B.A. (PROGRAMME) 1 YEAR MATHEMATICS

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1 Gdute Couse B.A. (PROGRAMME) YEAR MATHEMATICS ALGEBRA & CALCULUS PART B : CALCULUS SM 4 CONTENTS Lesson Lesson Lesson Lesson Lesson Lesson Lesson : Tngents nd Nomls : Tngents nd Nomls (Pol Co-odintes) : Cuvtue 4 : Asymptotes 5 : Singul Point 6 : Cuve Tcing 7 : Cuve Tcing (Pol Co-odintes) Edito: D. S.K. Vem SCHOOL OF OPEN LEARNING UNIVERSITY OF DELHI 5, Cvly Lne, Delhi-0007

2 Session 0-0 (000 Copies) School Open Lening Published By : Executive Diecto, School of Open Lening, 5, Cvly Lne, Delhi-0007 Lse Composing by : M/s Computek System

3 LESSON TANGENTS AND NORMALS INTRODUCTION One of the eliest pplictions of Diffeentil clculus ws the study of cuves by mens of thei tngents nd nomls. In this lesson we shll give geometicl intepettion of the diffeentil coefficient dy dx t point on cuve. Subsequently we will find out the equtions of the tngent nd noml to cuve whose eqution is given.. () Definition of the Tngent to Cuve t Given Point Let P be given point on cuve, nd Q nothe point on the cuve close to P. Then s Q tends to P, the secnt PQS in genel tends to stight line PT, sy, which is clled the tngent to the cuve t P. The tngent to cuve t point my, theefoe, be defined s the limit to which secnt though the point tends s the othe point of intesection with the cuve tends to the given point. Since the two points of intesection of the secnt nd the cuve tend to coincide, tngent is sometimes defined s stight line meeting the cuve into ultimtely coincident point.. (b) Geometicl Intepettion of the Diffeentil Coefficient Let P be ny point (x, y) on given cuve nd Q be neighbouing point (x + δx, y + δy) on the cuve so tht δx nd δy e smll quntities. Dw PL nd QM pependiculs on OX nd PK pependicul to QM. As is cle fom the digm tn QPK QK δy PK δ x, when Q P, δx 0, QPK TPK Ψ (sy) Poceeding to the limit s δx 0, we get nd equte the two esults. tn Ψ δy dy lim (t the point P). δx 0 δx dx

4 Thus the vlue of the diffeentil coefficient t point on the cuve is the tngent of the ngle which the tngent t tht point mkes with the X-xis i.e., the vlue of dy, t P is nothing but the slope of the dx tngent t P. Note. If dy, t point P on cuve is zeo, the tngent to the cuve t P is pllel to X-xis; dx convesely if the tngent t point P is pllel to the X-xis, dy is zeo t P. dx Definition of Noml The stight line pependicul to tngent t its point of contct is clled the noml to the cuve t tht point. Eqution of the Tngent nd Noml Let the eqution of the cuve be y f (x) nd let P (x, y) be ny point on it. Let the tngent PT t P to the cuve meet the X-xis in M. Let TMX Ψ, we hve Tn Ψ dy (t) P f (x).... () dx Eqution of the tngent PT is Y y dy (X x) f (x) (X x)... () dx which is the line though (x, y) hving slope i.e. m. Hee (X, Y) e the cuent coodintes i.e., the coodintes of ny point on the line () nd (x, y) e the coodintes of P, whee the tngent is dwn. 4

5 The noml PN being pependicul to the tngent PT t P, its eqution is given by i.e., ( x) dy X + Y y dx Y y ( X x ) dy dx ( ) 0... () Note. In cse dy does not exist t (x, y) but dx 0, then the tngent t (x, y) cn be obtined s dx dy follows: In this cse the tngent is pllel to the Y-xis nd hence the eqution of the tngent line is given by X x. Note. When the equtions of cuve e given in pmetic fom viz. then nd the equtions of the tngent nd noml cn be witten immeditely fom () nd () in the foms; g ( t ) Y g (t) X f ( t ), f t nd [X f (t)] + g f ( t) ( t) Note. The intecept of the tngent on X-xis is obtined by putting Y 0 in (). Intecept on the X-xis is given by x f (t), y g (t) dy ( ) dx dt g t dx f ( t) nd intecept of the tngetn on Y-xis is obtined by putting X 0 in (). ( ) Y g t 0 dy dt X x 5 ( ) y dy dx

6 Intecept on the Y-xis is given by Y y x dy dx. Exmple. Find the equtions of tngent nd noml to the pbol y 4x t the point (, ). Solution. Diffeentiting w..t.x we hve, At the point Tngent t (, ) is i.e., And the noml t (, ) is y 4x y dy dx 4 dy, y 0. dx y (, ), dy dx. y (x ), y x + y (x ) o x + y. Exmple. Find the tngent nd the noml to the cuve x (θ sinθ), y ( cosθ) t ny point θ. Solution. Hee Eqution of tngent is: Eqution of noml is dy dy dθ dx dx dθ θ θ sin.cos sin θ θ cot. ( cos θ) θ sin y ( cosθ) cot ( sin ) 6 θ θ θ x. θ θ θ x cot [y ( cosθ)] ( sin ) o [y ( cosθ). cot + x (θ sinθ) 0. Exmple. Show tht the line x cosθ + y sinθ p, will touch the cuve x m y n m+n if Solution. Let the line be the tngent to the cuve t the point (x, y ). Fom (), tking logithms, we hve Diffeentiting w..t.x, we get p m+n m m n n (m + n) m+n m+n cos m θ sin n θ. x cosθ + y sinθ p,... () x m y n m+n,... () m log x + n log y (m + n) log.

7 m n dt + x y dx dy t the point (x y ) my. dx nx Eqution of tngent t (x, y ) to () is 0. ( 0) y y my nx 7 dy dx (x x ) m y y n x o (my ) x + (nx ) y (m + n) x.y.... () If () is tngent to () t (x, y ) then () nd () must be identicl. Comping coefficients in () nd (), we get my nx cos θ sin θ m + n x y ( ) pm x ( m + n )cos θ pn nd y ( m + n )sin θ Also since (x, y ) is point on the cuve (), we get Substituting fo x, y, we hve o m m ( m n) cos ( m n) is the equied condition. x y m+n. m n m m n n p m p m. n n + θ + sin θ m+n p p m+n m m n n (m + n) m+n m+n cos m θ. sin n θ Exmple 4. Find the eqution of the tngent nd noml to the cuve y (x ) (x 4) x t the point whee it cuts the X-xis. Solution. y (x ) (x ) x y (x 5x + 6) x () The cuve () cuts the X-xis viz. y 0 whee 0(x ) (x ) x o x 7 The point of intesection is (7, 0). Diffeentiting both sides of () w..t. x we hve; dy dx At the point (7, 0); (x ) (x ) + y (x 5) 0 dy dx (7 ) (7 )

8 Eqution of tngent t the point (7, 0) is dy dx 0 y 0 nd eqution of the noml t the point (7, 0) is 0 y 0 0 (x 7) 0x + y 7 0. (x 7) x 0y 7 0 Exmple 5. Pove tht the sum of the intecepts on the co-odinte xes of ny tngent to the cuve x + y is constnt. Solution. Eqution of the cuve is x + y Diffeentiting both sides w..t. to x, we get dy +. 0 x y dx dy dx Eqution of tngent to the given cuve t (x, y) is Y y y x y x (X x) X y + Y x x y + y x + X y Y x x y ( x + y ) X y + Y x x y [Q x + y ] Intecept on the xis of X is obtined by putting Y 0 x y i.e., Intecept on X-xis y And Intecept on the xis of Y is obtined by putting X 0 in the eqution of tngent. x y Intecept on y-xis x Sum of the intecepts on co-odinte xes x + y. ( x + y ) x y Q is independent of x nd y. constnt. 8

