Cpter Differentition. Introduction In its initil stges differentition is lrgely mtter of finding limiting vlues, wen te vribles ( δ ) pproces zero, nd to begin tis cpter few emples will be tken. Emple..: Find te limit of te followings: (i) + + Lim + + 5 of limiting vlues (ii) (iii) Lim Sin Lim 0.. Definition: Te first differentil coefficient or derivtives of y wit respect to δy is Lim 0 were δ y is smll increses inn te dependnt vrible y, δ corresponding to smll increment δ in, nd y is some function of. Tis differentil coefficient is denoted by ny one of te symbols d, {f ()}, f (), Dy, were coefficient of y wit respect to is If y f(), it cn be redily sown tt known s te differentil coefficient. y f (). Hence te definition of te first differentil Lim 0 δy δ f ( + δ) f () Lim 0 nd te result is δ Note: Te following teorems cn be used.. Te derivtive of constnt is zero.. n d( ) n n. Te differentil of te sum is equl to te sum of te differentil coefficients.
. If y be function of z, were z is function of, ten 5. If u nd v re function of, ten d(uv) d(sin ) d(cos ) 6. cos nd sin dv u d(sin ) cos 7. Te slope of te curve y f() t (,y) is 8. d(tn ) d(sec ) 9. sec nd sec tn + du v d(cot ) d(cos ec) 0. cos ec nd cosec cot. Derivtive of eponentil function. Derivtive of log function d(loge ) d(e ) Derivtive of inverse trigonometric functions () (b) d cos (c) dz dz d( ) e nd loge d tn d sin d(sin ) d(cos ). Derivtives of te yperbolic functions () cos (b) sin d(cos ec) cosec cot (c) (e) d(sec ) + sec tn (d) d (tn ) sec d(cot ) (f) cos ec
5. Derivtives of te inverse yperbolic function. () d sin + (b) d cos ± (c) d tn (d) d sec ± (e) d cos ec ± + (f) d cot Emple..: Find te derivtives wit respect to of (i) sin ( ) (ii) cos + sin (iii) e sin + cos (iv) tn + ( + 5 + ) Applictions of differentition. Relted rts Suppose in prcticl problem two quntities nd y re relted by formul y f (), nd suppose tt nd y depend on vrible t, so tt g (t), y (t) (for us t usully represents time). Ten or y & & dt dt i.e. te rtes of cnge of z nd y w.r.t. t re relted by te cin rule. In tis section, use tis rule to solve relted rte problems using te following pproc:. Drw digrm nd introduce pproprite vribles.. Decide wt rte of cnge is sked for (e.g. y& )).. Find n eqution relting vribles (e.g. y f()).
. Use cin rule to y& to &. 5. Substitute pproprite vlues of vribles (if required). Emple.. Air is blown into spericl sop bubble t rte of 0 cm s -. Wen its rdius is.5cm, ow fst is () its rdius nd (b) its surfce re cnging wit time? Solution: Let te bubble ve volume V, rdius r nd surfce re S. We re given V &. We seek r&, we ve V πr so V& πr r& nd Given tt V 0 wen r.5, ten 0 0 9π V& r& πr r& 0.5cms. π(.5) (b) We seek S &. We ve S πr so S 8πrr& nd & 0 S& 8π.5 9π 0 cms t tt instnt
Emple.. A mn.8 m tll wlks t.5ms - wy from lmp m ig. How quickly is () is sdow growing, (b) te tip of is sdow dvncing? Let, nd y be te distnces sown. (i) We re given &. 5 nd we seek y&. Introduce ngle α s Ten so nd.8 tn α y + y.8.8 +.8y y y. 9 y& & 9.5.ms 9 ngle s sown. (ii) Te tip of te sdow is t distnce L & & + y& from te post so L& & + y&.5 +..7ms Emple.. A ligtouse one mile offsore rottes t rpm. How fst is te spot of ligt S moving long te bec wen it is miles from te nerest point N to te ligtouse? θ N bec S Let θ nd be te ngle nd distnce sown. We re given θ rpm πrd min 6πrd min We need &,. d Clerly tn θ, so tt tn & θ θ θ & & + Terefore & ( + ) θ& nd wen & 56π 0π 9.mp min 5655mp
. Smll vritions Let ( 0, y0 ) be point on curve y f(). Suppose tt wen 0 increses to +, y o increses to y, o y y + y. Since o y lim 0 t 0, we epect tt y for smll. Tus y wen is smll. Tis is essentilly sying tt y in te digrm. Some terminology: Te bsolute cnge in 0 is 0 Te percentge cnge in is 00 Emple.. : Estimte 6 0 Let y so tt y. we seek y wit 6. Coose 0 6 ten 0 nd y0 6, y0. 6 8 Hence y ( ) 0. 07 8 y nd 6 0.07. 958
Emple..: A metl spere cools so tt its rdius r decreses % Estimte te percentge cnge in its volume V. We ve so nd V πr dv V r π r r dr V V πr πr r r dv dr πr r V r But so tt 6% r 00 V r 00 Since f ( ), +, f ( ) f ( + ) nd y0 0 0 y 0 y we ve from y y0 + y f 0 (0 + ) f (0 ) + f ( ) f 0 0 0 ( smll, Writing 0 (so + ) we ve () f ( ) + f ( )( ) for ner. Tis is of term c + m( ), i.e strigt line, nd so clled te liner pproimtion to f() ner y 0 Emple..: Find te liner pproimtion to ner f () 5e + sin π π We ve f () f (0 ) + f (0 )( 0 ) so ere f () 5e + cos, f () 5 + 0, f () 5 nd f () 5 + ( ) +, i.e f() looks like te line y+ ner
