gu e u f g e u, where e s u. Notice lso tht u s. Combining the equtions for u nd gives f g e g e, so f gg e g f (g())e e e. Since e nd e go to zero s goes to zero, three of the four terms on the right vnish in the limit, leving lim f gg. This concludes the proof.

17 578 Appendices A5 Wht ou ll lern bout Circles Interiors nd Eteriors of Circles Prbols Ellipses Aes of n Ellipse Hperbols Asmptotes nd Drwing Reflective Properties Other Applictions... nd wh This section provides some bsic informtion bout equtions of conic sections nd how to sketch their grphs. Conic Sections Overview Conic sections re the pths trveled b plnets, stellites, nd other bodies (even electrons) whose motions re driven b inverse-squre forces. Once we know tht the pth of moving bod is conic section, we immeditel hve informtion bout the bod s velocit nd the force tht drives it. In this ppendi, we stud the connections between conic sections nd qudrtic equtions nd clssif conic sections b eccentricit (Pluto s orbit is highl eccentric while Erth s is nerl circulr). A5. Conic Sections nd Qudrtic Equtions Circles The Greeks of Plto s time defined conic sections s the curves formed b cutting through double cone with plne (Figure A5.). Tod, we define conic sections with the distnce function in the coordinte plne. Circle: plne perpendiculr to cone is Ellipse: plne oblique to cone is Prbol: plne prllel to side of cone Hperbol: plne cuts both hlves of cone () Point: plne through cone verte onl Single line: plne tngent to cone (b) Pir of intersecting lines Figure A5. The stndrd conic sections () re the curves in which plne cuts double cone. Hperbols come in two prts, clled brnches. The point nd lines obtined b pssing the plne through the cone s verte (b) re degenerte conic sections.

18 Section A5 Conic Sections 579 DEFINITION Circle A circle is the set of points in plne whose distnce from given fied point in the plne is constnt. The fied point is the center of the circle; the constnt distnce is the rdius. If, the eqution represents ll the points, in the plne whose distnce from the origin is. These re the points of the circle of rdius centered t the origin. If we shift the circle to plce its center t the point (h, k), its eqution becomes h k. Circle of Rdius Centered t (h, k) h k k Eterior: ( h) ( k) On: ( h) ( k) (h, k) EXAMPLE Finding Center nd Rdius Find the center nd rdius of the circle SOLUTION We convert the eqution to stndrd form b completing the squres in nd : ( 4 ( 4 ) ) ( ) ) 3 ( 4 ) 6 ( 6 ( 6 ) Strt with the given eqution. Gther terms. Move the constnt to the right-hnd side. Add the squre of hlf the coefficient of to ech side of the eqution. Do the sme for. The prentheticl epressions on the left-hnd side re now perfect squres Write ech qudrtic s 6 squred liner epression. With the eqution now in stndrd form, we red off the center s coordintes nd the rdius: h, k, 3 nd 4. Now tr Eercise 3. Interior: ( h) ( k) h Figure A5. The interior nd eterior of the circle h k. Interiors nd Eteriors of Circles The points tht lie inside the circle h k re the points less thn units from h, k. The stisf the inequlit h k. The mke up the region we cll the interior of the circle (Figure A5.).

19 58 Appendices The circle s eterior consists of the points tht lie more thn units from h, k. These points stisf the inequlit h k. EXAMPLE Interpreting Inequlities Inequlit Region Interior of the unit circle Unit circle plus its interior Eterior of the unit circle Unit circle plus its eterior Now tr Eercise 5. Prbols DEFINITION Prbol A set tht consists of ll the points in plne equidistnt from given fied point nd given fied line in the plne is prbol. The fied point is the focus of the prbol. The fied line is the directri. 4p Focus p The verte lies hlfw between directri nd focus. Directri: p p F(, p) Q(, p) P(, ) L If the focus F lies on the directri L, the prbol is the line through F perpendiculr to L. We consider this to be degenerte cse nd ssume henceforth tht F does not lie on L. A prbol hs its simplest eqution when its focus nd directri strddle one of the coordinte es. For emple, suppose tht the focus lies t the point F, p on the positive -is nd tht the directri is the line p (Figure A5.3). In the nottion of the figure, point P, lies on the prbol if nd onl if PF PQ. From the distnce formul, PF p p Figure A5.3 The prbol 4p. PQ p p. When we equte these epressions, squre, nd simplif, we get Directri: p Verte t origin Focus (, p) = 4p Figure A5.4 The prbol 4p. or 4 p 4p. Stndrd form () These equtions revel the prbol s smmetr bout the -is. We cll the -is the is of the prbol (short for is of smmetr ). The point where prbol crosses its is is the verte. The verte of the prbol 4p lies t the origin (Figure A5.3). The positive number p is the prbol s focl length. If the prbol opens downwrd, with its focus t, p nd its directri the line p, Equtions become or 4 p 4p (Figure A5.4). We obtin similr equtions for prbols opening to the right or to the left ( Figure A5.5 nd Tble A5.).

20 Section A5 Conic Sections 58 Directri p 4p 4p Directri p Verte Focus O F(p, ) Focus F( p, ) O Verte () (b) Figure A5.5 () The prbol 4p. (b) The prbol 4p. Tble A5. Stndrd-form equtions for prbols with vertices t the origin (p ) Eqution Focus Directri Ais Opens 4p, p p -is Up 4p, p p -is Down 4p p, p -is To the right 4p p, p -is To the left EXAMPLE 3 Finding Focus nd Directri Find the focus nd directri of the prbol. SOLUTION We find the vlue of p in the stndrd eqution 4p: 4p, so p 5 4. Then we find the focus nd directri for this vlue of p: Focus: p, ( 5, ) Directri: p or 5. Now tr Eercise 7. Ellipses Verte Focus Center Focus Verte Focl is Figure A5.6 Points on the focl is of n ellipse. DEFINITION Ellipse An ellipse is the set of points in plne whose distnces from two fied points in the plne hve constnt sum. The fied points re the foci of the ellipse. The line through the foci is the focl is. The point on the is hlfw between the foci is the center. The points where the focl is nd ellipse cross re the vertices (Figure A5.6).

