Geometrical Applications of Differentiation

Transcription

1 Geometricl Applictions of Differentition. INTRDUCTIN Though we hve some lgebric results which give useful informtion bout the grph of function nd the function rte of chnge over most if not ll of the functions. But to know the complete insight nd detils bout the grph of the curve in spce, we need to know first bout certin other things like Mxim-Minim problems, Estimting pproximtion Errors, Intermedite forms, Role s Theorem, men vlue theorem, Tlor s theorem, concvit, points of inflexion, sign of first derivtives, Asmptotes, etc. nd which in turn cn be known onl with the differentil co-efficient/derivtive of the function t point or over certin chnge. Derivtives re interpreted s slope of curves nd s instntneous rte of chnge. We know tht the first nd second derivtives together tell how the grph of the function is shped. Second derivtive helps in estimting the liner pproximtion of the function. Collectivel ll bove inference help in sketching the trce of the curve.. TANGENTS AND NRMALS I. Tngent nd its Eqution Let P(x, ) nd Q(x + δx, + δ) be neighbouring points on the curve f(x) with supposition tht the curve is continuous ner P. Eqution of n line through P(x, ) is m( x) () where, re the current coordintes of n point on this B line (Fig..). f( x) T Now, s Q P, the stright line PQ tends in generl to ( x + δx, + δ) Q definite stright line TPT', which is clled the tngent to the curve t P(x, ). (, x P ) Whence the slope of PQ, which in other words known s ( ) grdient of PQ becomes ( ) +δ δ nd in the limiting x +δx x δx δ cse when Q P, lt m tn Ψ δ x 0δx 97 A T Ψ Ψ+ δψ T Fig..

2 98 Engineering Mthemtics through Applictions Therefore, the eqn. () becomes eqution of the tngent t P(x, ), nd reduces to ( x) or tnψ( x) () Cor. : Intercepts of the tngent on x-xis (i.e. T): Putting 0 in eqn (), we get T ( ) x () Intercepts of the tngent of -xis (i.e. T"): Putting 0 in eqn (), we get T" ( ) x (4) f fx Cor. : If the given eqution of the curve be f(x, ) 0, then we know tht x f f nd hence eqn () reduces to fx ( x) or ( x)f x + ( )f 0 (5) f Cor. : If the given eqution of the curve is in prmetric form x φ(t), ψ(t), dt ψ' () t φ '() t nd whence the eqn () reduces to dt ψ' () t ψ () t ( φ() t ) φ' t (6) () Exmple : Find the eqution of the tngent t n point (x, ) to the curve x / + / /. Show tht the portion of the tngent intercepted between the xes is of constnt length. Solution: The given curve is x / + / /. n differentition with respect to x, we get / / x + 0 or Now eqution of the tngent t (x, ) is m( x), i.e. ( ) (/x) / ( x) Further, intercept on x-xis, i.e. 0 implies x x + x ( x + ) x x Intercept on -xis, i.e. 0 implies / x + x + x x / / x /. () ()

3 Geometricl Applictions of Differentition 99 (x / + / ) / / / () The squre root of sum of squre of the two intercepts ( ) ( x ) + 4/ / Thus, the portion of the tngent intercepted between the xis is which is constnt length. x Exmple : Prove tht + touches the curve be x/ t the point where the curve b crosses the xis of. Solution: The curve be x/ shll cross the -xis t point where x 0, i.e. be 0/ b Now t the point (0, b), Tngent t (0, b) is x/ b b b be ( ) 0 0, b ( 0, b) e ( 0, b) b or + b b 0 ( b) ( 0) or x + (Becuse reltion is true for ll x, ). b Exmple : Find the eqution of the tngent to the curve (x )(x ) x t the point where it cuts the x-xis. Solution: The curve cuts the x-xis t 0. But 0 implies x or x 7 Thereb mens tngent is tken t the point (7, 0) The eqution of the tngent t (7, 0), however, is given b f, where f x (x 5) nd f (x )(x ) f (7,0) 0 7, i.e is the desired eqution of tngent t (7, 0). 0 x ( ) ( x) or ( ) ( ) Exmple 4: Find the eqution of the tngent t n point θ on the curve whose equtions re x (θ + sinθ); ( cosθ). Solution: As the given eqution is in prmetric form whence the eqution of tngent t n point θ is given b ( θ) ( ) ( ()) θ ' ( θ ) x θ x'

4 00 Engineering Mthemtics through Applictions d θ + θ, sin d θ θ Here ( cos ) sinθ cosθ θ+ sinθ ( ) θ { ( )} ( + cosθ) θ θ sin cos θ or sin { θ sinθ} cos θ θ θ sin tn θ sinθ tn or ( ) θ θ θ sin tn ( θ) sin θ θ or ( θ )tn or ( x θ)tn (on chnging the current coordintes to generl coordintes). Exmple 5: If the stright line p x cosα + sinα touch the curve m cos α + bsin α p. tht ( ) m ( ) m m m m m m x +, prove b m m Solution: The given eqution of the curve is x + () m m b m m n differentiting with respect to x, we get x m m 0 m + m b m b x Eqution of the tngent t (x, ) becomes or Slope of the tngent t ( x, ) m b x m ( ) ( x) m m m m x x or + + m m m m, (using ()) () b b Now if the line xcos α + sin α p touch the given curve then it should be identicl to eqution (). Whence cosα sinα p m m x m m b m.

5 Geometricl Applictions of Differentition 0 m m x cosα bsinα nd p b p m m m m m m cosα bsinα x + + p p b whence the result. II Norml nd its Eqution Norml to curve f(x) t n point P(x, ) is the stright Ψ line (NP, s) through P(x, ) but it is perpendiculr to the T tngent t P. For m to be the slope of the tngent, (s m ) Fig.. m will be the slope of the norml (since we know tht for two perpendiculr lines, slope of one is the negtive reciprocl of the nother). With this notion, it becomes es to write the eqution of the norml, if the eqution of the curve is given, i.e. m ( ) ( x) or ( ) ( x) Cor. : If the eqution of the curve is f(x, ) 0, then reduces to f x f ( ) ( x) ( ) or ( x) f () fx so tht eqution of the norml f () f Cor. : If the eqution of the curve is in the prmetric form x φ(t), ψ(t) so tht dψ dt ψ't () d '() t, the eqution of the norml becomes φ φ dt ( ) ( x) or [ φ(t)] φ'(t) + [ ψ(t)]ψ'(t) 0 () ψ't () φ' () t III Lengths of Tngent, Norml, Subtngent, Subnorml, etc. Let f(x) be the eqution of the curve with PTT' s tngent t point P(x, ) meeting the x-xis nd -xis t T nd T' respectivel. Let RPN be the norml t point P(x, ) meeting the x-xis t N nd PM be the ordinte, Q be the perpendiculr from the origin to the norml (See Fig..). Let MTP be Ψ so tht tn Ψ nd MPN Ψ. Then we hve the following geometricl results: x N Norml P T

6 0 Engineering Mthemtics through Applictions. Length of the tngent ( TP) MP cosec Ψ + cot Ψ + (i.e. portion of the tngent between the curve nd x-xis).. Length of norml R 90 Px (, ) Ψ ( NP) MPsec Ψ + tn Ψ + (i.e. portion of the norml between the curve nd x-xis). Length of subtngent ( TM) cot Ψ. T Ψ Q T Ψ M N 4. Length of subnorml ( MN) tn Ψ. Fig.. 5. Length of intercept (of the tngent) on x-xis ( T) x (see cor. on tngents). 6. Length of intercept (of the tngent) on -xis ( T' ) x (see cor. on tngent). 7. Length of perpendiculr from (0, 0) on tngent ( ) + (since the eqution of the tngent ( ) ( x) m be rewritten s (x ) 0 comprble to + b + c 0. Therefore, length of perpendiculr x from origin on the tngent become Q + Q x comprble to 8. Length of perpendiculr from (0, 0) on the norml ( ) R x+ + c + b (since on rewriting the eqution of the norml ( ) ( x) s + (x + ) 0 which is comprble to + b + c 0, the perpendiculr distnce from (0, 0) becomes R x + + comprble to c + b bservtions: To usul conventions,, both re positive nd T is tken to the left of M nd further sign of TM is positive, otherwise, negtive. Similr interprettions cn be given for other vlues. Exmple 6: Find the eqution of the norml t n point t to the curve x (cost + t sint), (sint t cost). Verif tht these normls touch circle with its centre t the origin nd whose rdius is constnt..