9 Exmple 6. Show tht the length of the potion of the tngent to the cuve x cos θ, y sin θ intecepted between the co-odinte xes is constnt. Solution. We hve dx d θ dy d θ dy dy dθ dx dx dθ Eqution of tngent t the point θ is cos θ sinθ. sin θ cosθ sin θcos θ cos θsin θ y sin θ tn θ (x cos θ) tnθ x sin θ + y cos θ sin θ cos θ (sin θ + cos θ) x sin θ + y cos θ sin θ cos θ... () Tngent line () meets the co-odinte xes t the points A ( cos θ, 0) nd B(0, sin θ). [These points e obtined by putting y 0 nd x 0 in () espectively]. Thus the length of the potion of the tngent intecepted between the co-odinte xes is given by AB cos θ+ sin θ, which is constnt. Exmple 7. Show tht the tngent nd noml t ny point of the cuve e equidistnt fom the oigin. Solution. Hee nd dx d θ dy d θ dy dy dθ dx dx dθ x e θ (sin θ cos θ) y e 0 (sin θ + cos θ) e θ (sin θ cos θ) + e θ (cos θ + sin θ) e 0 sin θ e 0 (sin θ + cos θ) + e θ (cos θ sin θ) e θ cos θ θ e cos θ cos θ θ e sin θ sin θ Eqution of the tngent to the given cuve t ny point θ is cosθ sin θ x e y e θ (sin θ + cos θ) θ ( sin θ cos θ ) x cos θ y sin θ e θ (cos θ sin θ cos θ sin θ sin θ cos θ) 0 x cos θ y sin θ + e θ 0... () 9

10 And eqution of the noml t ny point θ to the given cuve is sin θ cosθ x e y e θ (sin θ + cos θ) θ ( sin θ cos θ ) x sin θ + y cos θ e θ (sin θ cos θ + cos θ + sin θ sin θ cos θ) 0 x sin θ + y cos θ e () Now the tngent nd the noml e equidistnt fom the oigin if the length of the pependicul fom the oigin up on the tngent () nd noml () e equl. The length of the pependicul fom (0, 0) upon the line () is p And the length of the pependicul fom (0, 0) upon the line () is [neglecting the ve sign] p Fom () nd (4), p p Hence the esult. θ 0.cos θ 0.sin θ+ e e θ... () cos θ + sin θ 0.sin θ 0.cos θ+ e sin Exmple 8. Show tht tngent t ny point of the cuve θ+ cos θ θ e θ +e θ... (4) x (t + sin t cos t), y ( sin t) mkes n ngle 4 Solution. Hee nd (π + t) with the x xis. dx dt dy dt [ + cos t. cos t + sin t ( sin t)] [ + cos t sin t] [ + cos t] cos t ( + sin t) cos t dy dx ( + ) sin t cos t cos t t t cos + sin + sin t cos t t t cos sin t t cos + sin t t cos sin 0

11 t + tn π t tn + t tn 4 If θ is the ngle which the tngent t ny point t mkes with the X-xis. then tnθ π t tn + 4 θ Hence the esult. π t + 4 ( ) 4 π + t. Exmple 9. Pove tht fo the cteny y c cos h x c the odinte upon the tngent is of constnt length., the pependicul dopped fom the foot of Solution. Hence y c cos h x c Eqution of tngent t ny point (x, y) is o Y cos h x c c o x x x X sin h Y + c cos h x sin h c c c dy dx x x c. sin h sin h c c c x c Y y sin h [ X x] x c sin h [ X x] 0 The foot of the odinte t the point (x, y) is (x, 0). The length of the pependicul fom (x, 0) on the tngent x x x xsin h 0 + c cos h x sin h c c c x + sin h c x c cos h c x cos h c c, i.e., constnt. x y Exmple 0. Pove tht + touches the cuve y b the y-xis. [Q y c cos h x c ] x be t the point whee the cuve cosses

12 Solution. y whee The cuve cosses y-xis t the point (0, b). x be cosses y-xis i.e., the line x 0 y be b Hence it is equied to find the tngent t (0, b). Now At (0, b) Eqution of tngent t (0, b) is dy dx be x dy dx b y b b (x 0) i.e., y b x x x i.e., + b Hence poved. Execise. Find the eqution of the tngent nd noml to the cuve x cosθ, y b sinθ t ny point θ.. Find the tngent nd noml to the cuve (i) x + y b t the point (, 0) (ii) x (θ + sinθ), y ( + cosθ) t θ π. (iii) y c cos h x c t the point (0, c). Show tht the line x cos θ + y sin θ c is tngent to the cuve x y (x + y ). [Hint: The eqution of the cuve cn be witten s + ] x y c x 4. Pove tht the stight line + whteve be the vlue of n. y b n x y touches the cuve + t the point (, b) b 5. At wht point of the cuve y x x + is the tngent pependicul to the line y x? n [Ans. (, 0)] 6. At wht point of the cuve y x + x 0x + 7 e the tngents pllel to the line y x? [Ans. (, ), (, )] 7. Pove tht the eqution of the tngent t ny point (4m, 8m ) of the semicubicl pbol x y 0 is y mx 4m nd show tht it meets the cuve gin t (m, m ), whee it is noml if 9m.

13 8. Show tht the noml t ny point of the cuve x cos θ + θ sin θ, y sin θ 0 cos θ is t constnt distnce fom the oigin. 9. The tngent t ny point on the cuve x + y cuts off lengths p nd q on the co-odinte xes, show tht p / + q / / /. Angle of Intesection of Two Cuves The ngle t which two cuve intesect t point is defined s the ngle between the tngents to the cuves t tht point. Applying the fomul tnθ m m fo the ngle θ between the two lines whose + mm gdients (slopes) e m nd m, the ngle of intesection of two cuves t point of intesection is esily found. When the ngle between two cuves t point of intesection is ight ngle, the cuves e sid to intesect othogonlly the conditionn fo which is m m. Exmple. Find the ngle of intesection of the pbols y 4x nd x 4byy t the point othe thn the oigin. Solution. The points of intesection of the pbols y 4x, nd x 4by e given by x 4 6b y 6b. 4x 64b x o x (x 64b ) 0 x 0 o x 4 / b / Substituting the vlue of x in x 4by, we get y x 6 b 4b 4b / 4 / (0, 0) nd (4 / b /, 4 / b / / ) e the two points of intesection. Now y 4x... (i) x 4by... (ii) Diffeentiting (i) w..t. x, we hve y dy dx 4 dy dx y 4 / b /.

14 dy dx / / / / t( 4 b, 4 b ) Diffeentiting (ii) w..t.x; we get, dy dx / / / / t( 4 b, 4 b ) 4 b / / b 4 / / x 4b dy dx 4 b b b / / / / dy x dx b. Thus if m, m e the slopes of the tngents to the two cuves, we hve m b / /, m m The ngle θ tn m + m m b. / / tn b b +. b b / / / / / /. / / b / / tn / / ( + b ) Exmple. Find the condition tht the cuves x + by nd x + b y should cut othogonlly. Solution. Let (x, y ) be the point of intesection. Then since the point lies on both the cuves, we hve x by 0 + x b y 0 + x b + b Now diffeentiting x + by, we get x + by dy dx 0 dy dx x by m y + b b b b b b nd b b dy dx x (, y ) x b y... ()

15 Similly diffeentiting x + b y, we get x + b y dy dx 0 m The two cuves will cut othogonlly if dy dx x b y m m dy dx x (, y ) x b y i.e., x x if by b y + x bb y 0 Substituting the vlues of x nd y fom (), we get, b b + bb b b b b ( ) + ( ) which is the equied condition. Execises 0 b b bb 0 b b + bb b b bb 0 b b Find the ngle between the following pis of cuves t ech one of thei points of intesection (i) x y, x + y (ii) Pove tht the cuves y x nd y x cut othogonlly when. (iii) Find the ngle of intesection of the following cuves: t the point (, ). x + xy y + x 0 y x 4 y + 0 Lengths of the Tngent, the Noml, the Sub-tngent nd the Sub-noml Let the tngents t ny point P(x, y) on cuve meet the X-xis in T, then PT is the length of the tngent intecepted between the point P nd the X-xis nd is often clled the length of the tngent. The pojection of this length on OX is clled the subtngent. Hee TM is the subtngent (whee PM is the pependicul on OX). 5