21 58 Appendices F P(, ) Figure A5.7 How to drw n ellipse. Focus F ( c, ) b Focus O Center F (c, ) F P(, ) Figure A5.8 The ellipse defined b the eqution PF PF is the grph of the eqution b. The quickest w to construct n ellipse uses the definition. Put loop of string round two tcks F nd F, pull the string tut with pencil point P, nd move the pencil round to trce closed curve (Figure A5.7). The curve is n ellipse becuse the sum PF PF, being the length of the loop minus the distnce between the tcks, remins constnt. The ellipse s foci lie t F nd F. If the foci re F c, nd F c, (Figure A5.8) nd PF PF is denoted b, then the coordintes of point P on the ellipse stisf the eqution c c. To simplif this eqution, we move the second rdicl to the right-hnd side, squre, isolte the remining rdicl, nd squre gin, obtining c. () Since PF PF is greter thn the length F F (tringle inequlit for tringle PF F ), the number is greter thn c. Accordingl, c nd the number c in Eqution is positive. The lgebric steps leding to Eqution cn be reversed to show tht ever point P whose coordintes stisf n eqution of this form with c lso stisfies the eqution PF PF. A point therefore lies on the ellipse if nd onl if its coordintes stisf Eqution. If b c, (3) then c b nd Eqution tkes the form b. (4) Eqution 4 revels tht this ellipse is smmetric with respect to the origin nd both coordinte es. It lies inside the rectngle bounded b the lines nd b. It crosses the es t the points, nd, b. The tngents t these points re perpendiculr to the es becuse d d b is zero if nd infinite if. Obtined from Eq. 4 b implicit differentition Aes of n Ellipse Verte ( 4, ) 6 9 (, 3) Verte (4, ) The mjor is of the ellipse in Eqution 4 is the line segment of length joining the points,. The minor is is the line segment of length b joining the points, b. The number itself is the semimjor is, the number b the semiminor is. The number c found from Eqution 3 s c b, is the center-to-focus distnce of the ellipse. Focus Focus ( 7, ) Center ( 7, ) (, 3) Figure A5.9 Mjor is horizontl. (Emple 4) EXAMPLE 4 Mjor Ais Horizontl The ellipse 6 9 (Figure A5.9) hs (5) Semimjor is: 6 4 Semiminor is: b 9 3 continued

22 Section A5 Conic Sections (, 4) Verte Focus (, 7) Center-to-focus distnce: c Foci: c, 7, Vertices:, 4, Center:,. Now tr Eercise 3. ( 3, ) Center Focus Verte (, 7) (, 4) Figure A5. Mjor is verticl. (Emple 5) (3, ) EXAMPLE 5 Mjor Ais Verticl The ellipse, 9 6 (6) obtined b interchnging nd in Eqution 5, hs its mjor is verticl insted of horizontl (Figure A5.). With still equl to 6 nd b equl to 9, we hve Semimjor is: 6 4 Semiminor is: b 9 3 Center-to-focus distnce: c Foci:, c, 7 Vertices:,, 4 Center:,. Now tr Eercise 7. There is never n cuse for confusion in nlzing equtions like (5) nd (6). We simpl find the intercepts on the coordinte es; then we know which w the mjor is runs becuse it is the longer of the two es. The center lws lies t the origin nd the foci lie on the mjor is. Stndrd-Form Equtions for Ellipses Centered t the Origin Foci on the -is: b Center-to-focus distnce: Foci: Vertices: Foci on the -is: b Center-to-focus distnce: Foci: Vertices: b c b c,, b c b, c, In ech cse, is the semimjor is nd b is the semiminor is. Hperbols DEFINITION Hperbol A hperbol is the set of points in plne whose distnces from two fied points in the plne hve constnt difference. The two fied points re the foci of the hperbol.

23 584 Appendices If the foci re F c, nd F c, (Figure A5.) nd the constnt difference is, then point, lies on the hperbol if nd onl if F ( c, ) O F (c, ) P(, ) Figure A5. Hperbols hve two brnches. For points on the right-hnd brnch of the hperbol shown here, PF PF. For points on the lefthnd brnch, PF PF. c c. (7) To simplif this eqution, we move the second rdicl to the right-hnd side, squre, isolte the remining rdicl, nd squre gin, obtining c. (8) So fr, this looks just like the eqution for n ellipse. But now c is negtive becuse, being the difference of two sides of tringle PF F is less thn c, the third side. The lgebric steps leding to Eqution 8 cn be reversed to show tht ever point P whose coordintes stisf n eqution of this form with c lso stisfies Eqution 7. A point therefore lies on the hperbol if nd onl if its coordintes stisf Eqution 8. If we let b denote the positive squre root of c, b c, (9) then c b nd Eqution 8 tkes the more compct form b. () The differences between Eqution nd the eqution for n ellipse (Eqution 4) re the minus sign nd the new reltion c b. From Eq. 9 Like the ellipse, the hperbol is smmetric with respect to the origin nd coordinte es. It crosses the -is t the points,. The tngents t these points re verticl becuse d d b Obtined from Eq. b implicit differentition is infinite when. The hperbol hs no -intercepts; in fct, no prt of the curve lies between the lines nd. Focus Vertices Center Focl is Focus Figure A5. Points on the focl is of hperbol. DEFINITION Prts of Hperbol The line through the foci of hperbol is the focl is. The point on the is hlfw between the foci is the hperbol s center. The points where the focl is nd hperbol cross re the vertices ( Figure A5.). Asmptotes nd Drwing The hperbol b ()

24 Section A5 Conic Sections 585 hs two smptotes, the lines b. The smptotes give the guidnce we need to drw hperbols quickl. (See the drwing lesson.) The fstest w to find the equtions of the smptotes is to replce the in Eqution b nd solve the new eqution for : b b b. hperbol for smptotes Stndrd-Form Equtions for Hperbols Centered t the Origin Foci on the -is: b Center-to-focus distnce: Foci: Vertices: Asmptotes: c b c,, b or b Foci on the -is: b Center-to-focus distnce: Foci: Vertices: Asmptotes: c b, c, b or b Notice the difference in the smptote equtions (b in the first, b in the second). DRAWING LESSON How to Grph the Hperbol b. Mrk the points, nd, b with line segments nd complete the rectngle the determine.. Sketch the smptotes b etending the rectngle s digonls. 3. Use the rectngle nd smptotes to guide our drwing. b b b b b b b b b

25 586 Appendices EXAMPLE 6 Foci on the -Ais The eqution () 4 5 is Eqution with 4 nd b 5 (Figure A5.3). We hve F( 3, ) F(3, ) Figure A5.3 The hperbol in Emple 6. Center-to-focus distnce: c b Foci: c, 3, Vertices:,, Center:, Asmptotes: or Now tr Eercise F(, 3) F(, 3) 5 Figure A5.4 The hperbol in Emple 7. EXAMPLE 7 Foci on the -Ais The hperbol, 4 5 obtined b interchnging nd in Eqution, hs its vertices on the -is insted of the -is (Figure A5.4). With still equl to 4 nd b equl to 5, we hve Center-to-focus distnce: c b Foci:, c, 3 Vertices:,, Center:, Asmptotes: or Now tr Eercise 3. Reflective Properties The chief pplictions of prbols involve their use s reflectors of light nd rdio wves. Rs originting t prbol s focus re reflected out of the prbol prllel to the prbol s is (Figure A5.5). Prbolic rdio wve reflector Prbolic light reflector Outgoing light prllel to is Filment (point source) t focus Incoming rdio signls concentrte t focus HEADLAMP RADIO TELESCOPE Figure A5.5 Two of the mn uses of prbolic reflectors.