7 Geometricl Applictions of Differentition 0 Solution: Eqution of the norml t n point t is given b ( ) ( x) where ' x' dshes represents derivtives with respect to t. or [ x(t)] x'(t) + [ (t)] '(t) 0 () Here x' () t ( sin t + sin t + t cost) t cos t dt t cost cost + tsin t tsin t dt Hence the eqution of the norml t t becomes nd () ( ) [ ] [ ] (cos t+ tsin t) tcos t+ (sin t tcos t) tsin t 0 cost + sint (cos t + sin t) 0 or xcost + sint () (on replcing current coordintes to generl coordintes) Now the perpendiculr distnce of this norml from (0, 0) (or in Fig..) is given b x + + ( + ) + ( ) cost tsin t sin t tcost tn t t + tn sin cos t t + cost, which is constnt sect Hence it touches circle of rdius hving its centre t (0, 0).??? som mi Exmple 7: Find the equtions of the tngent nd normls, the lengths of the tngent nd subtngents, length of the norml nd subnorml for the ellipse x cos t, b sint, t fixed point (x, ) for which t /4. Solution: The given eqution of the ellipse is x cost, b sint () b b Thus, we get sin t, bcos t, cot t, () dt dt See the geometr, here, PT is the tngent, MT is the subtngent, PN is the norml, NM is the subnorml. Now, eqution of the tngent t P(x, ) is ( ) m( x ) t 4 or i.e. bsin t cos t t t 4 t 4 4 b b bx + b 0 () Px (, ) N M Fig..4 /4 T

8 04 Engineering Mthemtics through Applictions Eqution of the norml t P(x, ) is or ( ) ( x) i.e. b b bsin t + cost b t or ( x b) t 4 ( b ) (4) Length of the tngent PT ( b t) b + sin + + b b (t t /4) t 4 b (t t /4) Length of subtngent MT ( bsint) + b (5) (6) Length of the norml, b t 4 b + b PN t t + ( b sin t) + 4 (7) b b Length of subnorml, MN t t ( bsin) (8) 4 t t 4 4 IV Angle of Intersection of Two Curves B ngle of intersection of two curves, we men the ngle between the tngents t common point of intersection. Let us suppose tht the eqution of tngent of the two curves AB nd CD be m x + c nd m x + c with m nd m being the slopes respectivel. Then the ngle of intersection of the bove two curves will be T T 4 D tn m m θ + mm () When m m, θ becomes 90, i.e. the two curves cut orthogonll wheres if m m then tnθ 0, i.e θ 0 mens the two tngents will be common one. Exmple 8: Find the ngle of intersection of the circle x + nd the rectngulr hperbol x. A T C P θ B Fig..5 T Solution: Intersection mens common point viz. tke the sum of the two equtions, impling ( + ) ( ) x ± + x For this vlue of x, we get ( ) or ( ) ± So tht common point re ± ( + ), ± ( ) / /

9 Geometricl Applictions of Differentition 05 For the given curves f( x, ) x + 0 nd φ(x, ) x 0 the eqution grdients will be fx x x φ x x x m f nd m φ x x m m x tnθ (with bove obtined x nd ) + mm x x x + whence θ. 4 Exmple 9: The curves x + b nd cx + shll cut orthogonll if. b c d Solution: Let P(h, k) be the point of intersection for given equtions x + b cx + Mens this common point must lie on both the equtions i.e., h + bk nd ch + dk () () h k n solving for (h, k), we get b + d c + d cb ( )/( ) ( )/( ) h d b d cb or k c d cb Differentiting eqution () with respect to x x b or m ( P) () t h (4) bk Likewise differentiting eqution () with respect to x cx ch or m dk (5) For orthogonl intersection, we must hve m m i.e., h ch bk dk or c h + bd k 0 or d b c bd c d b c + 0 or + 0 d cb d cb bd c or b d c whence the required condition for orthogonlit.

10 06 Engineering Mthemtics through Applictions ASSIGNMENT. At wht point is the tngent to the curve logx is prllel to the chord joining the points (0, 0) nd (0, ). n n x. The stright line + touches the curve x b + for ll vlues of n. Find b the point of contct. x. If p x cos α + sin α, touch the curve +, prove tht b p n (cos α) n + (bsinα) n. 4. Prove tht the condition for the line xcos α + sinα p to touch the curve x m n m + n, is p m + n m m n n (m + n) m + n m + n cos m α sin n α. 5. Show tht sum of the intercepts on the xes of n tngent to the curve x + is constnt. 6. For the curve x sin θ, cos θ, find the ngle which the perpendiculr drwn from the origin to the tngent t the point θ mkes with the xis of x. 7. Tngents re drwn from the origin to the curve sinx. Prove tht their points of contct lie on the curve x x. 8. If the tngent t (x, ) to the curve x + meets the curve gin in (x, ), show x tht +. x 9. If the tngent to the curve x / + / / t n point on it cuts the xes, t P, Q respectivel, prove tht P + Q. 0. Show tht the tngents t the points where the stright line x + b 0 meets the ellipse x + hx + b re prllel to the x-xis, nd tht the tngent t the points where the stright line hx + b 0 meets the ellipse re prllel to -xis. x/. Show tht the exponentil curve be, the subtngent is of constnt length nd the subnorml vries s the squre of the ordinte.. In the ctenr ccosh x, prove tht the length of the subtngent is ccosh x nd tht c c of the subnorml is csinh x c.. Find the length of the tngent, norml, sub-tngent nd sub-norml of the ccloid x (t + sint), ( cost). 4. For the curve x cos θ, sin θ, show tht the portion of the tngent intercepted between the point of contct nd the x-xis is cosec θ. Also find the length of the subnorml. θ 5. For the curve sinθ, x logcot cosθ, find the lengths of the sub-tngent nd sub-normls t the point θ /4. n n n n

11 Geometricl Applictions of Differentition Prove tht the subnorml to the curve x c vries s the cube of the ordinte. 7. Find the ngle of intersection of the curve x nd x Show tht the prbols 4x nd x intersect t n ngle of tn Prove tht the curves x + nd b x c + will cut orthogonll if b c d. d. TANGENTS AND NRMAL FR CURVES IN PLAR CRDINATES I. Angle between the Rdius Vector nd Tngent tnφ r d è dr Let P(r, θ) nd Q(r + δr, θ + δθ) be n two neighbouring points on the curve r f(θ). Let PT be the tngent t P(r, θ) nd φ be the ngle enclosed b this tngent with the rdius vector P. Join PQ nd drw PN Q, then from the right ngled tringle NP, NP r sinδθ, N rcos δθ, B Qr ( + δr, θ + δθ) NQ Q N N (r + δr) r cos δθ α δr + r( cosδθ) δθ Pr (, θ) δ r + () r sin r If NQP α, then NP rsinδθ tnα NQ δθ δ r + sin r () In the limiting cse, when Q P, i.e. δθ 0, the chord QP turns bout P nd becomes tngent t P nd thus resulting in α φ. rsinδθ tnφ Lt tnα Lt Q P δθ 0 δθ δ r+ sin r p δθ qθ T A f Fig..6 Lt δθ 0 sin r δθ δθ δθ sin δr r δθ + sin δθ δθ r r dr + r 0 dr ()

