b. moving away from the origin and towards its starting position

Transcription

1 Review Eercise, pp f 5, f d. d 77. v t t, t. vt 5t, t t 5. The upwrd velocit is positive for t.5 s, ero for t.5 s, nd negtive for t 7.5 s. Velocit (m/s) v(t) Time.. min: 5, m: min: 5, m: min:, m: 7.. m Yes, m eond the stop sign Stop signs re locted two or more metres from n intersection. Since the cr onl went m eond the stop sign, it is unlikel the cr would hit nother vehicle trvelling perpendiculr.. min is, m is i. $ ii. $5.5 iii. $.; $. i. $ ii. $. iii. $.; $. i. $5 ii. $.55 iii. $.; $. i. $75 ii. $.7 iii. $.99; $ moving w from its strting point t moving w from the origin nd towrds its strting position.. t Yes, since 7 for ll t 7, the prticle is ccelerting. 7. cm 7. cm for the se nd height.57 cm 5. length 9 m, width pproimtel m.. m. m. m 7. rdius. cm, height. cm. Run the pipe 7. km long the river shore nd then cross digonll to the refiner. 9. :5 p.m.. $ or $. The pipeline meets the shore t point C, 5.7 km from point A, directl cross from P...5 cm 7. cm. The two identicl rick sides should hve length 5 m; the fenced side nd the corresponding rick side should hve length m.. : p.m. 5.. km from point C.. solute mimum: f 7, solute minimum: f 5 solute mimum: f, solute minimum: f solute mimum: f 5 7, solute minimum: f 5 solute mimum: f 75, solute minimum: f m. m>s.7 s.. f f 5 f f f 9 f position:, velocit:, ccelertion: Q speed: R, position:, velocit:, 9 ccelertion:, speed: vt t t t, t 9 t t t t.9 m>s. m>s undefined. m>s Chpter Test, p... f 5 f 9.. v 57, v,.. vt t, t.5 m m>s etween t s nd t.5 s m> s.. min:, m: 7 min:, m: 5... s out.9 m. 5 m.7 m 7. mm mm 9 mm. $5>month Chpter Review of Prerequisite Skills, pp... or 7 or 5 or or.. 7 t or 7.. Answers

2 t t 7 As S q, f S q. As Sq, f S q. As S q, f S q. As Sq, f Sq... ; ; no verticl smptote ; ;.... i. no -intercept; (, 5) ii. (, ); (, ) iii. iv. Q ; no -intercept 5, R i. Domin: 5R, Rnge: 5R ii. Domin: 5R, Rnge: 5R iii. Domin: e R, f Rnge: e R f iv. Domin: 5R, Rnge: 5R Section., pp (, ), (, ) 5, ;, 5, incresing: ; decresing:, 7 incresing:, 7 ; decresing:, incresing: 7 ; decresing: 5. incresing:, 7 ; decresing:,. 7.., 9, c i. ii. iii. 5 (, ) 7 (, 5) 5 7.,,.5,.9.5,.,, 5,,.. If f n, where n is rel numer, then f n n.,, 5 If where k is constnt, f k,. A function is incresing when then f. f 7 nd is decresing when If k f g, then f. 5 k f g f g.. i., 7 If h f then ii. g, iii.,,, f g f g h i. i., 7 g, ii., 7 g. ii. iii.,, (, ) If f nd g re functions tht hve iii., i. derivtives, then the composite ii., function h f g hs iii. none derivtive given i., h f g # g. ii., If u is function of, nd n is iii. (, ) positive integer, then.. incresing:, 7 ; d decresing: d un nu ndu d. incresing:, 7 ; 9.. As S;q, f Sq. decresing: incresing:, 7 ; As S q, f Sq. decresing:, As Sq, f S q. Answers ( 5, ) (, )

3 i. ii. iii. i. ii. iii. 5, 7, 7. f c f Let f, then. If f, therefore the, function is decresing. If 7 f 7, therefore the, function is incresing.. f for, incresing: 7, decresing:, locl minimum:,. 5. Let f nd u g. Let nd e n two vlues in the intervl so tht. Since, oth functions re incresing: f 7 f () g 7 g () u f # g 5 results in f # g 7 f g The function u or f # g is strictl incresing.. strictl decresing Section., pp. 7. Determining the points on the grph of the function for which the derivtive of the function t the -coordinte is or is undefine.. Tke the derivtive of the function. Set the derivtive equl to. Solve for. Evlute the originl function for the vlues of. The (, ) pirs re the criticl points.,,,.. locl minim:,,,, locl mimum:, locl mimum:,., locl minimum: (,.) locl minimum:, 5, locl mimum: (, )..,,,,, f() g(),.,, (, ) h(t) locl minimum: (, ), locl mimum: (, 7), Tngent is prllel to the horiontl is for oth. (, ) neither mimum nor minimum, Tngent is prllel to the horiontl is. (5, ); neither mimum nor minimum, Tngent is not prllel to the horiontl is. locl minimum:,, Tngent is prllel to the horiontl is., nd (, ) re neither mim or minim. Tngent is not prllel to the horiontl is for either... t Answers

4 , locl mimum,, 5 locl minimum.. c 9,, p, q minimum; the derivtive is negtive to the left nd positive to the right.. k k k 7..,, c, d. f'() no criticl points f'() 7.., locl mimum, locl mimum,, locl minimum, locl minimum (, ) neither mimum nor minimum,, locl minimum. locl minim t nd ; locl mimum t 9. f() (, ) 5 (, ) f'() f'() 5..,, c, d 9, 9 locl minimum:, 7 nd, 9, locl mimum:, 9.. locl mimum: (, ) Answers

