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Calc III Ch12 Handouts (Shaw)

# Calc III Ch12 Handouts (Shaw) - 12.1 — Vector Valued...

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Unformatted text preview: 12.1 — Vector Valued Functions We’ve seen how a curve can be written as parameters: x = f(t) and y = g(t), well ﬁlm on may / I x; t: 2‘ ’ 2.4: this can be extended to the 3rd dimension with z = h 15:) ‘3? ti V_ {#12, Often we use vector valued functions for this: 2-dim: r(t) = £(C€)L + 5(Qj/ or 3-dim: r(t) = Hat 4 9650 J/m'j/kor f, g, & h are called Lem; Ami r' Ff/wr rmN 5 Properties of vector valued functions: 1. The domain 1s simply the Mag/MK ﬂavA/ Jed/C /QM(/WT [TV/V 5 5.5V FrosT‘ Ab\' 9‘2, I'M/S. Ex: What 1s the domain of r(t)= cos ti + \/—j _:6k FUN/7 77 MM: [0 00) {#0 / , , 2. The limit is simply the done by ( Qﬂd 93m 51 S limr t = I Ma 0 5;"; (l (t) b Jr Lva 03%. 4,, [w MHK \ c act 16 +4 Ex: Find the limit as t goes to 0 of the function above. [/M r(+): (UL + O + J— ) t --70‘* J (p K 3. A vector valued function is continuous at a if the limit exists and is r ( 0X). (Similar to continuity from calc l) limr(t)= rte) r—>a EX: For what values isr =( 1 -— sinﬁ + k continuous? t2 —4t ’6 ¥ 6 L/ / (P? 07 if} Sketching a curve: Remember to include an” WI’L Wt. Pm N t and hﬁlrfcfw . 2—Dim: One way is to convert it to a Cartesian equation by eliminating the parameter by fa Q or by jng’ I Lguﬂg \$ (as in calc 2). Ex: Sketch «0:51—43 r(t)=costi+3sintj /505 (ﬁat; J: (of: {a ’1 ‘3 f "t ._ Y z v.5?” ‘é .3: .— 5w 5 ‘3‘” (oszé #5/rzt:/ 1"”) )(va .12— l a _. :0 3-Dz'm: l. Recognize the graph (eg- a line). 2. Find a 3-D object on which it lies, then plot points. 3. Determine where it would lie in 2-D, then determine the z—value. Ex: Sketch r(t) = ti—2tj+(t—2)k 5K r(t)=costi+tj+§intk \3": (06% [VJ (“92\$ 9% Creating a vector-valued function: Often this is easy to do if you simply pick X = "t f— 1.199 Ex epresent the curve y—-_- —-x —x with a vector valued ﬁmction‘. f :5: t: t r(t):JcL - (1:3 NOTE. TJouuj’ can change the initial point and direction. A similar, but more intricate strategy can be used for more complicated sitiiations. Ex: p 838 #60, 66 “ﬁmwOfVR SILJL. Eff-K6 (Ufwe F} (34 mo. Z:X2+/L/ 2-"; l'i') X‘Z (0\$'b 2. .i a; ’(l'ly-Fj/Z: L/ 5:": 7 z x ( y (Z/aﬁf) 7Z \/ : L/ ! L5: ZS/A/é’ 7" / I”! 1 f/ w. HEX/4 fl; 2/05 6 t 25/” _ (INA—M ((0 / XV, Ll 1:4: (\V\ﬁocm/NT> K:_bﬂ V' t 12.2 — Calculus on Vector Valued Functions The technical deﬁnition is analogous to the deﬁnition of a function in Cale l: rm: Mm “HAW — Y W) (x j k) (“'20 At \n 70 Thankfully, differentiating and integrating a vector valued function can be done by components. .' . piece of cake. Ex: Find r'(t) for the following: r(t)=tanti+t3j r(t)=sinti+x/;j—lntk ' z \ , \ \ ~ _.:‘__\.