12 08 Engineering Mthemtics through Applictions II. Length of the Perpendiculr from Pole on the Tngent See the geometr of Fig..6, p be the length of the perpendiculr T from the pole to the tngent t P on the curve, then cosec φ ( + cot φ) (i) p rsinφ (ii) dr du + 4 (iii) u p r r +, where u. p r In PT, p r sin φ result (i) (4) p r r dr r + r dr + 4 r r result (ii) (5) Now for result (iii), given u r du d dr therefore, r r impling du dr 4 r or du p r u + du p using result (ii), result (iii) (6) These three reltions which involves p (length of the perpendiculr from the pole upon the tngent) nd r (the rdius vector) both, re known s pedl equtions of the curve. III. Polr Subtngent nd Polr Subnorml Let P(r, θ) be n point on the curve r f(θ) with P s the rdius vector nd let line drwn through the pole perpendiculr to the rdius vector P meets the tngent nd norml in T nd N respectivel. Then T is clled the polr subtngent nd N is clled polr subnorml. Since the ngle PT is φ, therefore, polr subtngent T r tnφ r r r dr dr Also in NP, polr subnorml, dr N r tn PN rcotφ r r dr (7)

13 Geometricl Applictions of Differentition 09 Length of polr tngent PT rsecφ r + tn φ r + r Length of polr norml dr PN rcosecφ r + cot φ dr r + r + r (8). (9) It is evident from the formul of subtngent tht it is mesured positivel to the right with supposition tht observer is sttioned t nd looking in the direction of P. Wheres negtive vlue of the subtngent shows tht T is to the left s shown in the Fig..8. A similr notion is pplied in cse of subnorml. Exmple 0: Show tht in the equingulr spirl r e θ cot α, the tngent is inclined t constnt ngle to the rdius vector. Solution: The given eqution r e θcotα dr θcotα n differentition gives, e cotα rcotα Fig..8 so tht tnφ r tnα or φ α dr Hence the ngle φ, the ngle between the tngent nd the rdius vector is constnt. Exmple : Prove tht in the prbol cosθ, r θ θ (i) φ (ii) p cosec (iii) Polr subtngent cosecθ (iv) p r N T φ θ T Fig..7 B 90 φ φ p Pr (, θ) A r θ Tking log on both sides, log logr + log( cosθ) Solution: Given ( cos ) or n differentition, dr sinθ 0 + r cosθ dr θ cot r θ tn dr r θ tn tn r r θ θ φ tn tn dr r

14 0 Engineering Mthemtics through Applictions impling θ φ Agin, we know tht, θ θ θ p rsinφ rsin sin sin θ θ sin cos impling or p cosecθ/ θ sin cosθ cosθ p r For polr subtngent, T (Fig..7) θ rtn φ tn cosθ θ sin cosce θ. θ θ sin cos sinθ Exmple : Show tht the ngle of intersection of the following curves r ( cosθ), r ( + cosθ) is /. Solution: Tking logs on both sides of Ist eqution, logr log( cosθ) n differentition, dr sin θ θ cot r cosθ θ θ tnφ r tn or φ dr cos Similrl from nd eqution, tnφ + θ cot θ sinθ θ or φ + Thus the ngle of intersection α between the two curves is given b α φ φ. tnα tn(φ ~ φ ) or ( ~ ) Exmple : Find the pedl eqution of the prbol 4( + x). Solution: Eqution of the tngent to the given prbol 4( + x) is written s, where in this cse () ( ) ( x)

15 Geometricl Applictions of Differentition The perpendiculr distnce of this tngent from (0, 0), i.e. Q (Fig..) p x x x impling x 4x 4 p 4x ( + ) ( + ) x 4 x () or p ( x) + Now, r x + x + (4x + 4 ) ( + x) p or r (using vlue of ) Therefore, the required pedl eqution, viz. reltion involving p (the length of the perpendiculr from the pole to the tngent) nd r (the rdius vector) for this problem is p r. Exmple 4: Find the pedl eqution of the ellipse x +. b x Solution: The eqution of the tngent to given ellipse + () b cn be written s: x + () b which is comprble to A + B + C 0 with its perpendiculr distnce from (, ), C p A + B x + b or x p b () or Now, r x + or r b x + ( b ) ( ) r b x b b b

16 Engineering Mthemtics through Applictions b x x x b (using ()) or x r b b (4) Similrl, r b b (5) n substituting the vlues of x nd b into eqn.(), we get or p b b b r b r ( rb b) + ( r) b + p b ( b r ) (6) whence b p + b r is the desired pedl eqution. Exmple 5: Prove tht the locus of the extremit of polr subnorml of the curve r f(θ) is r f (θ /); nd hence show tht the locus of the extremit of the polr subnorml of the equingulr spirl r e θcotα is nother equingulr spirl. Solution: Let the coordintes of N (Fig..7) be (r, θ), then dr r N polr subnorml f ( θ) () wheres θ N +θ or θθ () Now eliminting θ between eqn. () nd eqn. (), we get reltion between r nd θ s r f θ () Hence the locus of (r, θ), i.e., r f θ. dr f θ cotα e Now for the curve r e θ cotα θ cot, ( ) Hence the required locus is r f θ θ cotα or r cot e α

17 Geometricl Applictions of Differentition cot α. θ cot α cot e e be θ cot α where b cotα e / cotα is n rbitrr constnt. Whence the locus of (r, θ) is nother equingulr spirl. ASSIGNMENT. Show tht the tngent to the crdiod r ( + cosθ) t the point θ / is prllel to the initil line.. Show tht the ngle between the tngent t n point P nd the line joining P to the origin is the sme s ll the points of the curve log(x + ) k tn (/x).. Show tht in the curve r θ, the polr subnorml is constnt nd in the curve rθ, the polr subtngent is constnt. 4. Find the ngle of intersection of the curves r sin θ nd r cosθ 5. Show tht the curves r n n cosnθ nd r n b n sinnθ cut ech other orthogonll. 6. Prove tht the pedl eqution of the hperbol x b is r + b. b p 7. Show tht the reltion r p is the pedl eqution of the stroid x cos t, b sin t. 8. Find the pedl eqution of the curves: (i) r ( + cosθ) (ii) r sinθ (iii) r m m cos mθ 9. Show tht the pedl eqution of (i) the hperbol r cosθ is pr (ii) the lemniscte r cosθ is r p 4 r (iii) of the rchimedin spirl r θ is p. r + 0. Show tht the length of the perpendiculr from the pole on the tngent to the ellipse p ( ecos ) r + θ is given b l e + p e r..4 DERIVATIVE F ARC ds I. Crtesin Form: +. Let P(x, ) nd Q(x + δx, + δ) be two neighbouring points on the curve AB (i.e. f(x)) such tht rc AP s, AQ (s + δs) then rc PQ δs nd let chord PQ δc (Fig..9). Drw PL nd QM perpendiculr on the -xis nd PN perpendiculr on QM.