5 7. locl minimum:., 9., locl mimum:., 9. h f g Since f hs locl mimum t c, then f 7 for c nd f for 7 Since g hs locl minimum t c, then g for c nd g 7 for 7 h f g f g g f h g If c, f 7 nd g, then h 7. If 7 c, f nd g 7, then h. Since for c, h 7 nd for 7 c, h. Therefore, h hs locl mimum t Section., pp verticl smptotes t nd ; horiontl smptote t verticl smptote t ; horiontl smptote t. f g h Conditions for verticl smptote: h must hve t lest one solution s, nd lim f q. Sq Conditions for horiontl smptote: lim f k, where kr, or Sq lim f k, where kr. S q Condition for n olique smptote: The highest power of g must e one more thn the highest power of k q, q.. 5; lrge nd positive to left of smptote, lrge nd negtive to right of smptote ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote ; lrge nd positive to left of smptote, lrge nd positive to right of smptote hole t, no verticl smptote ; lrge nd positive to left of smptote, lrge nd negtive to right of smptote ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote ; lrge nd positive to left of smptote, lrge nd negtive to right of smptote ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote 5.. ; lrge negtive: pproches from ove, lrge positive: pproches from elow ; lrge negtive: pproches from elow, lrge positive: pproches from ove ; lrge negtive: pproches from ove, lrge positive: pproches from ove no horiontl smptotes lrge negtive: pproches from elow, lrge positive: pproches from ove lrge negtive: pproches from ove, lrge positive: pproches from elow 9.. 5; lrge nd positive to left of smptote, lrge nd negtive to right of smptote ; lrge nd positive to left of smptote, lrge nd positive to right of smptote ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote.. f() Answers 5

6 h(t) s(t) t t.. c d c.... = f''() = f''() 5. g hs verticl smptote t. As S, g Sq. As S, g S q. is n olique smptote for h. r hs verticl smptotes t nd. As S, r Sq. As S, r S q. As S, r S q. As S, r Sq. r hs horiontl smptote t. 9 5, 5.. lim lim Sq Sq q lim Sq lim Sq lim Sq q lim c Sq d lim Sq lim Sq g() 7. lim Sq f () s(t) t.. f nd r: lim f nd lim r Sq Sq eist. h: the highest degree of in the numertor is ectl one degree higher thn the highest degree of in the denomintor. h: the denomintor is defined for ll R. f hs verticl 7 Asmptotes t 7 nd. As S, f S q. As S, f Sq. As S 7, f Sq. As S 7, f S q. f hs horiontl smptote t. Mid-Chpter Review, pp decresing: q,, incresing:, q decresing: (, ), incresing: q,,, q incresing: q,,, q decresing: q,, incresing:, q Answers

7 . incresing: nd 7, decresing:. (, ) (, 5).. f is incresing. f is decresing... incresing: t.97, decresing: t 7.97 The velocit is lws decresing.. incresing: t 7.59, decresing: t iii. locl mimum t, locl minimum t, iv... ;, ; ;, ; ; 5.. locl mimum, locl minimum locl mimum, locl minimum. 7. ; locl minimum, incresing: 7, decresing:.. ; lrge nd positive to left of smptote, lrge nd negtive to right of smptote ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote ; lrge nd positive to left of smptote, lrge nd negtive to right of smptote ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote lrge nd positive to left of ; smptote, lrge nd negtive to right of smptote 5; lrge nd positive to left of smptote, lrge nd negtive to right of smptote 9.. ; lrge negtive: pproches from ove, lrge positive: pproches from elow ; lrge negtive: pproches from elow, lrge positive: pproches from ove.. 5; lrge nd positive to left of smptote, lrge nd positive to right of smptote no discontinuities ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote 5.. i. 7 ii. incresing: 7 7, decresing: 7 iii. locl minimum t 7 iv. 5 i., ii. incresing:, decresing:, 7 iii. locl minimum t, locl mimum t iv i.,, ii. incresing:, 7 ; decresing:, i., ii. incresing:, 7 ; decresing: iii. locl mimum t, locl minimum t iv... ; lrge nd positive to left of smptote, lrge nd negtive to right of smptote; ; As Sq, the function pproches the horiontl smptote from ov As Sq, the function pproches the horiontl smptote from elow. ; lrge nd positive to left of smptote, lrge nd positive to right of smptote; ; As Sq, the function pproches the horiontl smptote from ov As Sq, the function pproches the horiontl smptote from elow. ; lrge nd positive to left of smptote, lrge nd negtive to right of smptote; ; As Sq, the function pproches the horiontl smptote from ov As Sq, the function pproches the horiontl smptote from elow. ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote Answers 7

8 7.. q g. h. q Section., pp. 5.. A: negtive, B: negtive, C: positive, D: positive A: negtive, B: negtive, C: positive, D: negtive.. locl minimum: 5, 9, locl mimum:, locl mimum: Q, 5 R locl mimum:,, locl minimum: (, ) (, ) is neither locl mimum or minimum..., 5 Q, Q, 5 R R no points of inflection (, ).. ; ove ; ove 9 elow ; elow 7 ; 5.. i. concve up on, concve down on 7 ii. iii. i. concve up on or 7, concve down on ii. nd iii. = f(). For n function f, find the criticl points, i., the vlues of such tht f or f does not eist. Evlute f for ech criticl vlu If the vlue of the second derivtive t = f() criticl point is positive, the point is locl minimum. If the vlue of the second derivtive t criticl point is negtive, the point is locl mimum. 7. Use the first derivtive test or the second derivtive test to determine the tpe of criticl points tht m e present... i.,, (, ) ii. 9. i. Q,, 9 R ii.., 9, c (, ) (, 5) 7 f f f For possile points of inflection, we solve f : Q, 9 R or. The grph of f is prol with -intercepts nd. We know the vlues of f hve opposite signs when pssing through root. Thus, t nd t the concvit chnges s, the grph goes through these points. Thus, f hs points of inflection t nd. To find the -intercepts, we solve f or The point midw etween the -intercepts hs -coordinte. The points of inflection re (, ) nd Q, R... Answers m vr. For emple, there is section of the grph tht lies etween the two sections of the grph tht pproches the smptot. If n is odd ut not, then there is n inflection point t If n is even, there is not n inflection point. Section.5, pp.. A cuic polnomil tht hs locl minimum must lso hve locl mimum. If the locl minimum is to the left of the locl mimum, then f S q s S q nd f S q s S q. If the locl minimum is to the right of the locl mimum, then f S q s S q nd f S q s S q.. A polnomil of degree three hs t most two locl etremes. A polnomil of degree four hs t most three locl etremes. Since ech locl mimum nd minimum of function corresponds Answers

9 to ero of its derivtive, the numer of eros of the derivtive is the mimum numer of locl etreme vlues tht the function cn hv For polnomil of degree n, the derivtive hs degree n, so it hs t most n eros, nd thus t most n locl etremes... or no verticl smptotes.. 5 (5, 5) 5 (, 7) (, ) j (, 57) g. 5 (, ) h..... (, ) (, 5) i. (, ) (, 5) (, ) Answers 9

