Chapter 13 Differentiation and applications

Transcription

1 Differentiation and appiations MB Qd- 0 Capter Differentiation and appiations Eerise A Introdution to its Te series of numers is approaing 8. A ire n Te answer is B a As n gets arger, n approaes & n n a S 8 n represents te numer of terms. n 0 term 8 8 S 7 8 S n Te answer is C f() Tae of vaues f() f( ) Te answer is C 8 a f( ) d e f f( ) f( ) 0 f( ) f( ) f( ) 9 Assume graps are ontinuous, terefore sustitute vaue into funtion. Use grapia auator to ek: a ( ) 9 d (0 k) 0 8 k (7 p ) 7 p (9a ) 9 a 0 0 e ( ) f ( ) g ( ) ( ) i ( ) p p j ( ) p k ( ) 0 ( t t ) 0 t If f(), ten f ( ) is: 0 Sustitute 0 into f() f(0) 0 f( ) 0 7 Te answer is D ( ) ( ) Te Te answer is A a d

2 MB Qd- 0 Differentiation and appiations Eerise B Limits of disontinuous, rationa and yrid funtions Disontinuous graps are,, d and f. Grap is disontinuous at Grap is disontinuous at 0 Grap d is disontinuous at Grap f is disontinuous at a We annot sustitute into f() as f() 0 wi is 0 undefined. Must fatorise to simpify te funtion and remove te denominator. ( ), f( ) is te same as f() eept te point (, ) does not eist a Wen 0 f( ) 0 0 f (0) 0 Grap is undefined eause you annot divide y 0. Terefore wen 0, f() is disontinuous. ( ), d f( ) 0 f() an e rewritten as. e Wen graping, rememer grap is disontinuous wen 0. Wen 0, f(), so is disontinuous at point (0, ). f( ) ( ) f 0 0 a f() ( ) f(), 0 f() 8 ( ), f() ( ), 0 d f() ( )( ) f(), 7 e f() ( )( ) f(), 8 f f() ( ) f(), g f() ( )( ) f(), 7 f() ( ) 9 9, a ( ) d e 0 ( ) 0 ( ) ( ) ( ) f ( ) g ( ) ( ) ( ) a f() -interept (, 0) y-interept (0, ) (, ) not inuded f() [, ) inuded g() i Is a straigt ine wit a domain of < Find two points witin domain to grap ine g g g() i Is a straigt ine wit a domain of or < < Find two points witin domain to grap ine g 0 g () i straigt ine (, 0) wit a domain of < < 0 Find two points witin domain to grap ine 0 0 () A paraoa witin te domain 0 < d p() i is a paraoa wit a TP (0, ) witin te domain < Find end point, wen, p p() i Is a straigt ine witin te domain < < Two points on grap p p 8 a i ( ) i f ( ) does not eist (as eft it rigt it) i ( ) i ( ) g( ) does not eist (as eft it rigt it)

3 Differentiation and appiations MB Qd- 0 i i ( ) 0 0 d i ( ) i 9 a i d e i i i i i i i i ( ) f ( ) does not eist (as eft it rigt it) ( ) f ( ) does not eist (as eft it rigt it) ( ) ( ) f( ) ( ) ( ) 7 f ( ) does not eist (as eft it rigt it) ( ) ( ) f ( ) does not eist (as eft it rigt it) ( ) does not eist as ( ) ony eists for te domain < <. i ( ) f i ( ) i 0 ( ) 0 f( ) 0 0 a ( ) d i 7 7( ) 7 ( ) ( ) i f( ) 90 ( )( 0) ( 0) e ( ) 8 f i ( ) g i 0 ( ) 0 f ( ) does not eist (as eft it rigt it) 0 0 ( )( ) ( ) ( ) ( ) i j k ( )( ) ( ) ( ) ( )( ) ( ) ( ) 7 7 ( )( ) ( ) 8 ( )( ) ( ) Eerise C Differentiation using first prinipes f ( ) f( ) f () a 0 0 f () for 0. ( ) 0( ) 0 f () ( 0) 0 ( 0), 0 0 f () 0