26 Section A5 Conic Sections 587 F Figure A5.6 An ellipticl mirror (shown here in profile) reflects light from one focus to the other. F H F P Hperbol F E F H F E F This propert is used b flshlight, hedlight, nd spotlight reflectors nd b microwve brodcst ntenns to direct rdition from point sources into nrrow bems. Conversel, electromgnetic wves rriving prllel to prbolic reflector s is re directed towrd the reflector s focus. This propert is used to intensif signls picked up b rdio telescopes nd television stellite dishes, to focus rriving light in telescopes, nd to concentrte sunlight in solr heters. If n ellipse is revolved bout its mjor is to generte surfce ( surfce clled n ellipsoid ) nd the interior is silvered to produce mirror, light from one focus will be reflected to the other focus ( Figure A5.6). Ellipsoids reflect sound the sme w, nd this propert is used to construct whispering glleries, rooms in which person stnding t one focus cn her whisper from the other focus. Sttur Hll in the U.S. Cpitol building is whispering gller. Ellipsoids lso pper in instruments used to stud ircrft noise in wind tunnels (sound t one focus cn be received t the other focus with reltivel little interference from other sources). Light directed towrd one focus of hperbolic mirror is reflected towrd the other focus. This propert of hperbols is combined with the reflective properties of prbols nd ellipses in designing modern telescopes. In Figure A5.7 strlight reflects off primr prbolic mirror towrd the mirror s focus F P. It is then reflected b smll hperbolic mirror, whose focus is F H F P, towrd the second focus of the hperbol, F E F H. Since this focus is shred b n ellipse, the light is reflected b the ellipticl mirror to the ellipse s second focus to be seen b n observer. As pst eperience with NASA s Hubble spce telescope shows, the mirrors hve to be nerl perfect to focus properl. The berrtion tht cused the mlfunction in Hubble s primr mirror (now corrected with dditionl mirrors) mounted to bout hlf wvelength of visible light, no more thn 5 the width of humn hir. Ellipse Prbol Primr mirror Figure A5.7 Schemtic drwing of reflecting telescope. Other Applictions Wter pipes re sometimes designed with ellipticl cross sections to llow for epnsion when the wter freezes. The triggering mechnisms in some lsers re ellipticl, nd stones on bech become more nd more ellipticl s the re ground down b wves. There re lso pplictions of ellipses to fossil formtion. The ellipsolith, once thought to be seprte species, is now known to be n ellipticll deformed nutilus. Hperbolic pths rise in Einstein s theor of reltivit nd form the bsis for the (unrelted) LORAN rdio nvigtion sstem. (LORAN is short for long rnge nvigtion. ) Hperbols lso form the bsis for new sstem the Burlington Northern Rilrod developed for using snchronized electronic signls from stellites to trck freight trins. Computers bord Burlington Northern locomotives in Minnesot cn trck trins to within one mile per hour of their speed nd to within feet of their ctul loction.

27 588 Appendices Section A5. Eercises In Eercises nd, find n eqution for the circle with center Ch, k nd rdius. Sketch the circle in the -plne. Lbel the circle s center nd - nd -intercepts (if n) with their coordinte pirs.. C,,. C, 5, In Eercises 3 nd 4, find the center nd rdius of the circle. Then sketch the circle In Eercises 5 nd 6, describe the regions defined b the inequlities nd pirs of inequlities , 4 Mtch the prbols in Eercises 7 with the following equtions:, 6, 8, 4. Then find the prbol s focus nd directri Eercises 5 nd 6 give equtions of prbols. Find ech prbol s focus nd directri. Then sketch the prbol. Include the focus nd directri in our sketch Eercises 7 nd 8 give equtions for ellipses. Put ech eqution in stndrd form. Then sketch the ellipse. Include the foci in our sketch Eercises 9 nd give informtion bout the foci nd vertices of ellipses centered t the origin of the -plne. In ech cse, find the ellipse s stndrd-form eqution from the given informtion. 9. Foci:,. Foci:, 4 Vertices:, Vertices:, Eercises nd give equtions for hperbols. Put ech eqution in stndrd form nd find the hperbol s smptotes. Then sketch the hperbol. Include the smptotes nd foci in our sketch Mtch ech conic section in Eercises 4 with one of these equtions:, 4 9, 4,. 4 9 Then find the conic section s foci nd vertices. If the conic section is hperbol, find its smptotes s well... Eercises 3 nd 4 give informtion bout the foci, vertices, nd smptotes of hperbols centered t the origin of the -plne. In ech cse, find the hperbol s stndrd-form eqution from the informtion given. 3. Foci:, 4. Vertices: 3, Asmptotes: Asmptotes: The prbol 8 is shifted down units nd right unit to generte the prbol 8. () Find the new prbol s verte, focus, nd directri. (b) Plot the new verte, focus, nd directri, nd sketch in the prbol. 6. The ellipse 6 9 is shifted 4 units to the right nd 3 units up to generte the ellipse () Find the foci, vertices, nd center of the new ellipse. (b) Plot the new foci, vertices, nd center, nd sketch in the new ellipse.