18 4 Engineering Mthemtics through Applictions From the right ngle tringle PNQ, (PQ) (PN) + (NQ) (δc) (δx) + (δ) δc δ or + x () δ δx δs δs δc δs δ Now + δx δc δx δc δx () In the limiting cse s Q P, i.e. δc 0, we hve δs δs δ lt lt +, δs δx 0 δx δc 0δc lt δx δc 0 δc ds or + ( + ) then If s is mesured in such w tht ds ds + is positive, i.e. if x nd s increses together, () (4) Note: Strictl speking ds + holds onl for the cse when > 0 otherwise ds +. For this reson, in the generl cse, this formul is more correctl written s ds + Agin, if the eqution of the curve is x f(), then ds ds + + (5) Further, we know tht tn ψ, where ψ is the ngle which the tngent mkes with the initil xis. Hence ds + + tn ψ secψ or cos ds ψ (6) Likewise, tn ψ cosψ sin ψ ds ds (7) II. Prmetric Curves If the eqution of the curve is in prmetric form, viz. x f (t), φ(t), then T Fig..9 ψ B Q δc P δ A s δx N L M ds ds + + dt dt dt dt dt (8)

19 Geometricl Applictions of Differentition 5 III. Polr Formul: ds dr r + B N Qr ( + δr, θ + δθ) δs δc Let P(r, θ ) nd Q(r + δ r, θ + δθ) be n two neighbouring points on the curve AB, r f(θ) in such w tht AP s, rc PQ δs nd chord PQ δc. Drw perpendiculr PN on Q, then in PN, N r cosδθ, NP rsinδθ, nd Q QN NQ (r + δr) r cosδθ δ r + δθ rsin (0, 0) θ δθ r A T Fig..0 Pr (, θ) Ψ initl ixs θ 0 Now in right ngle tringle PNQ, (PQ) (PN) + (NQ) or δθ δ c ( rsin δθ ) + δ r + rsin δ s δ s δ c δ s δθ δc δθ δc sin r δθ rsin r δθ δ + + δθ δθ δθ δθ rsin δs sin δθ δr r δθ sin + + δc δθ δθ δθ In the limiting cse when Q P, i.e. δs 0 (or δθ 0), then ds dr dr r + + r 0 r + (9) (0) If s is mesured in such w tht ds is positive, i.e. s nd θ increses together, then ds dr r + () If the eqution of the curve is θ f(r), then ds ds r + dr + r dr dr dr dr () ds or + tn φ sec φ, since r tn φ dr dr dr or cos ds φ ()

20 6 Engineering Mthemtics through Applictions Likewise or Further, nd d d dr d θ r cosφ tn φ cosφ ds dr ds r dr r sinφ ds r (4) p rsinφ r r r ds ds (5) p r p cosφ sin φ dr ds r r (6) Exmple 6: For the ccloid x ( cosθ), (θ + sinθ), find ds, ds nd ds. Solution: Here sin, ( cos ), θ + θ therefore nd + sin θ+ + cosθ+ cos θ ds { ( cos )} / + θ / θ 4cos θ cos ds ds d θ cos θ cosec θ sinθ ds ds d θ cos θ sec θ. θ cos Exmple 7: Show tht in the curve r n n cosnθ, ( ) n ds sec nθ n. Further, ds (n )th power of r nd n dr n nr ds + 0. Solution: Given r n n cosnθ implies nlogr n log + logcosnθ ndr sinnθ dr n differentition, n or rtnn r cosnθ d θ θ vries s () () Now dr ds r + r + r tn nθ rsecnθ / n θ cosnθ θ θ hence the first result. n ( cos ) ( cos ) ( sec ) ( n n n n n n ) ()

21 Geometricl Applictions of Differentition 7 Agin, n ds n n n n r r rsecnθ r secnθ cosnθ secnθ ( constnt) n ds or n r mens ds vries s (n )th power of r dr We know cos ds φ which implies dr dφ dφ sin φ sinφ ds ds ds (4) Also tnφ r cotnθ tn nθ+ dr tn nθ, i.e. φ nθ+ (5) From the bove reltion, d φ n, sinφ sin(nθ + /) cosnθ. (6) From (4), ds n n n dr r r r cosnθ n n n n n or n dr ds n + nr 0 ASSIGNMENT. If c + s c for curve then show tht. Also show tht the perpendiculr c from the foot of the ordinte upon the tngent is of constnt length.. Find ds for the curves (i) x (ii) c cosh x/c. Find ds for the curves dt xsint + cos t f '( t) (i) x e t sint, e t cos t, (ii) xcost sin t f "( t) 4. Find ds for the curve (i) r cosθ (ii) r (+ cosθ) (iii) r (θ ) (iv) r θ (v) r θ (vi) r n n sinnθ ds r + 5. Prove tht in the hperbolic spirl rθ,. dr r 6. For the curve r e θcot α ds, prove tht (i) cot dr α (ii) s cr where c is constnt, s being mesured from the origin. ds 7. Prove tht for n curve rp d θ. 8. With usul menings for r, θ, s nd φ for the polr curve r f(θ), show tht dφ + r φ ds dr cosec 0. [Hint: Eqn (4) of Exmple 7]

22 8 Engineering Mthemtics through Applictions.5 CURVATURE AND RADIUS F CURVATURE A In Figure., curve PQ bends more shrpl thn the curve AB P The mesure of shrpness of bending of curve t prticulr point is clled the curvture of the curve t tht point. About Terms: Absolute Curvture, Averge Curvture;/Rdius of I II curvture Let AB be n curve with A s fixed point on it. Let P nd Q be two neighbouring points on the curve AB. B Q Fig.. Let the rc AP s, AQ (s + δs) so tht rc PQ δs (A being the fixed point on the curve, rcs re mesured from A.) Let the tngents to the curve t P nd Q, mkes ngle ψ B nd (ψ + δψ) with the initil xis respectivel. In moving from P to Q, through distnce δs the tngent Qx ( + δx, + δ) hs turned through n ngle of δψ. This ngle δψ is clled Px (, ) A s the totl bending or totl curvture of the rc PQ. (i) Thus the ngle δψ through which the tngent turns s its point of contct moves long the rc PQ is clled Ψ Ψ + δψ the totl bending/totl curvture. δψ Fig.. (ii) The rtio is clled the men curvture/verge δs curvture of the rc PQ. (iii) The limiting vlue of the men curvture when Q P is clled the curvture of the curve t the point P. δψ dψ Thus the curvture, K of the curve t point P lt. Q P δs ds (iv) The reciprocl of the curvture of the curve t P, provided this curvture of the curve is not zero, is clled the rdius of curvture t tht point P(s). This is usull denoted b ρ. Thus, ρ ds K d ψ The ngle δψ is lso clled the ngle of contingence of the rc PQ nd is mesured in rdins. From definition it is cler tht the curvture t point P depends onl on the position of the curve nd does not depend in n w on the sstem of co-ordintes. The reltion between the rc length s of curve (s AB) mesured from given fixed point (s A) on the curve nd the ngle ψ between the tngents t its extremities is clled the intrinsic eqution of the curve. The expression ds for rdius of curvture is suitble onl for those curves whose intrinsic dψ equtions re given. Formul for rdius of curvture for curves given in other different forms re discussed in succeeding section. Exmple 8: Find the rdius of curvture t the point (s, ψ) on the curve s log(tnψ + sec ψ) + tnψ secψ.

23 Geometricl Applictions of Differentition 9 Solution: ds ρ d ψ ψ+ ψ ψ tn ψ+ secψ ( sec sec tn ) secψ + secψ(sec ψ + tn ψ) secψ[ + tn ψ + sec ψ] sec ψ. ( sec sec tn sec tn ) + ψ ψ + ψ ψ ψ Rdius of Curvture (Crtesin, Prmetric nd Polr forms) I. Crtesin Form: Let f(x) be the eqution of the curve in implicit form, then the slope of the tngent t n point is given b m tn ψ () i.e. ψ tn dψ + + () ( ) Further, ds + + () ( ) + ds ds ρ dψ dψ which is the expression for rdius of curvture in crtesin form. (4) Convention of Signs: The positive root is tken in numertor of eqn (4), therefore, rdius of curvture, ρ, will be positive when is positive (i.e., when the curve is concve upwrds) nd negtive, when is negtive (i.e., when the curve is concve downwrds). In prctice, numericl vlue of ρ is tken. At point of inflexion is zero, therefore, curvture of the curve t the point of inflexion is zero. II Prmetric Form: Let x φ(t) nd ψ(t) so tht x dt nd dt (5) Now dt dt x nd d d dt ' d ' x'" 'x" dt x' x dt x' x' (6) (7)

