Conics and Polar Coordinates

Transcription

1 CHAPTER 11 Conics nd Polr Coordintes Ü11.1. Qudrtic Reltions We will see tht curve defined b qudrtic reltion between the vribles x is one of these three curves: ) prbol, b) ellipse, c) hperbol. There re other possibilities, considered degenerte. For exmple the grph of the eqution x we know to be circle, if. But if, the grph is just the point µ, nd if, there is no grph. Similrl the eqution x describes hperbol if, but if, we get the two lines x. First we list the stndrd forms of the bsic curves. These re stndrd in the sense tht n other curve given b qudrtic eqution is obtined from one of these b moving the curve in the plne b trnslting nd/or rotting. The Prbol. The stndrd form is one of these: (11.1) x x The sign of determines the orienttion of the prbol. We hve these four possibilities: x x x x Figure 11.1 Figure 11. Figure 11.3 Figure 11.4 Figure 11.1 Figure x 1 x 157

2 Chpter 11 Conics nd Polr Coordintes 158 Figure 11.3 Figure x x The mgnitude of determines the spred of the prbol: for ver smll, the curve is nrrow, nd s gets lrge, the prbol brodens. The origin is the vertex of the prbol. In the first two cses, the -xis is the xis of the prbol, in the second two cses it is the x-xis. The prbol is smmetric bout its xis. The Ellipse The stndrd form is (11.) x b 1 The vlues x cn tke lie between nd nd the vlues cn tke lie between b nd b. If b (s shown in figure 11.5), the mjor xis of the ellipse is the x-xis, the minor xis is the -xis nd the points µ re its vertices. Figure 11.5 Figure 11.6 b b b If b (s shown in figure 11.6), the mjor xis of the ellipse is the -xis, x is the minor xis, nd the points bµ re its vertices. Of course, if b, the curve is the circle of rdius, nd there re no specil vertices or xes. The Hperbol. The stndrd form is one of these: (11.3) corresponding to figures 11.7 nd x b 1 x b 1 b

3 Ü11.1 Qudrtic Reltions 159 Figure 11.7 Figure 11.8 b b The x-xis is the xis of the first hperbol. The points µ re the vertices of the hperbol; for x between these vlues, there corresponds no point on the curve. We similrl define the xis nd vertices of the hperbol of figure The lines (11.4) b x re the smptotes of the hperbol, in the sense tht, s x, the curve gets closer nd closer to these lines. We see this b dividing the defining eqution b x, nd consider wht hppens s x. For exmple, using the first eqution, we get (11.5) 1 1 b x 1 x Figure 11.9 b c which we cn rewrite s (11.6) b x 1 1 x so tht, s x gets lrge, the hperbol pproches the grph of (11.7) b x 1 which mounts to the two equtions bµx. Figure 11.9 shows these smptotes.

4 Chpter 11 Conics nd Polr Coordintes 16 Now, the generl qudrtic reltion between x nd is (11.8) Ax B Cx Dx E F If C, then b completing the squre in both x nd we re led to n eqution which looks much like one of the stndrd forms, but with the center removed to new point x µ. If C, the sitution is more difficult: rottion of the figure is lso required to get it into stndrd form. We will discuss this no further, nd consider onl the cse C. First, some exmples: µ Exmple 11.1 Grph the curve 3x 3x 73 We hve to complete the squre in x. We get (11.9) 3 x 1x 5µ which gives the stndrd form Figure 11.1 (11.1) 3 x 5µ Exmple 11. Grph the curve 9x 4 18x Completing the squres: 6 4 (11.11) 9 x x 1µ 4 4 4µ (11.1) x 1µ µ Figure Figure µ Exmple 11.3 the squres: Grph the curve 5x 3x 4 46 Completing (11.13) 5 x 6x 9µ 4 4µ (11.14) µ Ô5µ x 3µ 1 Proposition 11.1 The eqution (11.15) Ax B Dx E F cn be put into one of the following forms b completing the squre: ) (prbol): A x x µ if B The vertex of the prbol is t x µ, nd the xis is the line x x. b) (prbol): x x C µ if A The vertex of the prbol is t x µ, nd the xis is the line.

