text stringlengths 1 3.55k | source stringlengths 24 207 | emb listlengths 1.02k 1.02k |
|---|---|---|
6. 53 use equation 6. 18 to find the area of the surface of revolution obtained by rotating curve about the x - axis. surface integral of a scalar - valued function now that we can parameterize surfaces and we can calculate their surface areas, we are able to define surface integrals. first, let ’ s look at the surface... | openstax_calculus_volume_3_-_web | [
-0.013901746831834316,
-0.0037793132942169905,
0.011957775801420212,
-0.029087042436003685,
-0.0008976538083516061,
-0.015422297641634941,
0.01735542342066765,
-0.013257895596325397,
-0.01783018559217453,
0.060597848147153854,
0.06382489204406738,
-0.004220457281917334,
-0.0255202017724514,
... |
define a surface integral of a scalar - valued function, we let the areas of the pieces of s shrink to zero by taking a limit. definition the surface integral of a scalar - valued function of over a piecewise smooth surface s is ∞ again, notice the similarities between this definition and the definition of a scalar lin... | openstax_calculus_volume_3_-_web | [
-0.005568002350628376,
0.009326496161520481,
0.021873774006962776,
-0.01917707361280918,
0.00044672025251202285,
-0.030001958832144737,
0.017657652497291565,
-0.032401248812675476,
-0.008522977121174335,
0.05004854500293732,
0.07115481793880463,
0.0023702839389443398,
-0.02017010748386383,
... |
6. 6 • surface integrals 687 from the material we have already studied, we know that therefore, ∞ this approximation becomes arbitrarily close to ∞ as we increase the number of pieces by letting m and n go to infinity. therefore, we have the following equation to calculate scalar surface integrals : equation 6. 19 allo... | openstax_calculus_volume_3_-_web | [
-0.028540419414639473,
0.00823128130286932,
0.019146742299199104,
-0.022788135334849358,
-0.009315576404333115,
-0.004584694746881723,
-0.010457332246005535,
0.0016157726058736444,
0.016897395253181458,
0.049334004521369934,
0.060451705008745193,
-0.003124892245978117,
-0.014996937476098537,... |
6. 6 • surface integrals 689 figure 6. 72 calculating a surface integral over surface s. solution notice that s is not smooth but is piecewise smooth ; s can be written as the union of its base and its spherical top and both and are smooth. therefore, to calculate we write this integral as and we calculate integrals an... | openstax_calculus_volume_3_-_web | [
-0.009974207729101181,
0.031011050567030907,
0.04698502644896507,
-0.023174118250608444,
-0.006328728515654802,
0.007317682728171349,
-0.02406410500407219,
-0.011500890366733074,
0.019070645794272423,
0.03192611038684845,
0.024074112996459007,
0.01840030401945114,
-0.01029014028608799,
-0.... |
6. 6 • surface integrals 691 top and bottom. scalar surface integrals have several real - world applications. recall that scalar line integrals can be used to compute the mass of a wire given its density function. in a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density... | openstax_calculus_volume_3_-_web | [
0.015045953914523125,
0.015818174928426743,
0.031800203025341034,
-0.009360743686556816,
-0.026994282379746437,
-0.013109061866998672,
0.007968638092279434,
-0.024485282599925995,
0.035828977823257446,
0.04821500927209854,
0.06008746102452278,
-0.004759857431054115,
0.00407106988132,
-0.02... |
6. 56 a piece of metal has a shape that is modeled by paraboloid and the density of the metal is given by find the mass of the piece of metal. orientation of a surface recall that when we defined a scalar line integral, we did not need to worry about an orientation of the curve of integration. the same was true for sca... | openstax_calculus_volume_3_-_web | [
-0.02588161826133728,
0.004552065394818783,
-0.009961679577827454,
-0.015896888449788094,
-0.005177323240786791,
0.01087870355695486,
0.017738601192831993,
-0.017586413770914078,
0.007244677282869816,
0.05590642988681793,
0.03345438092947006,
0.010076146572828293,
-0.002102439757436514,
-0... |
the sphere, then the unit normal vectors vary continuously. this is called the positive orientation of the closed surface ( figure 6. 74 ). we also could choose the inward normal vector at each point to give an “ inward ” orientation, which is the 692 6 • vector calculus access for free at openstax. org negative orient... | openstax_calculus_volume_3_-_web | [
-0.018309984356164932,
0.02824144810438156,
0.00669126259163022,
-0.029833396896719933,
-0.003485234919935465,
0.029405295848846436,
-0.025512803345918655,
-0.00846932828426361,
0.00611178670078516,
0.05249008536338806,
0.019667576998472214,
0.029130982235074043,
-0.016690343618392944,
-0.... |
outward orientation of the cylinder ( figure 6. 75 ). figure 6. 75 if all the vectors normal to a cylinder point outward, then this is an outward orientation of the cylinder. | openstax_calculus_volume_3_-_web | [
0.00006615980964852497,
-0.015056712552905083,
0.005191261880099773,
-0.02026716247200966,
-0.006663935724645853,
0.00781963299959898,
0.012354102917015553,
0.01764434203505516,
-0.008892213925719261,
0.08276278525590897,
0.03853505477309227,
0.035781897604465485,
-0.04156692698597908,
-0.... |
6. 57 give the “ upward ” orientation of the graph of since every curve has a “ forward ” and “ backward ” direction ( or, in the case of a closed curve, a clockwise and counterclockwise direction ), it is possible to give an orientation to any curve. hence, it is possible to think of every curve as an oriented curve. ... | openstax_calculus_volume_3_-_web | [
-0.02300937846302986,
0.019432764500379562,
-0.020650437101721764,
-0.036387018859386444,
0.0038889299612492323,
0.015055578202009201,
0.027547426521778107,
-0.022095443680882454,
0.015149529092013836,
0.06172575429081917,
0.01827997714281082,
0.031700484454631805,
0.008361423388123512,
-0... |
analogous to the definition of the flux of a vector field along a plane curve. recall that if f is a two - dimensional vector field and c is a plane curve, then the definition of the flux of f along c involved chopping c into small pieces, choosing a point inside each piece, and calculating at the point ( where n is th... | openstax_calculus_volume_3_-_web | [
-0.008965573273599148,
0.04920540004968643,
0.026047907769680023,
0.028511064127087593,
-0.01503928005695343,
0.012775121256709099,
0.013190951198339462,
-0.045461442321538925,
0.001660733949393034,
0.06663946807384491,
0.0283198319375515,
0.006371499504894018,
0.001981022534891963,
-0.036... |
to approximate the mass of fluid per unit time flowing across ( and not just locally at point p ), we need to multiply by the area of therefore, the mass of fluid per unit time flowing across in the direction of n can be approximated by where n, and v are all evaluated at p ( figure 6. 78 ). this is analogous to the fl... | openstax_calculus_volume_3_-_web | [
-0.022160755470395088,
0.049502380192279816,
0.025166509672999382,
-0.008373291231691837,
0.002576600993052125,
0.01760573498904705,
0.022299019619822502,
-0.038081493228673935,
0.003751788754016161,
0.07266233116388321,
0.04036702215671539,
-0.008359014056622982,
0.003761719912290573,
-0.... |
6. 6 • surface integrals 695 figure 6. 78 the mass of fluid per unit time flowing across in the direction of n can be approximated by definition let f be a continuous vector field with a domain that contains oriented surface s with unit normal vector n. the surface integral of f over s is notice the parallel between th... | openstax_calculus_volume_3_-_web | [
-0.025275662541389465,
0.041990503668785095,
0.007835851982235909,
-0.004717301577329636,
-0.007602119818329811,
0.019243117421865463,
0.01828829012811184,
-0.0336306169629097,
0.006096845958381891,
0.06309806555509567,
0.02865055948495865,
0.01333351619541645,
0.006036080420017242,
-0.023... |
example 6. 64, the tangent vectors are and their cross product is notice that each component of the cross product is positive, and therefore this vector gives the outward orientation. 6. 6 • surface integrals 697 therefore we use the orientation for the sphere. by equation 6. 20, therefore, the mass flow rate is | openstax_calculus_volume_3_-_web | [