28 Section A5 Conic Sections The hperbol 6 9 is shifted units to the right to generte the hperbol. 6 9 () Find the center, foci, vertices, nd smptotes of the new hperbol. (b) Plot the new center, foci, vertices, nd smptotes, nd sketch in the hperbol. Eercises 8 3 give equtions for conic sections nd tell how mn units up or down nd to the right or left ech is to be shifted. Find n eqution for the new conic section nd find the new vertices, foci, directrices, center, nd smptotes, s pproprite. 8. 4, left, down 3 9., left, down , right, up , left, up Find the center, foci, vertices, smptotes, nd rdius, s pproprite, of ech conic section in Eercises Archimedes Formul for the Volume of Prbolic Solid The region enclosed b the prbol 4hb nd the line h is revolved bout the -is to generte the solid shown here. Show tht the volume of the solid is 3 the volume of the corresponding cone. h b, h 4h b 38. Compring Volumes If lines re drwn prllel to the coordinte es through point P on the prbol k, k, the prbol prtitions the rectngulr region bounded b these lines nd the coordinte es into two smller regions, A nd B. k () If the two smller regions re revolved bout the -is, show tht the generte solids whose volumes hve the rtio 4:. (b) Wht is the rtio of the volumes of the solids generted b revolving the regions bout the -is? 39. Perpendiculr Tngents Show tht the tngents to the curve 4p from n point on the line p re perpendiculr. 4. Mimizing Are Find the dimensions of the rectngle of lrgest re tht cn be inscribed in the ellipse 4 4 with its sides prllel to the coordinte es. Wht is the re of the rectngle? 4. Volume Find the volume of the solid generted b revolving the region enclosed b the ellipse bout the () -is, (b) -is. 4. Volume The tringulr region in the first qudrnt bounded b the -is, the line 4, nd the hperbol is revolved bout the -is to generte solid. Find the volume of the solid. 43. Volume The region bounded on the left b the -is, on the right b the hperbol, nd bove nd below b the lines 3 is revolved bout the -is to generte solid. Find the volume of the solid. Etending the Ides O A 44. Suspension Bridge Cbles The suspension bridge cble shown here supports uniform lod of w pounds per horizontl foot. It cn be shown tht if H is the horizontl tension of the cble t the origin, then the curve of the cble stisfies the eqution d w. d H Show tht the cble hngs in prbol b solving this differentil eqution subject to the initil condition tht when. B P Bridge cble b O

29 59 Appendices 45. Ripple Tnk Circulr wves were mde b touching the surfce of ripple tnk, first t point A nd shortl therefter t nerb point B. As the wves epnded, their points of intersection ppered to trce hperbol. Did the rell do tht? To find out, we cn model the wves with circles in the plne centered t nerb points lbeled A nd B s in the ccompning figure. P(t) 46. How the Astronomer Kepler Used String to Drw Prbols Kepler s method for drwing prbol (with more modern tools) requires string the length of T squre nd tble whose edge cn serve s the prbol s directri. Pin one end of the string to the point where ou wnt the focus to be nd the other end to the upper end of the T squre. Then, holding the string tut ginst the T squre with pencil, slide the T squre long the tble s edge. As the T squre moves, the pencil will trce prbol. Wh? r A (t) r B (t) A A B Focus F String P At time t, the point P is r A t units from A nd r B t units from B. Since the rdii of the circles increse t constnt rte, the rte t which the wves re trveling is d r d A t d rb d. t Conclude from this eqution tht r A r B hs constnt vlue, so tht P must lie on hperbol with foci t A nd B. Directri B Wht ou ll lern bout Ellipses nd Orbits Hperbols Focus-Directri Eqution... nd wh This section provides bsic informtion bout eccentricit of conic sections. c e F F A5. Clssifing Conic Sections b Eccentricit Ellipses nd Orbits We now ssocite with ech conic section number clled the eccentricit. The eccentricit tells whether the conic is circle, ellipse, prbol, or hperbol, nd, in the cse of ellipses nd hperbols, describes the conic s proportions. We begin with the ellipse. Although the center-to-focus distnce c does not pper in the eqution b, b for n ellipse, we cn still determine c from the eqution c b. If we fi nd vr c over the intervl c, the resulting ellipses will vr in shpe (Figure A5.8). The re circles if c (so tht b) nd fltten s c increses. If c, the foci nd vertices overlp nd the ellipse degenertes into line segment. We use the rtio of c to to describe the vrious shpes the ellipse cn tke. We cll this rtio the ellipse s eccentricit. F F 4 c 5 e 4 5 F c e F Figure A5.8 The ellipse chnges from circle to line segment s c increses DEFINITION Eccentricit of Ellipse The eccentricit of the ellipse b b is c e b.

30 Section A5 Conic Sections 59 Tble A5. Eccentricities of plnetr orbits Mercur. Venus. Erth. Mrs.9 Jupiter.5 Sturn.6 Urnus.5 Neptune. Pluto.5 The plnets in the solr sstem revolve round the sun in ellipticl orbits with the sun t one focus. Most of the orbits re nerl circulr, s cn be seen from the eccentricities in Tble A5.. Pluto hs firl eccentric orbit, with e.5, s does Mercur, with e.. Other members of the solr sstem hve orbits tht re even more eccentric. Icrus, n steroid bout mile wide tht revolves round the sun ever 49 Erth ds, hs n orbitl eccentricit of.83 (Figure A5.9). Mrs Erth Venus Mercur Sun Icrus Figure A5.9 The orbit of the steroid Icrus is highl eccentric. Erth s orbit is so nerl circulr tht its foci lie inside the sun. Hlle s comet Edmund Hlle (656 74; pronounced hw-le ), British biologist, geologist, se cptin, pirte, sp, Antrctic voger, stronomer, dviser on fortifictions, compn founder nd director, nd the uthor of the first cturil mortlit tbles, ws lso the mthemticin who pushed nd hrried Newton into writing his Principi. Despite these ccomplishments, Hlle is known tod chiefl s the mn who clculted the orbit of the gret comet of 68: wherefore if ccording to wht we hve lred sid [the comet] should return gin bout the er 758, cndid posterit will not refuse to cknowledge tht this ws first discovered b n Englishmn. Indeed, cndid posterit did not refuse ever since the comet s return in 758, it hs been known s Hlle s comet. Lst seen rounding the sun during the winter nd spring of , the comet is due to return in the er 6. The comet hs mde bout ccles so fr with bout the sme number to go before the sun erodes it w. EXAMPLE Finding the Eccentricit of n Orbit The orbit of Hlle s comet is n ellipse 36.8 stronomicl units long b 9. stronomicl units wide. (One stronomicl unit [AU] is 49,597,87 km, the semimjor is of Erth s orbit.) Its eccentricit is e b Now tr Eercise. EXAMPLE Locting Vertices Locte the vertices of n ellipse of eccentricit.8 whose foci lie t the points, 7. SOLUTION Since e c, the vertices re the points, where c e ,.8 or, Now tr Eercise 5.