__ r(e)%C+/,Jr3ezd CO5JW+Z£EA i/AZJc/j Determine the following integrals: f(ti+5j)dt f(3ﬁi+5j—k) 1v _ \ [2' §J+btj+\$r bf5¢J tk/ Cowwmewam’ a. 7L MJ , Z/C » Z /< 3 571": \X , "f (”365 ’qSIN‘l it 2(0’4 Kg; 3 makrc QC“ {5‘3 J ”(0N4 +6 gal/DD' L'(0\$(6>)} 35’ +6, : ZLHU: A3 —} O Q” o 4‘7 "1 C: t P " - OF CL. 01,15 5; % rrwwlc WW. Ra?" D ‘“ F a“ ﬁeﬁies of the Derivative: if»)! vawﬂ * 1. Constant: De ibr by} g C Y ’05) 2. Sum/Difference: ﬁt [(U') f \) 3)] I (/@> c 000:] : ( 'Lt) eUC’c) —l/<((t>b Uzi?) - ('LtQXUG’) 'l' FLOXU’GD 3. Product Rules: ____,______.__.____.H____————————-"'—'—' 4. Chain Rule: MEY- (f (13)]: r /( (’&D G/(E)> 5' Other: (G) a (0°) : A 3> F<t>e rIC-k>: O :>of‘rHoC/2/ Ex: Suppose r(t) = ti — 2tj + (t—2)k and u(t) = 3i—tj + tzk , determine the following: D,[r(t)5h(t)] D,[r(t)—5u(t)1 Home) Jr rt+>~u0c> (\L , a; + @m -6 my, (H — 2e; + t—22¢)-©L , 5 +2514) (3* ZJ‘ 4 ’61 “L [o,+ze r2345) #7 3+. 1 ¢ 3 '2' .4 tn D,[r(t)><u<t)] D,[||r(r)||] Va f . ﬂﬁﬂ ._ . {Q2} 13’4- L/{dJCl—E‘Iat J Maw Z Wt??? 2 13?- A vector value ﬁmction is smooth on an interval i-f Jf',_g ',and K1 are ﬁaﬂjﬁﬂ'kcﬁ'v 5 on the interval gnﬁg r‘(r)_ f 0 for any value in the interval. Ex: Where are the follow-ing'smooth‘? 1(9) =09 — sin 9)_i + cos 6’j f’é‘?) ~(l , ms é’jj » jaw/<4] 7’ Au. VLML ___ Q __ 2M r’ 12.3 — Velocity and Acceleration The velocity vector and acceleration vector are analogous to what we learned in Calc 1 (proofs are in the book). So, ._\ W) = r '(t) = X'(t)i + y'(t)j 30): Vl(i§) :. (”6%) speed = ||V(t)” = llr'(t)|| = é.” (/6 )J z +[y/Kfjj 2 D15 (M W D 3 Ex: Find the velocity, acceleration, and speed of the object at the given t—value. Sketch the velocity and acceleration vectors on the position graph. r(t)=ti—t2j att=2 r(t)=costi+3sintj att=7r/6 .. . - x: ”Izmétvtﬁ’wmﬁ , v04 a J w _.J.‘ £4? 2 3 UV 2 [:2 C Of course, if you’re given the acceleration/velocity and initial conditions, you can work backwards. (‘9 0 If a(t) = 3 j—k and the object starts at restrat the point (-7, 0, 8)[email protected]\ A ,/) I 0 / Q 7 . - t e t30 , v: 0 \(Ct): 3t5—tklc ‘7V‘, O'J * C) (L \L C W) " gt) ” 2’4 Z _ -3’61’ i L +(7. 73¢) 7. Z (“01), / Z __.__ _ .. Fa) : 3+ ' t Z __,___F r :— 5 E- k 4 C“ / z. i a . \ ' x'7" *?J z I v’lt +03t rap]: q, 4 O K + CLWQH . ---—"'*"‘_‘\ " Projectile Motion: We can now model projectile motion with vector valued functions assuming that gravity is the only force acting upon the particle after it’s launch. First, this is a2—D case. Why? W/VW/ \p a ‘ )t Second, we know from physics that (in vector notation), a: ‘ :i l (:4, 8 > (/1 g 7L” (9 rte): (“@569 ‘L 4” [ho+(v,énlé)t—‘/Z 51835 Conclusion: For a projectile, if (9 2 angle