Average Rate of Change (AROC) The average rate of change of y over an interval is equal to change in
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2 Averge Rte o Cnge AROC Te verge rte o cnge o y over n intervl is equl to b b y y cngein y cnge in. Emple: Find te verge rte o cnge o te unction wit rule 5 s cnges rom to & AROC Instntneous Rte o Cnge & st Principles I we look t te grp on te rigt, y nd wnted to clculte te rte o cnge t Point P, we ten clculte te grdient between P nd Q. I we bring te point Q closer nd closer to P ten te grdient will be pprocing te vlue o te tngent t P. PQ m I Q pproces P ten 0, te grdient pproces. Te instntneous rte o cnge o unction t point P on grp o y is equl to te grdient o te tngent to te grp t P. So, to ind te instntneous rte o cnge t point P, we evlute te derivtive o te unction t P. o Te instntneous rte o cnge o t is '. Emple: Find 0 lim or i lim ii 0 lim E9A,,, 4 LHS, 8 LHS
3 n Te derivtive o I n ten n ' n nd i n ten n ' n. Te derivtive o constnt I c ten ' 0. Emples: Find te derivtive o te ollowing:. y g y Remember to lwys to subtrct rom te power. Be creul wit + nd signs. Solutions. 8 d. Epnd irst: 6 tereore ' 8 d. 4g dg d 4. Simpliy irst: 6 7 tereore 6 d 5. y or 4 d d E9B,, 4, 5, 6
4 Te Grdient o Curve Te grdient o curve is not constnt. Te grdient o curve t certin point is equl to te grdient o te tngent to te curve t tt point. A tngent is line tt touces noter curve t one point only i.e.it does not cross it. T b T Te line T is tngent to te curve t point. Te line T is tngent to te curve t point b. Consider y d 4 8 nd its derivtive 6 Wt does ll tis men? y 4 8 is ormul tt gives te y-vlue o te curve t ny point. 6 is ormul tt gives te grdient o te curve t ny point. d Emple : Wt is te grdient o te curve Solution: 6 d t d d 80 6 y 4 8 t?
5 Emple : Wt re te co-ordintes o te points o te curve grdient is 4? Solution: 4 d Wen 6 4 or, y 4 Wen, y , 4 0, 7 y 4 8, were te E 9B 7,,,, 4, 6, 7 E 9C 4, 5, 6, 8, 0 Notes: {: > 0 } Mens: {: } Find te -vlues were Te grdient unction, > 0 tt is positive m = tn ɵ opp O tn dj A
6 Strictly incresing nd strictly decresing unctions A unction is sid to be strictly incresing wen < b implies <b or ll nd b in its domin. Te deinition does not require to be dierentible, or to ve non-zero derivtive, or ll elements o te domin. I unction is strictly incresing, ten it is one-to-one unction nd s n inverse tt is lso strictly incresing. I 0 or ll in te intervl ten te unction is strictly incresing. I 0 or ll in te intervl ten te unction is strictly decresing. Strictly Incresing Emple : Te unction : R R, is strictly incresing wit zero grdient t te origin. Te inverse unction : R R,, is lso strictly incresing, wit verticl tngent o undeined grdient t te origin. Emple : Te ybrid unction g wit domin [0, nd rule: 0 g is strictly incresing, nd is not dierentible t =. Emple : Consider : R R, H is not strictly incresing, But is strictly incresing over te intervl 0,.
7 Strictly Decresing A unction is sid to be strictly decresing wen < b implies b or ll nd b in its domin. A unction is sid to be strictly decresing over n intervl wen < b implies b or ll nd b in its intervl. Emple 4: Te unction : R R, e Te unction is strictly decresing over R. Emple 5: Te unction g : R R, g cos g is not strictly decresing. But g is strictly decresing over te intervl 0,. lso,, etc. nd E9B 8, 9, 0,,
8 y Sketcing te Grdient Function GRAPH OF THE ORIGINAL FUNCTION Were te grdient is lt i.e t ll sttionry points Were tere is positive grdient i.e. slopes Were tere is negtive grdient i.e. slopes Were te grdient gets ltter Were te grdient gets steeper At te steepest prt o ec section o te grp GRAPH OF THE GRADIENT FUNCTION Will cross te is Will be bove te is Will be below te is Gets closer to is Gets urter wy rom is Will ve pek Emple : Sketcing te Grdient Function y =
9 Emple : Sketc te grdient grp o: 4 y Solution:
10 Emple : Sketc te grdient unction o: y Solution: E9D cdei, cdegi,, 5, 6, 7
11 Cin Rule Te derivtive o unction n Te unction in unction rule or Composite Function rule. Emple : Find te derivtive o y 4 4. Solution: In words: ind te derivtive o te ting s wole, ten multiply it by te derivtive o te inside. 4 d [ 4 d d 8 ] In symbols: I y n u were u, ten d du du d Let u 4, 4 y u du d du 8, u 4 4 d 8 etc Emple : Find ' i 4 8. ' 8 ' Emple : I y ten ind. d y d d d E9E,,, 5 d g g'. ' g d Emple 4 d d. ' d d '. Emple 5
12 Dierentiting Rtionl Powers Emples: Find te derivtive o ec o te ollowing wit respect to or d d y y b cin rule E9F,,4, 6, 7
13 Derivtives o Trnscendentl unctions k Te derivtive o e k k In generl: I y e ten ke. I y e ten d Emple : Find te derivtives o: i y e ii iii y e y e 5 Solutions: i e d ii 5e 5 d iii e d d ' e Emple : Find ' given 4 e. Solution: Product Rule: ' e ' e HCF e e 4 4 E9G,,, 4, 5, 6
14 Derivtive o log e In generl, i y log e ten d ' I y log e ten ' d y log e log e, 0 I y log e, 0, 0, or R\{0}, 0 Emples: Find te derivtives o: i y log e ii y log e iii y log e Solution: i note: ny rule o te orm y log e k s derivtive o, 0 d ii, -, 0 d iii, 0 d E9H,,, 4, 5, 6, 7, 8
15 Derivtive o te Trigonometric Functions I y sink ten k cosk d I y cosk ten k sink d k or cos k y tnk ten k sec k d Emples: Find te derivtive o te ollowing: i y cos cos ii y sin iii y sin sin iv y tn v y cos Solution: i sin d ii cos d iii cossin d iv d 6 d.sec 6sec v sin E9I,,, 4, 5, 6 NOTE: Angle MUST be in RADIANS c o 80 e. g. sin d d o sin 80 o d o sin sin cos or cos d
16 Te Product Rule Te derivtive o te product o two unctions Emple : Find using te product rule i y. d Solution: In words: Te derivtive o te irst term multiplied by te second term, ADD te derivtive o te second term multiplied by te irst term. In symbols: I y u. v ten d du dv v. u. d d In te bove emple, u nd v du dv 6 nd d d 6 d d 8 d Emple : Find ' i 4. Solution: Let u du d d d d d 4 4 nd v 4 nd 4 4 dv d Let u e du e d e d e d.sin e nd v sin nd dv cos d. cos sin cos Emple : Find ' i e sin. Solution: E9J,,, 4, 5, 6, 7, 8
17 Quotient Rule Te derivtive o te quotient o two unctions Used wen you ve problem in rction orm. Emple : I y 4 7 ten ind d Solution: In words: Te derivtive o te top term, multiplied by te bottom term, subtrct te derivtive o te bottom term, multiplied by te top term, ll over te bottom term squred. du dv v. u. u In symbols: I y ten d d, v d v In te bove emple: u 4 top v 7 bottom du d dv d d d d Emple : I Solution: d y ten ind. d 4 e Emple : Find i y. d e Solution: d d e e. e e.e e e e e e e d e E9K ceg, de, 4, 5, 6, 7
