Chapter 16. 1) is a particular point on the graph of the function. 1. y, where x y 1

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1 Prctic qustions W now tht th prmtr p is dirctl rltd to th mplitud; thrfor, w cn find tht p. cos d [ sin ] sin sin Not: Evn though ou might not now how to find th prmtr in prt, it is lws dvisl to procd with prt nd ttmpt to writ th dfinit intgrl. () ; thrfor, point P hs th coordints (, ). V d d c d ( ) ( ) Solution Ppr tp d [ ] Solution Ppr tp Notic tht w chngd vril into vril, nd tht Solvr ccpts ll of th fturs from th clcultor s mnu with onl on vril prmtr to rprsntd s function in tht vril. W could hv usd th grphicl mod, ut, sinc thr is no mor thn on solution, w wr stisfid using Solvr. At th nd, w cn chc whthr th numricl rsult is spcil vlu which w could hv rcognizd. '. At th point (, ) th slop of th tngnt is: m '( ). Th tngnt cn found using th formul for th tngnt: f ' ( ) ( )+, whr (, ) is prticulr point on th grph of th function. ( )+ + Sinc th linr function hs no -intrcpt, it mns tht it psss through th origin. If w input th coordints of th origin into th qution, w gt tru sttmnt:. For th first trm, w nd to ppl th product rul. ( ) ' + + c Th shdd rgion cn split into two. Th first rgion is tringl oundd th tngnt lin, -is nd th vrticl lin. Sinc tht is right-ngld tringl, th r is clcultd s: A Tringl. In ordr to find th r of th scond rgion, w nd to vlut th following intgrl:

2 d ( ) + + Now, th totl r is th sum of thos two rs; thrfor, A +. Solution Ppr tp 5 i st () 8 + t t s () m ii v() t s' () t v() t 8t v() m s iii v() t 8t 8t t 8 s iv s () m Firstl, w nd to find th tim t which th p stops ftr touchdown: 5 v() t 8t t. 5 s 8 Now, w nd to find th distnc th p will trvl ftr touchdown: 5 s 5 5 s () m Using prt iv, th rmining runw gth is 5 m; thrfor, thr is nough runw to stop th p if it ms touchdown for point P. Solution Ppr tp 5 i nd ii iii nd iv To drw th function, w input th function into th clcultor nd thn us Tl to plot th points (or, ltrntivl, us th trc ftur).

3 Not: Prts nd cnnot solvd without using clcultor. c ( sin ) d cos + c, c Ar ( sin ) cos d cos cos ( cos )., corrct to thr significnt figurs. Othrwis: 7 Mthod I: From th dirction of th -is W split th shdd rgion into two: R, th rctngl nclosd th lins,, nd ; nd R, th rgion nclosd, nd +. Th r of th first rgion is: A ( ). To find th r of th scond rgion, w nd to do two things: firstl, w hv to find th point of intrsction of th lin nd th curv +. B inspction, w s tht th point of intrsction is,. Now, to l to us th dfinit intgrl, w nd to trnslt th grph vrticll units down. A + d d ( ) + + A A A Mthod II: From th dirction of th -is W nd to prss in trms of nd clcult th intgrl with rspct to th -vril. + A d d ( ) [ ] ( ) 8 i nd ii To drw th function, w input th function into th clcultor nd thn us Tl to plot th points (or, ltrntivl, us th trc ftur)..5.5 (, ) (.,.55) (.57, ) (, -.7)

4 cos, > cos c i S th digrm in prt. ii cos d d cos d sin + cos sin + sin cos sin ()+ () sin cos Or: ( sin() ) 7. f ( + ) sin ( + ) cos ( + ) sin ( ) cos ( ) f ( ), so th fundmntl priod of f is. B looing t th grph, w stimt tht th rng would [.,.]. c i f '( ) ( sin cos) cos + sin ( sin ) sin cos sin. W cn lso continu to prss th whol drivtiv in trms of sin onl: sin cos sin sin ( sin ) sin sin sin ii In this prolm, w us th first form of th drivtiv: f ' ( ) sin cos sin sin( cos sin ) Sinc th vlu of sin cnnot qul to t A, w cn conclud tht: cos sin cos cos cos co ( ) ( ) ( ) iii f sin cos m m m d f ( ) sin or cos, i sin ( ) cos ( ) d ( sin) d ( sin ) + cc ii f ( ) d sin sin ( ) s or or, so th -coordint of point B is. sin, sin ( ) f f " ( ) cos 7cos cos cos 7. Sinc th -coordint of C is lss thn, th scond fctor must qul to cos 7 cos cos rccos.