5 Ü11. Eccentricit nd Foci 161 c) (ellipse) x x µ µ b x µ, nd its xes re the lines x x. d) (hperbol): x x µ µ b 1 if A nd B re of the sme sign The center of the ellipse is t 1 or µ b x x µ 1 if A nd B re of different signs. The center of the hperbol is x µ, nd its xes re the lines x x. eµ If both A nd B re zero, the curve is line. The following degenerte cses m lso result: (11.16) A x x µ B µ : no grph or just the point x µ (11.17) A x x µ B µ : two lines crossing t x µ Exmple 11.4 Finll, just to illustrte the sitution of qudrtic whose coefficient of x is nonzero, we consider the curve x 1. This curve is smmetric bout the lines x, nd hs the smptotes x. This ppers to be hperbol with mjor xis the line x. In fct, if we mke the liner chnge of vribles x u v u v, this becomes the curve u v 1 in the new vribles. (This chnge of vribles represents rottion b 45 Æ, with slight chnge of scle.) Ü11.. Eccentricit nd Foci These curves re clled the conic sections becuse the cn be visulized s the intersection of cone with plne. We shll now consider nother definition, dting from the ncient Greeks, which leds to importnt properties of the conics. Fix point F nd line L in the plne such tht L does not go through F. Pick positive number e. We consider the locus C of ll points X in the plne such tht (11.18) XF exl where XY mens the distnce from X to Y. e is the eccentricit of C; F the focus nd L the directrix. Note tht the curve C is smmetric bout the line through the focus nd perpendiculr to the directrix. This is the xis of the curve. There is one point between F nd L on C which is on this xis; this point is the vertex of C. We now show tht if e 1, C is prbol, if e 1, C is n ellipse nd if e 1, C is hperbol. Let s tke the xis of C to be the x xis, nd plce the vertex t the origin, O. Then the focus is some point pµ; we tke p. Since OF p, from we find tht the directrix is the line x pe (see figure 11.14).

6 Chpter 11 Conics nd Polr Coordintes 16 Figure Figure µ µ m X n n em L X xµ V F p V F pµ L Now, for point X xµ on the curve, we hve (11.19) XL x pe nd XF Ô x pµ nd so eqution in coordintes is given b (11.) Ô x pµ e x peµ ex p Cse e 1. Squring both sides we get (11.1) x px p x px p simplifing to 4px This of course is the stndrd form of prbol. It lso loctes the focus nd the directrix of prbol. Proposition 11. The focus of the prbol x is 4 units on one side of the vertex of the prbol long the xis, nd nd the directrix intersects the xis 4 units on the other side. Exmple 11.5 Find the vertex, focus nd directrix of the prbol given b the eqution x 6x 4 First we put the eqution in stndrd form. Completing the squre, we hve (11.) x 3x or x Thus the vertex is t 31µ, the xis of the prbol is the line x 3 nd we hve 4p 1, so p 18. Thus the focus is t 3 1µ 18µµ 358µ nd the directrix is the line 38. Exmple 11.6 Find the eqution of the prbol whose vertex is t 4µ nd whose directrix is the line x 1. Find the focus of this prbol. Since the directrix is verticl line, the xis is horizontl, so the eqution hs the form (11.3) µ 4p x 4µ

7 Ü11. Eccentricit nd Foci 163 since the vertex is t 4µ. Now p is the distnce between the vertex nd the directrix, so p 1µ 5. Thus the eqution of the prbol is (11.4) µ x 4µ The focus is 5 units to the right of the vertex, so is t 9µ. Exmple 11.7 Find the eqution of the prbol whose focus is the origin nd whose vertex is t the point µ with. The prbol hs its xis the x-xis, nd since the vertex is to the right of the focus, the prbol opens to the left. Thus the eqution hs the form (11.5) 4p x µ where p is the distnce between focus nd vertex. But tht is, so the eqution is (11.6) 4 x µ Cse e 1.Squring both sides of 11. gives us (11.7) x px p e x px p µ which simplifies to (11.8) 1 e µx p 1 eµx Thus, if e 1, this is n ellipse, nd if e 1 this is hperbol. Notice, becuse of smmetr in the minor xis, ellipses nd hperbols hve two foci; one on ech side of the minor xis. We now wnt show how to locte the foci of n ellipse given in stndrd form. Thus we strt b putting 11.8 in stndrd form, nd then compre it to the formul of proposition 11.1c. Dividing eqution 11.8 b the coefficient of x gives us (11.9) x p x 1 e 1 e Now completing the squre, we come to (11.3) Compring this to (11.31) x p 1 e 1 e p 1 eµ x x µ µ b 1 we see tht the center of the ellipse is t p 1 eµµ nd p 1 eµ b 1 e µ. Let c be the distnce of the center from the focus. Then (11.3) c p 1 e p p e 1 e e