Emple..5: Find te liner pproimtion to f () sin ner 0 Now f () cos, f (0) 0, f (0) 0 so tt f () 0 + ( 0) nd sin ner 0 Emple..6: Find te liner pproimtion to α f () ( + ) ner 0 Here f () α( + ) α, f (0), f (0) α so tt f () + α( 0) nd ( + ) + α α ner 0. Curve sketcing To sketc curve of y f (), first we need to ceck sttionry (turning) point of f() in te domin, were tese points re clled For sttionry points: At point, curve y f() s locl mim if f () f (), close to mim nd minims. locl minim if nd s globl mim if globl minim if f () f () f () f (), close to f (), in te domin of f f (), in te domin of f Mimum nd minimum re referred to s etrem. Te curve y f () increses were > 0 nd decreses were < 0 so it is flt were 0. Tus 0 t mimum or minimum t wic is defined. A point t wic 0 is clled sttionry point(s P). Notes: Te fct tt 0 t point does not imply tt is mimum or minimum ( just tt te slope tere is zero).
Points to be defined for mmin:. points were 0. points were is not defined (i.e corners ). end points of te domin (if ny) Emple..: (i) y s min t 0, nd so tt 0 t te minimum. (ii) y s no minim or mim, but (iii) y s minim t 0, but so tt 0 ( nd te curve is flt tere) is not defined t 0. Emple..: Te curve y 6 + 8 s 0 t (were y-), nd t 0 (were y 0), but globl mimum ner y (t ) nd globl minimum y - (t - ). Note tt te test 0 s not detected te globl m min. We must ceck sttionry points nd ny oter criticl vlues of suc s -, ere. A test for mmin Ner sttionry point if < 0 for < nd > 0 for > ten is minimum. if > 0 for < nd < 0for > is minimum. slope slope - + - + - \ MIN - + - \ MIN
Find te etrems of y +. Here + ( + )( + )( ). Tere re sttionry points were 0, i.e t -,-, were y -9. 7, 9 respectively. nd no points were y is not defined. <- - >- - 0 + slope \ MIN <- - >- - 0 + \ MAX < > - 0 + \ MIN Emple..: Find te etrems of Here y 9 + 9 wic is ner zero, so tere re no sttionry points. ( in fct < 0 ( 9) so tt y is lwys decresing ). Note tt A test using te second derivtive y is not defined t, -. if y > 0 t ten te curve lies bove its tngent t. if y < 0 t ten te curve lies below its tngent t. Tese re respectively concve up nd concve down.
Hence y 0 nd y > 0t minimum t. y 0 nd y < 0 t mimum t. Emple..: Find te etrem of y +. (see emple.) Here + ( + )( + )( ) d y nd + Tere re sttionry points t -, -, y type - 08-7- >0 minmum - --<0 mimum +->0 minimum Points of infections Te inflection of y() s point of inflection t if y 0 t nd y 0 cnge sign s psses troug. Note: y need not be zero t point of inflection. A point of inflection cn be tougt s point were curve stops bending one wy nd strts bending te oter wy. A curve crosses its tngent t point of inflection. up up point of inflection down down
Emple..5: If + y y ten + nd y 6, so tere is point of inflection t 0 ( note tt y tere). Emple..6: If y ten y nd y. Here y 0 t 0 but y > 0 for >0 nd <0 so tere is no point of inflection ( in fct 0 is minimum)..5 Asymptotes An symptote is curve (e.g strigt line) wic noter curve pproces.. Te line y b is orizontl symptote to curve y() if y b s ndor y b s Emple.5.: (ii) (iii) ± y 0 s ± so y 0 is orizontl symptote. + + ± + y s ± so + + 5 6 + + 6 + 5 + 6 y + 7 7 + ± y is orizontl symptote. s ± so y is orizontl symptote.. Te line is verticl symptote to curve y() if eiter lim + y ± or lim y ± in prticulr, if p() y wit q() 0, p() 0 ten is verticl symptote. q() Emple.5.: Consider y ( )( )
Denomintor is zero t nd so y s verticl symptote t nd. Also ± y s ± so y is orizontl symptote. For, y > 0 nd y ± s. For, y < 0 nd y s +. Similrly, y s ±. Te line y m + c is n inclined symptote to curve if y() (m + c) 0 s or + Emple.5.: y + + s verticl symptote nd inclined symptote y +, As ±, y ± As ±, y ( + ) 0 Emple.5.: Sketc te curve y ( + 5)( ) ± We ve y ( + + 5)
so y 0 wen + 0, i.e t -, wen Hence, is sttionry point. 9 y 9 < - - > - + 0 - slope \ So sttionry point is mimum. Wen 0, y, so intercept is ( 0,. 5 5 We ve y 0 for ll so curve crosses is As ±, ± y 0, so curve s orizontl symptote y 0. Te denomintor of y is zero t, -5 so curve s verticl symptotes t, nd -5. As 5, y, s + 5, y : s, y : s +, y +
Emple.5.5: Sketc te curve y By polynomil division, we ve y + ± So y 0 s ± giving y s n inclined symptote. Denomintor is zero wen 0, so tere re verticl symptotes nd Wen 0, y 0 nd vice vers so cuts es only t (0,0). Also y ( ) ( ). ( ( 9) 9) so y 0 wen 0,+, - ( nd <- - >- + 0 - \ y 0, ± ). 9 <0 0 >0 < > - 0 + \ \ - 0 + \ slope Mimum Point of inflection Minimum.6 Optimistion Suppose y() is defined for [,b]. Ten y s bsolute mimum nd n bsolute minimum m in [,b]. Outline of pproc for optimistion problems:. Identify nd nme vribles: drw picture.. Decide vrible to be optimised (e.g. y).