5 5 5 + cos c In this prolm, if w us Solvr, w nd to us n stimtd vlu tht is furthr to th right of p, which w found s th first zro. Our stimt ws. So, th nswr is. 7. d S th digrm ov. Ar ( + cos ) d Ar ( + ) cos d [ + sin + cos ] ( + ) ( + + ) 78.. Or, using GDC, w gt: So, th nswr, corrct to si significnt figurs, is 7.8. ( ) + i p g( ) f ( ) ( + ) + 5 ii p' ( s.f.) i + ( ) ( ) ii f ( ) f () 5 5 ( ) c V + d Bonus: Evlut th intgrl in prt c: V ( + ) d + + d + + ( )

6 It is not possil to solv this qustion with GDC. ( ( ) ) ( (( ) ( ) + ) d + V ( + ) + d d ) u + u + d ( u) u du u u du du d d du u u c cc, dv ds dv ( s ) ( s + ) s + 7s v ds dt ds ( s ) s ( + ) ( s ) s 7 ( + ) 5 ( ) 7 5 V ( ) d d ( ) 5 5 Solution Ppr tp + + d ( + ) ( ) or, sinc >. Solution Ppr tp W will us Solvr on GDC. W could hv sil usd th grphing mnu too, in vr similr mnnr. Not: Th vlu of, shown in th finl scrn, is not rlvnt nd it cn n vlu. 7 W cn dduc tht () t t +, v(). So, w cn now procd with finding th distnc trvlld th trin. v() t t + dt t t c c + +, v() c v() t t + t + t d t t + t m

7 8 Firstl, w nd to find th zros of th prol. ( ) ( + ), Th r of th rctngl is A R h, whr h is th hight of th rctngl. Th r undr th prol is clcultd th following intgrl. AP ( ) d + Sinc th two rs must qul, w cn find th hight of th rctngl: h h So, th dimnsions of th rctngl r:. f ( ), > f '( ) +, > If th function is incrsing, th first drivtiv is positiv; thrfor: + > > >,, + Th qustion ss us to find th intrvl ovr which f() is incrsing; thrfor, th vlu of is nd th intrvl is: >,, +. c i f '( ) + ii f ( ) ( ) or So, th othr -intrcpt is: d Sinc th curv is low th -is, w nd to t th solut vlu of th intgrl. ( ) d To solv th first prt of th intgrl, w nd to us intgrtion prts. u du d d d d + c dv d v ( ) d ( ) So, th r nclosd th curv nd th -is is:. A (, ), m f '( ) + Eqution of tngnt: ( )+ f Th -intrcpt is, so th r of th tringl nclosd th tngnt nd th coordint s is: A ( ), which is twic th r nclosd th curv. g,,,,... To vrif th sttmnt, w r going to t two conscutiv -intrcpts, for nd + : + +. Th rtio is constnt nd thrfor th zros form gomtric squnc. 7

8 d d d + d d + c d + + rctn, rctn + c c rctn tn( ) mv dv d mv dv d mv dv m v + cc d,, v v m v mv + c c m v mv mv v ( ) + mv v m v v v m v ( ) ^() ^() c Givn tht th point of intrsction hs n -coordint qul to p, w notic tht th rctngl hs dimnsions p nd th tringl hs vrticl s of gth nd hight p. p ATRIANGLE < AREGION < ARECTANGLE p < AREGION < p AREGION p < < d A d REGION p ( ( )) p. 7 (corrct to four dciml plcs). In cs li this, it is dvisl to stor th coordints of th point of intrsction in th GDC s mmor nd thn wor with this mor ccurt vlu. So, th r of th rgion is.7 (corrct to thr significnt figurs). Notic tht th lst two prts of th qustion cnnot don without using clcultor. u du d cos d sin dv cos d v sin sin d sin cos + c sin + cos + c, c 8

i cos d sin + cos + sin cos + sin cos 8 A Th vlu of th intgrl is ngtiv, ut, s w nd to clcult th r, w simpl t th solut vlu of th intgrl, sinc th function is lws ngtiv for th givn intgrl.