16 If the noml t P to the cuve meets OX in N, then PN is clled the length of the noml nd its pojection MN on OX is clled the subnoml. These lengths e esily obtined fom the ight ngled tingles PTM nd PMN. We hve whee PT, the length of the tngent t PTM MPN Ψ, tn Ψ dy t P dx P PM cosec Ψ [Q PM PT sin Ψ PT PM cosec Ψ nd PM y] y + cot Ψ y dy y + + tn Ψ dx tn Ψ dy dx TM, the length of the sub-tngent Also PN, the length of the noml MN, the length of the subnoml PM cot Ψ PM sec Ψ dx dy + + dx y y dy dx dy PM y dx y tn Ψ dy dy dx Q PN y + tn Ψ PM cos ψ PN dy y + dx PM Q tn ψ TM N PM sec ψ 6

17 PM tn Ψ y dy dx, PM Q tn ψ TM Exmple. Show tht the subtngent t ny point of the exponentil cuve y e x/b is constnt nd the subnoml vies s the sque of the odinte. Solution. Hee Subtngent t (x, y) y dx dy dy dx x / b y. e b b.. b y y b which is constnt since b is independent of x, y. Subnoml dy y y which vies s the sque of the odinte y. dx b Exmple 4. Show tht fo the cuve by (x + ), the sque of subtngent vies s subnoml. Solution. Eqution of the cuve is Diffeentiting w..t. x, we get Length of subtngent Length of subnoml by dy dx by (x + ) (x + ) dy dx ( x + ) y dy dx y dy by by y ( ) x + ( x + ) by ( x ) + b. b x dx y. ( x + ) (x + ) b N ( + ) (x + ) T (sy) by Q y ( x + ) b (x + ) N (sy) b Sque of subtngent 4 9 (x + ) T N. Hence poved. T 4 b. N 9 8 b N 7 kn [whee k 8 b constnt] 7 Exmple 5. Show tht in the cuve y log (x ) the sum of the tngent nd the subtngent vies s the poduct of the co-odintes of the point. Solution. Hee y log (x ) 7

18 Length of Subtngent y dx dy dx x x x x. y( x ) dy x And length of tngent y dx + dy y Hence the sum of the tngent nd the subtngent Execises ( ) ( + ) x y + x ( ) 4 x + x 4 x y ( x ) ( + ) y x x + y x y x y x y y x + ( x + x + ) x x x x. xy ( xy) which vies s xy.. Find the lengths of the subtngent, subnoml, tngent nd noml fo the following cuves (i) x y 5 t (, ) (ii) x cos θ, y sinθ t θ. (iii) x (θ sin θ), y ( cos θ ) t θ π.. Find the lengths of the noml nd subnoml to the cuve y x / x / e e.. Show tht the subtngent t ny point of the cuve x m y n m+n vies s the bsciss. 4. Show tht in the pbols y 4x, the subnoml is constnt nd the subtngent vies s the bsciss of the point of contct. x y 5. Pove tht in the ellipse +, the length of the noml vies invesely s the b pependicul fom the oigin upon the tngent. 6. Fo the cteny y c cos h x, pove tht the length of noml is c 7. Show tht the subnoml t ny point of cuve y x (x ) vies invesely s the cube of the bsciss. + y c. 8

19 8. Show tht fo the cuve βy (x + α), the sque of the subtngent vies s the subnoml. 9. Show tht in the cuve y log (x ), the sum of the tngent nd the subtngent vies s the poduct of the co-odintes of the point. 9

20 LESSON TANGENTS AND NORMALS (POLAR COORDINATES) Angle between Rdius Vecto nd Tngent Let P be given point (, θ) on the cuve f (θ) nd Q neighbouing point on the cuve vey close to P whose coodintes e ( + δ, θ + δθ) so tht + δ f (θ + δθ). Let the tngent PT t P mkes TPM equl to φ with the dius secto OP. To find φ we obseve tht PT is the limiting position of the secnt PQ when Q P. Fom Q, dw QM on OP. Then tn QPM QM Now when Q P, then δθ 0 nd δ 0 But we know tht lim δθ 0 TPM lim QPM Q P ( ) PM ( ) + δ sinδθ + δ cos δθ tn φ lim tn QPM Q P sin δθ δθ δ lim δθ 0 δθ lim δθ 0 δθ 0 ( ) ( ) + δ sin δθ + δ cos δθ sin δθ ( + δ ) lim δθ cos δθ δ + cos δθ δθ δθ d d θ nd lim ( cos ) δθ 0 [Q PM OM OP OM ] δθ 0

21 Also lim δθ 0 cos δθ δθ cos δθ δθ tn φ Then ngle φ is the ngle which the positive diection of the tngent i.e., the diection n which θ inceses mkes with the positive diection of the dius vecto i.e., the diection in which inceses. This ngle φ lies between 0 nd π nd when φ is obtuse, tn φ is negtive. Pol Subtngent nd Pol Subnoml δθ sin δθ δθ sin δθ sin δθ δθ sin lim δθ lim sin δθ 0 δθ 0 δθ 0 0 lim δθ 0 dθ d d dθ Let O be the oigin nd OX the initil line. Let P (, θ) be ny point on the cuve whose pol eqution is given nd let the pependicul t O to the dius vecto OP meet the tngent t P nd T nd noml t P in N. Then OT is clled the pol subtngent t P nd ON the pol subnoml t P. Also PT is the length of tngent nd PN the length of the noml t P. Fom the ight ngled OPT, we hve pol subtngent OT OP tn φ. θ d d θ d d

22 Fom OPN, we hve Pol subnoml Length of tngent And length of noml ON OP cot φ tn φ θ d d dθ d PT OP sec φ + tn φ PN OP cosec φ dθ + d + cot φ dθ + d d + d θ Note: A negtive vlue of pol subtngent o the pol subnoml implies tht in the cuve t the point unde considetion deceses s θ inceses. Angle of Intesection in Pol Coodintes The ngle of intesection of two cuves is evidently the diffeence between the vlues of φ fo them t the point of intesection. If these ngles e φ, φ, the equied ngle is given by whee tn φ tn (φ φ ) tn φ tn φ + tn φ tn φ tn φ when φ φ i.e., i.e., when i.e., when dθ d dθ d. fo the fist cuve fo the second cuve π, the cuves e sid to intesect othogonlly. when tn φ tn φ, the two cuves intesect othogonlly. φ φ φ φ 0, the two cuves touch. tn φ tn φ the two cuves touch. Exmple. Fo the cdioid ( cos θ) pove tht (i) φ θ (ii) Pol subtngent sin θ θ tn.

23 Solution. Hee tn φ d d θ d θ d sin θ ( cos θ) sin θ θ sin θ θ sin cos tn θ φ θ. Pol subtngent d θ d θ d d ( cos θ), tn φ θ ( cos θ). tn Q φ θ θ θ sin tn Exmple. Find the ngle of intesection of the two cuve cos θ nd ( cos θ). Solution. The two cuves cos θ nd ( cos θ) intesect when cos θ cos θ cos θ cos θ θ nπ + π, whee n is ny intege Let us tke the vlue θ π. d Fo the cuve cos θ d θ dθ tn φ d Also fo the cuve tn φ, sin θ cos θ cot θ sin θ ( cos θ), d d θ dθ cos θ d sin θ sin θ

24 The cuves intesect, when θ θ sin θ θ sin cos θ tn tn (φ φ ) tn φ tn φ + tn φ tn φ tn (φ φ ) θ cot θ tn θ cot θ. tn π π π cot tn 6 π π cot tn 6. φ φ tn ( ) The equied ngle is π. π π π Exmple. Show tht the cuves ( + cos θ) nd b ( cos θ) intesect othogonlly. π Solution. The two cuves will intesect othogonlly when φ φ the condition fo which is tn φ. tn φ. o + tn φ tn φ 0 Fo the fist cuve ( + cos θ) we hve d dθ sin θ 4