31 59 Appendices Wheres prbol hs one focus nd one directri, ech ellipse hs two foci nd two directrices. These re the lines perpendiculr to the mjor is t distnces e from the center. The prbol hs the propert tht PF PD () for n point P on it, where F is the focus nd D is the point nerest P on the directri. For n ellipse, it cn be shown tht the equtions tht replce () re PF e PD, PF e PD. () Here, e is the eccentricit, P is n point on the ellipse, F nd F re the foci, nd D nd D re the points on the directrices nerest P (Figure A5.). Directri Directri e e b F ( c, ) O F (c, ) D D P(, ) b c e e Figure A5. The foci nd directrices of the ellipse b. Directri corresponds to focus F, nd directri to focus F. In ech eqution in () the directri nd focus must correspond; tht is, if we use the distnce from P to F, we must lso use the distnce from P to the directri t the sme end of the ellipse. The directri e corresponds to F c,, nd the directri e corresponds to F c,. Hperbols The eccentricit of hperbol is lso e c, onl in this cse c equls b insted of b. In contrst to the eccentricit of n ellipse, the eccentricit of hperbol is lws greter thn. DEFINITION Eccentricit of Hperbol The eccentricit of the hperbol b is c e b. In both ellipse nd hperbol, the eccentricit is the rtio of the distnce between the foci to the distnce between the vertices (becuse c c). distnce between foci Eccentricit distnce between vertices

32 Section A5 Conic Sections 593 In n ellipse, the foci re closer together thn the vertices nd the rtio is less thn. In hperbol, the foci re frther prt thn the vertices nd the rtio is greter thn. Directri Directri e e D P(, ) D EXAMPLE 3 Finding Eccentricit Find the eccentricit of the hperbol SOLUTION We divide both sides of the hperbol s eqution b 44 to put it in stndrd form, obtining or F ( c, ) O e F (c, ) With 6 nd b 9, we find tht c b 6 9 5, so c 5 e 4. Now tr Eercise 5. c e Figure A5. The foci nd directrices of the hperbol b. No mtter where P lies on the hperbol, PF e PD, nd PF e PD. As with the ellipse, it cn be shown tht the lines e ct s directrices for the hperbol nd tht PF e PD nd PF e PD. (3) Here P is n point on the hperbol, F nd F re the foci, nd D nd D re the points nerest P on the directrices ( Figure A5.). Focus-Directri Eqution To complete the picture, we define the eccentricit of prbol to be e. Equtions 3 then hve the common form PF e PD. DEFINITION Eccentricit of Prbol The eccentricit of prbol is e. The focus-directri eqution PF e PD unites the prbol, ellipse, nd hperbol in the following w. Suppose tht the distnce PF of point P from fied point F (the focus) is constnt multiple of its distnce from fied line (the directri). Tht is, suppose PF e PD, (4) where e is the constnt of proportionlit. Then the pth trced b P is () prbol if e, (b) n ellipse of eccentricit e if e, nd (c) hperbol of eccentricit e if e. Eqution 4 m not look like much to get ecited bout. There re no coordintes in it nd when we tr to trnslte it into coordinte form it trnsltes in different ws, depending on the size of e. At lest, tht is wht hppens in Crtesin coordintes. However, in polr coordintes, the eqution PF e PD trnsltes into single eqution regrdless of the vlue of e, n eqution so simple tht it hs been the eqution of choice of stronomers nd spce scientists for nerl 3 ers.

33 594 Appendices Given the focus nd corresponding directri of hperbol centered t the origin nd with foci on the -is, we cn use the dimensions shown in Figure A5. to find e. Knowing e, we cn derive Crtesin eqution for the hperbol from the eqution PF e PD, s in the net emple. We cn find equtions for ellipses centered t the origin nd with foci on the -is in similr w, using the dimensions shown in Figure A5.. EXAMPLE 4 Using Focus nd Directri Find Crtesin eqution for the hperbol centered t the origin tht hs focus t 3, nd the line s the corresponding directri. SOLUTION We first use the dimensions shown in Figure A5. to find the hperbol s eccentricit. The focus is c, 3,, so c 3. The directri is the line, so e. e D(, ) 3 6 P(, ) When combined with the eqution e c tht defines eccentricit, these results give c 3 e e, so e 3 nd e 3. Knowing e, we cn now derive the eqution we wnt from the eqution PF e PD. In the nottion of Figure A5., we hve F(3, ) PF e PD Eq e 3 Figure A5. The hperbol in Emple Now tr Eercise 9. Section A5. Eercises In Eercises 4, find the eccentricit, foci, nd directrices of the ellipse Eercises 5 8 give the foci or vertices nd the eccentricities of ellipses centered t the origin of the -plne. In ech cse, find the ellipse s stndrd-form eqution. 5. Foci:, 3 Eccentricit:.5 6. Foci: 8, Eccentricit:. 7. Vertices:, Eccentricit:.4 8. Vertices:, 7 Eccentricit:.

34 Section A5 Conic Sections 595 Eercises 9 nd give foci nd corresponding directrices of ellipses centered t the origin of the -plne. In ech cse, use the dimensions in Figure A5. to find the eccentricit of the ellipse. Then find the ellipse s stndrd-form eqution. 9. Focus: 5, 9 Directri: 5. Focus: 4, Directri: 6. Drw n ellipse of eccentricit 45. Eplin our procedure.. Drw the orbit of Pluto (eccentricit.5) to scle. Eplin our procedure. 3. The endpoints of the mjor nd minor es of n ellipse re,, 3, 4,, 7, nd, 4. Sketch the ellipse, give its eqution in stndrd form, nd find its foci, eccentricit, nd directrices. 4. Find n eqution for the ellipse of eccentricit 3 tht hs the line 9 s directri nd the point 4, s the corresponding focus. In Eercises 5 8, find the eccentricit, foci, nd directrices of the hperbol Eercises 9 nd give the eccentricities nd the vertices or foci of hperbols centered t the origin of the -plne. In ech cse, find the hperbol s stndrd-form eqution. 9. Eccentricit: 3 Vertices:,. Eccentricit: 3 Foci: 3, Eercises nd give foci nd corresponding directrices of hperbols centered t the origin of the -plne. In ech cse, find the hperbol s eccentricit. Then find the hperbol s stndrd-form eqution.. Focus: 4, Directri:. Focus:, Directri: 3. A hperbol of eccentricit 3 hs one focus t, 3. The corresponding directri is the line. Find n eqution for the hperbol. 5. Determining Constnts Wht vlues of the constnts, b, nd c mke the ellipse 4 b c lie tngent to the -is t the origin nd pss through the point,? Wht is the eccentricit of the ellipse? Etending the Ides 6. The Reflective Propert of Ellipses An ellipse is revolved bout its mjor is to generte n ellipsoid. The inner surfce of the ellipsoid is silvered to mke mirror. Show tht r of light emnting from one focus will be reflected to the other focus. Sound wves lso follow such pths, nd this propert is used in constructing whispering glleries. (Hint: Plce the ellipse in stndrd position in the -plne nd show tht the lines from point P on the ellipse to the two foci mke congruent ngles with the tngent to the ellipse t P.) 7. The Reflective Propert of Hperbols Show tht r of light directed towrd one focus of hperbolic mirror, s in the ccompning figure, is reflected towrd the other focus. (Hint: Show tht the tngent to the hperbol t P bisects the ngle mde b segments PF nd PF.) F ( c, ) 8. A Confocl Ellipse nd Hperbol Show tht n ellipse nd hperbol tht hve the sme foci A nd B, s in the ccompning figure, cross t right ngles t their points of intersection. [Hint: A r of light from focus A tht met the hperbol t P would be reflected from the hperbol s if it cme directl from B (Eercise 7). The sme r would be reflected off the ellipse to pss through B (Eercise 6).] A O P(, ) B F (c, ) P C Eplortions 4. The Effect of Eccentricit on Hperbol s Shpe Wht hppens to the grph of hperbol s its eccentricit increses? To find out, rewrite the eqution b in terms of nd e insted of nd b. Grph the hperbol for vrious vlues of e nd describe wht ou find.