10 . g. h. i. j., c, d, 7.. Answers m vr. For emple:. 9. Answers m vr. For emple: (,.).. is horiontl smptote to the right-hnd rnch of the grph. is horiontl smptote to the left-hnd rnch of the grph. nd re horiontl smptotes.. c d d d c d d For possile points of inflection, we d solve d : d The sign of chnges s goes from d vlues less thn to vlues greter thn Thus, there is point of. inflection t. At, d d c c Review Eercise, pp. 9.. i. ii. 7 iii. (, ) i.,, 7.5 ii.,.5 iii.,,.5,. No, counter emple is sufficient to justif the conclusion. The function f is lws incresing, et the grph is concve down for nd concve up for 7... (, ), locl minimum; tngent is horiontl (, 7), locl mimum; tngent is horiontl (, ), locl mimum; tngent is horiontl (, ), neither locl mimum nor minimum; tngent is horiontl Q,, locl minimum; R Q7,, locl mimum; tngents t R oth points re prllel (, ), neither locl mimum nor minimum; tngent is not horiontl.., 7 e c, d e c d 5.. ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote 5; lrge nd positive to left of smptote, lrge nd negtive to right of smptote hole t ; lrge nd positive to left of smptote, lrge nd negtive to right of smptote 5; lrge nd negtive to left of smptote, lrge nd positive to right of smptote. (, 5); Since the derivtive is t, the tngent line is prllel to the -is t tht point. Becuse the derivtive is lws positive, the function is lws incresing nd, therefore, must cross the tngent line insted of just touching it. 5 Answers

11 7. (, ).. (, ) (, ).. i. concve up: ; concve down:, ii. points of inflection:, iii. (, 9) (, 9) (., 5.) (.,.) i. concve up:.5, 5 concve down:.5, 5 ii. points of inflection:.5,, nd 5 iii... k, ; k There re three different grphs tht result for vlues of k chosen. k k 5 9.., Answers 5

12 For ll other vlues of k, the grph will e similr to the grph elow....,, ; incresing:, 7 ; decresing:,. locl mimum:.7, 7.5, locl minimum: (.7,.), solute mimum: (,.5), solute minimum:, p: olique smptote,.75 q: verticl smptotes t nd ; horiontl smptote t r: verticl smptotes t nd ; horiontl smptote t s: verticl smptote t Domin: 5R ; -intercept: ; verticl smptote: ; lrge nd negtive to the left of the smptote, lrge nd positive to the right of the smptote; no horiontl or olique smptote; incresing: 7.59; decresing:,.59; concve up:, 7 ; concve down: ; locl minimum t (.59, 7.5); point of inflection t,. If f is incresing, then f 7. From the grph of f, f 7 for 7. If f is decresing, then f. From the grph of f, f for. At sttionr point,. From the grph, the ero for f occurs t. At, f chnges from negtive to positve, so f hs locl minimum point ther If the grph of f is concve up, then f is positiv From the slope of f, the grph of f is concve up for... If the grph of f is concve down, then f is negtiv From the slope of f, the grph of f is concve down for. nd 7.. Grphs will vr slightl. 9. domin: 5R ; -intercept nd -intercept:, ; verticl smptote: ; lrge nd positive on either side of the smptote; horiontl smptote: ; incresing: ; decresing:, 7 ; concve down: ; concve up:, 7 ; locl minimum t,.5; point of inflection:,... Grph A is f, grph C is f, nd grph B is f. We know this ecuse when ou tke the derivtive, the degree of the denomintor increses on Grph A hs squred term in the denomintor, grph C hs cuic term in the denomintor, nd grph B hs term to the power of four in the denomintor. Grph F is f, grph E is f nd grph D is f. We know this ecuse the degree of the denomintor increses one degree when the derivtive is tken. Chpter Test, p... 9 or or or 7 9 or or 9,,,, (, ),,, f 7 or g.,,,.. or or Q locl mimum, 7 R: Q locl mimum, 5 R:, 5: locl minimum. (, 7) (, ) (, ) 5 Answers

13 . hole t ; verticl smptote t ; lrge nd negtive to left of smptote, lrge nd positive to right of smptote; ; Domin: 5R, 5.. There re discontinuities t nd. 7. lim f q S is verticl f q smptot lim S lim f q S is verticl f q smptot lim S The -intercept is nd -intercept 9 is 5. Q9, is locl minimum nd 9 R, is locl mimum. is horiontl smptot, c (, ) (, ) Chpter 5 Review of Prerequisite Skills, pp log 5 5 log log log 5 w log log T.. -intercept:, no -intercept.. r r 5.. p p p p 5p p g. p p h cos u, tn u 5 sin u 5, sin u, 5 cos u 5 cos u, tn u is undefined.. period: p, mplitude: period: p, mplitude: period:, mplitude: p period:, mplitude: 7 period: p, mplitude: 5 period: p, mplitude: 9.. tn u 5 p p p.. tn cot sec csc LS tn cot sin cos cos sin p p p 5p p 7p p sin cos cos sin cos sin RS sec csc cos sin cos sin Therefore, tn cot sec csc. p Answers 5

14 sin sin tn sec LS sin sin sin cos RS tn sec sin cos cos sin cos sin Therefore, sin tn sec... 5p p, 5p p, Section 5., pp.. You cn onl use the power rule when the term contining vriles is in the se of the eponentil epression. In the cse of e, the eponent contins vril.... e t 5 e e t e 5 e e e e e e e t e e te t e t e t.. e e e e 5... The nswers gree ver well; the clcultor does not show slope of ectl.5, due to internl rounding. e 7. e. (, ) nd, e 9. If 5 then e 5 e 5, 5 nd 5 e 5 5 e 5, 5 5 e 5 5 e 5 5 c 5 e 5 e 5 d 5.. d, d e d 9e, d d d 7e d n d n n n e.. d, d e d d e d, d e d d e e d, d e d d e.. e t 7 cteri> h t time t The numer of cteri is constntl decresing s time psses... Q e t R dv dt et e t From, v Q e t R, which gives e t v Thus,. v v. m/s out s, out 7. m.. i. e ii. e The limits hve the sme vlue ecuse s S q, S. 5.. e. m or m d 7.. d sinh d d c e e d cosh d d cosh e e sinh Since tnh sinh cosh, d sinh cosh d d tnh d cosh cosh e e e e e e sinh d d cosh cosh e e e e cosh e e cosh e e cosh cosh cosh.. Four terms:. Five terms:.7 Si terms:.7 Seven terms: Answers