24 0 Engineering Mthemtics through Applictions ' + ( + ) x' ( x' + ' ) ρ x'" 'x" x'" x"' x' is desired expression of rdius of curvture in the prmetric form. (8) bservtions (i) The curvture (or rdius of curvture) of the curve t n point is independent of the choice of nd xes since the curvture or the rdius of curvture is n intrinsic propert of the curve. Hence, x nd cn be interchnged in the formul of curvture (κ) nd rdius of curvture (ρ). (ii) When the tngent t point (where rdius of curvture is desired) on the curve is prllel to -xis, i.e. +, then the formul ρ which reduces to ρ, since + becomes inpplicble nd is given b ρ mens 0 ( defined quntit). III. Rdius of Curvture t the rigin (Newton s Formul) (i) If xis of x is tngent to the curve t the origin, x then ρ ( 0, 0) Lt 0 Since xis of x s tngent t (0, 0) mens 0 t (0, 0) x Lt x Lt 0 x 0 x Lt x 0 Also ( defined quntit) 0 form 0 form (0,0) ( 0, 0) Lt x ρ x 0 0 (0,0) ( 0, 0) (9) (0) (ii) If xis of is tngent to the curve t the origin, then ρ ( 0, 0) Lt x 0x (Its proof is similr to prt (i)) 0

25 Geometricl Applictions of Differentition (iii) In cse neither of the xes is tngent to the curve t the origin: Write the eqution of the curve s x f( x) f(0) + xf' (0) + f" (0) + [B Mclurin s series]! x px + q + (³ f(0) 0 in this cse) where d p f' (0) t x 0 nd q f" (0) d t x 0 + 0, 0 + p ρ ( 0, 0) q ( 0, 0) ( ) () bservtions (i) For scertining the tngents t the origin, equte to zero the lowest degree terms occurring in f(x, ) 0 since the eqution of the tngent is liner one, i.e. of the form x + b + c 0 nd if psses through the origin then c 0. (ii) When xis of x is tngent to the curve t the origin then ( 0), 0 (iii) When xis of is tngent to the curve t the origin then (x 0), 0 or. IV Some other Forms of ρ: (i) When x nd re given functions of the length of rc s. cosψ ds Differentiting with respect to s, sin d ψ ψ or ds ds ds ρ ds () Therefore ρ ds ds (ii) Similrl, strting with sin ψ, we hve ds ds ρ () (4)

26 Engineering Mthemtics through Applictions ρ ds ds (5) (iii) Squring nd dding the vlues ρ ds nd ρ ds, we get + ρ ds ds Some Stndrd Prmetric Curves: x φ(t), ψ(t) (6) S. No. Nme of the curve Crtisin Form Prmetric Form x x cos t'. Ellipse + b bsint, }. Circle x + x cos t,, sint t x, x + + t t, + t. Prbol 4x x t, t, x 4. Hperbol x sec t, b btnt x ct 5. Rectngulr x c c, t or Hperbol x } x cosht sinht 6. Hpoccloid 7. Astroid x + b x + cos, x t bsin t x cos t, sin t

27 Geometricl Applictions of Differentition 8. Cissoid Cissoid x x sin t ( x) x r cost x t t t x sin t sin t cost 9. Folium of Descrte s x + x Exmple 9: For the curve show tht Solution: Given nd i.e. Thus ρ x + x t x, + t, t + t x, if ρ is the rdius of curvture t n point (x, ) ( +x ). x + x () ( ) ( + x) + x x + x x Hence the result. + x x ( + x) 4 ( ) + x + ρ x 4 + x ρ x 4 x x ρ x x () Exmple 0: Appl Newton s formul to find the rdius of curvture t the origin for the ccloid x (θ + sinθ), ( cosθ).

28 4 Engineering Mthemtics through Applictions Solution: Here corresponding to θ 0, x 0 nd 0, nd hence the curve psses through the origin Further d sinθ θ ( + cosθ) so tht 0 Lt 0 θ 0 ( + ) Mens xis of x is tngent to the curve t the origin. x ρ ( 0, 0) Lt x 0 0 θ 0 θ 0 ( θ+ sinθ) ( θ) Lt cos ( θ+ θ )( + θ) sin cos Lt θ 0 sinθ Lt θ 0 0 form 0 0 form 0 ( θ+ sinθ)( sinθ ) + ( + cosθ )( + cosθ) ( )( ) cosθ Exmple : Show tht the rdii of curvture of the curve x ( + x)/( x) t the origin is. Solution: The eqution of the curve, ( x) ( + x)x psses through the origin. To see the nture of the tngent t the origin, equte to zero the lowest degree terms in x nd, i.e. x or ±x i.e., t the origin, neither of the xis re tngent to the given curve. Putting or ( ) x px + q + in the given eqution, we get x x px q x x ( ) + + ( + ) qx x p x + pq + + x + x 4 n compring the coefficients of x nd x, we get p p ± 4 x

29 Geometricl Applictions of Differentition 5 nd pq p q± ( ) + p ρ ( 0, 0) where p q x0 f'(0) nd q x0 f"(0) ( + ) ± ± Hence ρ(0, 0) is numericll. Alterntel: Eqution of the curve is ( + x ) x x ± x ( ) ( + x) ( x) x x ± x + x x ± x , [using ( + x)n + nx + ] x x x ± x i.e. So tht x x ± x x x ± x nd ± At (0, 0) ±, ± ( ) ( ) + + ρ ( 0, 0) ± ± ± Hence ρ(0, 0) is numericll.

30 6 Engineering Mthemtics through Applictions x Exmple : Prove tht for the ellipse +, b ρ, p being the perpendiculr b p distnce from the centre on the tngent t (x, ). [MDU, 005; NIT Kurukshetr, 007] x Solution: For the given ellipse + b, on differentiting with respect to x, we get b x () b x x d b x b Further, 4 b x + b Therefore, t n generl point P(x, ), the rdius of curvture 4 b () b x ( ) + + ρ 4 b / 4 b x b 4 b b x b The eqution of the tngent t (x, ) to the ellipse x x + b x + is given b b (comprble the generl form of tngent x + b + c 0) Now perpendiculr distnce of the tngent (4) from (0, 0) is () (4) x 0+ 0 b p x x + b b (5)

31 Geometricl Applictions of Differentition 7 Which comprble to c p + b n using eqn (5), rdius of curvture t P(x, ) becomes b ρ. p x cosφ Alterntel we cn use prmetric coordintes with bsinφ} Exmple : If ρ, ρ be the rdii of curvture t the extremities of two conjugte semidimeters of n ellipse, prove tht (ρ / + ρ / ) (b) / ( + b ) [NIT Kurukshetr, 006] Solution: Clerl in the geometr, CP nd CD re two conjugte semi-dimeters to ech other to the ellipse with nd b s the semi-mjor nd semiminor xis respectivel. Thus for P(x, ), x cosφ bsinφ} () We get x' sin φ nd ' b cosφ dφ nd x" cosφ nd " bsinφ dφ Rdius of curvture t P(x, ), ( ' ' ) ( ' ' ) () ( sin cos ) ( sin )( sin ) ( cos )( cos ) x + φ+ b φ ρ x" x" φ b φ φ b φ ( ) sin φ+ b cos φ cos φ + () b Now for position D rdius vector CD encloses n ngle (90 + φ) with the initil xis (insted φ s in cse of P) Therefore, rdius of curvture t D D, b sin φ + 90 (0, b) (90 φ) C (0, 0) P P D φ Fig.. ( cos φ, b sin φ) (, 0) Whence sin b cos +φ + +φ φ+ φ ρ b b ( ) ( b) ( ) sin φ+ b cos φ cos φ+ b sin φ ρ +ρ + i.e. (ρ / + ρ / ) (b) / ( + b ). ( b) ( cos b sin ) (4)

