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# Calculus MAT 021B

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This 20 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 021B at University of California - Davis taught by Qinglan Xia in Fall. Since its upload, it has received 17 views. For similar materials see /class/187344/mat-021b-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15

66 Work Work When a body moves a distance d along a straight line by a force F in the direction of motion what is the work done by the force on the body 39gt Work Done by a Constant Force The formula for the work is W F d where F is the constant force and d is the distance traveled SI International System units 0 force Newton N 0 distance meter m 0 work joule J L 1N m Work Done by 3 Variable Force along a line What happens if the magnitude of the force F is a continuous function of the position x Fx H a X lo Ax Axg Liana 5 Anal l l I I l I I I I I I I I v ann v r H II 3quot I7 On each small subinterval 11617 u the force is approximated by a constant force F Thus Wk g F xkAack Therefore the total work done from a to bis approximated by the Riemann sum W 2 Wk m 2 Hum k k Let the norm of the partition P approaches zero we have I W F 1011 That is work W done by the force is just the de nite integral of Example Spring WW ummI x0 iWMWfW L mm W stretched xlt 0 Here the force F depends on the posilion e What is the fonnulaforF39 Hooke s law for springs Fkac Tln c t 39 L quot CL 39 IIALC 1 I L quot I r 4 length N tenth o a gt U ifit is compressed and a lt 0 ifit is stretched ExampleHow much work does it take to change a spring from its nature length L to a length ofL e a A s n39ng has a natural length of 1771 A force of p E mple39 301v stretches the spring to a length of 3771 Example A 5 lb bucket is lifted from the ground into the air by pulling in 20 ft ofrope at a constant speed The rope weighs 0 llbft How much work was spent lifting the bucket and rope FIGURE 660 Li ing the bucket in Example 4 Example The conical tank in the gure is lled to within 2ft ofthe top with olive oil weighing a 57 lbftg How much work does it take to pump the oil to the rim ofthe tank FIGURE 661 The olive oil and ink in EVarnpleA 39 39 L 39 c iinuer m hi h and 39 quot It is full ofkerosene weighing 51 Zlbftg How much work does it take to pump the kerosene to the level of the top of the tank 84 Trigonometric Integrals Idea the general idea is to use identities to transform complicated trigonometric integrals into integrals that are easier to work with Products of powers of sines and cosines Evaluate sinm 05 cos xdx where m n are integers Key identity sin2 05 cos2 x 1 Case 1 If m ie the power of sin 05 is odd then use the substitution u cos 05 sinm 05 cos xdx sin2k1 05 cos xdx i 216 n i s1n 95 cos ac s1n xdx k l cos2 05 cos msm xdx 2 k n 39 l u u dubysettmgucosx So the integrand becomes a polynomial or a rational function Example 2 cos mdm sma More generally singkl1 cos x dx 1 cos2 00k f cos x sin xdx sin2kxf cos x sin xdx l u2k f u du by setting u cosx Example 3 300s5x 4cosx 4 s1 l 200s2x Zeosx Evaluate sinm 05 cos xdx Case 2 If n ie the power of cos x is odd then use the substitution u sin 05 sinm 05 cos xdx sinm x cosy 1 xdx sinmx l sin2 00k cos xdx um l u2k du by setting u sin 05 Example cos3xeSimdx Case 3 If both m and n are even that is m 2k 71 21 Then we substitute 2 l cost 2 lcos2x sin x andcos x 2 2 to reduce the integrand to one in lower powers of cos 2x sin2k 05 cos xdx l 020s205kl 020s2xldx Example sin4 05 cos2 xdx Eliminating square roots Example W Z de Integrals of powers of tan 1 and sec 391quot 0 Here the key identity is sech ltan2xor tauan seCZx 1 Example Evaluate tan3 05 sec xdx 82 Integration by Parts Example Evaluate xcosxdx xcosxdx xsinxdx my sin xdx is easy to evaluate Is there any relation between f as sin my dx and f my sin xdx Indeed by the product rule Note that It seems not that easy to solve it But 95 sinx 05 sinx x sinxy xcosxdx xsinxdx x sinx 05 sin x dx Therefore xsinx sinxdx xsinx cosx C The technique used here is called integration by parts In general the product rule says fx9fr ffr9 ffr995 Thus ltrltxgtgltxgtgtdx frrgrrdxfxgrrdrr fltxgtgltxgt frrgmdxfxgrrdrr Therefore fxgxdxfxgrr frrgrrd This is called the integrationbyparts formula Let u f and v g 95 then du f dx and dv g dx udvuv vdu Therefore Example Find f xlnmdx Example Find f In xdx To evaluate a complicated integral Fxdx using the method of