25 tn φ Similly fo the second cuve d dθ dθ + cos θ d sin θ b ( cos θ) tn φ b sin θ 5 dθ cos θ d sin θ cos cos tn φ tn φ + θ θ sin θ sin θ The cuves intesect othogonlly. Execises. Find the ngle φ fo the cuve (i) cos φ (ii) m m cos mθ ( cos θ ) sin θ (Q θ 0). Find the ngle of intesection of cuves sin θ + cos θ nd sin θ.. Show tht in the equingul spil e θcotα, the tngent is inclined t constnt ngle to the dius vecto. [Hint : Pove tht φ α] Also show tht the pol subtngent is tn α nd pol subnoml is cot α. 4. Show tht in the cuve θ, the subnoml is constnt nd in the cuve θ, the pol subtngent is constnt. 5. Fo the cdioid ( cos θ), pove tht (i) Pol subtngent sin (ii) Pol tngent sin (iii) Pol noml sin θ (iv) Pol subnoml sin θ. θ θ tn θ θ sec 6. Show tht the logithmic spil e bθ hs the lengths of its pol tngent, pol noml, pol subtngent nd pol subnoml ech popotionl to. 7. Pove tht the two cuves cos θ nd cut othogonlly. + cos θ

26 The Pependicul fom the Pole on the Tngent Let P be point on cuve, distnt fom the pole. Let ON be the pependicul fom O on the tngent t P. Then the length of ON denoted by p is n impotnt quntity nd cn sometimes be used s coodinte to define the position of P. Evidently fom the ight ngled ONP, we hve, Also we hve p sin φ tn φ p sin φ d θ d d d θ d + dθ... () p d + dθ 4 A net fom of this is obtined if in plce of, we used 6 + d θ 4 d

27 du d θ p u, d d θ du u + dθ Pedl Eqution The eltion between the quntities p nd of ny point on cuve is clled the pedl eqution of the cuve. Fo some cuves this eqution is vey simple. To obtin the pedl eqution of cuve whose eqution is given in ctesin fom we use the fomul The length of the pependicul fom the oigin on the tngent x + y... () Y y dy dx p (X x) is given by dy x y dx dy + dx [i.e., Length of fom (0, 0) to the stight line x + by + c 0 is The vlue of dy dx is obtined fom the eqution of the cuve viz. c + b ]... () f (x, y) 0... () Eliminting x, y fom (), () nd () the eqution obtined is clled Pedl eqution. Note: When the position of the pole is not mention then it is to be tken t the oigin of the ctesin xes of coodintes. To obtin the pedl eqution fom the pol eqution We use the eltion p f (θ) o f (, θ) 0... (4) + d θ 4 d The elimintion of θ between (4) nd (5) gives the pedl eqution. Exmple 4. Show tht the pedl eqution of the pbol y 4 (x + ) is p. Solution. Hee y 4 (x + ) Eqution of tngent t ny point (x, y) is y dy dx 4 dy dx y... (5) 7

28 Y y y (X x) o X yy + y x 0. The length of the pependicul p fom the oigin on the tngent is given by p y 4 x + y y 4 4 x ( ) + x + ( ) 4 x + x x x + x 4 4 ( + ) 4 x ( + ) ( x + ) ( + ) x x [Q y 4 (x +)] Also x + y x + 4 (x + ) (x + ) x + p (x + ) p is the equied pedl eqution. Exmple 5. Show tht the pedl eqution of the cuve cos θ sin p. Solution. Diffeentiting w..t. θ we hve: Also d d θ d d θ p cos θ sin θ sin θ + d θ 4 d Fom (), cos θ sin sin θ 6 θ sin θ cos θ 4 4 8

29 p is the equied pedl eqution. Execise. Obtin the pedl eqution of the following cuves (i) ( cosθ) (ii) cosθ (iii) n n sin nθ (iv) e θ cot α (v) cosθ. Fo the pbol θ cos θ, show tht pol subtngent is cosec θ nd p cosec. Deivtive of Ac Let the tngents t two neighbouingg points P nd Q on cuve meet in T, then we shelll ssume tht c PQ lim chod PQ Q P Deivtive of Ac in Ctesin Coodintes p p p i Let P be ny point (x, y) on cuve nd Q neighbouing point (x + δx, y + δy) on the cuve, vey close to P. Dw PM fom P on the odinte QL. 9

30 Chod PQ PM + QM ( δ ) + ( δ ) x y... () Let A be ny fixed point on the cuve of which PQ is n c. Let the length of the c AP s nd tht of c AQ be s + δs so tht c PQ δs. Then fom (), δs δs chod PQ. δx chod PQ δx Ac PQ. chod PQ ( δ x) + ( δy) x When Q P, then δx 0 nd c PQ lim Q P chod PQ Ac PQ. chod PQ δy + δx Tking limits s δx 0, we get ds dx δs lim δx 0 δx δy + δx The positive vlue of the dicl being tken if the convention is mde tht s is so mesued s to mke it incese with x incesing, s in the figue. Multiplying both sides of () by dx, we hve dy ds dx +, dy dy which is useful when the eqution of the cuve is given in the fom of x f (y). hve When the eqution of cuve is given in pmetic fom x f ( ), y g ( ) then Coolly: Putting dy dx ds dt dx dy + dt dt tn ψ, whee ψ is the ngle which the tngent mkes with the X-xis we Also ds dx dy + + tn ψ sec ψ dx cos ψ dx ds ds dx + + cot ψ cosec ψ dy dy dy ds sin ψ. Deivtives of the Ac in Pol Coodintes 0

31 Let P nd Q be two neighbouing points on cuve whee ctul distnces fom fixed point A sy on the cuve e s nd s + δs espectively. Let the pol coodintes of P nd Q be (, θ) nd ( + δ, θ + δθ). Join PQ. Though P dw PM pependicul to OQ. Then fom ight ngled PQM. PQ PM + MQ ( sin δθ) + [ + δ cos δθ] [Q MQ OQ OM ( + δ) OP cos δθ] δθ sin PQ sin δ + δθ + δθ δθ δθ ( ) PQ δθ sin δθ δ δθ + + sin δθ δθ δθ δθ [Q cos δθ sin ] Now Also δθ sin lim δθ δθ 0 δθ 0 lim δθ 0 PQ lim δθ δθ 0 sin δθ δθ δθ sin lim δθ. δθ 4 On tking limits s δθ 0, we get ds dθ, 0 PQ c PQ lim. c PQ δθ δθ 0 ds d θ lim δθ 0 d + dθ δ d δθ d θ chod PQ [Q lim. Q P c PQ ]

32 If the eqution of the cuve is then it cn be shown tht ds d θ θ f (), d + d θ ds d dθ + d Othe Fomule Since d θ d d d θ ds dθ ds d θ tn φ sin φ, cot φ d + dθ ( + cot φ) cosec φ cosec φ d θ ds + cot φ the positive oot being tken if the length of the c is mesued in the diection in which θ inceses. We hve sinφ, d ds d θ ds dθ. d d ds tn φ. d ds cos φ d ds sin φ tn φ cos φ [Q tn φ Exmple 6. Pove tht fo cuve e θ cot α, d is constnt, s being mesued fom the oigin. Solution. Eqution of the given cuve is Diffeentiting w..t. θ we get, e θ cot α lo log + 0 cot α d θ d ]

33 d θ d θ d cot α tn α i.e., tn φ tn α φ α o Integting, we get d ds ds d cos φ cos α sec α s sec α + c Assuming tht s 0 when 0, we get c 0 s sec α Exmple 7. Show tht fo the cuve θ cos, ds d k k is constnt. Solution. Diffeentiting the given eqution w..t. we get, d θ d s sec α constnt.. k k k k ( ) k k k k + k k + k k k k k

34 d dθ d θ d k (whee ± k) Also ds dθ d + dθ ds d θ + k 4 k k k k ds d ds dθ. dθ d k k k k, which is constnt. Exmple 8. Show tht fo cuve, whose pedl eqution is p f () Solution. Fo ny cuve, we hve ds d cos φ d ds, ds d p p sin φ, cos φ cos φ cos φ ( sin φ) p whee 0 (Q ± k)... (i)... (ii) [Q p sin φ] Execises. Find ds ds dx dy fo the following cuves. (i) y c cos h x c 4