-0.0019982620142400265,
0.04377766698598862,
0.010044638998806477,
-0.007232478819787502,
-0.026218699291348457,
0.026314690709114075,
0.006875736638903618,
-0.015547547489404678,
0.02972170151770115,
0.055580127984285355,
0.013329533860087395,
0.05593334883451462,
0.002261672168970108,
-0... |
6. 59 let m / sec represent a velocity field of a fluid with constant density 100 kg / m3. let s be the half - cylinder oriented outward. calculate the mass flux of the fluid across s. in example 6. 70, we computed the mass flux, which is the rate of mass flow per unit area. if we want to find the flow rate ( measured ... | openstax_calculus_volume_3_-_web | [
-0.023100774735212326,
0.05459293723106384,
0.012234369292855263,
0.0254746675491333,
0.0039463103748857975,
0.016980314627289772,
0.031193343922495842,
-0.02627549134194851,
-0.005897963419556618,
0.0637989193201065,
0.019369205459952354,
-0.0027341048698872328,
0.03020475246012211,
-0.05... |
object. to find the heat flow, we need to calculate flux integral notice that s is not a smooth surface but is piecewise smooth, since s is the union of three smooth surfaces ( the circular top and bottom, and the cylindrical side ). therefore, we calculate three separate integrals, one for each smooth piece of s. befo... | openstax_calculus_volume_3_-_web | [
0.005956222303211689,
0.021529456600546837,
0.00005548600893234834,
-0.00744727673009038,
0.0206298865377903,
0.015886686742305756,
0.019798096269369125,
-0.007977232336997986,
0.014265282079577446,
0.05651039257645607,
0.04679372161626816,
0.019275525584816933,
0.021286968141794205,
-0.05... |
6. 60 a cast - iron solid ball is given by inequality the temperature at a point in a region containing the ball is find the heat flow across the boundary of the solid if this boundary is oriented outward. section 6. 6 exercises for the following exercises, determine whether the statements are true or false. 269. if su... | openstax_calculus_volume_3_-_web | [
-0.0013698387192562222,
0.026412393897771835,
0.023382917046546936,
-0.030419526621699333,
0.009899755008518696,
0.04158651828765869,
-0.004860658664256334,
0.006236776243895292,
-0.0013510534772649407,
0.0539005845785141,
0.07137599587440491,
0.03468920662999153,
-0.011889890767633915,
-0... |
6. 6 • surface integrals 701 291. evaluate surface integral where and s is the portion of plane that lies over unit square r : 292. evaluate where is the surface defined parametrically by for 293. [ t ] evaluate where s is the surface defined by 294. [ t ] evaluate where is the surface defined by 702 6 • vector calculu... | openstax_calculus_volume_3_-_web | [
0.01304519735276699,
0.035078588873147964,
0.03863142058253288,
-0.03210369870066643,
-0.00522301159799099,
0.01293862983584404,
-0.0025997161865234375,
-0.0012617222964763641,
0.036944177001714706,
0.03969937190413475,
0.0556371845304966,
0.05315841734409332,
-0.032611340284347534,
-0.033... |
6. 6 • surface integrals 703 302. compute where and n is an outward normal vector s, where s is the union of two squares and 303. compute where and n is an upward pointing normal vector, where s is the triangular region of the plane in the first octant. 304. compute where and n is an outward normal vector s, where s is... | openstax_calculus_volume_3_-_web | [
-0.011081097647547722,
0.0020619395654648542,
0.024683527648448944,
-0.018149517476558685,
-0.007562866434454918,
0.013004082255065441,
-0.008595817722380161,
-0.031694069504737854,
0.027536699548363686,
0.06166442111134529,
0.05078720301389694,
0.04428613930940628,
-0.03312268108129501,
-... |
cone let s be the spherical shell centered at the origin with radius a, and let c be the right circular cone with a vertex at the origin and an axis of symmetry that coincides with the z - axis. determine the mass of the lamina if 320. a lamina has the shape of a portion of sphere that lies within cone let s be the sph... | openstax_calculus_volume_3_-_web | [
0.023846454918384552,
0.030391691252589226,
0.054014116525650024,
-0.03433996066451073,
0.0005527909379452467,
0.014052839949727058,
-0.007459414657205343,
-0.005773379001766443,
0.04114415869116783,
0.022896144539117813,
0.06928354501724243,
0.035363346338272095,
-0.021484190598130226,
-0... |
6. 6 • surface integrals 705 321. a paper cup has the shape of an inverted right circular cone of height 6 in. and radius of top 3 in. if the cup is full of water weighing find the total force exerted by the water on the inside surface of the cup. for the following exercises, the heat flow vector field for conducting o... | openstax_calculus_volume_3_-_web | [
-0.024240856990218163,
0.030849607661366463,
0.010981076397001743,
-0.030887678265571594,
-0.0027471950743347406,
0.030652422457933426,
0.007732780184596777,
-0.009293925017118454,
0.030812297016382217,
0.07299131155014038,
0.06930797547101974,
0.04516323283314705,
0.012463197112083435,
-0... |
6. 7 stokes ’ theorem learning objectives 6. 7. 1 explain the meaning of stokes ’ theorem. 6. 7. 2 use stokes ’ theorem to evaluate a line integral. 6. 7. 3 use stokes ’ theorem to calculate a surface integral. 6. 7. 4 use stokes ’ theorem to calculate a curl. in this section, we study stokes ’ theorem, a higher - dime... | openstax_calculus_volume_3_-_web | [
-0.013308689929544926,
0.003166083712130785,
0.00528718950226903,
-0.025376014411449432,
0.006790297571569681,
0.007073542568832636,
0.03487185016274452,
-0.019042866304516792,
0.0042220838367938995,
0.08579551428556442,
0.055471450090408325,
0.01769343949854374,
0.0030418250244110823,
-0.... |
we can state stokes ’ theorem. theorem 6. 19 stokes ’ theorem let s be a piecewise smooth oriented surface with a boundary that is a simple closed curve c with positive orientation ( figure 6. 79 ). if f is a vector field with component functions that have continuous partial derivatives on an open region containing s, ... | openstax_calculus_volume_3_-_web | [
-0.025716157630085945,
-0.0005691325641237199,
0.0019316555699333549,
-0.017640016973018646,
0.010186320170760155,
0.021049797534942627,
0.011224593035876751,
-0.00022431148681789637,
0.010039088316261768,
0.08903394639492035,
0.03009079024195671,
0.017200622707605362,
0.030229518190026283,
... |
6. 7 • stokes ’ theorem 707 why the theorem is true. let s be a surface and let d be a small piece of the surface so that d does not share any points with the boundary of s. we choose d to be small enough so that it can be approximated by an oriented square e. let d inherit its orientation from s, and give e the same o... | openstax_calculus_volume_3_-_web | [
-0.01644766703248024,
0.0025162212550640106,
0.02532431297004223,
-0.019325019791722298,
0.002343504223972559,
0.012383099645376205,
0.010221155360341072,
0.002745550824329257,
0.03542686998844147,
0.09100073575973511,
0.05586022511124611,
-0.0013506998075172305,
0.011625188402831554,
-0.0... |
is the sum of all the line integrals around the boundaries of approximating squares ) can be approximated by a line integral over the boundary of s. in the limit, as the areas of the approximating squares go to zero, this approximation gets arbitrarily close to the flux. figure 6. 80 chop the surface into small pieces.... | openstax_calculus_volume_3_-_web | [
-0.02017432451248169,
0.010547774843871593,
0.028489423915743828,
-0.0009571536211296916,
-0.035308852791786194,
-0.010171144269406796,
-0.028751052916049957,
0.021921996027231216,
0.01256338320672512,
0.0908605232834816,
0.041199274361133575,
0.02334192395210266,
0.010332346893846989,
0.0... |
mind that p, q, and r are all functions of x and y, we can evaluate line integral by clairaut ’ s theorem, therefore, four of the terms disappear from this double integral, and we are left with which equals | openstax_calculus_volume_3_-_web | [
-0.019565125927329063,
0.017634006217122078,
0.030434023588895798,
0.00344829517416656,
-0.06416234374046326,
-0.01206936314702034,
0.02384086698293686,
0.022838963195681572,
0.017099713906645775,
0.08286648243665695,
0.07239934056997299,
0.022282373160123825,
-0.02628270722925663,
-0.0315... |