4 MB Qd- 0 Differentiation and appiations f () 0 ( ) 8( ) ( 8 ) 0 ( 8), 0 0 f () 8 d f () 0 0 ( ) ( ) 0 ( ) ( ) 0 f () d f () f ( ) f( ) 0, 0 ( ) a f () 0 0 ( ) 0, 0 0 d f () ( ) ( ) d f () d f () 0 ( ) ( ) 0 0 ( ) (8 ), 0 8 d 9 ( ) e f () 0 d f f () ( ), 0 ( ) ( ) 0 0 ( ), 0 d ( ) ( ) 0 ( ) d a g () 0, 0 ( ) ( ) 0 0 g () Wen g () 0 0 a f () ( ), 0 ( ) ( ) 0 f () Wen gradient ± a 7( ) 7 f () f () 7 f () 0, 0 ( ) ( ) 0 0 ( ), 0

5 Differentiation and appiations MB Qd- 0 f () 0 f () ( ) ( ) 0 0 ( ), 0 0 f () ( ) d f () 0 0 ( ), 0 0 f () 7 C and E do not denote te gradient at any point. 7 C Te most aurate metod of finding te gradient of a funtion at a partiuar point is to find te derivative f () and evauate f (). Te answer is C 8 f() g() f () g () ( ) Te answer is A Eerise D Finding derivatives y rue a y d y 7 d y d d y d a y d y 8 7 d y d d y d e y d f y 7 d 7 a y 8 d 87 D y d B y 7 d E d y d F e y d G f y d 8 A g y p d p p C a y d y 7 d 0 y d d y 0 8 d 0 8 e y d 0 f y d 0 a f() d e f () 8 f() 8 f () f() 7 7 f () f() f () 9 f() f () f f() 7 8 f () 8 7 a f() ( ) f ()

6 MB Qd- 0 Differentiation and appiations f() ( ) f () f() ( ) 8 f () 8 d f() 9(8 ) 9( 8 9 ) 7 8 f () e f() ( ) 8 f () f f() ( ) f () 0 0 ( ) 7 a f() f (), 0 ( ) f() f (), 0 ( ) f() f() f (), 0 ( 7) d f() f () 0, 0 8 a f() f () f() 7 f () f() f () d f() 8 f () e f() f () 7 7 f f() f () g f() f () f() 9 9 f () i f() f () 0 j f() 0 f () 0 7 k f() f () 0 7 f() f () m f() f () n f() f () o f() f () p f() f () q f() r f() f () f() f () s f() f () 9 a f() f () 0 i f () f () 0 7 i f (0) f() f () i f () f () 8 i f (0) 0 f() f () i f () f () i f (0) d f() 7 8 f () 7 i f () 7 0 f () 7 9 i f (0) 7 0 y a -interept wen y 0 0 ( )( ) or d Sustitute in d Sustitute in d Gradient at Gradient at i d d i d

7 Differentiation and appiations MB Qd- 07 a y 0 ( )( ) -interepts our wen or Gradient of -interept points d Wen, d y d 7 Wen, d y d 7 Wen gradient 0 d 0 terefore, y ( ) ( ) Answer: (, ) 8 a y d Parae to -ais means d y d ( ) 0, Wen 0, y 0 (0, 0) Wen, y 8 (, ) Parae to y means d. 0 0 ( ) 0 ( ) Wen, y (, ) y a d Find wen Wen, y Equation of ine passing troug point (, ) wit gradient is: y m y Equation of norma: m point (, ) y m y y y Crosses y-ais at (0, ) Find gradient at tis point d Wen 0 d Tis is gradient of tangent. Hene gradient of norma Equation of norma: y or y 0 y d a At d y d 0 Terefore te gradient of te norma is infinite. Equation of norma: At d y d Terefore te gradient of te norma is. Te straigt ine passes troug, y Equation of norma: y ( ) y y 7 y d At, d y d () () 8 Gradient of norma Wen y () () () Equation of norma: y 7 ( ) y 7 y Eerise E Rates of ange f() a f() 9 9 f() Average rate of ange 9 Rate of ange f () Instantaneous rate of ange wen is f () f () 8 a V t t for t [0, 0] 8 i t 0, V m 8 t 0, V m Average rate of ange etween t 0 and t 0 Cange in V Cange in t m /s 8 V (t) t t t 8t i V (0) 0 m /s V () m /s i V (0) m /s y Average rate of ange to At, y 9 At, y Cange in y Average rate of ange Cange in t 9 7 Te answer is E f() f () f () () () f () 8 Te answer is C