35 . Find (ii) x y. ds d θ fo the following cuves (i) x cos θ, y b sin θ (ii) x (θ sin θ), y ( cos θ) (iii) cos θ. ds. Show tht fo ny cuve dθ p. 5

36 L]ESSON CURVATURE. Definitions Conside point moving on smll c PQ of cuve. The diection of motion t ny point being long the tngent to the cuve t the point, it is cle fom the figue tht the diection of motion chnges fom KPT to KQT while the point descibes the c PQ. Thus the diection tuns though n ngle TMT i.e., TMQ when the moving point tveses the c PQ. A mesue of the te of tuning (i.e., the te of TMQ chnge in diection) is. c PQ This mesue gives us n ide of the vege cuvtue of the c PQ. We shll now obtin pecise mesue of cuvtue t point P on cuve. Let A be fixed point on the cuve such tht c AP s nd the c AQ s + δs, so tht the length of the c PQ δs. Let ψ be the ngle which the tngent t P mkes with the positive diection of the x- xis. i.e., Let the ngle which the tngent K MQT t Q mkes with positive diection of the x-xis be ψ + δψ whee δψ is smll ngle depending on the cuve nd the position of Q eltive to P on it. The ngle though which the diection of the tngent chnges when c δs is tvesed is theefoe equl to TMT. An ppoximte mesue of the cuvtue of the c PQ is, theefoe δψ δs. Now let Q P, then δs 0 nd δψ 0, nd we hve δψ Lt δs δs 0 The ecipocl of cuvtue is clled the dius of cuvtue nd is genelly denoted by the geek ds lette ρ. Thus ρ. d ψ. Cuvtue of Cicle TKX ψ. δψ δs cuvtue t P 6

37 Let PQ be n c of cicle whose cente is C. If PCQ θ, i.e., the ngle subtended t the cente of the cicle by the c PQ of the cicle) nd the dius CP, then c PQ θ (θ being mesued in dins). The ngle between the tngents t P nd Q is lso θ since CPT 90 CQT. Hence the cuvtue of the c PQ is θ i.e., which is independent of the mgnitude of θ. ρ θ Thus the dius of cuvtue ρ t point on the cicle is (the dius of the cicle). We lso see tht in cicle the nomls t ny two points P nd Q meet t the point C clled the cente of cuvtue nd length of the noml t ny point P up to the cente of cuvtue, is the dius of cuvtue t P.. Ctesin Fomule fo ρ When the eqution of cuve is given in ectngul (ctesin) co-odintes viz y f (x), then dius of cuvtue is obtined s follows: We know tht enble us to expess ds d ψ, the fomul fo ρ, in tems of nd its deivtives. We hve Diffeentiting with espect to x, we get But dy dx d ψ ds tn ψ nd ds dx sec ψ y dy dx tn ψ d y d d d ψ tn tn dx dx d ψ dx y ( ψ ) ( ψ ) d ψ dψ ds dx ds dx sec ψ sec ψ. ρ ds nd dx sec ψ y sec ψ ρ. sec ψ ρ sec ψ ( + tn ψ ) / ρ 7

38 + y ρ ( ) / ( / + y ρ ) y... () Note. The dius of cuvtue of cuve is n intinsic popety of the cuve. It depends on the cuve itself nd not on the choice of the xes of co-odintes. The eltion between s nd ψ obtined by ds integting, is clled the intinsic eqution of the cuve. d ψ Note. If t ny point of cuve the tngent is pllel to the y-xis, dy does not exist thee, nd the dx fomul () bove becomes meningless. In this cse dx 0, t the point unde considetion nd we cn dy employ deivtives of x w..t. y. Fo exmple; stting with the diffeentition of the eltion with espect to y we get, s befoe, dx dy ρ cot ψ dx + dy d x dy /... () ds Note. Genelly s nd ψ e mesued tht one inceses with the othe, so tht the deivtive d ψ i.e., ρ is positive. Accodingly while extcting the sque oot involved in () nd (), tht sign is tken in the fomule () nd () which gives positive vlue to ρ. Solved Exmple. Pove tht the dius of cuvtue t ny point (x, y) on the cteny y c cos h x c is y c. Solution. We hve y cos h, x c Q So tht nd y ( / + y Theefoe ρ ) y x c x c cos h sin h y c sinh c x c c cosh x c x + sin h c x cos h c c sinh x c / 8

39 Exmple. Show tht fo ny cuve (i) (ii) d dy ρ dx ds y ρ + ( y ) / Solution. (i) We hve d dy dx ds Solved Exmple. Show tht fo ny cuve cos h 9 x c x y c cos h x c c c cos h c dy d dx dx ds dx dy d dx dx dy + dx d dx y + y + y. y y ( ) ( + y ) / y y + y + y + y y y y y ( + y ) / / ρ x y y x ρ ( x + y ) / Q x y c cos h c Q ds + dy dx dx when the eqution of cuve is given in pmetic fom viz., x f (t), y g (t) whee ccents denote diffeentition w..t. t i.e., x dx dt nd y dy dt. Solution. In the fomul ρ ( + y ) We hve to expess y nd y in tems of the deivtives of x nd y w..t. t. We know tht y /... ()

40 dy dy dx dt y dx x dt y i.e., y x d Hence y dy dy dx dx dx dy dt dx dt d y dt x x y x x y x ( ) x y x x y ( x ) Thus Solved Exmple 4. In the cycloid Pove tht ρ y x x y y + ( x ) x y x x y ( x ) + ( x ) / / x (t + sin t), y ( cos t), ρ 4 cos t Solution. x (t + sin t), y ( cos t) dx dt dy dt d x dt d y Now fom the solved exmple, we hve dt ρ ( + cos t), sin t sin t, cos t. x y y x ( x + y ) / 40

41 whee x i.e., ρ ρ dx dy d x d y, y, x, y dt dt dt dt ( + cos ) cos sin ( sin ) t t t t ( ) + cost + sin t cost + cos t + sin t cos t cost sin t [ cost + ] [ + cost] / [ + cost] ( ) / + cost + cost /. cos t / 4 cos t / ρ 4 cos t We cn lso find the vlue of ρ by using the fomule Now y ( / + y ρ ) y dy dy sin dt t dx dx t dt t t sin cos cos ( + cos ) tn t t / / y d y d t t dt tn sec. dx dx dx sec t sec. dx cos t dx t ( + ) t sec t 4 t. cos 4cos 4

42 ( + tn t ρ )/ 4 t t 4 cos sec 4 t 4 cos 4 cos t. Solved Exmple 5. If CP, CD be pi of conjugte semi dimetes of the cllipse x y + b ( CD pove tht the dius of cuvtue t P is ) b,. Solution. We know tht P ( cosθ, b sinθ) nd D ( sinθ, b cosθ) e the extemities of CP nd CD. Eqution of the ellipse is x y + b b x + y b Diffeentiting with espect to x, we get b x + yy 0 b x b x y y y b x y + b y xy b y y y y [By putting the vlue of y ] x y b b + 4 b y + b x b b y y y 4 ( ) ( ) y + Now ρ y y b x y b x + 4 / / b y b ( 4 b sin θ + b 4 cos θ At P ( cosθ, b sinθ), ρ )/ b 4 4 ( sin θ + cos θ) / b b 4 4 b 4