6. 7 • stokes ’ theorem 709 we have shown that stokes ’ theorem is true in the case of a function with a domain that is a simply connected region of finite area. we can quickly confirm this theorem for another important case : when vector field f is conservative. if f is conservative, the curl of f is zero, so since th... | openstax_calculus_volume_3_-_web | [
0.0012897559208795428,
0.012923323549330235,
0.03903737664222717,
-0.025415616109967232,
0.019644109532237053,
-0.0019035584991797805,
0.017074331641197205,
-0.004560426343232393,
0.0055207861587405205,
0.08012708276510239,
0.0658692866563797,
0.017808321863412857,
0.013452108949422836,
-0... |
6. 61 verify that stokes ’ theorem is true for vector field and surface s, where s is the upwardly oriented portion of the graph of over a triangle in the xy - plane with vertices and applying stokes ’ theorem stokes ’ theorem translates between the flux integral of surface s to a line integral around the boundary of s... | openstax_calculus_volume_3_-_web | [
-0.03114709071815014,
-0.0021406272426247597,
0.014352425932884216,
-0.04275748133659363,
0.029093973338603973,
-0.00258050044067204,
0.0012571129482239485,
-0.020614799112081528,
0.023748908191919327,
0.07077405601739883,
0.052425265312194824,
0.008884822949767113,
0.012031923979520798,
-... |
6. 7 • stokes ’ theorem 711 solution note that to calculate without using stokes ’ theorem, we would need to use equation 6. 19. use of this equation requires a parameterization of s. surface s is complicated enough that it would be extremely difficult to find a parameterization. therefore, the methods we have learned ... | openstax_calculus_volume_3_-_web | [
-0.02394033782184124,
0.023035218939185143,
0.013595051132142544,
-0.03404784947633743,
0.008848979137837887,
0.0050123995169997215,
0.025744574144482613,
-0.007115913089364767,
0.00004840774636249989,
0.07852106541395187,
0.05959327518939972,
0.0012791447807103395,
0.014321181923151016,
-... |
depends on the boundary of the path only ; it does not really depend on the path itself. analogously, suppose that s and s ′ are surfaces with the same boundary and same orientation, and suppose that g is a three - dimensional vector field that can be written as the curl of another vector field f ( so that f is like a ... | openstax_calculus_volume_3_-_web | [
-0.04153835028409958,
0.023910148069262505,
-0.008495807647705078,
-0.01546997670084238,
0.0033588348887860775,
0.017079157754778862,
0.029775027185678482,
-0.0035667186602950096,
-0.014003301970660686,
0.07708784192800522,
0.03203393891453743,
0.028926998376846313,
0.0003384642768651247,
... |
6. 62 use stokes ’ theorem to calculate surface integral where and s is the surface as shown in the following figure. the boundary curve, c, is oriented clockwise when looking along the positive y - axis. example 6. 75 calculating a line integral calculate the line integral where and c is the boundary of the parallelog... | openstax_calculus_volume_3_-_web | [
-0.0012894602259621024,
0.0037715863436460495,
0.015404111705720425,
-0.03990379348397255,
-0.009701763279736042,
0.003113986225798726,
0.03894355893135071,
-0.01669333502650261,
0.029377000406384468,
0.07320433109998703,
0.05849142745137215,
0.010579395107924938,
0.0030322351958602667,
-0... |
6. 7 • stokes ’ theorem 713 recall that if c is a closed curve and f is a vector field defined on c, then the circulation of f around c is line integral if f represents the velocity field of a fluid in space, then the circulation measures the tendency of the fluid to move in the direction of c. let f be a continuous ve... | openstax_calculus_volume_3_-_web | [
-0.011737367138266563,
0.02418985776603222,
0.03294633328914642,
-0.009718605317175388,
0.007117022760212421,
0.017692212015390396,
0.022066107019782066,
-0.020014842972159386,
-0.0013816147111356258,
0.09955733269453049,
0.055670760571956635,
0.013782639987766743,
0.017813077196478844,
0.... |
the direction of curlf. this justifies the interpretation of the curl we have learned : curl is a measure of the rotation in the vector field about the axis that points in the direction of the normal vector n, and stokes ’ theorem justifies this interpretation. 714 6 • vector calculus access for free at openstax. org f... | openstax_calculus_volume_3_-_web | [
-0.016957249492406845,
0.006848741788417101,
-0.021876104176044464,
-0.03786372393369675,
-0.0009670714498497546,
0.000740990974009037,
0.07273917645215988,
-0.04358236491680145,
-0.0009527162183076143,
0.08650682866573334,
0.0489426925778389,
0.033404745161533356,
0.01633348874747753,
-0.... |
of the integral form. by stokes ’ theorem, we can convert the line integral in the integral form into surface integral since then as long as the integration of the surface does not vary with time we also have therefore, to derive the differential form of faraday ’ s law, we would like to conclude that in general, the e... | openstax_calculus_volume_3_-_web | [
-0.03750740736722946,
0.012793416157364845,
0.019523143768310547,
-0.0301968976855278,
-0.016707012429833412,
0.010369127616286278,
0.023910026997327805,
-0.010922993533313274,
0.03064514696598053,
0.08137186616659164,
0.04433019459247589,
0.02499539591372013,
0.01869424432516098,
-0.04598... |
6. 7 • stokes ’ theorem 715 is not enough to conclude that the integral symbols do not simply “ cancel out, ” leaving equality of the integrands. to see why the integral symbol does not just cancel out in general, consider the two single - variable integrals and where both of these integrals equal so however, analogous... | openstax_calculus_volume_3_-_web | [
-0.01983991451561451,
0.007932006381452084,
0.01545084547251463,
-0.02171516604721546,
-0.013632792979478836,
-0.02467752993106842,
0.06470663845539093,
-0.008443606086075306,
-0.023767489939928055,
0.0903838723897934,
0.08033949136734009,
0.03093268722295761,
0.02474335767328739,
-0.03580... |
6. 64 calculate the curl of electric field e if the corresponding magnetic field is ∞ notice that the curl of the electric field does not change over time, although the magnetic field does change over time. section 6. 7 exercises for the following exercises, without using stokes ’ theorem, calculate directly both the f... | openstax_calculus_volume_3_-_web | [
-0.02316432259976864,
0.028316468000411987,
0.0036002276465296745,
-0.05518639460206032,
0.016266698017716408,
-0.0018799839308485389,
0.01874581165611744,
-0.02444768138229847,
0.02596353180706501,
0.06401946395635605,
0.05927858129143715,
0.02529965527355671,
0.00019174421322531998,
-0.0... |
6. 7 • stokes ’ theorem 717 337. [ t ] use a cas and stokes ’ theorem to approximate line integral where c is the intersection of the xy - plane and hemisphere traversed counterclockwise viewed from the top — that is, from the positive z - axis toward the xy - plane. 338. [ t ] use a cas and stokes ’ theorem to approxi... | openstax_calculus_volume_3_-_web | [
-0.005871979519724846,
-0.010805939324200153,
0.04730259254574776,
-0.04035850614309311,
0.018603490665555,
0.0010745751205831766,
-0.023890463635325432,
-0.024845130741596222,
0.01270224992185831,
0.09813734889030457,
0.0643911138176918,
0.017100326716899872,
-0.029222644865512848,
-0.042... |
where s is surface c is boundary circle and s is oriented in the positive z - direction. 348. use stokes ’ theorem for vector field where s is that part of the surface of plane contained within triangle c with vertices ( 1, 0, 0 ), ( 0, 1, 0 ), and ( 0, 0, 1 ), traversed counterclockwise as viewed from above. 349. a ce... | openstax_calculus_volume_3_-_web | [
0.009810938499867916,
0.011132894083857536,
0.02632058970630169,
-0.03461417928338051,
0.00891825556755066,
0.016223164275288582,
-0.027781177312135696,
-0.009179194457828999,
-0.00736259575933218,
0.06791423261165619,
0.04261786490678787,
0.01879667304456234,
-0.007243467960506678,
-0.022... |
6. 7 • stokes ’ theorem 719 353. use stokes ’ theorem to evaluate where and c is a triangle with vertices ( 0, 0, 0 ), ( 2, 0, 0 ) and oriented counterclockwise when viewed from above. 354. use the surface integral in stokes ’ theorem to calculate the circulation of field f, around c, which is the intersection of cylin... | openstax_calculus_volume_3_-_web | [