9 i cos d sin + cos + sin cos + sin cos 8 A Th vlu of th intgrl is ngtiv, ut, s w nd to clcult th r, w simpl t th solut vlu of th intgrl, sinc th function is lws ngtiv for th givn intgrl. 5 5 ii cos d sin + cos sin cos 8 + sin cos 7 iii 5 cos d sin + cos sin cos sin cos 8 A Agin, th intgrl ws ngtiv, so, for th r, w t th solut vlu. c Th rs nclosd th givn oundris form n rithmtic squnc with first trm u nd common diffrnc d. Thrfor, th sum of th first n trms is givn : nn n n n n Sn n ( ) nn ( + ), n + v() t t t sin t t t or, Using th rstrictd domin, w cn clcult th vlus of t : t or t or t. i In ordr to void discussion of th positiv or ngtiv vlus of th prts of th intgrl, w will simpl us th solut vlus. Totldistnctrvlld t dt ii So, th totl distnc trvlld is.5 m (corrct to thr significnt figurs). Not: If not using clcultor, w should split th intgrl into two prts, from to nd from to, whr th lst on hs ngtiv vlu nd w t its opposit vlu. Th nti-drivtiv cn found using intgrtion prts. 5 Distnc trvlld v() t dt dt + t rctn t rctn rctn rctn. 5 m

10 dv t () dt t ( + t ) d d u + + d udu u + c + du d + ( ) + cc, c + ( + ) d ( + ) ( + ) This prt cn solvd dirctl using GDC. 5 ( ) Agin, th vlu of is irrlvnt for this clcultion. To corrct, w nd to s tht th vlu of must within th domin of th function. In our cs, th domin is th st of ll rl numrs., distnc ( ) ( ) 7 v() t t t t t tdt t t 8 dt + t t dt ( t t ) + ( t t ) ( )+ ( + ) m Th curv is prol which opns upwrds, with th zros t nd ; thrfor, th function is ngtiv from to nd positiv from to. Th totl distnc trvlld is.85 m (corrct to thr significnt figurs). dt dt dt dt ( T ) dt dt T T T t + c T t c t, c T + A, A + + To find th constnts A nd, w nd to solv th simultnous qutions formd from th givn informtion. i T () + A A 78 A 78 A 78 5 T ( 5) 7 + A A

11 d d 8 t 8 t 5 5 ii T t 5 t t 5 t 5 d d d + rctn ( ) + cc, d +, rctn() + c c rctn( ) rctn( ) tn ( + ) u + ( u ) u u + u 8 d du d du u du u + u 8 u 8 ( + ) 8 du u+ u + + c ( + )+ + + u u u + + c, c 5 f() g() 5 i Th logrithmic function g ( ) ( + ) hs vrticl smptot:. ii -intrcpt: g ( ) (). -intrcpt: ( + ) ( + ) +. c d i Rfr to th digrm in prt. ii 5. (( ( ) ) ( + ) ) d

12 iii So, th shdd rgion hs n r of. (corrct to thr significnt figurs). Not: W wr l to stor th -coordint of th intrsction in th GDC s mmor sinc tht ws th lst clcultion don for w found th intgrl. To find th mimum distnc, w nd to form nw function: h( ) f ( ) g ( ). So, th mimum distnc twn f() nd g() is. (corrct to thr significnt figurs). θ d d θ d d d + d d d + d θ θ, d θ θ ( + ) d d ( + ) d d c c + ( + )+, c ( ) + d d + ( )+ c θ ( θ + )+ c, c ( ) + θ, + c c θ θ ( + )+ θ θ +

Integration Continued. Integration by Parts Solving Definite Integrals: Area Under a Curve Improper Integrals

Integration Continued. Integration by Parts Solving Definite Integrals: Area Under a Curve Improper Integrals Intgrtion Continud Intgrtion y Prts Solving Dinit Intgrls: Ar Undr Curv Impropr Intgrls Intgrtion y Prts Prticulrly usul whn you r trying to tk th intgrl o som unction tht is th product o n lgric prssion