43 ( sin θ + b cos θ )/ b Also CD ( ) ( ) 0 + sin θ + 0 b cos θ sin θ + b cos θ ( sin θ + b cos θ ) Hence ρ ( CD ) Solved Exmple 6. Pove tht fo the ellipse fom the cente upon the tngent t ny point (x, y). Solution. Fom solved Exmple 5, ρ x y + b b, ρ ( 4 y + b 4 x ) / b 4 4 The eqution of the tngent to the given ellipse t (x, y) is Xx + b. x X y Y y ( x) Yy b Xx Yy + b x 0 y + b The pependicul distnce fom the cente upon the tngent is Substituting this vlue in (), we get Execises I p 4 4 y + b x ρ / b p b x y y + b x b b p 4 4 b b. 4 4 p b p 6 6 Poved.. Find the dius of cuvtue t ny point (x, y) on the following cuves (i) y 4x (ii) y x (iii) xy c b., ρ being the pependicul Q y b. x y... (). Show tht the dius of cuvtue t the point ( cos θ, sin θ) on the cuve x / + y / / is sinθ cosθ. [Hint: The pmetic equtions of the cuve e 4

44 x cos θ y sin θ Use the fomul obtined in solved exmple ]. Find the dius of cuvtue t the specified point on the following cuves (i) x + y, t the point,, 4 4 π (ii) y 4 sin x sin x, t the point x, (iii) x + y xy t the point,, 4 4. ρ Fo Pedl Equtions The eltion between p nd fo points on cuve is clled the pedl eqution of the cuve. We shll now obtin n expession fo ρ suitble fo cuves given by pedl equtions. Fom the digm it is evident tht nd d θ ds ψ θ + φ d ds sin φ cos φ p sin φ ρ ds d ψ... (i)... (ii)... (iii)... (iv)... (v). We shll now eliminte θ, φ, s nd ψ fom these eltions nd obtin the vlue of ρ in tems of p nd ρ d ψ ds d dθ dφ θ + φ + ds ds ds ( ) sin φ dφ d sin φ dφ cosφ d ds d 44

45 p d p d p + φ + d d. ( sin ) ρ d dp p p dp dp d d Solved Exmple 7. Find the dius of cuvtue t the point (p, ) fo the ellipse + p b b Solution. Diffeentiting the given eqution w..t. p, we get i.e., Execise II p d dp d., b dp b p ρ d dp b p. Find the dius of cuvtue t ny point on the following cuves: (i) p (ii) p (iii) p (iv) p. In the cuve p the dius vecto. n+ n (pbol) (hypebol) (cdioid) x. Pove tht fo the ellipse upon the tngent t (x, y). 5. Pol Fomul Fo ρ (lemniscte) show tht the dius of cuvtue vies invesely s the (n )th powe of y + b, ρ b p whee p is the pependicul fom the cente To obtin the fomul fo the dius of cuvtue when the eqution of cuve is given in pol coodintes we mke use of the following esults we hve d ds d ψ ds dθ cos φ, ψ θ + φ, tn φ d dψ dθ ds dθ... () 45

46 Also, d ψ d θ φ + d d θ... () Diffeentiting w..t. θ, we hve sec φ. ds d θ d dθ dθ tn φ d d dθ d φ d θ d d dθ dθ d dθ d φ d θ d d dθ dθ d dθ + d dθ (Q sec φ + tn φ) Thus fom (), we get Putting the vlues of d ψ d θ nd ds d θ d ψ d θ d d dθ dθ d + dθ + + d φ d θ in (), we obtin d ψ ds d d dθ dθ d + dθ d d θ θ + d d d + dθ d + dθ / 46

47 i.e., ρ If we wite d d θ, d dθ ρ d + d dθ dθ d + dθ / d + d dθ dθ the fomul becomes Solved Exmple 8. Show tht fo the cdioid Also pove tht ρ is constnt. Solution. Hee ρ d + dθ / / d + dθ d d + dθ dθ ( ) / + ρ +. ( + cosθ), ρ 4 θ cos. d d θ sinθ, cosθ. / ( + ) ( + ) ( ) (Q ρ d) + cosθ + sin θ / ( + cosθ ) + sin θ + cosθ ( + cosθ) + cos θ + cosθ + sin θ + cos θ + cos θ + sin θ + cosθ + cos θ ( + θ) / cos + cos θ + cos θ / θ 4 θ d cos cos 47

48 Squing, we get ρ 6 θ cos ( + cosθ) 8 9 ρ constnt. Solved Exmple 9. Fo the cuve m m cos mθ, pove tht Solution. ρ m ( m + ) m m cos mθ Diffeentiting both sides with espect to θ, we get m m log m log + log cos mθ m d d θ m sin mθ cos mθ tn mθ tn mθ sec mθ. m tn mθ m sec mθ [Q tn mθ] ( ) / + ρ + / + tn mθ + θ θ + θ tn m tn m m sec m sec m θ sec m sec mθ + m sec mθ m + θ ( ) m m. m + cos mθ ( m + ) ( m + ) ( ) m m Solved Exmple 0. If ρ nd ρ be the dii of cuvtue t the extemities of ny chod of the cdioide ( + cosθ) which psses though the pole, then ρ + ρ Solution. Fo ( + cosθ) d d θ d dθ sinθ cosθ ( ) / + Now ρ

49 ( ) + cosθ + sin θ ( + cosθ ) + sin θ ( + cosθ)( cosθ) / + cosθ + cos θ + sin θ + cos θ + cos θ + sin θ + cosθ + cos θ ( θ) / + cos + cosθ ( + θ) ( + θ) cos cos + cos θ cos 4 θ cos Let PQ be ny chod of the cuve pssing though the pole. If the vectoil ngle t P is θ, then vectoil ngle t Q is π + θ. Now ρ At P, ρ nd t Q, ρ Hence ρ + ρ 4 θ cos 4 θ cos 4 π + θ cos 4 θ sin θ / 6 θ 6 θ cos + sin Solved Exmple. Show tht in the cuve sin θ, the tngent tuns thee times s fst s the dius vecto nd tht the cuvtue vies s the dius vecto. Solution. sin θ... (i) Fom (i) d d θ d θ d cos θ cosθ /... (ii) 49

50 Also tn φ φ θ d θ d tn θ (Q sin θ) cos θ ψ θ + φ θ + θ θ dψ dθ, which shows tht the tngent tuns thee times s fst s the dius vecto. ds ds d ψ ds d θ ds ds dψ ds dθ We hve ds d θ d + d θ o ds d ψ d ψ ds 4 cos θ + + cos θ sin θ + cos θ ds d θ k, whee k constnt i.e., cuvtue vies s the dius vecto. 6. Rdius of Cuvtue fo Tngentil Pol Equtions A eltion between p nd ψ on cuve is clled tngentil pol eqution. We mke use of the following known fomule nd ρ d ds, ρ, p sin φ dp d ψ d ds cos φ The two expessions fo ρ when equted give d ds dp d ψ 50

51 dp d ψ Eliminting φ using p sin φ, we hve p dp + dψ d ds cos φ (sin φ + cos φ). d dp Also ρ ( ) d dp d p dp dp + dψ d dp dψ. p +. dψ dψ dp p + dp... d p dψ dψ dψ dp ρ p + d p p + ψ d d p dψ ψ Q dp. d ψ d dp is the equied tngentil pol fomul. Solved Exmple. Find the dius of cuvtue t ny point on the cuve p sin ψ cos ψ. Solution. dp d ψ d p dψ ρ p + Solved Exmple. Pove tht fo ny cuve Solution. Now p sin ψ. cos ψ ρ ρ (cos ψ sin ψ) cos ψ. sin ψ sin ψ. cos ψ 4 sin ψ cos ψ d p dψ sin ψ cos ψ p. sin φ d φ +. d θ ψ θ + φ, dψ d ds ds sin ψ cos ψ 4 sin ψ cos ψ d θ ds sin φ, d ds cos φ (θ + φ) 5