18 Continuous unctions nd Dierentible Functions Te grp o continuous unction is one witout breks. It is usully smoot unbroken curve, owever it my ve srp corners. I te derivtive o unction eists t point on curve tis unction is sid to be dierentible t tis point. Te derivtive eists t point i it is possible to drw tngent t tt point. i.e. te curve must be smoot nd continuous. y y = -5-0 At, te grp is continuous nd dierentible. At, te grp is continuous BUT NOT dierentible. y At, te grp is neiter continuous or dierentible. -4 E9L,, 4 E9M,,, 5 Note: No derivtive eists t: CUSP point END-POINT open or closed HOLE point
19 Finding te Eqution o Tngent nd Norml y= y P,y P, y is point on te curve y. Te Norml nd Tngent re t rigt ngles to ec oter. Norml t P I m = grdient o te curve t P, ten te grdient o te tngent t Point P = m Te grdient o te norml t point P is Tngent t P m Te eqution o te Tngent is: y y m. Te eqution o te Norml is: y y. m How would you ind m i you knew te eqution o te curve? Find nd substitute te - coordinte o P into it. d Emple : Find te eqution o te tngent nd o te norml to te curve point 0,. y 8 8 Solution: 9 8 d At 0, , so mtngent t = 0 = 8 & mnorml t = 0 = d 8 Eqution o te tngent: y 8 c 0, 80 c c y 8 9 t te Eqution o te norml: y c 8 0, 0 c 8 0 c y 8 E0A,,, 6, 8c, 9bc, 4, 6
20 Rtes o Cnge Wt is rte? I you work nd ern $ n our your rte o py = $ per our = $/r. Tis is linked wit clculus by I P = totl Py $ & t = time worked r Te P t dp & dt dp rte o cnge o P wit respect to t. dt dp $ For te unit o, $ per our,. dt r I you ve to ind Te rte o cnge o volume wit respect to te rdius Te rte o increse o cost o production o dolls w.r.t te number o dolls Te rte o cnge o circumerence w.r.t eigt Te rte o decrese o mount o wter in drining tnk Coose letters or te vribles V= volume r = rdius C = cost n = no. o dolls C = circumerence = eigt V = volume t = time Te rte you need is dv dr dc dn dc d dv dt So you ll need n eqution relting V nd r C nd n C nd V nd t Unit o rte is cm cm $ doll mm mm m min In te lst cse, wt s missing? w.r.t nd vrible ssumes it is time. Solving rte problem is very, very similr to solving m/min prob.. need wt rte? no second vrible ssume time. ind ormul.. ormul must be in terms o one vrible only, i not reltionsip between te vribles by oter ino. From question. 4. ind te rte. 5. substitute given vlue o second vrible, include units 6. nswer ll questions. I rte is positive, it is incresing, i te rte is negtive, it is decresing.
21 Emple : A spericl blloon is being inlted. Find te rte o increse o volume wit respect to te rdius wen te rdius is 0cm. Solution: dv. dr. 4 V r. dv 4. 4r dr 5. wen r = 0 cm dv cm / cm dr 6. volume o te spere is incresing t rte o 400 cm / cm. Emple : Te mount o wter in tnk A litres t ny time seconds is given by te rte o cnge o A wen Solution: da. need dt. A t. A t t 5s. da 4. t dt da dt t 5. wen t 5 da l / s dt A is cnging t rte o l / s 5 OR A is decresing t rte o l / s 5, wen t 5, wen t 5 A. Find t
22 Emple : A blloon develops microscopic lek. It s volume V cm t time, t s is: t V 600 0t, t 0 00 i At wt rte is te volume cnging wen t = 0 seconds? ii Wt is te verge rte o cnge o volume in te irst 0 seconds? iii Wt is te verge rte o cnge o volume in te time intervl rom t 0 to t 0 seconds? Solution: i need dv t t 0 dt dv t t 0 0 dt dv 0 t t 0, 0 0. cm dt 50 i.e. te volume is decresing t rte o 0. / s cm / s. ii verge rte o cnge o V: V t V t t 0, V 600 nd 0, V A. R. O. C 0. cm / s Te volume is decresing t n verge rte o 0. iii verge rte o cnge o V: t 0, V 499 nd t V t V t A. R. O. C 0. cm / s Te volume is decresing t n verge rte o 0. t 0, V 96 cm / s. cm / s. Prt i bove is n INSTANTANEOUS rte o cnge, Prt ii & iii is n AVERAGE rte o cnge, i.e. nd verge o number o instntneous rtes.