35 596 Appendices Wht ou ll lern bout Qudrtic Curves Cross Product Term Rotting Aes to Eliminte B Possible Grphs of Qudrtic Equtions Discriminnt Test Technolog Appliction... nd wh This section provides bsic informtion bout generl qudrtic equtions in two vribles nd their grphs. F ( 3, 3) O 9 P(, ) /4 Focl is F (3, 3) Figure A5.3 The focl is of the hperbol 9 mkes n ngle of p4 rdins with the positive -is. ' (, ) P (', ') ' A5.3 Qudrtic Equtions nd Rottions Qudrtic Curves In this section, we emine one of the most mzing results in nltic geometr, which is tht the Crtesin grph of n eqution A B C D E F, () in which A, B, nd C re not ll zero, is nerl lws conic section. The eceptions re the cses in which there is no grph t ll or the grph consists of two prllel lines. It is conventionl to cll ll grphs of Eqution, curved or not, qudrtic curves. Cross Product Term You m hve noticed tht the term B did not pper in the equtions for the conic sections in Section A5.. This hppened becuse the es of the conic sections rn prllel to (in fct, coincided with) the coordinte es. To see wht hppens when the prllelism is bsent, let us write n eqution for hperbol with 3 nd foci t F 3, 3 nd F 3, 3 (Figure A5.3). The eqution PF PF becomes PF PF 3 6 nd When we trnspose one rdicl, squre, solve for the remining rdicl nd squre gin, the eqution reduces to 9, () cse of Eqution in which the cross product term is present. The smptotes of the hperbol in Eqution re the - nd -es, nd the focl is mkes n ngle of p4 rdins with the positive -is. As in this emple, the cross product term is present in Eqution onl when the es of the conic re tilted. Rotting Aes to Eliminte B To eliminte the -term from the eqution of conic, we rotte the coordinte es to eliminte the tilt in the es of the conic. The equtions for the rottions we use re derived in the following w. In the nottion of Figure A5.4, which shows counterclockwise rottion bout the origin through n ngle, OM OP cos u OP cos u cos OP sin u sin (3) MP OP sin u OP cos u sin OP sin u cos. Since OP cos u OM nd OP sin u MP, the equtions in (3) reduce to the following. O M M' Figure A5.4 A counterclockwise rottion through ngle bout the origin. Equtions for Rotting Coordinte Aes cos sin sin cos (4)

36 Section A5 Conic Sections 597 ' ' 9 ' 9 ' EXAMPLE Chnging n Eqution The - nd -es re rotted through n ngle of p4 rdins bout the origin. Find n eqution for the hperbol 9 in the new coordintes. SOLUTION 3 3 /4 9 Since cos p4 sin p4, we substitute, from Equtions 4 into the eqution 9, obtining Figure A5.5 The hperbol in Emple ( nd re the new coordintes). ( )( ) See Figure A5.5. Now tr Eercise 35. If we ppl Equtions 4 to the qudrtic Eqution, we obtin new qudrtic eqution A BC DEF. (5) The new nd old coefficients re relted b the equtions AA cos B cos sin C sin BB cos C A sin CA sin B sin cos C cos (6) DD cos E sin ED sin E cos FF. Equtions 6 show, mong other things, tht if we strt with n eqution for curve in which the cross product term is present B, we cn find rottion ngle tht produces n eqution in which no cross product term ppers B. To find, we set B in the second eqution in (6) nd solve the resulting eqution, B cos C A sin, for. In prctice, this mens determining from one of the two equtions. cot A C B or tn. (7) B A C EXAMPLE Eliminting Cross Product Term The coordinte es re to be rotted through n ngle to produce n eqution for the curve 3 tht hs no cross product term. Find suitble nd the corresponding new eqution. Identif the curve. continued

37 598 Appendices 3 Figure A5.6 This tringle identifies cot 3 s p3. (Emple ) ' ' 4 ' Figure A5.7 The conic section in Emple. ' SOLUTION The eqution 3 hs A, B 3, nd C. We substitute these vlues into Eqution 7 to find : cot A C. B 3 3 From the right tringle in Figure A5.6, we see tht one pproprite choice of ngle is p3, so we tke p6. Substituting p6, A, B 3, C, D E, nd F into Equtions 6 gives A 5, B, C, DE, F. Eqution 5 then gives 5, or. 4 The curve is n ellipse with foci on the new -is (Figure A5.7). Now tr Eercise. Possible Grphs of Qudrtic Equtions We now return to the grph of the generl qudrtic eqution. Since es cn lws be rotted to eliminte the cross product term, there is no loss of generlit in ssuming tht this hs been done nd tht the eqution hs the form A C D E F. (8) Eqution 8 represents () circle if A C (specil cses: the grph is point or there is no grph t ll); (b) prbol if Eqution 8 is qudrtic in one vrible nd liner in the other; (c) n ellipse if A nd C re both positive or both negtive (specil cses: circles, single point, or no grph t ll); (d) hperbol if A nd C hve opposite signs (specil cse: pir of intersecting lines); (e) stright line if A nd C re zero nd t lest one of D nd E is different from zero; (f) one or two stright lines if the left-hnd side of Eqution 8 cn be fctored into the product of two liner fctors. See Tble A5.3 (on pge 6) for emples. Discriminnt Test We do not need to eliminte the -term from the eqution A B C D E F (9) to tell wht kind of conic section the eqution represents. If this is the onl informtion we wnt, we cn ppl the following test insted. As we hve seen, if B, then rotting the coordinte es through n ngle tht stisfies the eqution cot A C () B will chnge Eqution 9 into n equivlent form A C DEF () without cross product term.