15 The epression for e in prt is specil cse of e!!!! p. in tht it is the cse when. Then e e e is in fct e e!!!! 5! p. The vlue of is. Section 5., p... ln ln.. t5 ln n 5nn ln ln ln ln 5 5 ln t t t ln t ln ln out. ers out 9.% > er 7.. In 97, the rte of increse of det pments ws $9 7 > nnum compred to $ 5 > nnum in 9. The rte of increse for 97 is 7. times lrger thn tht for 9. The rte of increse for 99 is 7. times lrger thn tht for 9. Answers m vr. For emple, dt from the pst re not necessril good indictors of wht will hppen in the futur Interest rtes chnge, orrowing m decrese, principl m e pid off erl.. 9. v(t) From the grph, the vlues of vt quickl rise in the rnge of out t 5. The slope for these vlues is positive nd steep. Then s the grph ners t, the steepness of the slope decreses nd seems to get ver close to. One cn reson tht the cr quickl ccelertes for the first units of tim Then, it seems to mintin constnt ccelertion for the rest of the tim To verif this, one could differentite nd look t vlues where v t is incresing. Section 5., pp solute m: out.9, solute min: solute m: out., solute min: out f : m:.9, min: ; m: m: out, min: out 59 The grphing pproch seems to e esier to use for the functions. It is quicker nd it gives the grphs of the functions in good viewing rectngl The onl prolem m come in the second function, m, ecuse for.5, the function quickl pproches vlues in the negtive thousnds... 5 squirrels squirrels (5.9, ) P(t) P grows eponentill until the point of inflection, then the growth rte decreses nd the curve ecomes concve down. t t.. items 5 items 5. 5 units. 7.%;. h 7.. C(t) Cpitl investment from U.S. sources ($ million) Yers since 7 The growth rte of cpitl investment grew from million dollrs per er in 97 to. illion dollrs per er in % C dollrs, dc.9 9 dollrs> er dt Sttistics Cnd dt shows the ctul mount of U.S. investment in 977 ws.5 9 dollrs. The error in the model is.5%. C dollrs, dc.975 dollrs> er dt ;. min fter the drug ws introduced.7 min fter the drug ws introduced 9. h of stud should e ssigned to the first em nd h of stud for the second em.. Use the lgorithm for finding etreme vlues. First, find the derivte f. Then find n criticl points setting f nd solving for. Also find the vlues of for which f is undefine Together these re the criticl vlues. Now evlute f for the criticl vlues nd the endpoints nd. The highest vlue will e the solute mimum on the intervl, nd the lowest vlue will e the solute minimum on the intervl... f is incresing on the intervls q, nd, q. Also, f is decresing on the intervl,. t Answers 55

16 .. no mimum or minimum vlue min: e., no m min: e.7, no m no m or min. out... d(t) incresing when t nd ln decresing when t 7 ln out.5 > s t 7 s t 5. The solution strts in similr w to tht of question 9. The effectiveness function is Et.5Q te t R The derivtive simplifies to E t.5e t t This epression is ver difficult to solve nlticll. B clcultion on grphing clcultor, we cn determine tht the mimum effectiveness occurs when t. h... P Numer of cells (thousnds).q9 5 te 5t R.e 5t 5 t Ds fter. ds, 5 The rte of growth is slowing down s the colon is getting closer to its limiting vlu Mid-Chpter Review, pp e e 7 e e e e e e e.. 5e 5t.5... e, e e e, e e e e, e e 5.. ln 5.ln. ln 9ln 55 ln ln Qe t R decresing out 5 rits/month 55 t The grph is constntl decresing. The -intercept is (, 55). Rit popultions normll grow eponentill, ut this popultion is shrinking eponentill. Perhps lrge numer of rit predtors, such s snkes, recentl egn to pper in the forest. A lrge numer of predtors would quickl shrink the rit popultion. 7. t out. h. The originl function represents growth when ck 7, mening tht c nd k must hve the sme sign. The originl function represents dec when c nd k hve opposite signs mm Hg. mm Hg.7 mm Hg. A e. ; 5% per er. f e f e e e So, e This mens tht the function is incresing when At. t A t. t ln. A $5.7, A 5 $77.9, A $.5 No 5 Answers

17 A ln., A A 5 ln., A5 A ln. A All the rtios re equivlent (the equl ln., which is out.5 7), which mens tht A t is constnt. At 5. ce ce e ce ce Section 5., pp cos sin cos sin g. cos sin ln cos sin e cos e h. 9 cos p i. sin j... cos sin cos sin sin cos cos e cos cos sin sin cos.. cos p p p p.. One could esil find f nd g to see tht the oth equl sin cos. However, it is esier to notice fundmentl trigonometric identit. It is known tht sin cos. So, sin cos. Therefore, f is in fct equl to g. So, ecuse f g, f g. f nd g re ech others negtiv Tht is, f sin cos, while g sin cos. sin t cos t 5.. v t t sin t sin tcos t v t cos t sin t h sin sin cos sin sin cos sin sin cos m cos sin cos.. solute m:, solute min: solute m:., solute min: 5. solute m:, solute min: solute m: 5, solute min: 5 p 7.. t p pk for positive pk, integers k.. f () p, p p 9. csc csc cot, sec sec tn. p.. t p pk, pk for positive integers k minimum: mimum:. u p. u p. First find. A cos kt B sin kt ka sin kt kb cos kt k A cos kt k B sin kt So, k k A cos kt k B sin kt k A cos kt B sin kt k A cos kt k B sin kt k A cos kt k B sin kt Therefore, k. Section 5.5, p... sec sec sec tn sec tn p p sec p tn p sec tn sec 5tn 5 cos 5 sin 5 sec 5.. p.. cos sec sin tn sec tn cos sec cos sin tn cos sec sin sin tn cos sin sec etn sec.. cos sec sin cos sec tn 5., p, nd p. 7. p,.57 sec tn sin cos cos sin cos d d cos sin sin cos cos sin sin cos cos sin sin cos sin cos The denomintor is never negtiv sin 7 for p since p, sin reches its minimum of t p Since the derivtive of the. originl function is lws positive in the specified intervl, the function is lws incresing in tht intervl.. p 9. Write tn sin nd use the cos quotient rule to derive the derivtive of the tngent function.. csc. f csc cot Answers 57