23 Prticulr Cse Displcement Velocity Accelertion Symbol Units Deinition Displcement, t, st, d m, km, Te distnce rom ied point O Velocity Accelertion d ds m/s, ms -, v,, dt dt km/, dv d m/s, ms -,,, dt dt km/ Te rte o cnge o displcement Te rte o cnge o velocity Originl displcement/velocity/ccelertion occurs t t = 0 NOTE: I you were sked to ind te verge rte o velocity, it would be done s n verge rte o cnge i.e. using te displcement vlues not te velocity vlues. I te velocity vlues t t were used ten you get te verge ccelertion! E0B,, 4, 8, 0,,
24 Finding te Sttionry Points o Curve Emple : Sketc te grp o nd determine te coordintes o ll turning points d.p.. Solution:.Find nd y intercepts: X-Int y = 0 Y-Int = 0 0,, Sttionry Points 0: d 5 6 ' ' ,.54 qudrtic ormul y , 6.06 &.54, Type o Sttionry Points: o Locl Minimum o Locl Mimum; o Point o Inlection Emple : Using te bove emple, determine te nture o te Turning Points. Solution: 0 ~0. ~.54 ' Slope / - \ - / Nture o T.P Locl Mimum Locl Minimum Consider te grps: y y y Grps similr but dierent number o sttionry points. E0C LHS,,, 5, 7, 0; E0D ce, d, 4, 0,,, 7, 8,, 4, 5, 6 NOTE: nd derivtives cn be used, < = M, > 0 = Min, = 0 = inconclusive e.g. rom bove = 6 8, 0 = -8, 0. = -ve, = - nd = 0
25 Mim/Minim Problems Solving mimum/minimum problem SETP : Need to Mimise/minimise wt? Cll it A STEP : Write te ormul or A =, mking up vribles were necessry, mybe digrm could elp. STEP : Cn you write noter eqution? A = must be written s A = wit only vrible on te Rigt Hnd Side. da da STEP 4: Dierentite A = i.e. nd equte to zero, 0 d d nd solve or. STEP 5: Test or te type o sttionry point obtined. da d Are ny nswers impossible i.e. negtive lengt STEP 6: Answer te question in words. Emple : Four squre corners re removed rom seet o crd o dimensions cm by 0 cm. Te seet is olded to orm n open rectngulr continer. Find te dimensions to d.p., suc tt te totl volume o te continer is mimum. Solution: using te 6 steps rom te potocopy seet.. Need to mimise te volume o te continer, V.. H cm H 0 cm L W V = L W H. Rigt Hnd side s vribles must be in terms o only vrible L = 0 H W = H V = 0 H HH
26 4. Mimise let te derivtive = 0 i.e. 0. d Epnd V = V 60 60H 4H 4H H V 60 0 H 4H V 60H 0H dv 60 04H H dh 605 4H H 0 H 4H 05 0 Qud. Formul... H 4H 7 79 H, H 4., H.9 V= 0 H HH Must ignore H =.9 wy? int: wt is te implied domin o H? b H = 4. cm H dv dh / - \ Tereore locl mimum L 0 H W H 7 L 0 L.9cm W W.9cm Te mimum volume is obtined wit te dimensions: H = 4. cm L =.9 cm & W =.9 cm Te mimum/minimum vlue o unction DOES NOT NECESSARILY OCCUR AT A TURNING POINT. It depends on te esible Domin cused by te Pysicl constrints. E0F,,, 4, 6, 7,, 4, 6, 7 Workseet
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28 Absolute Mimum/Minimum Problems Emple: Let A be te unction tt models te totl enclosed re wen 00 cm piece o wire is cut into two pieces, were one piece is used to orm te perimeter o squre, nd te oter piece is used to orm te circumerence o circle. 00 Sow tt A cn be modelled by, A: 0,00 R, A, were 6 4 cm is te lengt o te piece o wire used to orm te perimeter o te squre. b For wt vlues o, is A is mimum nd minimum? c Wt is te minimum re? Solution: Squre: Circle : C r 00 r r A r 4 00 A : 0,00 R, A A', cm minimum, A 50.cm Mimumt 0, A 795.8cm E0E,, 5, 7, 9, 0,,, 4, 6 Clcultor use: m, 0 00
29 Fmilies o unctions Emple : Consider te mily o unctions o te orm b, were nd b re positive constnts wit b >. Find te derivtive o wit respect to. b Find te coordintes o te sttionry points o te grp o y. c Sow tt te sttionry point t, 0 is lwys locl minimum. d Find te vlues o nd b i te sttionry points occur were = nd = 4. Solution: b b b b 7 4, 7 4,0 0, 0 b b b b b b c look t grp, <, 0, nd i 0, ten b d Since < b, we must ve = nd 9 4 b b Emple : Te grp o y, is trnslted by units in te positive direction o te -is nd b units in te positive direction o te y-is were nd b re positive constnts. Find te coordintes o te turning points o te grp y. b Find te coordintes o te sttionry points o its imge. Solution: 4, 4, 0,0 0 0, 0, y y d
30 b Te turning points o te imge re:, b nd +, b - 4. E0G,,, 5, 7, 9
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