32 8 Engineering Mthemtics through Applictions Exmple 4: Show tht the rdius of curvture t P(x, ) on the ellipse given b ρ CD b where CD is the semi-dimeter conjugte to CP. x + is b Solution: In the previous problem ρ t P(x, ) is ( ) sin φ+ b cos φ ρ b Now the distnce CD with C(0, 0) nd D (cos(90 + φ), bsin(90 + φ)), i.e. D( sinφ, b cosφ) is given b CD b sin φ+ cos φ CD Now clerl ρ. b Exmple 5: If ρ nd ρ be the rdii of curvture t the ends of focl chord of the prbol 4x, then show tht ρ / + ρ / () /. [MDU, 006; KUK 008] Solution: The given prbol 4x pssing through n generl point P(x, ) in its prmetric form is given s follows: x t, t } So tht x t, x, 0 () ( x ) ρ t P(x, ) + ( ) 4t + 4 t + x' x t 0 If ρ t P(x, ) is denoted b ρ, then ( ) / / ( ) ( ) ( t ) ρ + () + t Further, the prmetric coordintes of point Q t the nd end of the focl chord would be x nd (4) t t The generl eqution of the line pssing though P(t ) nd Q(t ) with prmetric vribles t nd t, (t + t ) x + t t θ (0, 0) S (, 0) Fig..4 () Px (, ) But if it pss through S(, 0) where x, 0, we get (t + t ) 0 + t t, i.e. t t With bove rguments, ρ t Q if denoted b ρ, then ( ) t ρ ( ) t + + (5) t Q

33 Geometricl Applictions of Differentition 9 Adding (4) nd (5), we get t ρ +ρ + t + t + Hence the result. ( ) ( ) Exmple 6: If x, re given s functions of the rc s, show tht + ρ ds ds Hence show tht for ctenr x c log s+ s + c nd s +c, ρ is s +c. c Solution: We know tht tnψ (stndrd result) ds tn sec which implies + + ψ ψ cos ds ψ nd tnψ cosψ sinψ ds ds Further sin d ψ ψ ds ds dψ nd cosψ ds ds Whence on squring nd dding expressions under (), we get dψ + ds ds ds ρ. () () () Now Further, x clog s + s + c nd s + c d c s+ s + c ds s+ s + c ds c + s s+ s + c s + c c s + c c d c cs s ds ds s + c ( s + c ) ( s + c ) (4)

34 0 Engineering Mthemtics through Applictions s ds s + c Further d s c ds ds s + c Now using (4) nd (5), in (), we get + ρ ds ds ( s + c ) (5) or ρ + ds ds cs c + ( ) s + c ( s + c ) s + c cs + c c. 4 ( s + c ) Exmple 7: Show tht for ccloid x (θ sinθ), ( cosθ), rdius of curvture t n point is twice the portion of the norml intercepted between the curve nd the x-xis. Solution: Here x (θ sin θ), ( cos θ) () x' ( cosθ), ' sinθ x" sinθ, " cosθ () Thus ρ ( ) x' + ' x'" 'x" { ( cosθ )} + ( sinθ) ( ) cosθ cosθ sinθ sinθ ( ) cos θ cosθ cos θ+ sin θ ( ) ( ) ( cosθ) ( cos ) θ

35 Geometricl Applictions of Differentition ( ) cosθ Q θ ( cos ) sin θ θ 4sin Now the length of the norml intercepted between the curve nd the x-xis, p + d + θ s sinθ ( cosθ ) + θ θ sin + cot ( cosθ) θ θ θ i.e., p From eqn () nd (4), the rdius of curvture ρ t n generl point is twice the length of the norml (p). sin cosce sin (4) x Exmple 8: Prove tht the rdius of curvture for the ctenr ccosh is equl to the c portion of the norml intercepted between the curve nd the -xis nd tht it vries s the squre of the ordinte. Solution: Eqution of the curve is ccosh x c () d x sinh c () nd d x cosh c c () ( ) x x + + sinh cosh c c x Thus ρ ccosh cosh x cosh x c (4) c c c c Now portion of the norml intercepted between the curve nd the -xis is n + x cosh sinh x c c + c using ()

36 Engineering Mthemtics through Applictions ccosh x cosh x ccosh x (5) c c c Clerl from eqn (4) nd (5) we see tht ρ (rdius of curvture) n (length of the norml) x ccosh, c c using () ρ vries s squre of the ordinte. ASSIGNMENT 4. Find the rdius of curvture t the point (s, ψ) on the following curves: (i) s 8sin ψ (Crdioid) (ii) s 4 sinψ (Ccloid) 6 (iii) s c log sec ψ (Trctrix) (iv) s (e mψ ) ( ) x +. Show tht for the rectngulr hperbol x c, ρ. [MDU, 004] c θ. In the ccloid x (θ + sinθ), ( cosθ), prove tht ρ 4cos. 4. Show tht the rdius of curvture t point ( cos θ, sin θ) on the curve x / + / / is sinθ cosθ. 5. Find the rdius of curvture t the point (i) (t, t) of the prbol x 4x, (ii) (0, c) of the ctenr ccosh. c 6. Show tht the rdius of curvture t (/4, /4) on the curve x + is. x x 7. Prove tht the rdius of curvture of the ctenr ( e e ) + is nd tht of the ctenr of uniform strength c log sec(x/c) is csec(x/c). 8. Find the rdius of curvture t the origin for (i) x + x (ii) x x + x + x Show tht the rdius of curvture of the lemniscte (x + ) (x ) t the point where tngent is prllel to -xis is. 0. The coordintes of point on curve re given b x sint bsin t, b cost bcos t. Show tht the rdius of curvture t the point with prmeter t is b b b 4 sin t. + b b. Prove tht the rdius of curvture t n point of the stroid x / + / / is three times the length of the perpendiculr from the origin to the tngent t tht point. [Hint: p x / / (x / + / ) / (x) / ]

37 Geometricl Applictions of Differentition x. Show tht for n ellipse +, the rdius of curvture t the end of the mjor xis b is equl to the semi-ltus-rectum of the ellipse. b [Hint: ρ t (,0) (the semi-ltus-rectum)]. Prove tht the curvture t point on the curve f (x) is given b cos ψ. V Rdius of Curvture for Polr Curve r f(θ). With usul nottions, from the geometr ψ θ + φ n differentition with respect to s, () dψ dφ + ρ ds ds ds i.e., dφ d dφ + θ + ds ds ds Also we hd derived the result r tnφ r dr r r φ tn r dφ d r so tht r d r θ + r r rr rr r r r ( + ) p P φ r ψ θ T N Fig..5 () Also r rr + () ( r r ) ds r r d θ + (4) Using () nd (4), Hence r rr r + r rr + ρ + ( r + r ) r r ( r + r ) ( ) r + r ρ r r rr +

38 4 Engineering Mthemtics through Applictions Corollr If eqution of curve is given in the form u f(θ), where u or even in cse of vice vers r i.e., du d u du r u,then r, r + u u u u then ( ) r + r ρ r r rr + reduces to where du + u u u + u ρ du d u du u u u u u du d u u u,. ( ) u ( u u ) 4 (5) VI Rdius of Curvture for Pedl Curve p f(r) From the (Fig..5), we see tht ψ θ + φ nd d ψ d θ d φ + ρ ds ds ds Also in tringle PN, p r sinφ dp dφ so tht sinφ+ rcosφ dr dr (6) dp dr dφ r + r dr ds ds dr dr Hence ρr. dp dp dφ r r + dr ds ds ρ (7) (8) Exmple 9: Find the rdius of curvture for the following curves (i) r n n cosnθ (ii) r e θ cotα (iii) r cosθ Solution (i) Given r n n cosnθ Tking logs on both sides nlogr n log + log cos nθ ()