18 Review Eercise, pp. 5.. e t.. ln ln 5 5 ln 5 ln ln 5 ln 5.. cos sin sec sin cos.. The function hs horiontl tngent t (, e). So this point could e possile locl m or min. 5.. The slope of the tngent to f t the point with -coordinte is e e e 5e 5 e e t Now, sec tn e sin cos cos sin e e 5 e e d d e e e e e e e e e e e e e e e e e e e e e d d. ln 9.. out.9 m per unit of time.. t After ds, out.5 mice re infected per d. Essentill, lmost mice re infected per d when t... c c.. 9e e e e e e 5e 5 e ln 5.7 ln.7 5 ln 5 5 ln e ln ln cos cos sin cos p cos sin cos sin sin cos sin. 7. p v ds dt ; Thus, v cos pt p p cos pt The ccelertion t n time t is dv dt d s dt. Hence, psin pt p p sin pt. Now, d s dt p s p sin pt p sin pt.. displcement: 5, velocit:, ccelertion: p 9. ech ngle rd, or 5..5 m..5 m. 5.9 ft.. f sin cos f sec sec tn Chpter 5 Test, p... e ln e e cos 5 sin 5 sin cos sec., The tngent line is the given lin... t v t ke kt ke kt kvt Thus, the ccelertion is constnt multiple of the velocit. As the velocit of the prticle decreses, the ccelertion increses fctor of k. cm> s ln k ; 5k 5.. f sin cos f csc cot csc sin. solute m:, solute min: 7... minimum: Q, no mimum e R 9.. p, 5p, p incresing: 5p p ; 5p decresing: p nd p p locl mimum t p,.59 locl minimum t 5p,.59 Cumultive Review of Clculus, pp ln.. ms > 5 ms >. f.. 9. ms > 9. ms > 5.55 ms > p p p p p p p p 5 Answers

19 fish> er fish> er.. i. ii. iii. iv. No, lim f does not eist. In order S for the limit to eist, lim f S nd lim f must eist nd the S must e the sm In this cse, lim f q, ut S lim f q, so lim f S S does not eist. 7. f is discontinuous t lim f 5, ut lim f. S S p t t people per er.. f 5 5 ; f f f ; f f ; 5 f 5; f 5.. mimum:, minimum: mimum: 9 minimum:, e mimum:, minimum: e mimum: 5, minimum:.. vt 9t t, t t sttionr when t or t, dvncing when vt 7, nd retreting when vt t.5 t.5.5 t m r. cm, h. cm 9. r. cm, h 7.5 cm cm ;.7 cm.7 cm. cm.. $7 or $. $.. d, d is criticl numer, Increse:, Decrese: 7 d, d is criticl numer, Increse: 7, Decrese: d d, ; re criticl numers, Increse:, 7, Decrese: d The function hs d. no criticl numers. The function is decresing everwhere it is defined, tht is,. 5.. is horiontl smptot ; re the verticl smptotes. There is no olique smptot Q, is locl mimum. 9 R There re no horiontl smptotes. ; re the verticl smptotes. is n olique smptot, is locl mimum,, is locl minimum... = e 5 e ln cos e sin. 9. e e. 5 ds 7.. cos 5 sin 5 cos sin cos sin cos cos. out. m. out.5 m Chpter sec tn sec sin cos cos = Review of Prerequisite Skills, p AB 9.7, B.5, C 5.5 A 97.9, B 9.7, C 5.. Z 5, XZ 7., YZ.7 5. R, S, T. 5. km 7.. km.. km cm Answers 59

20 Section., pp Flse; two vectors with the sme mgnitude cn hve different directions, so the re not equl. True; equl vectors hve the sme direction nd the sme mgnitud Flse; equl or opposite vectors must e prllel nd hve the sme mgnitud If two prllel vectors hve different mgnitude, the cnnot e equl or opposit Flse; equl or opposite vectors must e prllel nd hve the sme mgnitud Two vectors with the sme mgnitude cn hve directions tht re not prllel, so the re not equl or opposit. The following re sclrs: height, temperture, mss, re, volume, distnce, nd spee There is not direction ssocited with n of these quntities. The following re vectors: weight, displcement, force, nd velocit. There is direction ssocited with ech of these quntities.. Answers m vr. For emple: A rolling ll stops due to friction, which resists the direction of motion. A swinging pendulum stops due to friction resisting the swinging pendulum.. Answers! m! vr.! For emple:!!!. AD DE! BC AD! EB! ; AB DC ; AE EC ;!! AE! CB! AC! CE! ; AB! CD! ; & DB! ; ED ; AE! EB & EB! 5. B H D E A. AB! AB! CD! AB! EF! EF! ut C G F GH! AB! AB! JI! I!! AB EF J. Drwings not to scl. cm cm cm cm.5 cm 7.. km> h, south 5 km> h, west km> h, northest 5 km> h, northwest km> h, est.. km> h, due south 7 km> h, southwesterl km> h southesterl 5 km> h, due est 9.. i. Flse; the hve equl mgnitude, ut opposite direction. ii. True; the hve equl mgnitud iii. True; the se hs sides of equl length, so the vectors hve equl mgnitud iv. True; the hve equl mgnitude @ The tngent vector descries Jmes s velocit t tht moment. At point A, his speed is 5 km> h nd he is heding north. The tngent vector shows his velocit is 5 km> h, north. Jmes s speed The mgnitude of Jmes s velocit (his speed) is constnt, ut the direction of his velocit chnges t ever point. C.5 min southwest.. or.,, (, ) Section., pp A C BA! A C! A C CB! A C CA! B B B B Answers

21 .. c 5.. PS! R RQ Q km> h + + c! S RS PS PR P PQ Q km/h 7 km/h km/h.. c + c c c c + c ( + c) + ( + c) ( + ) c c.! 7..!!!!!!!! g.!!!!!!! h... S!! MR! RS!! t! MS! ST! SQ! TQ! so!!! t! MS! MQ! SQ! See the figure in prt. for the drwn vectors.!!!!!! nd so!!!! cos u!!, 9.. km> h km/h 7 km/h RQ PS A B u RS R PQ P km/h D u C.. ship f f + f The vectors form tringle!! withside f f f using the cosine cos f cos u. 7 km h, N. W.!! > 5, u The digonls of prllelogrm isect ech other. So, nd ED! EB! EA! EC!. Therefore, EA! EB! EC! ED!!. 5. Multiple pplictions of the Tringle Lw for dding vectors show tht RM!!! TP! (since oth! re equl to the undrwn vector ), nd tht RM!!! SQ! TM (since oth re equl to the undrwn vector RQ! ). Adding these two equtions gives RM! RM!!.!! TP!! nd! SQ!!! TP! SQ!! represent the digonls of prllelogrm with sides! nd the onl prllelogrm with equl digonls is rectngle, the prllelogrm must lso e rectngl u f ( + )+c The resultnt vectors re the sm The order in which ou dd vectors does not mtter.!! c!!! c! Answers