8 MB Qd- 08 Differentiation and appiations d 7 (Add one to raised power and divide y tis) 7 y y 7 or y 7 Te answer is C (t) 8t t a (t) 8 t i () 8 8 m/s () m/s i () 8 8 m/s Wen t te a stops rising. It reaes its greatest eigt and its instantaneous veoity is zero. It starts to fa after tat. d Rate of ange of eigt wen a reaes a eigt of m (t) 8t t 8t t 0 t 8t 0 t 8t 0 (t t ) 0 (t )(t ) t or t Now (t) 8 t t is wen it first reaes m. () 8 8 m/s 7 (t) t 0t a Rate of ange of dispaement (veoity) at any time t is d d t 0 i At t d m/s At t 9 d m/s i At t d 0 0 m/s Te ift anged diretion. d Rate of ange 0 wen d 0 0 t 0 t 0 t 0 t 0 s m 8 N 00 t 00 t t [0, ] a i t N N N 000 t N N N 000 Average rate of ange etween t and t Cange in N Cange in t peope/our Instantaneous rate N (t) 000 t 00 i At t 0 N (0) peope/our At t N () peope/our i At t N () peope/our iv At t N () peope/our d Most peope arrive after te gates open, tat is, oser to starting time. 9 W 80 t 0 t 0 < t < 0 a Weigt of foa at irt ours wen t 0 W W 80 kg dw W (t) t 0. t i At t dw 0. 9 kg/week At t 0 dw 0. 0 kg/week i At t dw 0. 9 kg/week d Rate of foa s weigt t 9 kg/week t 0 kg/week t kg/week Rate of ange of foa s weigt is dereasing e Wen does foa weig 00 kg? W 80 t 0 t t 0 t t t t 0t 00 0 or (t 0t 00) 0 (t 0)(t 0) 0 (t 0) 0 t 0 So te foa wi weig 00 kg after 0 weeks. 0 P.n n a P (n). dn n. n..n i If n P ().... P (). So te rate of ange of profit is. undred doars for empoyees.. 00 doars per empoyee $7.0 If n P () So te rate of ange of profit is. undred doars for te empoyees. 00 doars per empoyee $9.8 i If n P ()... 7.

9 Differentiation and appiations MB Qd- 09 So te rate of ange of profit is undred doars for te empoyees. 00 doars per empoyee $ dn..n 0 wen. n. n n 9 Te rate of ange of profit is zero for nine empoyees. V 000 0t 00 t V (t) 0 00 t V (t) 0 0 t a i V (0) m /s V (0) m /s i V (00) m /s No, eause te gas is esaping from te yinder. a A πr (area of a ire rue) Rate of ange of A wit respet to te radius d A dr dr πr i wen r 0 m π 0 dr 0π m /m wen r 0 m π 0 dr 00π m /m i wen r 00 m π 00 dr 00π m /m d Yes, eause d A is inreasing dr (Voume of a spere π r ) a V π r Rate of ange of V wit respet to r is d V dr dr π r π r i Wen r 0. m π (0.) dr π π m /m r 0. m π (0.) dr π.0 0.π m /m i r 0. m π (0.) dr 0.π m /m a engt (ase) eigt wi ase eigt V L W H V V Rate of ange of V d V d d i wen d m /m wen d 8 m /m i wen d 08 m /m a To epress in terms of : sin 0 sin 0 V area end engt (Base engt of triange tan(0 ) o ) d i Wen d Wen y a Gradient of sope d y d d i At 0 d At 00 d i 0 d iv At 0 d Gradient 0. d y d d Using any program to sove a quadrati equation wit a , 0.0, 0. gives 0 or 0 d Te gradient must not e greater tan 0.. Sketing y and y 0. gives te vaues aove 0 and 0. If 0, y If 0, y 7.