8 Chpter 11 Conics nd Polr Coordintes 164 nd c e b. Summrizing Proposition 11.3 If n ellipse is in stndrd form (11.31), with b, then the foci of the ellipse re on the mjor xis, c units w from the center where (11.33) c b The eccentricit of the ellipse is given b the equtions (11.34) b 1 e µ or c e The sme rguments for the cse e 1, the hperbol, led to Proposition 11.4 If hperbol is in stndrd form (11.35) x x µ µ b 1 then the foci of the hperbol re on the mjor xis, c units w from the center where (11.36) c b The eccentricit of the hperbol is given b the equtions (11.37) b e 1µ or c e Exmple 11.8 Find the foci of the conic given b the eqution x 4 First, we complete the squre to get the eqution in stndrd form: x 8 (11.38) x 1µ 3 3µ 1 This conic is n ellipse centered t (1,), with mjor xis the x-xis, nd 9 b 94. Thus c b 9 34µ, so c 3µ Ô 3. This is the distnce of the foci from the center (long the mjor xis), so the foci re t 1 3µ Ô 3µ. Exmple 11.9 Find the foci of the conic given b the eqution Complete the squres, nd get the stndrd form x 4x 13 (11.39) x µ This is hperbol with center t µ, nd mjor xis the line x. We hve c b 18, so c 3 Ô is the distnce of the foci from the center long the line x. Thus the foci re t 3 Ô µ. The vertices re t 3µ. Exmple 11.1 Find the eqution of the ellipse centered t the origin, with focus t µ nd vertex t 3µ.

9 Ü11.3 String nd Opticl Properties of the Conics 165 (11.4) The eqution of n ellipse centered t the origin is x b 1 We re given 3 c. Thus b c 5, nd the eqution is (11.41) x Ü11.3. String nd Opticl Properties of the Conics We hve seen tht the prbol cn be defined s the locus of points X equidistnt from given point F nd given line L. The ellipse nd the hperbol hve similr definitions. Proposition 11.5 Given two points F 1 nd F nd number greter thn the distnce between F 1 nd F, the locus of points X such tht (11.4) XF 1 XF is n ellipse with foci t F 1 nd F nd mjor xis of length. Given n ellipse in stndrd form, we cn verif 11.4 b length lgebric computtion. To show tht 11.4 leds to the eqution of n ellipse is nother lgebric computtion beginning this w. Choose coordintes so tht the points F 1 nd F lie on the x-xis, nd the origin is midw between the points. Then F 1 hs coordintes cµ, nd F hs coordintes cµ for some c. Let X hve the coordintes xµ. Then 11.4 becomes (11.43) Ô x cµ Ô x cµ Eliminte the rdicls to verif tht we end up with qudrtic eqution which is tht of n ellipse. We hve similr description of the hperbol: Proposition 11.6 Given two points F 1 nd F nd positive number, the locus of points X such tht (11.44) XF 1 XF is hperbol with foci t F 1 nd F. Actull, this is just the brnch of the hperbol which wrps round the focus F ; the other brnch is given b the eqution (11.45) XF XF 1 The opticl properties of the conics follow from these string chrcteriztions. Let s strt with the prbol. Suppose tht the prbol is coted with light-reflecting mteril. The rs of bem of light originting fr w long the xis of the prbol will pproch the prbol long lines prllel to