22 7. Q P Let point M e defined s shown. Two pplictions of the tringle lw for dding vectors show tht GQ! GR! QM! RM! MG! MG!!! Adding these two equtions gives GQ! QM! MG! GR! RM!! From the given informtion, MG! nd QM! GP! RM!! (since the re opposing vectors of equl length), so GQ! GP! GR!!, s desire Section., pp. 9. A vector cnnot equl sclr.. Drwings not to scl. cm G M R. Drwings not to scl Answers m vr. For emple:. v! v! v! v! + cm 9 cm v! v! 5. Drwings not to scl.. Answers m vr. For emple:. cm. E5 N descries direction tht is 5 towrd the north of due est. N5 E nd ering of 5 oth descrie direction tht is 5 towrd the est of due north. + + Answers

23 7.. Answers m vr. For emple: m, n, infinitel mn Answers m vr. For emple: d, e, f ; not unique.!! 9. es. 9. MN! QR!!!!! Notice tht MN!! QR!!! We cn conclude tht is prllel to MN! nd QR! QR MN!. A B.. + +!!!! c E ( + ) ( + c) ( + )+ c = = + + c + (+ c).. colliner not colliner not colliner colliner.. is vector with length unit in!!! the sme direction s is vector with length unit!!. in the opposite direction of!.. m..!!!!!. 5 or., from towrd! u.! 5..9, 9.9 from! towrd!.!!!! `!!! `!!! 7. AD! AD! c! AD!! CD! But CD! c! BD!! BD! CD! BD!. So, or AD! AD!!! c, c!!. C Answers m vr. For emple:. nd!! u! nd nd u! ED! v!! u! AD! v!! u! AB! CD AC! v AE nd v! DB AD!.. n m m.. CD!, BE!! n!!!! CD!! The two re, therefore, prllel (colliner) Appling the tringle lw for dding vectors shows tht AC! AD! DC! The given informtion sttes tht AB! DC! AB! DC! B the properties of trpeoids, this gives, nd since AE! EC! AC! AE! EC!, the originl eqution gives AE! AD! AE! AB! 5 AE! AE! 5 AD! 5 AB! Section., pp. 7..! AD! AB! D. Answers m vr. For emple: 5... Yes, the digonls of rectngulr prism re of equl length. 7.!.. i! 9! 7i! 7j! c! i! j! 5k! j! k! 7k! 9.! 5,!!. k( + ) k PQ! RQ! SR! TS! PT! RQ! SR! TS! PT! SR! SQ! RQ! PS! PT! TS! PS! PQ! SQ! EC!!!!!!!!!!!!!!!!!!!!! k + Answers

24 .. AG!!! c!, AG!!!!! c! c! AG! BH!. Appling the tringle lw for dding vectors shows tht TY! TZ! ZY! The given informtion sttes tht TX! ZY! TX! ZY! B the properties of trpeoids, this gives, nd since TO! OY! TY! TO! OY!, the originl eqution gives BH!!! c!, CE! DF!!!!! c! c!, BH! TO! TO! TZ! TX! TO! TZ! TX! TO! TZ! TX! Mid-Chpter Review, pp. 9..!! AB!,, AD! DC! BC! BA, CB! CD DA! There is not enough informtion to determine if there is vector equl (prllelogrm)!.. RV! PS! RU! PS PQ!.. 5. t or t 5. In qudrilterl PQRS, look t ^PQR. Joining the midpoints B nd C cretes! vector BC! tht is prllel to nd hlf the length of PR!. Look t PR^ SPR. Joining the midpoints A nd D cretes! vector AD! tht is prllel to! nd hlf the length of. is prllel to AD! PR! PR BC nd equl in length to AD!. Therefore, ABCD is prllelogrm... 7 u! v! u! v! u! v! m! 9. BC! m!,, BD!! DC! AC!!!. Construct prllelogrm with sides OA! nd OC!. Since the! digonls isect ech other, is the digonl equl to. Or OB! OA! OA! OC!OB AB! nd AB! AC! So, OB! OA!. And AC!. Now OB! OC! AC! OA! OA!, OC! OA! Multipling gives OB! BD! OA! BC!!!! OC!.!. km h, south PT! > PT! SR! PS! PQ! QS! RS!! QS!!! QR! + Section.5, pp.. No, s the -coordinte is not rel numer... We first rrnge the -, -, nd -es (ech cop of the rel line) in w so tht ech pir of es re perpendiculr to ech other (i., the - nd -es re rrnged in their usul w to form the plne, nd the -is psses through the origin of the -plne nd is perpendiculr to this plne). This is esiest viewed s right-hnded sstem, where, from the viewer s perspective, the positive -is points upwrd, the positive -is points out of the pge, nd the positive -is points rightwrd in the plne of the pg Then, given point P(,, c), we locte this point s unique position moving units long the -is, then from there units prllel to the -is, nd finll c units prllel to the -is. It s ssocited unique position vector is determined drwing vector with til t the origin O(,, ) nd hed t P. Since this position vector is unique, its coordintes re uniqu Therefore,,, nd c... 5,, nd c. 5,,. This is not n cceptle vector in I s the -coordinte is not n integer. However, since ll of the coordintes re rel numers, this is cceptle s vector in R. 5. (,, ) A(,, ) (,, ) (,, ) (,, ) O(,, ) (,, ) (,, ) Answers