52 Hence the esult. Execise III ρ dθ dφ + ds ds d θ ds + d φ ds dφ dθ sin φ +. dθ ds dθ dφ sin φ +. ds dθ d sin φ + sin φ φ d θ sind φ d φ + d θ. Find the dius of cuvtue t ny point (, θ) on the following cuves: (i) θ (ii) cos θ (ii) cos θ (iv). Show tht the dius of cuvtue t ny point on the cdioid ( cos θ) is.. Estblish the fomul ( u + u ) ρ nd dshes denote diffeentition w..t.θ. 7. Newton s Method ( + ) u u u 5 / whee u If cuve psses though the oigin the dius of cuvtue t the oigin is obtined by specil method which is due to Newton. Fo cuve pssing though the oigin nd hving X-xis s the tngent theet, the dius of cuvtue t the oigin is given by ρ 0 x lim x 0 y Similly, fo cuve pssing though the oigin nd hving Y-xis s the tngent theet the dius of cuvtue t the oigin is given by the fomule ρ 0 y lim y 0 x Exmple. Find the dius of cuvtue t the oigin fo the following cuves (i) x y (ii) y 4x Solution. Fo the cuve x y, X-xis is the tngent t the oigin

53 By Newton s fomul ρ 0 x y lim lim x 0 0 y x y Fo the second cuve y 4x, Y-xis is the tngent t (0, 0) By Newton s fomul ρ 0 y 4x lim lim x 0 0 x x x 8. Cente of Cuvtue The cente of cuvtue t ny point on cuve is the limiting position of the point of intesection of the noml t P nd the noml t neighbouing point Q vey close to P on the cuve. Let us find the coodintes of the cente of cuvtue t ny point (x, y) on the cuve y f (x). Let Q (x + δx, y + δy) be point on the cuve close to P. The equtions of the noml t P nd Q e espectively (Y y) f (x) + X x 0... (i) [Y (y + δy)] f (x + δx) + X (x + δx) 0... (ii) Subtcting (i) fom (ii) X is eliminted nd we get: O (Y y) f (x) + [Y y δy] f (x + δx) δx 0 (Y y) [f (x + δx) f (x)] δx + δy f (x + δx) Dividing thoughout by δx nd poceeding to the limit s δx 0, we get (Y y) d dx [f (x)] + f (x). f (x) lim f ( x ) i.e., (Y y) + [f (x)] ( ) + Y y + f x f ( x) Q δ x 0 δy δx Putting the vlue of Y in (i) o (ii) we get, dy + dx y + d y dx X x (Y y) dy dx dy dy + x dx dx d y dx The coodintes of the cente of cuvtue e given by (iii) nd (iv).... (iii)... (iv) 5

54 9. Cicle of Cuvtue nd Chod of Cuvtue The cicle hving the cente of cuvtue fo its cente nd the dius of cuvtue fo its dius, is clled the cicle of cuvtue fo the point unde considetion. The cente nd the dius being known, the eqution of the cicle of cuvtue cn be witten down. The locus of centes of cuvtue of cuve is clled the evolute nd the given cuve is clled involute. Any chod of the cicle of cuvtue t point on the cuve pssing though the point is clled the chod of cuvtue. Solved Exmple 9. Show tht the cente of cuvtue t the oigin lying on the pbol x y mx + is given by Hence show tht the eqution of the cicle of cuvtue is α m ( + m ), β ( + m ) x + y ( + m ) (y mx) Solution. y. x + Hee y m + x (y ) 0 m y (y ) 0, x / / ( ) ( + y + m ) ρ y ( ) / + m The coodintes of the cente of cuvtue t ny point (x, y) e given by the fomule d y dy dy + dx α x dx dx dy + dx β y + d y dx... (i) 54

55 At the oigin, α 0 Eqution of the cicle of cuvtue is (x α) + (y β) ρ ( + m ) m m + m β 0 + ( + m ) i.e., ( ) ( ) x + m + m + y + m ( ) 4 + Simplifying we get, x + y ( + m ) (y mx) ( m) m ( ) / Q ρ + m 55

56 LESSON 4 ASYMPTOTES. Infinite Bnches of Cuve Cuves like the pbol o the hypebol e unlimited in extent nd it is of get impotnce to know how they behve s one o moe of thei bnches tend to infinity. It my be thtt n infinite bnch might be ppoximted by stight line in the pt of the x y plne vey emote fom the oigin of coodintes. In ou study of the cuves, we e concened with points nd lines, etc. lying in the plne contining the coodinte xes of x nd y. We sy tht point P lies in the finite pt of the plne, if the distnce of P fom the oigin is finite. Similly stight line, lying on the plne, is sid to be in the finite pt of the plne, if the pependicul distnce of the line fom the oigin is finite. By point, eceding (o tending) to infinity on cuve, is ment point moving on the cuve such tht its distnce fom the oigin tends to infinity. Difinition. A stight line is sid to be n symptote of n infinite bnch of cuve if it is t finite distnce fom the oigin nd if the pependicul distnce, fom this line, of point moving on the cuve, tends to zeo, s the point ecedes to infinity long the infinite bnch.. Asymptote to Genel Cuve Let the eqution of ny cuve be φ (x, y) 0... () In Figue the symptote y mx + c is ppoched when x long the cuv. Then n symptote of the cuve, not pllel to the y-xis hs n eqution of the fom y mx + c... () when m nd c both e finite. The pependicul distnce fom the line () of point (x, y) on the bnch of the cuve to which the line (), is n symptote is given by the fomul. p y my c + m... () 56

57 whee p must tend to zeo when x + o x, o x + nd both. Suppose tht the line () is n symptote, when x +. Since m is finite, theefoe finite nd non-zeo (in fct gete thn ). Hence p tends to zeo if lim ( y mx c ) 0 x + whence, on dividing by x we see tht lim x + y c m x x 0 c Since c is finite, lim x x 0 n nd hence, we get y m lim x x whee the eltion between x nd y is φ (x, y) 0, since the point (x, y) lies on the given cuve (). Hving obtined m fom (5), the eltion (4) then shows tht c lim( y mx ) x +m is... (4)... (5) In Figue, the symptote y mx + c is ppoched when x long the cuve. Fo exmple, let the eqution of cuve be y e x nd let us find its symptote. Let the eqution of the symptote be y mx + c. Hee Also y e m lim lim x x x x x y e [Note: tht in this cse lim lim is not finite]. x x x x c lim( ) x x 0. y mx lim lim x y e 0. x x 57

58 Hence the pt of line y 0 (whee x is positive) is n symptote of the cuve. On the othe hnd, if the eqution of the cuve is y e x, y lim x x lim e x x x is not finite. But nd m lim x y x lim x c lim ( ) x e x x 0 x y mx lim e 0. x Hence, the pt of the line y 0 (whee x is negtive) is symptote to the cuve y e x. In this cse, it is towds the negtive side tht the x-xis is n symptote of the cuve, whee, s in the peceding cse the x-xis is n symptote towds the positive side. Lstly, if the cuve is + sin x y mx + c +... (A) x y then m c + sin x lim lim m + + x x x x x s x + o x in both the cses the limit is m. Also y mx tends to c s x + o. Hence the symptote to the cuve (A) is the line y mx + c whethe x o x. 58

59 . Asymptotes of n Algebic Cuve of the nth Degee IN X nd Y An lgebic cuve of the nth degee in x nd y, cn be lwys witten in the fom. n n n n n n n n 0x + x y + x y n y b x b x y b x y bn y + n n cx c n y (Ax + By) + M (α) in which the fist bcket contins tems ech of which is of the nth degee in x nd y togethe (i.e., homogeneous in x nd y of degee n), the second bcket similly contins tems ech of which is of (n )th degee in x nd y togethe, similly othe bckets contins tems of lowe degee in x nd y togethe. in y x Let us ewite the tem in the fist bcket in the fom n x x x x n y y y n n n y x φ n x whee y φ stnds fo the lgebic sum of numbe of tems in which the highest degee tems n x does not exceed n. The eqution (α) cn theefoe, be witten s n y n y n y y x φ n + x φ n + x φ n xφ + M 0 x x x x Note: Hd the tems in the fist bcket consisted of only tems, viz., x x y x y n n n 0 + +, the goup of nth degee tems could hve been expesed s n y y x x x... (α )... () in which the highest degee of y x tht if () is expessed s within the box bcket [ ] is (nd not n). We notice in this cse n y x φ n then y φ n x x is goup of tems in which the highest degee of y is, which is supposed to be less thn n. We x e expessing this goup of thee tems not s y φ but s y φ, the eson being tht y n φ is n x x x the co-efficient of x n n. This is the justifiction fo using the symbol y x φ n fo the tems ech of which x is of the nth degee is x nd y. A simil intepettion is tue fo ech tem of (α ). Let us find out the symptotes of (α ) which e not pllel to the y-xis. We ledy know tht such n symptote is of the type y mx + c, whee m nd c e finite quntities, we lso ecll in ou mind tht y m lim lim y mx x x x nd ( ) These limits e to be found fom the eqution of the given cuve itself. 59

60 Fom (α ), by dividing both sides by x n, we get y y y y M φ + φ + φ φ + 0 n x x n x x n x x n x x n Let x, then we hve fom (α ) y φ n x 0... (α ) i.e., φ n (m) 0... (β) y [By witing lim x x m] Let the oots of (β) be m, m, m,... Conside the vlue m m nd let the coesponding vlue of c denoted by c then c lim (y m x x) To find the vlue c fom the given eqution of the cuve (α ) let us poceed s follows: Put y m x p whee p is function of x. [Note: Fom the eqution of the given cuve (α ) y is to be looked upon s n implicit function of x nd thus y m x is function of x, x nd y being the co-odintes of point on (α )] When x, lim p c. Fom y m x p we find y x m + p x. Substituting this vlue of y x (α ), we find n p n p n p p x φ n m + + x φ n m + + x φ n m xφ m + + M 0 x x x x Expnding the functions in the eqution of the cuve p p p φ n m +, φ n m +, φ n m +,... with the help of Tylo s theoems in scending x x x p powes of, we get x φ ( ) + φ ( ) +. φ ( ) φ ( ) + φ ( ) +. φ ( ) +... n p p n p p n n n n n n x m m m x m m m x! x x! x n p p p + x φ ( ) ( ) ( ) ( ). n m + φ n m + + x φ m + φ m + φ ( m ) M 0 x x! x Renging the tems in decending powes of x, we get p x m x p m m x m p m m! n n n φn + φ n + φn + φ n + φ n + φn ( ) ( ) ( ) ( ) ( ) ( ) Since m is oot of p p + x φ m + φ m + p φ m + φ m!! n n n n n ( ) ( ) ( ) ( ) + [tems contining lowe powes of x] + M () φ n (m) 0. We get 60

61 () becomes φ n (m ) 0 p x p m m x m p m m! n φ n + φn + φ n + φ n + φn n ( ) ( ) ( ) ( ) ( ) n p p + x φ ( ) ( ) n m + φ n m + pφ n ( m ) + φn ( m )... ()!! M 0 Dividing both sides of () by x n, we get p p φ m + φ m + φ m + p φ m + φ m x! ( ) ( ) ( ) ( ) ( ) n n n n n p p + φ ( ) ( ) n m + φ n m + pφ n ( m ) + φn ( m ) x!! M (4) n x Let x. Since ccoding to ou ssumption lim ( ) ( ) n n p c x, we get fom (4) c φ m +φ m 0... (5) c Hence when m m we get the symptote φ n φ n y m x ( m ) ( ) Similly when m m, we get s befoe n symptote y m x m, povided φ ( ) φ φ n φ n n φ n ( m ) ( ) m, if φ ( ) ( m ) ( ) 0 n m. 0 n m. m, if φ ( ) 0 n m. It is supposed hee tht the oots of (β) [viz. φ n (m) 0] e ll el nd unequl i.e., no two of them e equl. Cse I If in (5) φ n ( m ) 0, but φ ( ) n m 0 no vlue of c cn be obtined. In this cse thee is no symptote of the cuve fo the vlue m m. If, howeve, both φ n ( m ) 0 nd ( ) φ n m 0, the eqution (5) becomes n identity. To detemine the vlue of c in this, let us poceed fom (4) which is, theefoe, witten on multiplying by x s 6

62 p ( ) ( ) ( ) p p φ + φ + φ + φ ( ) + φ ( ) + φ ( ) + φ ( ) n m p n m n m n m n m p n m n m! x!! M n x Now, let x. Then since lim p c x, we get fom (4 ) (4 ) c φ ( ) n m + cφ n ( m ) + φn ( m ) 0... (A)! which is qudtic in c to detemine two vlues of c [povided φ ( ) 6 0 n m ]. Let the oots of (A) be el unequl viz., c nd c. In this cse we get two pllel symptotes. y m x + c. y m x + c Note: We get two pllel symptotes fo the vlue m m only when φ ( ) But if lso ( ) φ n φ n ( m ) 0 nd lso ( ) φ n m 0 0 n m. m 0 then the vlue of c is obtined s bove fom cubic eqution in c viz., c c φ ( ) ( ) n m + φ n m + cφ n ( m ) + φn ( m ) 0... (B)!! povided φ ( ) 0 n m. In this cse, if the thee oots e el nd distinct we hve thee pllel symptotes fo the vlue m m viz. whee c, c nd c e the oots of (B). y m x + c y m x + c y m x + c Deductions mde fom the Eqution B viz., φ n (m) 0 () We notice tht expession φ n (m) is obtined fom the nth degee tems of the lgebic cuve (α) viz., the tems of n y x φ n by putting x nd y m. x Since, s ledy pointed out the degee of y φ in y is neve gete thn n (i.e., is eithe less n x x thn n o t the most equl to n. The degee of φ n (m) is t the most n. Hence the eqution φ n (m) 0 hs t the most n oots in m. Since the eqution φ n (m) 0 detemines the diection of symptotes not pllel to the y-xis, we conclude tht in lgebic cuve of the nth degee given by (α) cn hve t the most n symptotes not pllel to the y-xis. () Since the detemintion of m nd c (fo symptotes not pllel to the y-xis) depends in genel upon the equtions.

63 (i) φ n (m) 0 (ii) φ ( ) +φ ( ) c m m 0 n n nd since φ n (m) [nd hence φ n ( m ) ] nd ( ) φ n m n depend upon y 6 n x φ n nd x φ x n y. x We conclude tht in genel the detemintion of symptotes not pllel to the y-xis, of n lgebic cuve of the nth degee depends upon the nth degee nd (n )th degee tems of the lgebic cuve. Solved Exmple. Find the symptotes of the cuve y x y xy + x 7xy + y + x + x + y + 0 Solution. Putting x, y m in the thid nd second degee tems septely, we get Now But c φ (m) m m m + (m ) (m m ) (m ) (m + ) (m ), φ (m) 7m + m + φ (m) m 4m φ (m) m m m + 0 gives m, o. ( ) ( ) Fo m, c Fo m, c Fo m, c The thee symptotes e, theefoe (i) y x, (ii) y x, nd (iii) y x. Solved Exmple. Find the symptotes of the cuve x + x y 4y x + y + 0. Solution. The given eqution is y y y x x + 0 x x x i.e., of the fom y y x φ + xφ + 0 x x φ m m + m + φ m m m 7 4 ( ) ( ) ( ) ( ) φ φ φ ( ) ( ) φ φ φ ( ) ( ) Thus φ (m) 4m + m +, φ (m) 0. nd φ (m) m.

64 The gdients m of the symptotes e given by φ (m) 0 o 4m m 0 o (m ) (4m +4m +) 0 o (m ) (m + ) 0. Thus thee e thee symptotes, one hving gdient m nd the othe two hve the sme gdient m nd theefoe e pllel. To find c, we hve the eqution ( ) ( ) cφ m +φ m 0... () But φ (m) m + nd φ (m) 0 when m, φ () 9 nd when m φ Thus when m then c φ c φ + 0. ( ) ( ) 0 nd the symptote is y x. When m, then the eqution detemining c viz., the eqution () becomes n identity. We do not get ny vlue of c fom this eqution. To detemine the vlue of c, we hve to conside the eqution c φ ( m) + cφ ( m) + φ ( m ) 0! c m c m 0, i.e., ( 4 ).0 + ( ) Thus when m,we get 6c 0 o c 4 c ±. Hence the othe two symptotes e y o y + x ±. Execise I. Find the symptotes of the following cuves: x ± 64

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