10 Chpter 11 Conics nd Polr Coordintes 166 Figure T β α xµ γ X ds α dx d L L F cµ F β x its xis. According to the phsics of the sitution, the ngle of reflection off the prbol is equl to the ngle of incidence. The opticl propert of the prbol is tht these reflected rs ll meet t the focus. Proposition 11.7 Let X be point on the prbol, nd T the tngent line to the prbol t X. Let L F be the line from the focus to X, nd L the line through X prllel to the xis of the prbol. Then the ngle between T nd L F is equl to the ngle between T nd L. Wht we wnt to show, referring to figure 11.15, is tht γ α. From the figure we see tht this mounts to showing tht β α α. Let us think of the prbol s being trced out b prticle moving to the right t constnt velocit 1. This expresses the coordintes xµ of the point X s functions of rc length s. The string propert of the prbol tells us tht (11.46) Ô x cµ x c Differentiting with respect to rc length gives us (11.47) x cµ dx ds d ds Ô x cµ dx ds which simplifies to (11.48) x c dx Ô x cµ ds d Ô x cµ ds dx ds Now we do little trigonometr: (11.49) cosβ x c Ô x cµ sin β Ô x cµ cosα dx ds sinα d ds

11 Ü11.3 String nd Opticl Properties of the Conics 167 Figure α β dx ds d α F 1 cµ β β 1 β 1 F cµ so (1) becomes (11.5) cosβ cosα sinβ sin α cosα or cos β αµ cosα, from which we conclude β α α s desired. The opticl propert of the ellipse is tht r of light emnting from one focus reflects off the ellipse so s to pss through the other focus. Proposition 11.8 Let X be point on the ellipse, nd T the tngent line to the ellipse t X. Let L 1 be the line from the focus F 1 to X, nd L the line from the other focus F to X. Then the ngle between T nd L 1 is equl to the ngle between T nd L. Wht we wnt to show, referring to figure 11.16, is tht β α β 1 α. We strt with the string propert, written in the coordintes s shown in the figure: (11.51) Ô x cµ Ô x cµ We now differentite with respect to rc length, nd rrive t (11.5) x c Ô x cµ dx ds Ô x cµ d ds x c Ô x cµ dx ds Ô x cµ d ds We now mke the substitutions with the trigonometric functions, but here we hve to be creful: in our picture d nd x c re negtive, so since the sine nd cosine re rtios of lengths, we hve x c (11.53) cosβ 1 sinα d d Ô x cµ ds ds Thus our eqution becomes (11.54) cosβ cosα sinβ cosαµ cosβ 1 µcosα sinβ 1 sinαµ or (11.55) cosβ cosα sinβ cosαµ cosβ 1 cosα sinβ 1 sinαµ

12 Chpter 11 Conics nd Polr Coordintes 168 which is cos β αµ cos β 1 αµ, so β α β 1 α s desired The opticl propert of the hperbol is tht r of light emnting from one focus reflects off the opposite brnch of the hperbol so s to pper to hve come from the other focus. Proposition 11.9 Let X be point on the hperbol, nd T the tngent line to the ellipse t X. Let L 1 be the line from the focus F 1 to X, nd L the line from the other focus F to X. Then the exterior ngles between T nd L 1 nd between T nd L re equl Figure 11.17: T α L α 1 L F 1 F Ü11.4. Polr Coordintes r θ Figure x Often problem cn be seen s tht of understnding the motion of prticle in the plne reltive to fixed point. In such sitution it is desirble to be ble to describe position in terms of the length nd the direction of the line between the two points. These re the polr coordintes of the point. We consider the fixed point s the origin of these coordintes, nd tke the positive x-xis s the zero direction. Then n other direction is described b the ngle between it nd the positive x xis, which we denote s θ. The distnce of point on this line from the origin is denoted r. These equtions relte the crtesin coordintes xµ with the polr coordintes rθ: (11.56) x r cosθ r sin θ r Ô x θ rctn x See figure Polr coordintes hve two peculrities which we need to get used to. Ever vlue of rθµ determines point in the plne. However, if r, the point is the origin, nd θ doesn t mke sense. Secondl, the vlues rθµ nd rθ πµ, nd in fct, rθ nπµ for n n give the sme point. This mbiguit is sometimes of vlue: for exmple, when discussing the motion of prticle, n tells us how mn times the prticle hs wound round the origin in the counterclockwise sense. Finll, it is lso of convenience to let r tke negtive vlues, mening distnce of r in the opposite direction of the r θ. Thus rθµ nd rθ πµ determine the sme point. We now consider the grphs of equtions in polr coordintes. Exmple The eqution r, for is stisfied b ll points of distnce from the origin, so is polr eqution of the circle of rdius centered t the origin.

13 Ü11.4 Polr Coordintes 169 Figure Figure 11. θ θ Exmple 11.1 The eqution θ θ is the line which mkes n ngle of θ with the x-xis. Exmple r θ describes the motion of point which rottes round the origin t ngulr velocit 1 while moving out long the r t velocit. This is the Archimeden spirl nd is shown in figure Exmple r e θ is nother spirl, however, the point moves out long the r t rte exponentil in the rte of rottion. This is the logrithmic spirl nd is shown in figure 11.. Exmple The eqution r cosθ is the circle of dimeter with center on the x-xis which goes through the origin. For, if we multipl b r we get r r cosθ, which cn now be written in crtesin coordintes (using (11.56)) s (11.57) x x or x 4 Given n eqution of the form r r θµ, we cn often trce out the grph b just studing the behvior of the function r θ µ. Let s redo exmple this w. We hve this tble θ π π 3π π r Ô Figure 11.1 Ô Figure 11. µ X xµ θ F r θ V d L

14 Chpter 11 Conics nd Polr Coordintes 17 It will be useful for ou to follow the following discussion long the curve in figure Between nd π the point is in the first qudrnt, nd s the ngle increses it moves towrd the origin, reching there t θ π. Then for θ between π nd π, the point is in the fourth qudrnt (becuse r µ, stedil moving w from the origin until we rech the point we ve strted with. This looks like circle, nd the rgument bove (in exmple 11.15) shows tht it is. Note tht s θ moves from π to π the circle is retrced. Exmple with center on the r of ngle θ. This mounts to the ssertion tht n eqution of the form Similrl, the eqution r cos θ θ µ is the circle through the origin of rdius (11.58) r cosθ bsinθ is circle with the origin the endpoint of one of its dimeters (see prctice problem 1.1). Exmple If we re given the eqution of curve in crtesin coordintes, we cn find its eqution in polr coordintes through the substitution x r cosθ r sinθ. For exmple (11.59) Eqution of line: r c cosθ bsinθ For, the generl eqution of line is x b c. After substitution this becomes (11.6) r cosθ br cosθ c which becomes the bove when we solve for r. Exmple The polr eqution of conic of eccentricit e, focus t the origin nd directrix the line x d is (11.61) Eqution of Conic: r ed 1 ecosθ To show 11.61, we strt with the defining reltion XF exl, referring to figure 11.. In polr coordintes this gives us (11.6) r e d xµ e d r cosθµ Solving for r brings us to (11.61). If the figure is rotted b θ, we just replce θ with θ θ Exmple r cosθ. We first construct the tble: π 3π 4 π θ π 4 r Follow this discussion long the grph shown in figure This time the curve strts (t θ ) t r, nd decreses to zero b θ π4. Between π4 nd π r is negtive, so the curve is in the third qudrnt, nd s θ rottes counterclockwise, r moves w from the origin finll to r for θ π. As θ increses from π the point continues to move towrd the origin (in the fourth qudrnt), rriving there t θ 3π. Moving on, r becomes positive, so we enter the second qudrnt with the distnce from the origin stedil incresing until, t θ π we re t r. Since cosθ is n even function, s we move from π to π (or wht is the sme, from π to ), we just get the sme curve, reflected in the x-xis. This is the four-petlled rose shown in figure 11.4.

15 Ü11.4 Polr Coordintes 171 Figure 11.3 Figure 11.4 Exmple 11. r cos3θ is three-petlled rose. Construct the tble of importnt vlues between nd π nd rgue s in exmple The tble is π 3 π π 3 5π 6 π θ 6 π r Tht completes the rose; s we proceed from π to π we trverse the rose gin. See figure Figure 11.5 Figure 11.6 We conclude Proposition 11.1 The grph of the eqution r cos nθµ or r sin nθµ is n-petlled rose if n is even, nd n n petlled rose if n is odd (trversed twice). Limçons. These re the curves defined b the eqution r bcosθ. First, we consider the cse: b. We hve the tble pi 3π θ π π r b b b leding to the grph shown in figure As b gets closer nd closer to, the vlue of r for θ π goes to zero. Thus when b, we get the grph shown in figure 11.7, clled the crdioid.

16 Chpter 11 Conics nd Polr Coordintes 17 Figure 11.7 Figure 11.8 Then s b goes beond, r becomes negtive s θ gets ner π, nd there is n inner loop of the limçon Exmple 11.1 r 4cosθ. Our tble is this: θ pi 3π π π r 6 6 When cosθ 1, tht is, for θ π3, the vlue of r is zero, nd between these two vlues r is negtive. We get the grph shown in figure We hve drwn the curve so tht it is tngent to θ π3 for those vlues of θ. This is correct, s we will show in the next section. Finll, it is importnt to note tht if the function cosθ is replced b cosθ the curve is reflected in the -xis, nd if it is replced b sinθ, it is rotted b right ngle. Ü11.5. Clculus in polr coordintes Arc length Consider the curve given in polr coordintes b the eqution r r θµ. We cn clculte the differentil ds of rc length b the differentil tringle in polr coordintes using this digrm. Figure 11.9 dr rdt ds dt r r r tµ The length of the rc of the circle of rdius r subtended b the ngle dθ is rdθ. The differentil tringle is thus right tringle with side lengths dr nd rdθ. B the pthgoren theorem (11.63) ds dr r dθ

17 Ü11.5 Clculus in polr coordintes 173 Exmple 11. Find the length of the curve r θ from to π. This curve is spirl whose distnce from the origin increses s the squre of the ngle. We hve dr θdθ, so (11.64) ds dr r dθ 4θ dθ θ 4 dθ θ 4 θ µdθ nd thus the length is (11.65) π ds π θ Ô 4 θ dθ θ 3 π π Are To cculte the re enclosed b curve given, in polr coordintes, b r r θµ, we clculte the differentil of re, using figure The re is dθ r Figure 11.3 rdθ (11.67) Are 1 r r θµ π The re of the wedge given b the increment dθ is 1µr dθ. To see this, we strt with the re of the circle of rdius r : A πr. Now n ngle α subtends segment of the circle which is the (απ)th prt of the full circle, thus the re of tht segment is 1µr α. Thus, for α dθ, we get (11.66) da 1 r dθ Exmple 11.3 Find the re enclosed b the crdioid r 3 1 sinθµ. 3 1 sinθµ dθ 9 π 1 sinθ sin θµdθ Now, we know tht the integrl of sinθ over n entire period is zero, so we cn neglect the middle term. We now use the double ngle formul for the lst term, nd drop the integrl of cos θµ for the sme reson: (11.68) Are 9 π 1 1 cos θµ dθ 9 π 3 7 dθ π Exmple 11.4 Find the re inside one petl of the rose r sin3θ. At θ we hve r, but then s the ngle rottes, r increses to its mximum t 3θ π, nd then decreses bck to zero for 3θ π. Thus one petl is spnned s θ rnges from to π3. We now clculte; (11.69) Are 1 π3 sin 3θµdθ 1 π3 1 cos 6θµ dθ 1 θ cos 6θ µ π3 π 1 1

18 Chpter 11 Conics nd Polr Coordintes 174 Figure r r θµ φ α θ dθ tn line θ Tngents Given the polr eqution r r θµ of curve, we cn find the tngent t n point s follows. First of ll, the crtesin coordintes re given b x r θµcosθ r θµsin θ. If m is the slope of the tngent line, we hve, b the chin rule (11.7) m d dx ddθ dxdθ r cosθ sinθ dr dθ r sinθ cosθ dr dθ Notice tht, s r, the right hnd side pproches tnθ. Thus, if θ is vlue for which r, then the curve pproches the origin long the r θ θ. Exmple 11.5 Wht is the slope of the tngent to the inner loop of the limcon (11.71) r 5cosθ t the origin? We find the vlues of θ for which r : (11.7) 5cosθ or cosθ 5 so tht θ 63π rdins or 1136 Æ.