25 (,, ) O(,, ) (,, ) B(,, ) (,, ) (,, ) (,, ) (,, ) 9.. A(,, ) C(,, ) B(,, ) (,, ) E(,, ) (,, ) (,, ) (,, ) (,, ) (,, ) O(,, ).. A,, is locted on the -is. B,, C,,, nd D,, re three other points on this is. OA!,,, the vector with til t the origin O,, nd hed t A. 7.. Answers m vr. For emple: OA!,,, OB!,,,. O(,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) C(,, ) OC!,, 5 Yes, these vectors re colliner (prllel), s the ll lie on the sme line, in this cse the -is. A generl vector ling on the -is would e of the form OA!,, for n rel numer. Therefore, this vector would e represented plcing the til t O nd the hed t the point (,, ) on the -is... (,, ) (,, ) C(,, ) (,, ) (,, ) (,, ) O(,, ) (,, ) (,, ) (,, ) O(,, ) (,, ) (,, ) A(,, ) (,, ) (,, ) (,, ) B(,, ) (,, ) O(,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) O(,, ) (,, ) F(,, ) (,, ).. OA!,, (,, ) (,, ) O(,, ) (,, ) OB!,, A(,, ) (,, ) OA (,, ) (,, ) (,, ) B(,, ) (,, ) OB (,, ) (,, ) (,, ) (,, ) O(,, ) E(,, ) B(,, ) A(,, ) F(,, ) D(,, ) C(,, ) (,, ) (,, ) O(,, ) (,, ) D(,, ) (,, ) (,, ) (,, ) OC!,, (,, ) (,, ) (,, ) C(,, ) (,, ) (,, ) (,, ) OC O(,, ) Answers 5

26 OD!,, (,, ) (,, ) O(,, ) (,, ) OE!,, (,, ) E(,, ) (,, ) (,, ) (,, ) D(,, ) (,, ) OD (,, ) (,, ) (,, ) (,, ) O(,, ) OE vectors ssocited with these points joining O (til) to the given point (hed). Tht is, the plne contins the vectors nd OP! OM! ON!,,. 5.. A,,, B,,, C,,, D,, 7, E,, 7, F,, 7 OA!,,, OB!,,, OC!,,, OD!,, 7, OE!,, 7, OF!,, 7 7 units Ever point contined in rectngle BCEP hs -coordinte equl to, nd so is of the form (,,), where nd re rel numers such tht nd 7... O(,, ) M(,, ) F(,, 5) O(,, ) OF!,, O(, ) J(,, ) (,, ) (,, ) (,, ) (,, ) F(,, ) OF O(,, ) (,, ) (,, ) D(, ) P(, ) 7. O(,, ).. c 5 Since P nd Q represent the sme point in R, the will hve the sme ssocited position vector, i., OP! OQ!. So, since these vectors re equl, the will certinl hve equl mgnitudes, i., OP! OQ!.. -plne, ; -plne, ; -plne,... Ever point on the plne contining points M, N, nd P hs -coordinte equl to. Therefore, the eqution of the plne contining these points is (this is just the -plne). The plne contins the origin O(,, ), nd so since it lso contins the points M, N, nd P s well, it will contin the position O(, ) C(,, 5) O(,, ) (,, ) (,, ) O(,, ) (,, ) (,, ) (,, ) (,, ) (,, ). First, OP! OA! OB! the tringle lw! of vector ddition,! where OA 5,,, OP! nd OA! OB,,, re drwn in stndrd position (strting from the origin O (,, )), nd OB! is drwn strting from the hed of OA!. Notice tht OA! Answers

27 9. lies in the -plne, nd OB! is perpendiculr to the -plne (so is!!! perpendiculr to OA ). So, OP, nd OB! OA, form right tringle nd, the Pthgoren theorem, OP! OA! Similrl, OA! OB!!! the tringle lw of vector ddition, where! 5,, nd!,,, nd these three vectors form right tringle s well. So, OA!!! 5 5 Oviousl OB!, nd so sustituting gives OP! OA! OB! 5 5 Section., pp.... OP! 5 5, 5, AB!,, BA!, A(, ) BA OA!, AB!. OB!., BA! AB B(, 5) A(, ) 5 O(, ) OA OA! nd OA! OA!, nd OA!. 5.., !,!,!!!!...., 7, 7. (, ). 7i! i! j! 5j! 9i! j!.. out. out.7 out. out 9.. AB!.,, EF!, CD!, AB!, GH!,,, EF!.7 CD! 5,,.. OC! 7. GH!.7 7,, BA! 5 5, 9, BC!, OA!, BC! so oviousl we will hve OA!, BC!... C(, ) AB!,,, AC! AB!, CB! CB! 5 5, AC!., AB! O(, ) (, 5) 9 9 (, 5) 5 (, ) 5 (, ) B(, ) A(, ) Since CB! AC! AB!, the tringle is right tringl.. A(, ) X(, ) A(, ) X,, Y,, Z,.. 7,,... B(, 9) Becuse ABCD is rectngle, we will hve BC! AD!,, 9,,, 9, 9 So, nd 7, i., C, P, Q,. 9, 7.. out.9 out 5. C(, ) C(, ) C(, ) B(7, ) Y(, ) D(, ) A(, ) Z(, ) B(7, ) Answers 7

28 Section.7, pp... i! j! k! out. OB!.5,,, OB!...., OA!,, OB!, OP!.57 AB!, AB! 5,,, AB!.7 represents the vector from the tip of OA! to the tip of OB!. It is the difference etween the two vectors. 5..,, 7,,,,,,.. i! i! j! 9i! j! k! 9i! j! k! j! k! k! 7.. out 5. out. out 5.9. out!,! i!. i! j! j! k! k! 9.. The vectors OA!, OB!, nd OC! represent the -plne, -plne, nd -plne, respectivel. The re lso the vector from the origin to points,,,,, c, nd,, respectivel. OA!, c, OB! i!, OC! i! j! i! j! k! j! ck! ck! OA!, OB! c, OB!, AB! c,, c is direction vector from A to B... 7 (5,, 9).9 5,, 9.9. In order to show tht ABCD is prllelogrm, we must show tht AB! DC! or BC! AD!. This will show the hve the sme direction, thus the opposite sides re prllel. AB!,,! DC,, We hve shown nd BC! AD! AB! DC!, so ABCD is prllelogrm.., 7, c.. OB O OA OC V,,, V,, 5, V,,, V, 5,, V 5,,, V, 7,, V 7, 9,, V,, 5. (,, ) 5.. Section., pp.. The re colliner, thus liner comintion is not pplicl. It is not possile to use! in spnning set. Therefore, the remining vectors onl spn R.. The set of vectors spnned (, ) is m,. If we let m, then,.!m,. i spns the set m,,. This is n vector long the -is. Emples: (,, ),,,. 5. As in question, it isn t possile to use! in spnning set.. 5,,,, 5,,,, 5,,, re ll the possile spnning sets for with vectors. 7.. i! 7i! j! R j! k! k!. 5,,,,, :,,,,,,,,,,,, 5,,,,,,,,,,,,,,,,, A B C 9.. It is the set of vectors in the -pln,,,, B prt., the vector is not in the -pln There is no comintion tht would produce numer other thn for the -component. It would still onl spn the -pln There would e no need for tht vector..,, c.,,, 5..,,,,, 5 9,,, 7,,.. The sttement,,,,,, does not hve consistent solution.,, 5,,,, m, n ; Non-prllel vectors cnnot e equl, unless their mgnitudes equl.. Answers m vr. For emple: p nd q, p 5 nd q, p nd q 7. As in question 5, non-prllel vectors. Their mgnitudes must e gin to mke the equlit tru m m m m m, m m m m m, So, when m, their sum will e. Review Eercise, pp. 7.. flse; Let, oth represent the lengths of the digonl of prllelogrm,! the first with sides! nd nd the second with sides! nd c! ; since oth prllelogrms hve! s side nd digonls of true; Sutrcting from oth sides shows tht! c!!. Answers