39 Geometricl Applictions of Differentition 5 Differentiting, n dr 0 nsinn r + cos nθ θ, i.e. r rtnnθ () Agin differentiting, r rnsec nθ r tnnθ rnsec nθ + rtn nθ () ( ) r + r ( ) r + r tn nθ ρ r + r rr r + r tn nθ+ r nsec nθ r tn nθ r n r n r sec θ sec θ n + r sec nθ n + n + cosnθ ( ) ( ) n n r or ρ n n ( n+ ) r ( n+ ) r In other words, here ρ vries s the (n )th power of the rdius vector. Alterntel: Chnge the given eqution into its pedl form nd then find dr Here in this problem, r rtnn d θ θ dr r dp r tnφ r cotnθ tn + nθ dr r tnnθ or φ + nθ Now n + r r p rsinφ rsin + nθ rcosnθ r n ( n+ ) r n n Differentiting it, dp dr n n dr ρ r r n n dp ( n+ ) r ( n+ ) r (ii) so tht r e θcotα, dr θcotα cot dr αe θ, cot α e r dr θcotα ( cot ) + e + α ( θ cot α ) e cosec α cotα e θcotα cosec α, θcotα ρ e cosec α r cosec α. θcotα e cosec α Extension: Show tht in equingulr spirl r e θcotα, rdius of curvture subtends right ngle t the pole. r Here tnφ r tnα, φ α (Fig..6) dr r cotα, n n

40 6 Engineering Mthemtics through Applictions QP Now cosecα sec( 90 α ) sec PQ nd P whence PQ 90 (iii) Given r cosθ () Differentiting () with respect to θ, we get Q 90 φ P dr r sin d θ θ () Further differentiting with respect to θ, we get 90 θ r Ψ T Fig..6 dr r dr + cosθ or dr r dr r () Now ρ r dr r + d θ dr d r + r r r dr + dr dr + + r + dr r + n squring nd dding () nd (), we get (4) 4 dr 4 r + r or or r r + dr 4 r dr + d θ r (5) Using (5), we get ρ r

41 Geometricl Applictions of Differentition 7 Corollr: Rdius of curvture of the lemniscte r cosθ t the point where tngent is prllel to -xis is. Here in this curve tngent nd the rdius vector coincides t θ ±/4. Therefore ρ θ 4 r. Exmple 0: If ρ, ρ re the rdii of curvture t the extremities of focl chord of the conic l ( + ecos θ), prove tht when e, ρ + ρ / l /. r Solution: Let r u () so tht for l/r ( + e cosθ) we get nd u ( + ecos θ)/ l u esin θ/ l u ecos θ / l () nd Now r u dr d u r u u implies ( ) r u u u + u () ρ ( ) r + r + r r rr u + u u u u u + + u u u u u / ( u + u ) u ( u+ u ) (4) Now ρ + ecosθ e sin θ + l l + ecosθ + ecosθ ecosθ l l l ρ + e + ecosθ l + ecosθ l l (5)

42 8 Engineering Mthemtics through Applictions l r cos, prbol for e n ellipse for e < The generl eqution of the conic ( e ) hperbol for e > ( ) l ( + cosθ ) ( + cosθ) l Thus for e, + cos θ ρ l cos θ (6) Now if ρ t P is termed s ρ nd ρ t Q is termed s ρ, then ρ / l / cos θ/ P( θ) θ θ θ 0 Q( θ + ) θ / Generl Geometr of the conic Fig..7 nd θ+ θ ρ l cos l sin Add the two, ρ / + ρ / l / Exmple : If φ be the ngle which the rdius vector of the curve r f(θ) mkes with the r dφ tngent prove tht sinφ +, where ρ is the rdius of curvture of the curve. Also ρ ppl the result to show tht ρ for the circle r cosθ. Solution: See the geometr given in Fig..8. In the PT, ψ θ + φ ψ is the ngle which the tngent to the curve mkes t with the initil xis θ is the ngle which the rdius vector P mkes with the initil xis; φ is the ngle which the rdius vector encloses with the tngent t P(r, θ). Eqution () implies d ψ d θ d φ + B ds ds ds i.e. dφ + ρ ds ds or dφ + ρ ds () In the PNQ, for the limiting rc when Q pproches to P(bck), i.e. when δθ 0 δ c ds Lt nd dr r + δθ 0 δs Thus Lt tn α tn θ, tn θr dr δθ 0 ds ds dr dr A δθ θ (0, 0) T s φ r ψ () α Qr ( + δr, θ + δθ) δs N δc Pr (, θ) Fig..8

43 Geometricl Applictions of Differentition 9 r dr + dr r + dr tn φ+ i.e. ds dr secφ or cos dr ds φ () Now sinφ tnφ r or r dr cosφ dr i.e. sinφ cos r dr φ dr r ds dr Using () or sin φ r ds (4) Now on using (4), () becomes r dφ sinφ + ρ (5) Further, the circle r cosθ implies r sinθ Thus, r cosθ tnφ cotθ tn +θ r sinθ i.e. φ +θ dφ nd r cosθ ρ dφ + sinφ ( + )sin +θ cosθ cosθ (6) VII Rdius of Curvture for Tngentil Polr Eqution p f(ψ). A reltion between perpendiculr p from the origin on n tngent to curve nd ψ which this tngent mkes with -xis is clled the tngentil polr eqution of the curve. Let p be the length of perpendiculr L drwn from the origin on the tngent to the curve t the point P(x, ), then L mkes n ngle (ψ /) with the positive direction of -xis. The eqution of the tngent PT p cos(ψ /) + sin(ψ /) sinψ cosψ () where, re the coordintes of n point on the tngent. Px (, ) p ψ / L / Fig..9 T ψ

44 40 Engineering Mthemtics through Applictions As P(x, ) lies on (), therefore p x sinψ cosψ () Differentiting both sides of () with respect to ψ, dp xcosψ+ sinψ + sinψ cosψ dψ dψ dψ ( xcosψ+ sin ψ ) + sinψ cosψ dψ dψ ds ds ( xcosψ+ sinψ ) + sinψ cosψ ds dψ ds dψ (xcosψ + sinψ) + sinψ ρ cosψ cosψ ρ sinψ (xcosψ + sinψ) + 0 () Differentiting () gin with respect to ψ, dp xsin ψ + cosψ + cosψ + sinψ dψ dψ dψ dp ( xsinψ + cosψ ) + cosψ + sinψ dψ dψ ( xsin cos ) (cos sin ) ψ + ψ +ρ ψ + ψ Q ρcos Ψ, ρsin Ψ dψ dψ dψ dp p+ρ dψ dp or ρ p+ (4) d ψ Alterntel: Using the reltion between p nd ψ. dp dp dr ds dψ dr ds dψ dp cosφ ρ dr dp dr cos r, dr φ dp As dr ρ r dp rcosφ dp p + dψ r sin φ+ r cos φ r, (As p r sinφ) (5) n differentiting (5) with respect to p, dp d p dψ dr dp p+ r d d or p+ ψ ψ dp dp dψ ρ (6) Hence the result.

45 Geometricl Applictions of Differentition 4 Exmple : Find the rdius of curvture for the hperbol p cos ψ b sin ψ Solution: The curve is p cos ψ b sin ψ () or ( + cosψ ) b ( cosψ) p or p ( b ) + ( + b ) cosψ () n differentiting both sides with respect to ψ, dp p ( + b ) sinψ dψ dp + sin ψ dψ p ( b ) () Also from (), ( p b ) ( + b ) cosψ (4) n squring nd dding () nd (4), dp ( ) ( ) ( + p p b p b b ) dψ 4 4 dp dψ 4 n dividing both sides b p, we get 4 or p p p ( b ) ( b ) ( b ) dp + p b b ( ) dψ p Differentiting both sides with respect to ψ, dp d p dp b + p dp dψ dψ dψ p dψ dp b or + p (on cncelling dp dψ p dψ throughout) b ρ p+ p numericll. (5) ASSIGNMENT 5. Find the rdius of curvture t point (r, θ) on the following curves: (i) r( + cosθ) (sine spirl) (ii) r m m sinmθ (iii) r (cosθ ) (iv) r cosθ (v) θ (r ) / cos (/r). Find the rdius of curvture to the curve r ( + cosθ) t the points where tngent is prllel to the initil line.