10 MB Qd- 0 Differentiation and appiations So te range of eigts not permitted is. < y < 7. 7 A 90t t a 80t 9t i Wen t Wen t i Wen t iv Wen t v Wen t vi Wen t v Wen t Te rate of urning inreases in te first 0 ours and ten dereases to zero in te net 0 ours. Tis an e seen from d A 80t 9t 9t(0 t) Tis is positive from 0 < t < 0, giving a maimum rate of urning midway etween t 0 and t 0, tat is, at t 0. d Te fire is spreading, te area urnt out does not derease. e Te fire stops spreading. It is put out or ontained to te area area urnt. is negative after t 0, onfirming tat te fire stops spreading. f Rate is equa to 7 etares per our 80t 9t 7 80t 9t 9t 80t 7 0 or 9(t 0t 8) 0 9(t )(t ) 0 t or t At t and t ours. Eerise F Soving maimum and minimum proems y. 0.0 a d 0.0 Let d y 0 for maimum eigt. d so To verify tis is a maimum, et < 0, 0 d (positive) Let 0, d 0 Let > 0, 0 d (negative) So Stationary point is a maimum. Maimum eigt reaed: sustitute 0 into y. 0.0 y y. 0 0 y. Maimum eigt reaed is. metres V 00.t 0.08t for te domain 0 t a To find te time for minimum voume, find te derivative and equate it to zero..t 0.t 0.t 0.t 0 0.t(0 t) 0 So 0.t 0 or 0 t 0 t 0 or t 0 (t 0 eause sower is turned on and we require a minimum after tat) So t 0 minutes. To verify tis is a minimum, et t < 0, t.t 0.t. 0. (negative) et t 0 d V 0 Let t > 0, t.t 0.t (positive) So stationary point is a minimum Minimum voume is found y sustituting in origina equation t 0. V 00.t 0.08t V itres Minimum voume 0 itres d If t 0, V 00 itres So t 0.08t or 0.t 0.08t 0 t (. 0.08t) t 0 or. 0.08t 0 t 0 or t So wen t minutes te tank wi e fu again. (t) t t a To find te greatest eigt reaed y te a and vaue of t for wi it ours, find te derivative and equate it to zero. d 0t 0 0t t 0 t. seonds For maimum eigt reaed, sustitute t. in origina equation: (t)..... m. To verify tis is a maimum: Let t <., t d 0t 0 (positive) Let t, d 0

11 Differentiation and appiations MB Qd- Let t >., t 0 0 (negative) Te stationary point is a maimum If 8 and y, ten y 8 aso. 0 0 dw (negative) Let te wi of te fene e metres. Ten te engt wi e ( ) metres. A ( ) 0 for maimum area d 0 and so Nature of te stationary vaue: Let < (say ) (positive) d Let, d A d 0 Let > (say ) 0 (negative) d So te stationary point is a maimum. Te argest area ( 8) m Let te first numer e and te seond numer y. a Ten y So y If P is te produt of te two numers ten P ( ) and d P For P to e a maimum d 0 d So 0 8 Consider nature of stationary point. Let < 8, 7 d (positive) Let 8 d P d 0 Let > 8, say 9 8 d (negative) It is a oa maimum at 8 7 Perimeter 0 m a P L W 0 L W L 0 W L 0 W L 0 A L W A (0 ) A 0 For maimum area d 0 0 d d For maimum area dimensions are wi m engt 0 m So engt and wi m e Maimum area LW m P 0 m If L engt and W wi a P L W 0 L W or LW 0 L 0 W aso A LW A W(0 W) For maimum area, find d A dw and equate to zero. A 0W W So d A 0 W dw 0 0 W W 0 Cek to see if stationary point is a maimum Let W <, W dw 0 (positive) Let W, d A dw 0 Let W >, W 0 At W tere is a maimum turning point Sustitute W into L W 0 L 0 L So engt and wi m Maimum area L W m 8 C $(0.n ) a C ost, n numer of toasters Toasters sod for $0 ea Revenue 0n P revenue ost P 0n (0.n ) P 0n 0.n Numer for maimum profit dn 0 0.n dn So 0 0.n.n 0 n 0. n Verify tat tis is a maimum Stationary point Let n <, n dn 0 8 (positive) Let n, d P dn 0 Let n >, n dn 0 7 (negative) Terefore a maimum vaue ours at n. For maimum daiy profit, Sustitute n into P. P 0n 0.n P Maimum daiy profit $00 9 Inome $( n 0n ) Wages 70 n $70n