29 true; Drw the prllelogrm formed RF! nd SW!. FW! nd RS! re the opposite sides of prllelogrm nd must e equl. true; the distriutive lw for sclrs flse; Let nd let c! Then,! d!!!. nd ut c!!! so!!!!..!! c!! c!!.. XY!!! c! c!! d! c! c! d! c!,,, OA!, OA! OB! OB!,, 9,, 7. u. AB!., BC! 59, CA! AB! 5 CA!! BC.5.,,.. XY! 7 7, 7, 7,,,, 7, 7, 7 + c c 5 9.,, 7,,,, 7,,,,, 7.. where P(,, ) is the point. (,, ) nd,,..,,, 7.5 c, c.. es es.. AB! 9, Since AB! AC! AC!, BC! BC! the tringle is right-ngle.. DA! nd DC!, BC! nd AD! AB! EB! CE!, ED!, But AC! DC!, EA! DB! AC! Therefore, AD! DC! DB! 5.. C(,, 5), P(,, 5), E(,, 5), F(,, ) DB!, CF!,, 5,, km S. W.. An pir of nonero, noncolliner vectors will spn R. To show tht (, ) nd (, 5) re noncolliner, show tht there does not eist n numer k such tht k,, 5. Solve the sstem of equtions: k k 5 Solving oth equtions gives two 5 different vlues for k, nd so, (, ) nd (, 5) re noncolliner nd thus spn R. m 77, n 9.. Find nd such tht 5, 9,,,,, 5, 9,,,,, 5, 9,,, i. 5 ii. 9 iii. Use the method of elimintion with i. nd iii. 5 B sustitution,.!! lies in the plne determined nd c! ecuse it cn e written s liner comintion of! nd c!. If vector is in the spn of nd then!!! c!, cn e written s liner comintion of! nd c!. Find m nd n such tht,, m,, n,, m, m, m n, n, n m n, m n, m n Solve the sstem of equtions: m n m n m n Use the method of elimintion: m n m n m n n n B sustitution, So, vector is in the spn of nd c!! m.!... (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ) (,, ),,,, (,, ). 7.. AB!, BC!, CA! If A, B, nd C re vertices of right tringle, then BC! CA! AB! BC! CA! 5 AB! So, tringle ABC is right tringl..!!!!! c!!!! c! c! Answers 9

30 . 5..!!!!!. Cse If nd re colliner, then!! is lso colliner with oth! nd. But is perpendiculr! to nd! c! c! c!! 9! c!, so!! is perpendiculr to. Cse If! nd c! c re not colliner, then spnning sets,!! nd c! spn plne in R!, nd is in tht pln If! c is perpendiculr to! nd c!, then it is perpendiculr to the plne nd ll vectors in the pln So, is perpendiculr to! c!!. Chpter Test, p.. Let P e the til of nd let Q e the hed of The vector sums!! c!! c!. nd!! c! cn e depicted s in the digrm elow, using the tringle lw of ddition. We see tht!! PQ! c!!! c!. This is the ssocitive propert for vector ddition. P.. (,, ) ( + ) PQ= ( + ) + c = + ( + c),, ( + c). 9..!!!,!! 5!,, 5..! nd! 5 c spn R, ecuse n vector (, ) in R cn e written s liner comintion of! nd!. These two vectors re not multiples of ech other. p, q..,, 9,, 7, r!, cnnot e written s liner comintion of nd In other words, r! p! q!. does not lie in the plne determined p! nd q!. 7., u.; 7.9 from! towrd! c Q. DE! DE! CE!!! CD! Also, BA! CA! CB! BA!!! Thus, DE! BA! Chpter 7 Review of Prerequisite Skills, p. 5. v km> h N 7. E. 5.9 units W. N...,, 7; l 7.7 9,, ; l.9,, ; l.,, 9; l on the -plne on the -plne on the -plne.. i! i! 7j! i! j! i! 7.. i! j! j! 5i! j! i! j! k! k! k! j! k! k! Section 7., pp... N is melon, 5 N is chir, N is computer Answers will vr... N N C(,, ) B(,, ) A(,, ) D(,, ) N. line long the sme direction. For three forces to e in equilirium, the must form tringle, which is plnr figur 5.. The resultnt is N t n ngle of N. E. The equilirnt is N t n ngle of S. W. The resultnt is 5 N t n ngle of S.9 W. The equilirnt is 5 N t N.9 E... es es no es 7. Arms 9 cm prt will ield resultnt with smller mgnitude thn t cm prt. A resultnt with smller mgnitude mens less force to counter our weight, hence hrder chin-up.. The resultnt would e.7 N t.7 from the N force towrd the N forc The equilirnt would e.7 N t 5. from the N force w from the N forc N 5 from given force,.95 N perpendiculr to 9. N force. 9 N directed up the rmp.. 7 N. pproimtel 7. N 5 south of est.. 7 N The ngle etween f nd the resultnt is The ngle etween f!.. nd the equilirnt is.7... N N N N N For these three equl forces to e in equilirium, the must form n equilterl tringl Since the resultnt will lie long one of these lines, nd since ll ngles of n equilterl tringle re, the resultnt will e t ngle with the other two vectors. Since the equilirnt is directed 7 Answers