of elevation, g = gravity, h = initial height, and v0 2 initial speed, then r(t) = EX;- P854 #28 "60nd" Latin {5'9 ---.... SWL (T 3/286: Fi— ____ _ ﬂeww ‘tOGt F®><<15fs (’05 )3 ‘5 L>+ C6 + (73,35/M/6” —- @593 J KM 7%ar0214°@_hz7)9 y: 5} 73.35/N/5’°(A27/) /é[/Z7/)z a; also +-t P855 #38 mm 1M7 Ma éi/z‘ ‘ Vi : .? lo: 0 .rx— 150) x, o \ my: V, (5522}; «L [Q Mm/A/IZ”)t 'é-(E'ZFz-J e Z K:['§b ‘3: (p V:/;: 9K“ ~‘ ‘0: 0 ‘2 I50 : V0 CoS|Z°t OfV ﬁlMlSOb'KOb \I: 50 0 ~ V70 5/3/57 e - wt 0 (“3'30 a) [05/th ' D WW3": -- {‘50 WW |Z° 12.4 4 Tangent Vectors and Normal Vectors Based on the last section, how should we deﬁne the unit tangent vector? T(t)= ﬂ :2 ll Hit)” . . I What is necessary for the tangent vector to ex1st? “f (15' ) H ii 0 How do we know this is satisﬁed for a speciﬁc vector function? fowl/7" Ex: Find the unit tangent vector: r(t)=3ti+t2j ‘ ‘ r(t)=costi+23intj+k VG); 5‘”th r’(e)' «ﬁ/A/éd‘ +Zco>t J' +ole I :UNW T I! We)“ = . r‘ / ,. Ex: P864#28 C' L ” of ""W W“ TU):— xteh‘. + get—)5 36%) In practice, we can short—cut the normal vector ' 2-D by using our knowledge about x e) perpendicular lines from algebra: / M 3 __v_ K I , ‘ C K N(t): 1/ (0t 4“)» -— 36c», + x603 The principal normal vector is the one that points toward the concave side. . is ovjfr'g co \$36172. if Ltwp/1 Wu 156,; .t i su‘aﬂé/ Ex: Find the principal normal vector for r(t) : 3ti + t2 j , >< mﬁoz, ‘/ ray): 3 ,L 1,, 2t #0:): ZHZW fler? “fife. n N(t)” : W a :LZ—EJ ,2. c M) m " m Wj““l””“ as .7 Tangent and normal components of acceleration: Last section we noticed how the velocity vectors and acceleration vectors of an object with constant speed were .J— . However, if the speed is variable, then this may not be true. This acceleration vector has tangential and normal components, similar to i and j. They Theorem: If r(t) is sfhﬁlfand N(t) \1 = , ,1/ \f- T‘HT“ T 4 T" \x l. \ .., \ aN‘I: 6i , M:: //\/////7‘///-f’—' ”VXQ H \Zd“ EX: P864 #36, 40, 54 ,—-- \9 a) 44:30, rug): {‘C ’r 2+1 15:3»! " «EL E @ 0.5: d' 7' =9 4%):r”(f):ZI Z: a Li. +.——)- ‘ W):“'(*)= “4L - 45 + Zk 4G): 0 _ r') on —w +Z\g 2 Z - J. Tug-Iii”- I L1: :‘é—L'gd +514 T(Z>:-gZ-u—_Z_‘o +J—k 5W / 3 3 "" “ WW)” 12. 5— Are Length and Curvature [019), Recall from Ch 10 that are 1ength,:; =i:lx W10 4_ y 3(6); d 6 Thus, this applies to this chapter as wellznaeemw 5 In 3—D, this becomes s = f I Q. (x (a); WW 42(0) ”/75 '""///r 2% Ex: Find the arclength of r(t): tzi— tk Ffam (0/7 z) (waif ty/VQWZ 1* / J/f' W, P\ 'f?)/ 96, 5%.”; How about 1'0): (,3: 2cost, 25in: M [0 ZrTj ("5/ 31¢yjzﬂzi4'V/aﬁz" 9’1! 