. Determine te domin.. Find te set of equtions relting l vribles. 5. Reduce tis set to one eqution (e.g. y f() ). 6. Find mimum or minimum. Emple.6.: Wt is te size of te cylinder of lrgest volume tt cn be cut from spere of rdius?. Let cylinder ve eigt nd rdius r. Ten te volume of te cylinder is r V π, were 0. But, by Pytgors s Teorem, r r + So π π V nd. dr V d, dr dv π π V s s sttionry point were 0 dr dv so, (ignore negtive root s 0 ). Ten r nd V π π
d V At sttionry point < 0 So sttionery point is mimum. ( Also V 0 wen dr so sttionry point gives bsolute mimum). Emple.6.: A womn in rowing bot is miles from sore. Se wnts to get ome, 0 miles long te sore, s soon s possible. Se cn run t 0 mp nd row t 8 mp. Were sould se coose lnding point A so tt er route from er position B to er ome His quickest, nd wt is te minimum time? O 0 A H B y river Let OA nd BA y, wit 0 0. Ten AH 0- nd y +. Time for y 0 te trip is T + + + (0 0) 8 0 8 0 so dt 8 + 0 d T ( ) nd + ( > 0) dt 8 + ( ) 8( + ) ( + ) + At sttionry point dt dt 0 8 + 0 8 + 00 6 + 6 6 6 56 6 8 miles 0
d T As > 0 tis is minimum. dt Wen 0, we ve 0 T +.5 ours r 8 0 nd 5min Wen, 0, we ve 0 T + 0.7 ours r 8 nd 6.5min Wen 8 6 8 69, we ve T + + ours r nd 9 min 8 9 0 60 8 Hence bsolute minimum is T r 9 min wit. Emple.6.: A metl plte of sides by s squres cut from its corners, nd is ten folded up to mke n open bo. Wt size is te bo tt s mimum volume? Let te squre ve sides, wit 0 < <. Ten te volume of te bo is V() ( ) ( + ) V ( 6 + ), V ( ). dv At v sttionry point, 0 so, 6 ± but only 6 6 ± 6 ± d V stisfies 0<<. Here < 0, so it is mimum. Also, V(0) 0 V() 0 so is te bsolute mimum. Required bo s sides
+ by by nd 8 V m 9 Emple.6.: : An open bo is to be mde of seet metl. It is to ve squre bse nd is to ve cpcity V. Wt size of bo needs lest metl for its construction? l l Let bo ve sides ll ( were l < ). Te mount used is proportionl to te re of te bo, nmely But te volume Hence S l S l + l V l is fied, so V + l l l + V l V l ds V so 0 l l V l (V) dt l d S 8V Also + > 0 dl l At tis l so te sttionry point is minimum, Ceck end points : S s l 0 ( tll tin bo ) nd s l (: sortest wide bo ) so l (V) is te bsolute minimum.
Te desired bo s l (V) nd V l (V).7 L Hopitl Rule We cn use differentition to elp evlute limits of function. Let g() f () wit () g()0 nd () 0 ( so tt f is indeterminte t ) nd suppose we seek Lim f () g() g() g() Now by definition g () Lim Lim nd so similrly Now divides tese two, () () () Lim Lim g () Lim () g () () () 0 Repetedly until te result is not. Tis is clled L Hopitl Rule. 0 Emple.7.: Evlute Lim sin 0 d sin sin Lim 0 Lim 0 Lim 0 cos Emple.7.: Evlute Lim Lim Lim 0 e e e ln Emple.7.: Lim Lim Lim Lim 0 6 6