0.015929872170090675,
0.017155734822154045,
0.03188081458210945,
-0.045742981135845184,
0.023880233988165855,
0.01453244686126709,
-0.004801567178219557,
-0.01623091846704483,
0.01965380646288395,
0.07046236097812653,
0.07432345300912857,
0.011350132524967194,
-0.01313717570155859,
-0.0464... |
maximum value? for the following application exercises, the goal is to evaluate where and s is the upper half of ellipsoid 365. evaluate a surface integral over a more convenient surface to find the value of a. 366. evaluate a using a line integral. 367. take paraboloid for and slice it with plane let s be the surface ... | openstax_calculus_volume_3_-_web | [
0.02952566370368004,
0.03124312311410904,
0.020550057291984558,
-0.012324215844273567,
-0.004707531537860632,
0.008303437381982803,
0.013591652736067772,
-0.00370399234816432,
0.01802666112780571,
0.061584360897541046,
0.02710181474685669,
0.034839533269405365,
0.035021159797906876,
-0.022... |
6. 8 the divergence theorem learning objectives 6. 8. 1 explain the meaning of the divergence theorem. 6. 8. 2 use the divergence theorem to calculate the flux of a vector field. 6. 8. 3 apply the divergence theorem to an electrostatic field. we have examined several versions of the fundamental theorem of calculus in h... | openstax_calculus_volume_3_-_web | [
-0.024164656177163124,
0.014072393998503685,
0.004055672325193882,
-0.00025025507784448564,
-0.014021293260157108,
0.0061276680789887905,
0.01228330284357071,
-0.013629255816340446,
0.000892257783561945,
0.10282008349895477,
0.018068131059408188,
0.03094499744474888,
0.020187100395560265,
... |
for free at openstax. org since and divergence is a derivative of sorts, the flux form of green ’ s theorem relates the integral of derivative divf over planar region d to an integral of f over the boundary of d. 5. stokes ’ theorem : if we think of the curl as a derivative of sorts, then stokes ’ theorem relates the i... | openstax_calculus_volume_3_-_web | [
-0.04002900794148445,
0.0035102074034512043,
0.020689962431788445,
-0.026136508211493492,
-0.0022953643929213285,
0.019588211551308632,
0.03942272067070007,
-0.004054322838783264,
0.0008856665808707476,
0.10164563357830048,
0.03210081905126572,
0.044038545340299606,
0.03526897355914116,
-0... |
planes inside e ( figure 6. 88 ). let the center of b have coordinates and suppose the edge lengths are and ( figure 6. 88 ( b ) ). the normal vector out of the top of the box is k and the normal vector out of the bottom of the box is the dot product of with k is r and the dot product with is the area of the top of the... | openstax_calculus_volume_3_-_web | [
0.016114478930830956,
0.010729873552918434,
0.036563582718372345,
-0.04434377700090408,
-0.004674104042351246,
-0.0255579873919487,
-0.034869235008955,
0.010155675932765007,
-0.0584307536482811,
0.06069262698292732,
0.027448007836937904,
0.037267573177814484,
-0.01651684381067753,
-0.04029... |
6. 8 • the divergence theorem 723 figure 6. 88 ( a ) a small box b inside surface e has sides parallel to the coordinate planes. ( b ) box b has side lengths and ( c ) if we look at the side view of b, we see that, since is the center of the box, to get to the top of the box we must travel a vertical distance of up fro... | openstax_calculus_volume_3_-_web | [
0.004096320364624262,
0.02071266621351242,
0.050363149493932724,
-0.048623714596033096,
0.024460287764668465,
-0.021288933232426643,
0.004394586198031902,
-0.002145160920917988,
-0.0009278662037104368,
0.09850683063268661,
0.05044364184141159,
0.05113207548856735,
0.00714331679046154,
-0.0... |
as the volumes of the approximating boxes shrink to zero, this approximation becomes arbitrarily close to the flux over s. example 6. 77 verifying the divergence theorem verify the divergence theorem for vector field and surface s that consists of cone and the circular top of the cone ( see the following figure ). assu... | openstax_calculus_volume_3_-_web | [
0.0003233939642086625,
0.02988894097507,
0.041038308292627335,
-0.03202322870492935,
0.008998501114547253,
-0.007296400144696236,
-0.009026297368109226,
-0.009138580411672592,
0.006064144894480705,
0.08819802850484848,
0.051187947392463684,
0.022485069930553436,
0.020036282017827034,
-0.03... |
6. 65 verify the divergence theorem for vector field and surface s given by the cylinder plus the circular top and bottom of the cylinder. assume that s is oriented outward. recall that the divergence of continuous field f at point p is a measure of the “ outflowing - ness ” of the field at p. if f represents the veloc... | openstax_calculus_volume_3_-_web | [
-0.023988282307982445,
0.02571280486881733,
0.028435034677386284,
-0.02403511106967926,
0.00435120752081275,
-0.010635549202561378,
0.010563269257545471,
-0.020872754976153374,
0.0013341817539185286,
0.08081547170877457,
0.05145309865474701,
0.032374609261751175,
0.00692354142665863,
-0.03... |
for the bottom, and one for the side. furthermore, each integral would require parameterizing the corresponding surface, calculating tangent vectors and their cross product, and using equation 6. 19. by contrast, the divergence theorem allows us to calculate the single triple integral where e is the solid enclosed by t... | openstax_calculus_volume_3_-_web | [
-0.016767606139183044,
0.028227735310792923,
0.02755620703101158,
-0.04534238204360008,
-0.046198513358831406,
-0.007653150707483292,
-0.001277883886359632,
0.0011110324412584305,
0.004685928579419851,
0.09178494662046432,
0.03346449136734009,
0.04376548156142235,
0.0029118163511157036,
-0... |
6. 8 • the divergence theorem 727 figure 6. 90 vector field solution the flow rate of the fluid across s is before calculating this flux integral, let ’ s discuss what the value of the integral should be. based on figure 6. 90, we see that if we place this cube in the fluid ( as long as the cube doesn ’ t encompass the... | openstax_calculus_volume_3_-_web | [
-0.038697972893714905,
0.03367053344845772,
0.014351519756019115,
-0.03084857016801834,
-0.011073869653046131,
-0.013908755034208298,
0.02086767368018627,
-0.011313353665173054,
-0.007220592815428972,
0.09534550458192825,
0.025047866627573967,
0.053176552057266235,
-0.00044584082206711173,
... |
6. 67 let be the velocity field of a fluid. let c be the solid cube given by and let s be the boundary of this cube ( see the following figure ). find the flow rate of the fluid across s. 728 6 • vector calculus access for free at openstax. org example 6. 79 illustrates a remarkable consequence of the divergence theore... | openstax_calculus_volume_3_-_web | [
-0.04397033154964447,
0.03764459118247032,
0.018744733184576035,
-0.01636846736073494,
-0.00969117134809494,
-0.013055761344730854,
0.02223089523613453,
-0.020661288872361183,
-0.00555534427985549,
0.09597668796777725,
0.024479424580931664,
0.03948724642395973,
0.01737494394183159,
-0.0307... |
positive, outward orientation. since the radius of is small and f is continuous, the divergence of f is approximately constant on that is, if is any point in then let denote the boundary sphere of we can approximate the flux across using the divergence theorem as follows : as we shrink the radius r to zero via a limit,... | openstax_calculus_volume_3_-_web | [
-0.00397785659879446,
0.02223561331629753,
0.0159092266112566,
-0.020272912457585335,
-0.015697108581662178,
-0.006881581153720617,
0.028550347313284874,
-0.002967358101159334,
-0.009920628741383553,
0.08569750934839249,
0.04923737421631813,
0.017863404005765915,
-0.004620423540472984,
-0.... |
6. 8 • the divergence theorem 729 application to electrostatic fields the divergence theorem has many applications in physics and engineering. it allows us to write many physical laws in both an integral form and a differential form ( in much the same way that stokes ’ theorem allowed us to translate between an integra... | openstax_calculus_volume_3_-_web | [
-0.02541419304907322,
0.01878722570836544,
0.014332344755530357,
-0.0187165979295969,
-0.025554364547133446,
-0.009749582037329674,
-0.003452486591413617,
-0.01800013706088066,
-0.004279519431293011,
0.09487929940223694,
0.02091439627110958,
0.04222848266363144,