12 MB Qd- Differentiation and appiations a Profit inome wages P n 0n 70n P 800 0n 0n For maimum weeky profit dn 0 0 0n dn 0 0 0n 0n 0 n 0 0 Verify if tis is a maimum Let n <, n 0 0 dn (positive) Let n dn 0 Let n >, n dn (negative) n gives a maimum profit For maimum daiy profit, sustitute n into P. P Maimum weeky profit $0 0 Let first numer and y seond numer Sum y y 0 0 y Sum of squares S S y S (0 y) y S 00 0y y y S y 0y 00 For sum to e a minimum 0 y 0 So y 0 0 y 0 y Sustitute for y in y 0 0 So ot numers are. a Ea side must e greater tan 0. Aso te engt of ea side is. > 0 and so <. Te range of vaues of is 0 < < i Heigt Lengt of o i Wi of o Voume of o V L W H V ( ) ( ) ( ) ( 8 ) 8 d For maimum voume 0 d 9 d 9 0 for maimum or ( 8 ) 0 ( 8 ) 0 ( )( ) 0 or If, te o wi not eist as L 0 and W 0 So (no o at a) Consider te vaue Let <, 9 d 0 (positive) Let, d V d 0 Let >, 9 d (negative) So, at we ave a maimum vaue Maimum possie voume of o Sustitute in V V m Te maimum possie voume of te o is 8 m. Voume L W H V (0 ) (0 ) V ( ) V For a maimum voume d V d d So ± a a ( 0) ± 00 0 ± 8.. or 7. Sustituting. into te engt and wi of te o gives a negative wi. Te domain for tis question is 0 < < 0 and so. is disarded. Consider 7. for SP. Let < 7., d (positive) Let 7. d 0 zero Let > 7., d (negative) At 7. we ave a maimum point Dimensions of o are Lengt m Wi m Heigt 7. m Te maimum voume L W H m Te maimum voume is. m a Let eigt of o Lengt 0 Wi 0

13 Differentiation and appiations MB Qd- a Voume L W H V V But V If o is open at te top A area (ase side ak front) A A But A A 0 Dimensions of o for surfae area A to e a minimum: d A 0 d A 0 A 0 0 d 0 d Now d A 0 d 0 0 Mutipy troug y To verify a minimum vaue at 8: Let, < 8, 0 d (negative) Let 8 0 d Let > 8, d (positive) So at 8 we ave a minimum vaue. Dimensions of o are: Lengt 8 m Wi 8 m Heigt m d Te minimum area: sustitute 8 into A A 0 A A 8 A 9 m V L W H V V But V () Surfae area (S) S ase and top sides S () Sustitute () into () S 000 S 000 For minimum amount of seet meta used 0 d S 000 S d 000 d Now d S 0 d 000 So 0 Mutipy troug y Verify tat tis is a minimum point. Let < 0, d 8. d 0. (negative) Let 0 0 d Let > 0, 000 d (positive) At 0 we ave a minimum vaue. So dimensions for minimum surfae area are: Lengt 0 m Wi 0 m 000 Heigt m 00 0 Bo is a ue m Cost v Doar per our (distane 900 km) a Cost if v 00 km/ Cost per our (00) Time for journey 900 or ours 00 Cost for ours 00 $700 Cost in terms of v ost ost/r numer of ours C (00 00 v ) time time Distane Speed 900 v v So C (00 ) v C 9v v Te most eonomia speed and minimum ost ours wen dc 0. dv C 9v v C 0000v 9v dc dv 0000v v If d C 0 dv v or 9 v

14 MB Qd- Differentiation and appiations v v v ± 9 v ±00 But v annot e negative, so v 00 km/ Verify tat tis is a minimum. Let v < 00, say 00 dc dv (00) 9 7 (negative) Let v 00 dc dv 0 zero Let v > 00, v 00 dc dv (00) 9.7. (positive) At v 00 we ave a minimum vaue. Minimum speed 00 km/ Minimum ost: C 9v v $700 So, te most eonomia speed is 00 km/ and minimum ost is $700. Capter review ( 7) 7 8 Te answer is E 9 7 ( ) 0 Te answer is A ( ) ( )( ) ( ) ( ), 7 Te answer is B Read off grap wen 0, f() Te answer is A 7 Read off grap wen, f() Te answer is B 8 Undefined eause grap is disontinuous at. Te answer is C 9 a f() ( )( ) ( ), ( ) 0 a Wen 0, f(0) Wen f() 0, f ( ) is undefined eause 0 as 0, f() approaes ut as 0, f() approaes. If f() f ( ) f( ) 0 ( ) ( ) 0 0 ( ) 0 Te answer is A f ( ) f() f () 0 Te answer is B Gradient of tangent of f() at f ( ) f() 0 Te answer is D f() f ( ) f( ) 0 ( ) 0 0 f() f( ) ( ) ( ) f ( ) f( ) 0 0 ( ) 0 ( ) 0 If g() 8 7 g () Te answer is E

15 Differentiation and appiations MB Qd- 7 y 0 7 d 0 7 Te answer is D 8 f() ( )( ) f () Te answer is B 9 y 0 d 0 Te answer is A 0 y d Te answer is C y y d Te answer is A f() 7 f () 7 f () 7 Te answer is E y 7 d Te answer is B g() 7 8 a g () 7 8 g () 7 8 i g () g () 8 0 Wen g () ( 8)( ) 8 or Wen 8 y Te gradient is zero at (8, ) and (, ). y d a Parae to -ais, d y 0 d 0 Wen y Gradient is zero at (, 9 ) 8 Parae to te ine y. d wen y 0 Gradient at (, 0) g() g () 0 0 ( )( ) or Wen y 0 7 Wen y (, 7) and (, ) f() 7 f () 0 If f () 0 0 f () Te answer is E 7 v t 7t 0 Wen t v 7 0 Wen t v Average rate of ange ange in V ange in t 0 8 Te answer is D 8 (t) t 8t a Initia position t 0 (0) 0 0 (0) m Rate of ange of dispaement at any time (t) (t) t 8t (t) t 8 m/s v d t 8 Wen t v m/s d 0 t 8 wen v 0 t t 8t At t 8 8 m Veoity is zero wen t and m e t v 8 8 Negative veoity means it is moving eft f v 0 at t and so te partie anges diretion ere. (0) () It moves 8 m to te rigt in te first seonds. () 9 It moves from to 9, tat is, m in te tird seond. Tota distane 8 0 m 9 A y 7 Has oa maimum and minimum B y Loa minimum ony; quadrati C y 7 Point of infetion at 0 D y ( ) Point of infetion at E y Quadrati, minimum TP. Te answer is A 0 V t t, t [0, ] a Rate of ange d V t At t, d V itres/our Maimum voume d V 0 t 0 t So maimum voume ours after ours. P 80 m Let L engt of retange W wi of retange P perimeter of retange A area of retange

16 MB Qd- Differentiation and appiations Now P L W 80 (L W) or L W 0 so W 0 L and A LW Sustitute for W into A: A L (0 L) A 0L L a To find area of argest retange dl 0 0 L dl So 0 0 L L 0 If L 0 W 0 0 W 0 And A 0L L m So area of argest retange is 00 m. C 80 m Now C πr 80 πr R 80 π R.7 m A πr A π (.7) A A 09. m Yes a ire of irumferene 80 m woud ave a arger area (eing 09. m as against 00 m for te retange.) Modeing and proem soving Tota surfae area of a yinder is πr πr, were r is te radius of te yinder and is te eigt of te yinder. If te surfae area is 00π square units, ten 00π πr πr πr(r ) 00 π r π r 00 r r Voume of a yinder is auated as Area of ase eigt πr ( 00 r) r Voume πr 00 r r V(r) 00πr πr Maimum voume ours wen V (r) 0 V (r) 00π πr 0 π(00 r ) (Differene of two squares) r 0 (ignore te negative soution) Voume 00πr πr 000π 000π 000π ui units a y d At eginning of trai 0 (0 km) Gradient 0. At end of trai ( km) Gradient ( ) 0.8 Gradient Wen y ( ) y.0 Gradient 0 at (,.0) d Maimum eigt of pat.0 km e Minimum eigt wen 0 terefore y.8 km Let R radius of ire Now πr 00 πr 00 R 00 π R (0 ) π R 0 π Area of ire πr 0 π π (0 ) π π (0 ) π Tota area of two sapes A square ire πr (0 ) π π π ( π ) π ( π ) Now A π ( π ) A π π π ( π ) 00 d π π Let d A d 0 ( ) 00 0 π π π ( π ) 00 π π 00 ( π ) So and te tota area is a minimum. Sustitute into A. Minimum area 0 m g(0) and g() 0 Gradient is zero wen. It is negative on eft and side and negative on rigt and side of tis point, so it is a stationary point of infetion. Two points on grap as stated aove. Hyrid funtion 0 [0, ) f( ) ( ) [, ] were is te orizonta distane in metres from te start of te side. a Wen 0, f () 0 (0) 0 metres. Wen, f () ( ) ( ) 0. metres.

17 Differentiation and appiations MB Qd- 7 e For te straigt setion f() 0 f () gradient of straigt setion For te urved setion f() ( ) ( ) d Wen, f () ( ) ( ) metres Using Pytagoras teorem et L engt of te straigt setion. L 8 L 80 L 80 metres. 8 f () Wen, f () gradient of urved setion at join. Side is smoot at te join.