7'" J zrf / / W3 Mi Arc Length Parameter: So far we have used time and angles as parameters, now we’re going to use the arc length, 5. If we have a smooth curve given by r(t) deﬁned on [a, b], then the arc length function is L , 8602/ I Z I ‘ Z ,. . “7 Z— « 4 o’ a) «a «a e w 2 a «( ~ / my 4+ 4 It follows from the FTC, that s’(t) = _. , ‘ «2 tzx’vrnwneqz/F Ex: Find s(t) for r(t): L 6“) fWZ’; fi/m «m If: it” wEl fie it ‘ f0.» '5 (1&3): ZWL—éd From this example, what 1s AT%(S) ?/ :W I (“73; F) [J] Theorem: If we have a smooth curve deﬁned by r(s) en "r ‘(s)||—___ = Furthermore, if ”r '(t)“=l, then t must be the 4 KC @ff [7’ . Curvature: Simply stated, curvature is a measure of how . . . 51-1/4 PP 4 c u ”v: Tng 5 ._-a— low 6&4?- T How can we determine this? TM V€LTOF Deﬁnition: If we have a smooth curve given by r(s), the curvature ﬁfhﬁ Ex: Whatisthecurvatureofr(t)-t/2i—tj? r/({—)= Zb L ’J Hr’(k)\l=J'-lez+ I I - ' 2m- ZelL4t1+6égt ’ T145); ?— L .- r7415); ﬁl£Zf// L 4.32:» 17'4sz I , z I “Tin :6 07 WJ ﬂz KW +)/z it'd f“ I M “(We)“ W: WW: 20/61 31,2394]? /£1’__— J W (”‘52 *’)/ LH‘+I What 1s the reﬁagre' of any circle? V-W FROG: f/OS 9L «J'CS/A/QJ K’KMLMWM‘ + rms 6‘, ”1"”: {.sz V fl Lit 21? Theorem: If we have a smooth curve given by r(t), K = I/ f I (f) l/ 3 Ex: P876#38 f: 24:21; + #5 +3Z—chk K we): qua +3 + "t /< I) )er: WI Natl-H +15“ + ”(15): Hi +k ‘ taro) "|)<“lt ~ng H40?!) u Theorem: If we have a smooth curve given by f(x), then K =-——-]——Lé—-L-—" 3 (a L OZ) /L Ex- y: 3x2—1 a K="“—"‘z ‘3, H— y . . 2. V :(t b. atx=3? L0 ——————-’ 3 c: .00 10;“! [x + (to)? " Acceleration and Curvature: The tangential component of acceleration is the change in 6 [3 IEE D , which therefore is a function of &rc L65? m L5 > NOTE: The tangential component is independent of C0 r \n M ‘C/ . The normal component of acceleration is acting memenddigﬂar to direction of motion, and Is the F0 (Y E you feel when driving in a curve. Thus, it makes sense this normal component is a ﬁmction of BOTH 28550 and A; An rwr e A WHY? 3 Application: We know from Physics that F = ma = m (qt 4 d") = WI 4 tT 'I' . Wt ‘1 N \mmbﬂI N ?s’b"‘r 1 Theorem: We have a smooth curve given by r(t), then a(t)— — 61‘, T + 6'\ 5N: ’I/ X Ex: If r(t)= 4ti— t 3j—étzk, use the above to determine aT and a”. f’ a): ”I *thJ " tic. rI/(t): ’(ﬂtj _K 3 j k 4 = q __5 114:1 " O wlﬂ : L(3t2— oe’)— J‘GH- ﬁlmy-62%) _‘_' ”352 4 +4' F th [Q '1— 51th! 4_‘“a +57‘1t1 I a ; «Pub/“\$7044- «I “o + 6m ”.5: N The force of friction' IS the force required to keep the object on the cuwefwb‘) and is equal to the M 3-— ”? I fl) Ex: Suppose a 80 kg pilot is ﬂying in a circle with radius of 1km at a speed of M: 1000km/hr. What is the force of the pilot on the seat of his ILpltine ( :he s 47*);‘7/ experiencing “G- force' ’)? Ui ”M 319605 (wow IIW M f”? a”: '3’” ' ' 2277 7704/5 701:3 Q/Mom) 277'7 0/0: é/7ZN ...
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