0.020005451515316963,
-0.04... |
because we say that electrostatic fields obey an inverse - square law. that is, the electrostatic force at a given point is inversely proportional to the square of the distance from the source of the charge ( which in this case is at the origin ). given this vector field, we show that the flux across closed surface s i... | openstax_calculus_volume_3_-_web | [
-0.04037191718816757,
0.017310380935668945,
0.006080749444663525,
-0.011929788626730442,
-0.008361068554222584,
-0.01001456193625927,
0.014327527955174446,
-0.00925854966044426,
-0.022168880328536034,
0.0856037512421608,
0.025239545851945877,
0.030863933265209198,
0.0304238460958004,
-0.04... |
of and since the divergence of is zero, we can immediately apply the divergence theorem and find that is zero. now suppose that s does encompass the origin. we cannot just use the divergence theorem to calculate the flux, because the field is not defined at the origin. let be a sphere of radius a inside of s centered a... | openstax_calculus_volume_3_-_web | [
-0.037680841982364655,
0.001070060534402728,
0.011497991159558296,
-0.04372239112854004,
-0.02172781527042389,
-0.0007476085447706282,
-0.009433194063603878,
-0.01149169635027647,
-0.02550230361521244,
0.08376012742519379,
0.028269480913877487,
0.036678213626146317,
0.02462761476635933,
-0... |
6. 8 • the divergence theorem 731 therefore, and we have our desired result. now we return to calculating the flux across a smooth surface in the context of electrostatic field of a point charge at the origin. let s be a piecewise smooth closed surface that encompasses the origin. then if s does not encompass the origi... | openstax_calculus_volume_3_-_web | [
-0.02514895424246788,
0.03378795459866524,
0.029298918321728706,
-0.007622682023793459,
-0.025700021535158157,
-0.027458960190415382,
0.029801174998283386,
-0.006279740482568741,
-0.006548140197992325,
0.09125399589538574,
0.009318012744188309,
0.04218463972210884,
0.01856834813952446,
-0.... |
6. 68 work the previous example for surface s that is a sphere of radius 4 centered at the origin, oriented outward. 732 6 • vector calculus access for free at openstax. org section 6. 8 exercises for the following exercises, use a computer algebraic system ( cas ) and the divergence theorem to evaluate surface integra... | openstax_calculus_volume_3_-_web | [
-0.037818145006895065,
0.00606573885306716,
0.04122207313776016,
-0.05327900871634483,
-0.0035973170306533575,
0.013965710997581482,
-0.021229146048426628,
-0.025455331429839134,
0.013105180114507675,
0.06299017369747162,
0.05131693556904793,
0.05304693058133125,
-0.023934325203299522,
-0.... |
6. 8 • the divergence theorem 733 389. [ t ] use a cas and the divergence theorem to calculate flux where and s is a sphere with center ( 0, 0, 0 ) and radius 2. 390. use the divergence theorem to compute the value of flux integral where and s is the surface of the solid bounded by 391. use the divergence theorem to co... | openstax_calculus_volume_3_-_web | [
-0.016705254092812538,
0.02964824065566063,
0.03222914785146713,
-0.01885228604078293,
0.02387169748544693,
0.0012196613242849708,
0.00032666660263203084,
-0.028616491705179214,
0.003081038361415267,
0.08340851217508316,
0.02081340178847313,
0.037782035768032074,
-0.008586262352764606,
-0.... |
in the first octant between planes and 402. let use the divergence theorem to calculate where s is the surface of the cube with corners at oriented outward. 403. use the divergence theorem to find the outward flux of field through the cube bounded by planes 404. let and let s be hemisphere together with disk in the xy ... | openstax_calculus_volume_3_-_web | [
-0.008250376209616661,
0.01440494041889906,
-0.004567082971334457,
-0.032105036079883575,
-0.007696887478232384,
0.012799987569451332,
0.03005075454711914,
-0.03633086755871773,
0.007034081965684891,
0.07702624052762985,
0.015675541013479233,
0.05399879813194275,
-0.004633190575987101,
-0.... |
6. 8 • the divergence theorem 735 407. use the divergence theorem to evaluate where and s is sphere with constant 408. use the divergence theorem to evaluate where and s is the boundary of the cube defined by 409. let r be the region defined by use the divergence theorem to find 410. let e be the solid bounded by the x... | openstax_calculus_volume_3_-_web | [
-0.021488331258296967,
0.028841432183980942,
0.029785804450511932,
-0.024184593930840492,
0.012277350760996342,
0.01621505431830883,
0.0068565900437533855,
-0.027740079909563065,
0.008474788628518581,
0.0891304612159729,
0.039622992277145386,
0.042863644659519196,
-0.01949736662209034,
-0.... |
the flux of across surface s, where s is the boundary of the solid bounded by hemispheres and and plane 421. use the divergence theorem to evaluate where and s is the surface consisting of three pieces : on the top ; on the sides ; and on the bottom. 422. [ t ] use a cas and the divergence theorem to evaluate where and... | openstax_calculus_volume_3_-_web | [
-0.025030093267560005,
0.03897182643413544,
0.011165388859808445,
-0.028484072536230087,
-0.017610246315598488,
0.030996980145573616,
0.017993614077568054,
-0.02343699336051941,
0.010311686433851719,
0.0890551507472992,
0.019740387797355652,
0.050169698894023895,
-0.00045649585081264377,
-... |
6. 8 • the divergence theorem 737 chapter review key terms circulation the tendency of a fluid to move in the direction of curve c. if c is a closed curve, then the circulation of f along c is line integral which we also denote closed curve a curve for which there exists a parameterization such that and the curve is tr... | openstax_calculus_volume_3_-_web | [
-0.020294025540351868,
0.0026537086814641953,
0.015905730426311493,
0.00506081897765398,
-0.002835208550095558,
0.007165215909481049,
0.033581145107746124,
-0.02718123234808445,
-0.012727237306535244,
0.10818837583065033,
0.021583348512649536,
0.027296319603919983,
0.004983352962881327,
-0... |
grid lines in a coordinate plane heat flow a vector field proportional to the negative temperature gradient in an object independence of path a vector field f has path independence if for any curves and in the domain of f with the same initial points and terminal points inverse - square law the electrostatic force at a... | openstax_calculus_volume_3_-_web | [
-0.02307509444653988,
0.012005086988210678,
-0.020835228264331818,
0.007091857958585024,
-0.0002983231970574707,
0.02697419561445713,
0.0020300045143812895,
-0.02321162074804306,
-0.029298976063728333,
0.07911095023155212,
0.03393978253006935,
0.015083208680152893,
-0.01562442909926176,
-0... |
a function along a curve c with respect to arc length is the integral it is the integral of a scalar function along a curve in a plane or in space ; such an integral is defined in terms of a riemann sum, as is a single - variable integral simple curve a curve that does not cross itself simply connected region a region ... | openstax_calculus_volume_3_-_web | [
-0.036370087414979935,
0.025810664519667625,
0.002586810849606991,
-0.008733825758099556,
0.011367741040885448,
0.00518087949603796,
0.04032426327466965,
-0.040172278881073,
-0.013188190758228302,
0.07080161571502686,
0.060106247663497925,
0.01566162332892418,
-0.013188214972615242,
-0.047... |
theorem, circulation form where c is the boundary of d green ’ s theorem, flux form where c is the boundary of d green ’ s theorem, extended version curl divergence divergence of curl is zero curl of a gradient is the zero vector scalar surface integral flux integral stokes ’ theorem divergence theorem key concepts 6. ... | openstax_calculus_volume_3_-_web | [
-0.018570397049188614,
0.010589811950922012,
0.017364250496029854,
-0.04582361504435539,
0.008936600759625435,
0.013220107182860374,
0.04349887743592262,
-0.023259969428181648,
-0.006932936143130064,
0.07893086224794388,
0.04313110187649727,
0.017133554443717003,
0.017260143533349037,
-0.0... |
6. 2 line integrals • line integrals generalize the notion of a single - variable integral to higher dimensions. the domain of integration in a single - variable integral is a line segment along the x - axis, but the domain of integration in a line integral is a curve in a plane or in space. • if c is a curve, then the... | openstax_calculus_volume_3_-_web | [
-0.015886245295405388,
0.009471483528614044,
0.020830443128943443,
0.010973574593663216,
-0.019076354801654816,
-0.008933154866099358,
0.032380830496549606,
-0.04152323678135872,
0.005964923650026321,
0.09033716470003128,
0.04293469712138176,
0.03207497298717499,
-0.008974548429250717,
-0.... |
6. 3 conservative vector fields • the theorems in this section require curves that are closed, simple, or both, and regions that are connected or simply connected. • the line integral of a conservative vector field can be calculated using the fundamental theorem for line integrals. this theorem is a generalization of t... | openstax_calculus_volume_3_-_web | [
-0.03231476619839668,
0.01755918189883232,
0.015393600799143314,
0.00019647044246084988,
0.020528322085738182,
-0.005793331656605005,
0.03194410726428032,
-0.005099582951515913,
0.001917287241667509,
0.10562854260206223,
0.02641334757208824,
0.0054436721839010715,
0.01636255346238613,
-0.0... |
6. 4 green ’ s theorem • green ’ s theorem relates the integral over a connected region to an integral over the boundary of the region. green ’ s theorem is a version of the fundamental theorem of calculus in one higher dimension. • green ’ s theorem comes in two forms : a circulation form and a flux form. in the circu... | openstax_calculus_volume_3_-_web | [
-0.040464576333761215,
-0.004730813670903444,
0.0165056511759758,
-0.006453523878008127,
-0.005052591674029827,
-0.0013621289981529117,
0.03181624040007591,
-0.013005243614315987,
-0.003008234780281782,
0.09962303191423416,
0.04252230376005173,
0.01986263133585453,
0.03443865850567818,
-0.... |
6. 6 surface integrals • surfaces can be parameterized, just as curves can be parameterized. in general, surfaces must be parameterized with two parameters. • surfaces can sometimes be oriented, just as curves can be oriented. some surfaces, such as a mobius strip, cannot be oriented. • a surface integral is like a lin... | openstax_calculus_volume_3_-_web | [
-0.0170615091919899,
-0.007661212235689163,
0.005265053827315569,
-0.04408324882388115,
-0.014711559750139713,
0.0077684186398983,
0.02013091742992401,
-0.018592311069369316,
0.008632482029497623,
0.07217273861169815,
0.06383564323186874,
0.01216807495802641,
-0.0036392465699464083,
-0.026... |
6. 8 the divergence theorem • the divergence theorem relates a surface integral across closed surface s to a triple integral over the solid enclosed by s. the divergence theorem is a higher dimensional version of the flux form of green ’ s theorem, and is therefore a 6 • chapter review 741 higher dimensional version of... | openstax_calculus_volume_3_-_web | [
-0.019779402762651443,
0.005746848415583372,
0.034143272787332535,
-0.012674622237682343,
0.019969148561358452,
0.010614781640470028,
0.011785841546952724,
-0.017355307936668396,
-0.0062683625146746635,
0.09721652418375015,
0.04818495362997055,
0.029455360025167465,
0.010096759535372257,
-... |
##6. over cube defined by 447. where is bounded by paraboloid and plane 448. find the amount of work performed by a 50 - kg woman ascending a helical staircase with radius 2 m and height 100 m. the woman completes five revolutions during the climb. 449. find the total mass of a thin wire in the shape of an upper semici... | openstax_calculus_volume_3_-_web | [
0.005339567083865404,
0.016156034544110298,
0.02821371704339981,
-0.023268267512321472,
-0.0012524640187621117,
0.007627974264323711,
0.027322981506586075,
0.026484161615371704,
0.04371669515967369,
0.06504166126251221,
0.05588126927614212,
0.06382332742214203,
0.00012878798588644713,
-0.0... |
7. 4 series solutions of differential equations introduction we have already studied the basics of differential equations, including separable first - order equations. in this chapter, we go a little further and look at second - order equations, which are equations containing second derivatives of the dependent variabl... | openstax_calculus_volume_3_-_web | [
0.0010585663840174675,
0.044666364789009094,
0.037608321756124496,
-0.017307810485363007,
-0.04633694887161255,
-0.003331267973408103,
-0.020595096051692963,
-0.009615597315132618,
0.04188692942261696,
0.08623743057250977,
-0.003159501124173403,
0.046611107885837555,
0.02805253304541111,
-... |
7. 1 second - order linear equations learning objectives 7. 1. 1 recognize homogeneous and nonhomogeneous linear differential equations. 7. 1. 2 determine the characteristic equation of a homogeneous linear equation. 7. 1. 3 use the roots of the characteristic equation to find the solution to a homogeneous linear equat... | openstax_calculus_volume_3_-_web | [
0.012696684338152409,
0.061191532760858536,
0.006373875308781862,
-0.037244610488414764,
-0.03908080607652664,
0.008467492647469044,
-0.010952816344797611,
0.01476070936769247,
0.026784004643559456,
0.07816636562347412,
-0.000009953122571459971,
0.030700474977493286,
0.006804474629461765,
... |
are real - valued functions and is not identically zero. if — in other words, if for every value of x — the equation is said to be a homogeneous linear equation. if for some value of the equation is said to be a nonhomogeneous linear equation. media visit this website ( http : / / www. openstax. org / l / 20 _ secondor... | openstax_calculus_volume_3_-_web | [
-0.006072566378861666,
0.039595719426870346,
0.014037013985216618,
-0.005866382736712694,
-0.0010987085988745093,
-0.005877206102013588,
-0.012415314093232155,
-0.009483986534178257,
-0.00540782930329442,
0.0855167880654335,
0.021620605140924454,
0.048337798565626144,
0.005102104507386684,
... |
f. this equation is linear. rewriting it in standard form gives with the equation in standard form, we can see that so the equation is nonhomogeneous. g. this equation looks like it ’ s linear, but we should rewrite it in standard form to be sure. we get this equation is, indeed, linear. with it is nonhomogeneous. h. t... | openstax_calculus_volume_3_-_web | [
0.028714003041386604,
0.022321350872516632,
0.022067494690418243,
0.0008800445939414203,
-0.009630946442484856,
-0.0029742002952843904,
-0.013111284002661705,
0.02322070114314556,
0.0034948699176311493,
0.09182706475257874,
0.011908440850675106,
0.03369934484362602,
0.008566608652472496,
-... |
7. 1 classify each of the following equations as linear or nonlinear. if the equation is linear, determine further whether it is homogeneous or nonhomogeneous. a. b. later in this section, we will see some techniques for solving specific types of differential equations. before we get to that, however, let ’ s get a fee... | openstax_calculus_volume_3_-_web | [
0.010751714929938316,
0.034356117248535156,
0.017885301262140274,
-0.031830813735723495,
-0.03335558995604515,
-0.0025454990100115538,
-0.019522204995155334,
0.023909341543912888,
0.03606696426868439,
0.09850798547267914,
0.022707052528858185,
0.04051940515637398,
0.016472505405545235,
0.0... |
7. 2 show that is a solution to the differential equation although simply finding any solution to a differential equation is important, mathematicians and engineers often want to go beyond finding one solution to a differential equation to finding all solutions to a differential equation. in other words, we want to fin... | openstax_calculus_volume_3_-_web | [
0.016631612554192543,
0.03718973696231842,
0.01206942554563284,
-0.005639474838972092,
-0.04564836621284485,
-0.02627870813012123,
-0.018742162734270096,
-0.00350961578078568,
0.05175168812274933,
0.10006310045719147,
0.015348073095083237,
0.026253115385770798,
0.003575370879843831,
0.0131... |
7. 3 consider the differential equation given that and are solutions to this differential equation, show that is a solution. unfortunately, to find the general solution to a second - order differential equation, it is not enough to find any two solutions and then combine them. consider the differential equation 748 7 •... | openstax_calculus_volume_3_-_web | [
0.026255201548337936,
0.03648946061730385,
0.027327679097652435,
0.006337563041597605,
-0.04106113314628601,
-0.021382726728916168,
-0.020113641396164894,
0.005074501037597656,
0.03596344590187073,
0.09969702363014221,
0.0092705599963665,
0.03940417245030403,
0.006625879090279341,
0.003317... |
for all x over the interval of interest. then, now, since we stated that and can ’ t both be zero, assume then, there are two cases : either or if then so one of the functions is identically zero. now suppose then, and we see that the functions are constant multiples of one another. theorem 7. 2 linear dependence of tw... | openstax_calculus_volume_3_-_web | [
0.018947793170809746,
0.05588396266102791,
0.02465401589870453,
0.007365722209215164,
-0.03123250976204872,
-0.0121468435972929,
-0.02179400436580181,
0.004248099867254496,
0.0050346702337265015,
0.08314362913370132,
0.0255904383957386,
0.027250684797763824,
-0.011161758564412594,
-0.04325... |
7. 4 determine whether the following pairs of functions are linearly dependent or linearly independent : if we are able to find two linearly independent solutions to a second - order differential equation, then we can combine them to find the general solution. this result is formally stated in the following theorem. th... | openstax_calculus_volume_3_-_web | [
0.019162455573678017,
0.04294108599424362,
0.018894540145993233,
0.0010918529005721211,
-0.04029010236263275,
-0.005080027971416712,
-0.008924019522964954,
0.002578077372163534,
0.055474910885095596,
0.0917242243885994,
0.006654560100287199,
0.04032306745648384,
-0.018109014257788658,
0.00... |
7. 5 if and are solutions to what is the general solution? second - order equations with constant coefficients now that we have a better feel for linear differential equations, we are going to concentrate on solving second - order equations of the form where and are constants. since all the coefficients are constants, ... | openstax_calculus_volume_3_-_web | [
0.0417746976017952,
0.0505862720310688,
0.036464523524045944,
-0.028301220387220383,
-0.033251743763685226,
-0.02403990365564823,
0.021610474213957787,
0.03151192516088486,
0.01892188936471939,
0.09027379006147385,
0.054495371878147125,
0.005090589169412851,
0.020570719614624977,
-0.009305... |
characteristic equation has a repeated real root, in this case, we know is a solution to equation 7. 1, but it is only one solution and we need two linearly independent solutions to determine the general solution. we might be tempted to try a function of the form where k is some constant, but it would not be linearly i... | openstax_calculus_volume_3_-_web | [
0.04071269556879997,
0.03831217437982559,
0.024706916883587837,
-0.01727321371436119,
-0.02928447537124157,
-0.01347438059747219,
0.0012957985745742917,
0.011406202800571918,
0.024427583441138268,
0.08925559371709824,
0.09468387067317963,
0.019891519099473953,
0.02266603522002697,
-0.00971... |
7. 1 • second - order linear equations 751 where and are constants. for example, the differential equation has the associated characteristic equation this factors into which has a repeated root therefore, the general solution to this differential equation is complex conjugate roots the third case we must consider is wh... | openstax_calculus_volume_3_-_web | [
0.011070183478295803,
0.042669590562582016,
0.04047928750514984,
-0.03193444013595581,
-0.01785079762339592,
0.008297593332827091,
-0.003907471429556608,
0.0011956961825489998,
0.021058082580566406,
0.0892830565571785,
0.04572233185172081,
0.031780317425727844,
0.022925537079572678,
-0.023... |
a little bit of algebraic manipulation, we can find two linearly independent, real - value solutions to equation 7. 1 and express our general solution in those terms. we encountered exponential functions with complex exponents earlier. one of the key tools we used to express these exponential functions in terms of sine... | openstax_calculus_volume_3_-_web | [
0.031247448176145554,
0.03881119191646576,
0.027797002345323563,
-0.03519269451498985,
-0.025646332651376724,
-0.02131730504333973,
0.00670202961191535,
0.020092010498046875,
0.03633693605661392,
0.0842980146408081,
0.06290895491838455,
0.01348903402686119,
0.037321917712688446,
-0.0281322... |
the characteristic equation or use the quadratic formula to find the roots. 4. determine the form of the general solution based on whether the characteristic equation has distinct, real roots ; a single, repeated real root ; or complex conjugate roots. example 7. 6 solving second - order equations with constant coeffic... | openstax_calculus_volume_3_-_web | [
0.028708836063742638,
0.04291002079844475,
0.0448983833193779,
-0.035901989787817,
-0.01146426796913147,
0.0015370656037703156,
0.01480054296553135,
0.020136939361691475,
0.032491497695446014,
0.08560016751289368,
0.052683811634778976,
0.02800123766064644,
0.029523620381951332,
-0.02779303... |
7. 1 • second - order linear equations 753 c. the characteristic equation is ( step 2 ). this factors into so the characteristic equation has a repeated real root ( step 3 ). then the general solution to the differential equation is d. the characteristic equation is ( step 2 ). this factors into so the roots of the cha... | openstax_calculus_volume_3_-_web | [
0.042069241404533386,
0.039126746356487274,
0.03716369718313217,
-0.04233931750059128,
-0.01721344329416752,
-0.005560921970754862,
0.006767801940441132,
0.021795526146888733,
0.030880313366651535,
0.0879557728767395,
0.028569180518388748,
0.025791382417082787,
0.038377609103918076,
-0.019... |
7. 6 find the general solution to the following differential equations : a. b. initial - value problems and boundary - value problems so far, we have been finding general solutions to differential equations. however, differential equations are often used to describe physical systems, and the person studying that physic... | openstax_calculus_volume_3_-_web | [
0.004198538139462471,
0.04996570199728012,
0.03754941001534462,
-0.028102098032832146,
-0.03854113817214966,
-0.01406074222177267,
-0.015176614746451378,
0.02223242074251175,
0.015054174698889256,
0.08234161138534546,
-0.002100676530972123,
0.007367119193077087,
0.022120269015431404,
0.012... |
conditions are provided. boundary - value problems, however, are not as well behaved. even when two boundary conditions are known, we may encounter boundary - value problems with unique solutions, many solutions, or no solution at all. example 7. 7 solving an initial - value problem solve the following initial - value ... | openstax_calculus_volume_3_-_web | [
0.006404637359082699,
0.025336209684610367,
0.03728463873267174,
-0.016179658472537994,
-0.03740531578660011,
-0.045087993144989014,
-0.02566670812666416,
0.03473043069243431,
0.010118106380105019,
0.07880675047636032,
0.013177716173231602,
0.020522572100162506,
0.02235778607428074,
0.0044... |
7. 1 • second - order linear equations 755 example 7. 9 initial - value problem representing a spring - mass system the following initial - value problem models the position of an object with mass attached to a spring. spring - mass systems are examined in detail in applications. the solution to the differential equati... | openstax_calculus_volume_3_-_web | [
-0.014320265501737595,
0.03811197355389595,
0.028864583000540733,
-0.026157420128583908,
-0.010144978761672974,
0.016797557473182678,
0.00946794357150793,
0.021654320880770683,
0.03239543363451958,
0.05903980880975723,
-0.0004841775808017701,
0.01757238060235977,
0.0078970892354846,
-0.005... |
7. 9 suppose the following initial - value problem models the position ( in feet ) of a mass in a spring - mass system at any given time. solve the initial - value problem and graph the solution. what is the position of the mass at time sec? how fast is it moving at time sec? in what direction? example 7. 10 solving a ... | openstax_calculus_volume_3_-_web | [
0.002314965706318617,
0.05110582709312439,
0.00778424134477973,
-0.005518496036529541,
-0.03031040169298649,
-0.005521544720977545,
-0.00596570922061801,
0.025497693568468094,
0.015176498331129551,
0.07336895912885666,
-0.006757709197700024,
0.023611783981323242,
0.02202305570244789,
-0.00... |
12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. solve the initial - value problem. 31. 32. 33. 34. 35. 36. | openstax_calculus_volume_3_-_web | [
0.005551021546125412,
0.009129933081567287,
0.026887763291597366,
-0.01712881587445736,
-0.03017611615359783,
-0.002807608339935541,
-0.018422478809952736,
0.017507826909422874,
0.029122252017259598,
0.05872452259063721,
0.021468011662364006,
0.022611422464251518,
0.019532902166247368,
-0.... |
7. 1 • second - order linear equations 757 37. 38. solve the boundary - value problem, if possible. 39. 40. 41. 42. 43. 44. 45. 46. 47. find a differential equation with a general solution that is 48. find a differential equation with a general solution that is for each of the following differential equations : a. solv... | openstax_calculus_volume_3_-_web | [
0.009999837726354599,
0.04904455319046974,
0.02134052850306034,
-0.009128575213253498,
-0.012968303635716438,
-0.008041509427130222,
0.006949730217456818,
0.009856508113443851,
0.03908929601311684,
0.0942324697971344,
-0.019501596689224243,
0.035537660121917725,
0.03279687464237213,
-0.011... |
7. 2 nonhomogeneous linear equations learning objectives 7. 2. 1 write the general solution to a nonhomogeneous differential equation. 7. 2. 2 solve a nonhomogeneous differential equation by the method of undetermined coefficients. 7. 2. 3 solve a nonhomogeneous differential equation by the method of variation of param... | openstax_calculus_volume_3_-_web | [
0.019880201667547226,
0.0391266904771328,
0.005591719876974821,
-0.013240817934274673,
-0.034599144011735916,
0.005463052075356245,
-0.007438921369612217,
0.005456084385514259,
-0.00002294642399647273,
0.08071988075971603,
0.02991861291229725,
0.050740957260131836,
-0.016260743141174316,
0... |
. 3 ) ( 7. 4 ) 7. 2 • nonhomogeneous linear equations 759 solution the complementary equation is which has the general solution so, the general solution to the nonhomogeneous equation is to verify that this is a solution, substitute it into the differential equation. we have then so, is a solution to | openstax_calculus_volume_3_-_web | [
0.01587075926363468,
0.016961952671408653,
0.011833230033516884,
-0.018532024696469307,
0.006423640064895153,
0.002586086979135871,
0.003632059320807457,
0.0035478209611028433,
0.0192726943641901,
0.07220426201820374,
0.02157881110906601,
0.06312881410121918,
-0.002643055282533169,
0.01154... |
7. 10 given that is a particular solution to write the general solution and verify that the general solution satisfies the equation. in the preceding section, we learned how to solve homogeneous equations with constant coefficients. therefore, for nonhomogeneous equations of the form we already know how to solve the co... | openstax_calculus_volume_3_-_web | [
0.01680169254541397,
0.0445893257856369,
0.007537537254393101,
-0.03874480724334717,
-0.036651428788900375,
-0.0014345652889460325,
-0.007663929369300604,
0.027568425983190536,
-0.007500064559280872,
0.07524892687797546,
0.056937459856271744,
0.04053870961070061,
0.0015193934086710215,
-0.... |
- order differential equations access for free at openstax. org present in example 7. 13 undetermined coefficients when is an exponential find the general solution to solution the complementary equation is with the general solution since the particular solution might have the form then, we have and for to be a solution... | openstax_calculus_volume_3_-_web | [
-0.02683333493769169,
0.010291635990142822,
0.008835905231535435,
-0.01794118992984295,
-0.022374585270881653,
-0.012066442519426346,
-0.005585359875112772,
0.037966445088386536,
-0.004539363086223602,
0.05600414425134659,
0.025057479739189148,
0.04055999219417572,
0.010821267031133175,
0.... |
7. 11 find the general solution to in the previous checkpoint, included both sine and cosine terms. however, even if included a sine term only or a cosine term only, both terms must be present in the guess. the method of undetermined coefficients also works with products of polynomials, exponentials, sines, and cosines... | openstax_calculus_volume_3_-_web | [
0.0244801826775074,
0.03220182657241821,
0.01734619215130806,
-0.05012621730566025,
-0.0655350387096405,
-0.012130542658269405,
0.010143744759261608,
0.012591474689543247,
-0.009931180626153946,
0.0866667851805687,
0.07062896341085434,
0.038702402263879776,
0.002469773404300213,
-0.0312328... |
7. 2 • nonhomogeneous linear equations 761 keep in mind that there is a key pitfall to this method. consider the differential equation based on the form of we guess a particular solution of the form but when we substitute this expression into the differential equation to find a value for we run into a problem. we have ... | openstax_calculus_volume_3_-_web | [
-0.009216981008648872,
0.0398392453789711,
-0.004078793805092573,
-0.01856977492570877,
-0.03344832360744476,
-0.015109763480722904,
-0.004720885306596756,
0.026131808757781982,
-0.00023271942336577922,
0.07113298028707504,
0.04427522420883179,
0.042553819715976715,
0.01605982519686222,
-0... |
based on the form of our initial guess for the particular solution is ( step 2 ). none of the terms in solve the complementary equation, so this is a valid guess ( step 3 ). now we want to find values for and so substitute into the differential equation. we have so we want to find values of and such that therefore, thi... | openstax_calculus_volume_3_-_web | [
0.020582158118486404,
0.018929747864603996,
0.01634998433291912,
-0.023719482123851776,
-0.016596069559454918,
-0.006711054127663374,
0.015466867946088314,
0.03764476254582405,
0.01363761629909277,
0.08260221034288406,
0.04744406044483185,
0.0415460430085659,
0.014315217733383179,
-0.02232... |
7. 2 • nonhomogeneous linear equations 763 find values for and so we substitute into the differential equation. we have and so we want to find values of and such that therefore, this gives and so ( step 4 ). putting everything together, we have the general solution d. the complementary equation is which has the general... | openstax_calculus_volume_3_-_web | [
0.024596668779850006,
0.0411602184176445,
0.016698209568858147,
-0.019202319905161858,
-0.016185959801077843,
-0.0034369472414255142,
0.020052500069141388,
0.02906448021531105,
0.01610865816473961,
0.0829816609621048,
0.04735717549920082,
0.050932593643665314,
0.014506653882563114,
-0.0010... |
7. 12 find the general solution to the following differential equations. a. b. variation of parameters sometimes, is not a combination of polynomials, exponentials, or sines and cosines. when this is the case, the method of undetermined coefficients does not work, and we have to use another approach to find a particula... | openstax_calculus_volume_3_-_web | [
0.012486050836741924,
0.03488650545477867,
0.025357212871313095,
-0.032738957554101944,
-0.06659138947725296,
0.006543333176523447,
-0.0338289812207222,
0.03609905764460564,
0.002866567810997367,
0.07958522439002991,
0.06169804185628891,
0.050751883536577225,
-0.03136391192674637,
0.007531... |
case, the solution is given by example 7. 15 using cramer ’ s rule use cramer ’ s rule to solve the following system of equations. solution we have 7. 2 • nonhomogeneous linear equations 765 then, and thus, in addition, thus, | openstax_calculus_volume_3_-_web | [
0.02541322633624077,
0.03682137280702591,
0.02984338253736496,
0.007977285422384739,
-0.06527896970510483,
0.011904499493539333,
-0.03605372831225395,
0.02313605137169361,
0.032721951603889465,
0.0910925567150116,
0.06622269749641418,
0.04574846476316452,
-0.054613031446933746,
0.003412912... |
7. 13 use cramer ’ s rule to solve the following system of equations. problem - solving strategy method of variation of parameters 1. solve the complementary equation and write down the general solution 2. use cramer ’ s rule or another suitable technique to find functions and satisfying 3. integrate and to find and th... | openstax_calculus_volume_3_-_web | [
0.012075522914528847,
0.038680531084537506,
0.014240753836929798,
-0.00598753010854125,
-0.05996953696012497,
0.0130851399153471,
-0.026706986129283905,
0.033387407660484314,
0.025000611320137978,
0.06484212726354599,
0.050697628408670425,
0.05118550732731819,
-0.03858962282538414,
0.00315... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.