46 4 Engineering Mthemtics through Applictions. Find the rdius of the curve r sinθ t the point where the rdii vector is perpendiculr to the tngent. [Hint: such points re (, /4). First find generl vlues of ρ(r, θ) nd then tke θ /4] 4. Find the rdius of curvture t (r, θ) for the polr curve r θ. ( ) u ( u u ) ( 4) ( ) Hint: θ u + u θ + u, ρ r + θ θ + 5. Find the rdius of curvture t the point (r, θ) of the curve cos θ sin θ u +. b 6. Find the rdius of curvture t the point (p, r) on the following curves: (i) r p (Crdiod); (ii) r p (Lemniscte) (iii) pr (Hperbol); (iv) p n r n+ (Sine spirl) (v) 4 r r p (Archimedin spirl) (iv) + (Ellipse) ( r + ) p b b 7. Find the rdius of curvture for the curves (i) p ( + sinψ) (ii) p cos ψ + b sin ψ 8. Show tht for the Epiccloid p sinbψ, ρ vries s p..6 CENTRE F CURVATURE, CIRCLE F CURVATURE, EVLUTE AND CHRD F CURVATURE I Centre of Curvture Centre of curvture C( x, ) for n point P(x, ) of curve is the point on the positive direction of the norml to the tngent t P t distnce ρ from it (Figure.0). Angle NCP 90 Angle NPC Angle NPT ψ x M L ML L NP x ρsinψ () ( ) + x Qtn ψ, sin ψ, cosψ + ( + ) x () nd MC MN + NC LP + NC + ρ cosψ () ( ) + Q + + (, x ) C P N Ψ + + Ψ ( + ) T M L + (4) x x Fig..0 II Circle of Curvture The circle with its centre t the centre of curvture C nd rdius equl to ρ is clled the Circle of Curvture of the curve t the point P nd its eqution t P is ( ) ( ) x x + ρ. ρ(, x )

47 Geometricl Applictions of Differentition 4 III Evolute P 6 The locus of the centre of curvture of the given curve is C 6 C 5 P clled its evolute nd in turn the curve is termed s n C 5 4 C involute of its evolute. P 4 C If C 0, C, C,, etc. re the centre of curvture of the C curve f (x) corresponding to the points P 0, P, P,, C P 0 etc. respectivel, we see tht in moving from P 0 to P, P,, etc. Center of curvture moves long the curve P C 0 C C in turn which is clled the evolute of f (x). P 0 P As x,, ρ nd ψ depends upon s, therefore the eqution () nd () m be treted s prmetric eqution of the evolute. Fig.. Two Importnt Properties of Evolutes () The norml t n point of curve is the tngent to its evolute t the corresponding centre of curvture. (figure.) (b) The length of the rc of the evolute between n two points is equl to the difference between the rdii of curvture t the corresponding points of originl curve i.e. S S ρ ρ, where S S length of the rc C C nd ρ ρ difference between rdii of curvture t point P nd P. (Fig..). Evolute C C Cx (, ) C 0 Px (, ) P P 0 P Fig.. Fig.. IV Chord of Curvture The length intercepted b the circle of curvture of the curve t P, on stright line drwn through P in n given direction is clled Chord of curvture through P in tht direction. Thus, if the chord of curvture PQ (Fig..4) mkes n ngle φ, with the norml PCD, then its length PQ is given b PQ PDcosφ ρcos φ () [³ DQP being in semi-circle is right ngle] D Q φ C P T Fig..4

48 Geometricl Applictions of Differentition 7 ( x + ) ± 8 x + 4 n neglecting the term with negtive sign before the rdicl on the right side, becuse if is negtive will become imginr. ( ) 8 4 x + + x + n differentiting the bove eqution, we get 6x x (, 0) 4x+ 4 8x + 90 x Now 0 8x implies 4x 4 8x + or x ± 8 Fig..4 Hence the tngent to the curve re prllel to the xis of t x ± nd, re thus 8 the extreme vlues of (i.e. point of mxim-minim). Further the tngents re prllel to -xis t x nd x which cn be verified b shifting the origin from (0, 0) to (x, 0) nd equting to zero the lowest degree in x nd for tngent t x. Similrl, for x b replcing x b (x + ). (5) Regions: For < x < nd < x <, is negtive, wheres for < x< 0 nd 0 < x <, first increses nd then diminishes to zero. Exmple 6: Trce the curve ( x ) x. Solution:. Smmetr: The given curve is smmetricl bout xis onl.. rigin: It psses through the origin nd t the origin, the tngent to the curve is cusp, viz. x 0 (i.e. -xis).. Asmptotes (i) n equting to zero the coefficient of highest powers of x, we get 0, hence -xis itself is the smptote to the curve. (ii) Agin, on equting to zero the coefficient of highest power of, we get x 0, i.e., x nd x re the two smptotes prllel to -xis. (iii) It hs no oblique smptote. 4. Points: The curve does not meet the xes except t the origin nd lso there re no such points where tngents re prllel to either of the xes except t the origin. x (0, 0) Fig..5 x

49 7 Engineering Mthemtics through Applictions. Smmetr: The curve does not hve n smmetr.. rigin: It does not pss through the origin s eqution of the curve hs constnt term.. Asmptote: The curve does not possess n smptote. 4. Points: The curve intersects with the -xis (i.e. x 0) t,,. i.e. It intersects the -xis (i.e. 0) t x 6. n simplifiction, x nd 0 implies for + 0 ± 44 4 or ± (.6,.4) for.6, x , x 0.84 Hence the tngent prllel to the xis of re t points (.84,.6) nd (.84,.4) respectivel. Exmple 6: Trce the curve (x + ) (x ) (i.e. r cosθ in polr form) Solution:. Smmetr: The given curve is smmetricl bout both nd -xes.. rigin: The curve psses through the origin nd t the origin the tngent to the curve re ±x (node).. Asmptotes: The given curve hs no smptotes. 4. Points: () Intersection with xes: (i) Intersection with -xis (i.e., 0) is t the points x 0, ± (ii) Intersection with -xis (i.e., x 0) is t the point 0. Hence the given curve intersects the xis t (0, 0), (, 0), (, 0). (b) Points where 0 or. Rewrite the given eqution fter simplifiction s: 4 + (x + ) + (x 4 x ) 0 or ( x ) ( x ) 4( x 4 x ) + ± + ( x 4 x 4 4 x 4 x 4 4 x ) + ± ( 6, 0) (0, ) (.84,.6) (0, ) (.84,.4) (0, ) Fig..

50 Geometricl Applictions of Differentition 7 {( ) ( ) ( ) ( ) } ± x x + x + x + x x ( x) ( x) + ( + x) x ± + ( + x) ( x) ( ) ( + x) x ( x) + ( + x) x ± + ( + x) ( x) ( ) ( + x) x ( x)( x) ( + x) ( x) + + ± ( ) x + x + x x ± ( ) ( ) + x x ( x ) ± + x x ( ) ( ) + ( ) ( + ) / / x x Now 0 implies x 0, i.e. x ±. (Leving the vlue x, since for < x <, is imginr). Thus, x is the point of mxim nd t this point, the tngent is prllel to the xis of. Similrl implies + x 0 nd x 0, i.e. x Hence the tngent to the curve is prllel to -xis t (, 0) nd lso t (, 0) which is behving s n smptote. Regions: For ± x ( x) + x (i) If x <, is imginr (ii) If < x < 0, decreses from to 0 (i.e., defined entit) (iii) If 0 < x <, is positive nd it first increses nd then decreses to zero. (iv) If x >, is imginr., x x (, 0) (, 0) x x (, 0) Exmple 60: Trce the curve x ( )( )( ). Fig.. Solution: