3.4 Conic sections. In polar coordinates (r, θ) conics are parameterized as. Next we consider the objects resulting from

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1 3.4 Conic sections Net we consier the objects resulting from + by + cy + + ey + f 0. Such type of cures re clle conics, becuse they rise from ifferent slices through cone In polr coorintes r, θ) conics re prmeterize s k 0 r e cosθ e is clle the eccentricity k is n oerll constnt Trnsform into Crtesin coorintes: k > 0, e 0 With r cosθ n y r sinθ r k + e r cosθ r k + e r cosθ) + y k + e) e k/e + ) Circles belong to specil clss of cures clle conic sections. Other such cures re the ellipse, prbol, n hyperbol. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 / 40 Geometricl interprettion of + y e k/e + ) : D k e P, y) F0, 0) FP edp FP + y DP + k e Thus conic is escribe by ll the points P, such tht the istnce to fie point F is fie rtio to line k e, clle the irectri. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 3 / The ellipse e < ) Mnipulting + y e k/e + ) gies Trnsforming the ribles: e ) ke e ) + y k e X ke e y Y yiels the norml form of the ellipse X + Y b with : k n b : k e e. We cn lso epress k, e in terms of, b We shifte the focus by k b e b ke e : c e. e 0 is circle of rius k. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 4 / 40 Collecting eerything gies 3.4. The hyperbol e > ) Similrly s for the ellipse e e b but now e ) < 0. e ) k X + e ) k Y Therefore the norml form of the hyperbol becomes /e /e The istnce is clle the mjor is n the istnce b is clle the minor is.the foci re t ±e. The two irectrices re t ±/e. We cn prmetrise the ellipse by or with rtionl function by Xφ) cosφ Y φ) bsinφ Xt) t + t Y t) b t + t. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 5 / 40 with : k n b : k e e. Epressing k, e in terms of, b k b X Y b e + b nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 6 / 40 e /e /e The shortest istnce between the two sections of the cure is clle the mjor is, equlling. The two irectrices re t ±/e. We cn prmetrise the hyperbol by e The prbol e ) Now we he + y k + ) such tht y k + k Trnsforming the ribles: X + k y Y yiels the norml form of the prbol Y 4X X ± φ) ±coshφ Y φ) bsinhφ with : k/. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 7 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 8 / 40

2 Now we he only one focus t, 0) The irectri is t. The cure prbol cn be prmeterise by X t Y t Emple 3.4.: Determine the foci n irectices of the ellipse ) 5 y + 3) +. 6 We compre with X + Y b. To trnsform in this wy we must he X Y y b 4. lso b e ) implies tht e 3 5. Therefore the centre of the ellipse is t, 3), the mjor is hs length 0 n the minor is hs length b 8. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 9 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 0 / 40 The foci lie on the mjor is t istnce e 3 from the centre. So the foci re 5, 3), 3). Directrices re perpeniculr to the mjor is n t istnce e 5 3 from the centre. So the irectrices re Emple 3.4.: n ellipse hs foci t, 5) n 8, 5) n eccentricity e 4. Fin its Crtesin eqution. The centre is miwy between the foci, so lies t 5, 5). The istnce from the centre to ech focus is e 3, n so. Therefore b e ) 35. From this we see tht the eqution is gien by 5) 44 y 5) nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 / 40 Lecture 4 4. Clculus I: Differentition 4. The eritie of function Suppose we re gien cure with point lying on it. If the cure is smooth t then we cn fin unique tngent to the cure t : If the tngent is unique then the grient of the cure t is efine to be the grient of the tngent to the cure t. The process of fining the generl grient function for cure is clle ifferentition. B b c Consier the chor B. s B gets closer to, the grient of the chor gets closer to the grient of the tngent t. Here the cure in ) is smooth t, but the cures in b) n c) re not. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 3 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 4 / 40 For y f ), the grient function is efine by ) δy ) f + ) f ) lim lim ,y+ δy),y) We enote the grient function by or f ), n cll it the eritie of f. This is not the forml efinition of the eritie, s we he not epline ectly wht we men by the limit s 0. But this intuitie efinition will be sufficient for the bsic functions which we consier. δy Emple 4..: Tke f ) c, constnt function. t eery the grient is 0, so f ) 0 for ll. Or f + ) f ) Emple 4..: Tke f ). c c 0. t eery the grient is, so f ) for ll. Or f + ) f ) + ). nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 5 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 6 / 40

3 Emple 4..3: Tke f ). Now we nee to consier the secon formultion, s we cnnot simply re the grient off from the grph. f + ) f ) + ) + + ) + ) +. The limit s tens to 0 is, so f ). Emple 4..4: Tke f ). f + ) f ) + ) + ) ) + ) ) + ) + ). The limit s tens to 0 is, so f ). nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 7 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 8 / 40 Emple 4..5: Tke f ) n with n N n n >. Recll tht n so n b n b) n + n b + n 3 b + + b n ) n b n b n + n b + n 3 b + + b n where the sum hs n terms. s b we he n b n ) lim lim n + n b + n 3 b + + b n ) nb n. b b b If + n b then ) f + ) f ) lim lim 0 b Hence f ) n n. n b n b ) nb n n n. Emple 4..6: f ) sin. We use the ientity for sin + sin B. n so f + ) f ) sin f + ) f ) sin ) ) cos + ) cos + ). We nee the following fct which we will not proe here): n so sin θ lim θ 0 θ f sin ) ) lim cos + ) cos). 0 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 9 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 0 / 40 Some stnr erities, which must be memorise: Lecture 5 4. Differentition of compoun functions f ) f ) k k k e e ln sin cos cos sin tn sec cosec cosec cot sec sec tn cot cosec Some of these results cn be erie from the results in the following sections, or from first principles. Howeer it is much more efficient to know them. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 / 40 Once we know few bsic erities, we cn etermine mny others using the following rules: Let u) n ) be functions of, n n b be constnts. Function Deritie Sum n ifference u ± b u ± b Prouct u u + u u Quotient u u u Composite u)) z. z where z ). The finl rule boe is known s the chin rule n hs the following specil cse u + b) u + b) For emple, the eritie of sin + b) is cos + b). nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 / 40 Emple 4..: Differentite Emple 4..: Differentite y y +. + ) ) 4 + ) + ). Emple 4..3: Differentite y ln + 3). ln + 3) Emple 4..4: Differentite y e 5. Set z 5, then z z ez 5 5e 5. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 3 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 4 / 40

4 Emple 4..5: Differentite y 4 sin + 3). Set z + 3, then z 4 cosz) 8 cos + 3). z s we he lre note, some of the stnr erities cn be euce from the others. Emple 4..6: Differentite y tn sin cos. cos cos sin sin ) cos cos + sin cos cos sec. Emple 4..7: y cosec sin. sin.0). cos sin cos sin cosec cot. Emple 4..8: y ln + + ), i.e. y ln u where u + +. u u n so ) u + + ). ) nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 5 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 6 / 40 Emple 4..9: y. We he y e ln ) e ln ), i.e. y e u where u ln. u eu e ln ln) + ) ln) + ). 4.3 Higher erities The eritie is itself function, so we cn consier its eritie. If y f ) then we enote the secon eritie, i.e. the eritie of with respect to, by y or f ). We cn lso clculte the higher erities n y or f n) ). n Emple 4.3.: y ln + ). Let z. + y z + ). ) + ) ) + ). Emple 4.3.: Show tht y e sin) stisfies y + + 5y 0. e sin + e cos e cos sin ) y e cos sin ) + e 4 sin cos ) e 3 sin 4 cos ). nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 7 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 8 / 40 Writing s for sin n c for cos we he y + y + 5y e 3s 4c s + 4c + 5s) 0. Now 4 + ) ) Emple 4.3.3: Elute ) 3 + ) + 3) t 0. We coul use the quotient rule, but this will get complicte. Inste we use prtil frctions. y ) + 3) + + B + ) + C + 3. y + ) ) 3 3 y ) ) 4 n substituting 0 we obtin tht 3 y 0) We obtin check!) 0, B, n C 3. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 9 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn / 40 Lecture 6 Generlly it is hr to gie simple formul for the nth eritie of function. Howeer, in some cses it is possible. The following cn be proe by inuction. Emple 4.3.4: y e. We cn show tht e n n y n n e. y e. Emple 4.3.5: y sin). We cn show tht y cos) sin + π ) y sin) sin + π) y 3 cos) 3 sin + 3π ) y i) 4 sin) 4 sin + π). n y n n sin + nπ ). nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 3 / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn 00 3 / 40

5 4.4 Differentiting implicit functions Sometimes we cnnot rerrnge function into the form y f ), or we my wish to consier the originl form nywy for emple, becuse it is simpler). Howeer, we my still wish to ifferentite with respect to. Gien function gy) we he from the chin rule gy)) gy)). Emple 4.4.: + 3y y 4. Therefore we he + 3y y 4 ) ) y ) y 4 ) 0 + 3y + 3 y ) 4y 3 + 3y + 6y 3 4y 0 0. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn / 40 Emple 4.4.: + 3 y. Therefore we he + 3 y ) ) ) 3 y y Differentiting prmetric equtions Sometimes there is no esy wy to epress the reltionship between n y irectly in single eqution. In such cses it my be possible to epress the reltionship between them by writing ech in terms of thir rible. We cll such equtions prmetric equtions s both n y epen on common prmeter. Emple 4.5.: t 3 y t 4t +. lthough we cn write this in the form y the prmetric ersion is esier to work with. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn / 40 To ifferentite prmetric eqution in the rible t we use Emple 4.5.: Continue.) n so t 4 n t 4 3t.. 3t Emple 4.5.: Fin the secon eritie with respect to of We he Therefore Now y sin θ y cos θ. θ cos θ sin θ. θ sin θ 4 sin θ. cos θ ) 4 sin θ) θ 4 cos θ 4 sin θ) 4. θ cos θ nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn / 40 Note: The rules so fr my suggest tht erities cn be trete just like frctions. Howeer y y t in generl. Moreoer y ). Emple 4.5.: Continue.) We he n Therefore θ y θ 4 cos θ 4sin θ cos θ) ) θ ) θ θ sec θ) sec θ tn θ. y θ θ 4 tn3 θ 4 tn θ 4 y. nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn / 40 nres Fring City Uniersity Lonon) S05 Lecture 3-6 utumn / 40