38 Section A5 Conic Sections 599 Now, the grph of Eqution is (rel or degenerte) () prbol if A or C; tht is, if AC; (b) ellipse if A nd C hve the sme sign; tht is, if AC; (c) hperbol if A nd C hve opposite signs; tht is, if AC. It cn lso be verified from Equtions 6 tht for n rottion of es, B 4AC B 4AC. () This mens tht the quntit B 4AC is not chnged b rottion. But when we rotte through the ngle given b Eqution, B becomes zero, so B 4AC 4AC. Since the curve is prbol if AC, n ellipse if AC, nd hperbol if AC, the curve must be prbol if B 4AC, n ellipse if B 4AC, nd hperbol if B 4AC. The number B 4AC is clled the discriminnt of Eqution 9. Discriminnt Test With the understnding tht occsionl degenerte cses m rise, the qudrtic curve A B C D E F is () prbol if B 4AC, (b) n ellipse if B 4AC, (c) hperbol if B 4AC. EXAMPLE 3 Appling the Discriminnt Test () represents prbol becuse B 4AC (b) represents n ellipse becuse B 4AC 4 3. (c) 5 represents hperbol becuse B 4AC 4. Now tr Eercise 5. Technolog Appliction How Some Clcultors Use Rottions to Evlute Sines nd Cosines Some clcultors use rottions to clculte sines nd cosines of rbitrr ngles. The procedure goes something like this: The clcultor hs, stored,. ten ngles or so, s sin, sin,, sin, nd. twent numbers, the sines nd cosines of the ngles,,,. To clculte the sine nd cosine of n rbitrr ngle u, we enter u (in rdins) into the clcultor. The clcultor substrcts or dds multiples of p to u to replce u b the ngle between nd p tht hs the sme sine nd cosine s u (we continue to cll the ngle u).

39 6 Appendices NOT TO SCALE (cos, sin) m 3 3 s m s The clcultor then writes u s sum of multiples of (s mn s possible without overshooting) plus multiples of (gin, s mn s possible), nd so on, working its w to. This gives u m m m. m s (, ) The clcultor then rottes the point, through m copies of (through, m times in succession), plus m copies of, nd so on, finishing off with m copies of (Figure A5.8). The coordintes of the finl position of, on the unit circle re the vlues the clcultor gives for cos u, sin u. Figure A5.8 To clculte the sine nd cosine of n ngle u between nd p, the clcultor rottes the point, to n pproprite loction on the unit circle nd displs the resulting coordintes. Tble A5.3 Emples of qudrtic curves A B C D E F A B C D E F Eqution Remrks Circle 4 4 A C; F Prbol 9 9 Ellipse Qudrtic in, liner in A, C hve sme sign, A C; F Hperbol A, C hve opposite signs One line (still conic section) -is Intersecting lines Fctors to (still conic, section) so, Prllel lines Fctors to (not conic 3 3, section) so, Point The origin No grph No grph

40 Section A5 Conic Sections 6 Section A5.3 Eercises Use the discriminnt B 4AC to decide whether the equtions in Eercises 6 represent prbols, ellipses, or hperbols In Eercises 7 6, rotte the coordinte es to chnge the given eqution into n eqution tht hs no cross product term. Then identif the grph of the eqution. (The new equtions will vr with the size nd direction of the rottion ou use.) Find the sine nd cosine of n ngle through which the coordinte es cn be rotted to eliminte the cross product term from the eqution 4 6 6,37 7. Do not crr out the rottion. 8. Find the sine nd cosine of n ngle through which the coordinte es cn be rotted to eliminte the cross product term from the eqution Do not crr out the rottion. The conic sections in Eercises 7 6 were chosen to hve rottion ngles tht were nice in the sense tht once we knew cot or tn we could identif nd find sin nd cos from fmilir tringles. The conic sections encountered in prctice m not hve such nice rottion ngles, nd we m hve to use clcultor to determine from the vlue of cot or tn. In Eercises 9 34, use clcultor to find n ngle through which the coordinte es cn be rotted to chnge the given eqution into qudrtic eqution tht hs no cross product term. Then find sin nd cos to two deciml plces nd use Equtions 6 to find the coefficients of the new eqution to the nerest deciml plce. In ech cse, s whether the conic section is n ellipse, hperbol, or prbol The Hperbol The hperbol is one of mn hperbols of the form tht pper in science nd mthemtics. () Rotte the coordinte es through n ngle of 45 to chnge the eqution into n eqution with no -term. Wht is the new eqution? (b) Do the sme for the eqution. 36. Writing to Lern Cn nthing be sid bout the grph of the eqution A B C D E F if AC? Give resons for our nswer. 37. Writing to Lern Does n nondegenerte conic section A B C D E F hve ll of the following properties? () It is smmetric with respect to the origin. (b) It psses through the point,. (c) It is tngent to the line t the point,. Give resons for our nswer. 38. When A C Show tht rotting the es through n ngle of p4 rdins will eliminte the -term from Eqution whenever A C. 39. Identifing Conic Section () Wht kind of conic section is the curve? (b) Solve the eqution for nd sketch the curve s the grph of rtionl function of. (c) Find equtions for the lines prllel to the line tht re norml to the curve. Add the lines to our sketch.

41 6 Appendices 4. Sign of AC Prove or find counteremples to the following sttements bout the grph of A B C D E F. () If AC, the grph is n ellipse. (b) If AC, the grph is hperbol. (c) If AC, the grph is hperbol. Eplortions 4. 9 Rottions Wht effect does 9 rottion bout the origin hve on the equtions of the following conic sections? Give the new eqution in ech cse. () The ellipse b b (b) The hperbol b (c) The circle (d) The line m (e) The line m b 4. 8 Rottions Wht effect does 8 rottion bout the origin hve on the equtions of the following conic sections? Give the new eqution in ech cse. () The ellipse b b (b) The hperbol b (c) The circle (d) The line m (e) The line m b 44. Degenerte Conic Section () Decide whether the conic section with eqution represents prbol, n ellipse, or hperbol. (b) Show tht the grph of the eqution in prt () is the line A Nice Are Formul for Ellipses When B 4AC is negtive, the eqution A B C represents n ellipse. If the ellipse s semi-es re nd b, its re is pb ( stndrd formul). Show tht the re is lso p4ac B. (Hint: Rotte the coordinte es to eliminte the -term nd ppl Eqution to the new eqution.) 46. Other Rottion Invrints We describe the fct tht B 4AC equls B 4AC fter rottion bout the origin b sing tht the discriminnt of qudrtic eqution is n invrint of the eqution. Use Equtions 6 to show tht the numbers () A C nd (b) D E re lso invrints, in the sense tht ACA C nd D E D E. We cn use these equlities to check ginst numericl errors when we rotte es. The cn lso be helpful in shortening the work required to find vlues for the new coefficients. Etending the Ides 43. Degenerte Conic Section () Decide whether the eqution represents n ellipse, prbol, or hperbol. (b) Show tht the grph of the eqution in prt () is the line 3.

42 Section A6 Hperbolic Functions 63 A6 Wht ou ll lern bout Bckground Definitions Identities Derivtives nd Integrls Inverse Hperbolic Functions Identities for sech, csch, coth Derivtives of Inverse Hperbolic Functions; Associted Integrls... nd wh This section provides bsic informtion bout hperbolic functions nd their derivtives nd integrls. Pronouncing cosh nd sinh Cosh is often pronounced kosh, rhming with gosh or guche. Sinh is pronounced s if spelled cinch or shine. Tble A6. Identities for hperbolic functions sinh sinh cosh cosh cosh sinh cosh cosh sinh cosh cosh sinh tnh sech coth csch Hperbolic Functions Bckground Suspension cbles like those of the Golden Gte Bridge, which support constnt lod per horizontl foot, hng in prbols (Section A5., Eercise 44). Cbles like power line cbles, which hng freel, hng in curves clled hperbolic cosine curves. Besides describing the shpes of hnging cbles, hperbolic functions describe the motions of wves in elstic solids, the temperture distributions in metl cooling fins, nd the motions of flling bodies tht encounter ir resistnce proportionl to the squre of the velocit. If hnging cble were turned upside down (without chnging shpe) to form n rch, the internl forces, then reversed, would once gin be in equilibrium, mking the inverted hperbolic cosine curve the idel shpe for self-stnding rch. The center line of the Gtew Arch to the West in St. Louis follows hperbolic cosine curve. Definitions The hperbolic cosine nd sine functions re defined b the first two equtions in Tble A6.. The tble lso defines the hperbolic tngent, cotngent, secnt, nd cosecnt. As we will see, the hperbolic functions ber number of similrities to trigonometric functions fter which the re nmed. Tble A6. Hperbolic cosine of : Hperbolic sine of : Hperbolic tngent: Hperbolic cotngent: The si bsic hperbolic functions cosh e e sinh e e sinh tnh c os h e e e e coth c osh e e sinh e e Hperbolic secnt: sech cos h e e Hperbolic cosecnt: csch sin h e e See Figure A6. for grphs. Identities Hperbolic functions stisf the identities in Tble A6.. Ecept for differences in sign, these re identities we lred know for trigonometric functions. Derivtives nd Integrls The si hperbolic functions, being rtionl combintions of the differentible functions e nd e, hve derivtives t ever point t which the re defined (Tble A6.3 on the following pge). Agin, there re similrities with trigonometric functions. The derivtive formuls in Tble A6.3 led to the integrl formuls seen there.

43 64 Appendices Tble A6.3 Derivtives nd compnion integrls d sinh u cosh u d u d d d cosh u sinh u d u d d d tnh u sech d u du d d coth u csch d u du d d sech u sech u tnh u du d d d csch u csch u coth u du d d sinh u du cosh u C cosh u du sinh u C sech u du tnh u C csch u du coth u C sech u tnh u du sech u C csch u coth u du csch u C 3 e cosh 3 coth 3 sinh 3 e 3 e e 3 coth tnh 3 () The hperbolic sine nd its component eponentils. (b) The hperbolic cosine nd its component eponentils. (c) The grphs of tnh nd coth /tnh. cosh sinh sech csch (d) The grphs of cosh nd sech /cosh. (e) The grphs of sinh nd csch /sinh. Figure A6. The grphs of the si hperbolic functions.

44 Section A6 Hperbolic Functions 65 sinh EXAMPLE Finding Derivtive d tnh t d t sech t d t d t t sech t t Now tr Eercise 3. EXAMPLE Integrting Hperbolic Cotngent sinh ( sinh ) coth 5 d c osh sinh 5 5 d 5 d u u u sinh 5, du 5 cosh 5 d 5 ln u C 5 ln sinh 5 C Now tr Eercise () = cosh, cosh ( cosh, ) (b) sech ( sech, ) sech, (c) Figure A6. The grphs of the inverse hperbolic sine, cosine, nd secnt of. Notice the smmetries bout the line. 3 EXAMPLE 3 Using n Identit to Integrte Evlute sinh d. SOLUTION Solve Numericll To five deciml plces, NINTsinh,,,.467. Confirm Anlticll sinh d cosh d Tble A6. cosh d [ sinh ] sin h Now tr Eercise 47. Inverse Hperbolic Functions We use the inverses of the si bsic hperbolic functions in integrtion. Since dsinh d cosh, the hperbolic sine is n incresing function of. We denote its inverse b sinh. For ever vlue of in the intervl, the vlue of sinh is the number whose hperbolic sine is ( Figure A6.). The function cosh is not one-to-one, s we cn see from the grph in Figure A6.. But the restricted function cosh,, is one-to-one nd therefore hs n inverse, denoted b cosh. For ever vlue of, cosh is the number in the intervl whose hperbolic cosine is (Figure A6.b). Like cosh, the function sech cosh fils to be one-to-one, but its restriction to nonnegtive vlues of does hve n inverse, denoted b sech. For ever vlue of in the intervl,, sech is the nonnegtive number whose hperbolic secnt is ( Figure A6.c).

45 66 Appendices The hperbolic tngent, cotngent, nd cosecnt re one-to-one on their domins nd therefore hve inverses, denoted b tnh, coth, csch (Figure A6.3). tnh tnh coth coth csch csch () (b) (c) Figure A6.3 The grphs of the inverse hperbolic tngent, cotngent, nd cosecnt of. EXPLORATION Viewing Inverses Let t t, t t, t t, t cosh t, 3 t t, 3 t t.. Grph the prmetric equtions simultneousl in squre viewing window tht contins 6, 4. Set tmin, tm 6, nd t-step.5. Eplin wht ou see. Eplin the domin of ech function.. Let 4 t t, 4 t cosh t. Grph nd compre 3, 3 nd 4, 4. Predict wht ou should see, nd eplin wht ou do see. Tble A6.4 Identities for inverse hperbolic functions sech cosh csch sinh coth tnh Identities for sech, csch, coth We use the identities in Tble A6.4 to clculte the vlues of sech, csch, nd coth on clcultors tht give onl cosh, sinh, nd tnh. Derivtives of Inverse Hperbolic Functions; Associted Integrls The chief use of inverse hperbolic functions lies in integrtions tht reverse the derivtive formuls in Tble A6.5.