integration by parts one needs to express F xdx udv Key part Find suitable u and dv if possible Some general hints on how to make the correct choice of u and dv 1 Choose u f to be a function that becomes simpler after differentiation or at least does not become more complicated 2 Choose dv to be the complicated part but still be something you can integrate 3 f vdu should be simpler or at least not more complicated than f udv 1 xexdx 0 Example A useful order ln x polynomial 617sinx or cos 95 When the integrand is the product of them pick the later one into the dv term Example ex cos xdm Example f x2 sin xdx Example Use integrationbyparts to derive the reduction formula xnerdx x7161 nxnilexdx Here we reduce the power of the variable in the integrand eg x4erdx x463 4f361d35 x463 4 361 30961de 461 4361 12 261 2BId 461 4361 12261 24 61 erdx 461 435361 1295261 243561 2461 C Example xge dx Example sec3 xdm Example ln 9 64 Moments and Centers of Mass Center of Mass Suppose two masses m1 m2 are located on a rigid line supported by a fulcrum m m1 x 2 39 X1 X 2 I1 To get balanced Where shall We put the supporter oIfm1 m2then l1 l2 0 If m1 74 m2 eg ababy and an adult on a seesaw then mlgll m2 7l2 Where g is the acceleration of gravity That is mill mglg What happens When there are a lot of masses m1 m2 mm on the line m 7 m1 3 x m 4 m2 X x1 x3 l x4 2 We usually want to know where to put the fulcrum to make the system balance that is at what point 9 to place it to make the torques add to zero For each mass mk the torque of mk about a is 0 0am The system torque about 0 equals to 2k 05k mkg Thus The system is balanced at a ltgt System torque about a 0 ltgt 2105c aka 0 ltgt Zltk a mk 0 ltgt Zkak mek0 ltgt ZkakEka izmkmkiMO lt2 ETWTW M0 ka k Mka i the moment of the system about the origin Here the number is called the moment of the system about the origin is the system s total mass and 93 n M the system s total mass is called the center of mass Example Find the center of mass for the following masses mass 2 II 6 System39s total mass rs System moment rs Therefore the center ofmass rs Wrres emmrrr Eods Home ndme center of mass of arod or athm smp ofmeta17 q mm m Idea Approximation cm r mto small preees of mass Am by apamuon ofthe rmerval Then the m1 mass ofthe system rs approximatedby 2AM Z 1kA1k e e where 6 is the density function of the rod at 05 Also the system moment is approximated by Z xkAmk Z k6ltkAQ k k Is Now we let the norm of partition P approaches 0 then b M 6xdx b M0 x6xdx and the center of mass is b 7 M0 fa x6xdx x b M fa 6xdx Example Show that the center of mass of a straight thin strip or rod of constant density 6 C lies halfway between its endpoints FIGURE 630 The center of mass of a straight thin rod or strip D39f cunstant density lice halfway between its ends Example I Exampte Fmd u center ofmass M a rod oftlensxty m E wet m 1H 1 HGURE 631 We can treat a re of variable thickness as a rod ofvariablc densin Example 2 Masses dtsttntuted over a Mane regxon o moment about 7axls mkyk o moment about yiaxxs mm Systenn moment about 0 wraxts M2 mea k FIGURE mnmcm nbom cm 532 Each mass m has a 39 a h axis My 2mm k W11 ma Hh n mt Each mass mk has a moment about each axis I System mass is The center ofmass lance me FIGURE 633 Alwudimensicnal army ofnlasscs balances on us cenler ofmaxsv Thin al plates How to nd the center ofmass ofa thin at plate Idea Cut it into strips parallel to one of the axes Then for each strip nd its mass Amk and center 5mm So smp of mass Am 0 Moment about the y axis b M ZAmkik H idm k u o Moment about the x axis b GURE 634 A plalc cm Into hm slrlps 1 4z parallel to me J39raxlsr The momcm exerted a moment Ils mass Am Ivonld exen If Mass concentramd a le smp39s ccnlcr or mass b i M dm 2 Center of mass b 2 7 My 7 fa zdm i b M fa dm b N g 7 M1 7 fa ydm i b M fa dm Thus to calculate the center of mass we need 92 g and dm 0 Sketch a strip parallel to one of the axes 0 nd the strip s mass dm and center ofmass 92 3 0 Use the formula to nd M1 My M 2737 Example Find the center of mass of a thin plate of constant density 6 covering the region bounded above by the parabola y 4 7 x and below by the ziaxis 2 1r 4 a x We cut the region into small pieces of vertical strips The typi N Cemer of mass cal vertical strip has 4 J 0 center of mass is 1 0 mass is 2 2 32 M dm 64 x2dx 6 72 2 3 2 MI gjdm 72 2 4 2 m 64 952dx 2 2 2 6 4 x22dx 0 2 256 6 16 8m2m4dm 0 15

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