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by: Kennith Herman


Kennith Herman
GPA 3.54


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This 35 page Class Notes was uploaded by Kennith Herman on Friday October 23, 2015. The Class Notes belongs to MA 213 at University of Kentucky taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/228156/ma-213-university-of-kentucky in Mathematics (M) at University of Kentucky.

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Date Created: 10/23/15
January 57 2009 Intersection of 3 c 39 Car Eberhart Some Calculus III Notes Extreme values gt withplots execute this line first 1 least squares t If you have 4 points in space m y 2i7 i 14 there probably isn t a plane which contains all of them However7 you can nd the plane 2 ax by c which best ts the point in the sense that ga7 b7 0 211 a xi by c 7 202 is as small as possible There is only one critical point for this function It is the solution of the system 3 linear equations in ab and c ya 07 9b 07 go 0 gt dataoo22o4238o35 gt matrixdata OMMO OJOJOO CHOOHgtM make up the function g gt g c 2quot2 2ac 4quot2 2a3bc 8quot2 3bc 5quot2 g 67222a67422a3b67823b6752 nd its critical point Since g is a sum of squares7 it will have a minimum at the critical point gt solsolvediffga diffgb diffgcabc 5 7 sola17bg7c1 If we put these values for a7 lo7 and c into g and then compute sqrtgabc4 we get a measure of the goodness of t gt goodnessoffitsqrtsubssol g4 goodnesso t i this says on average the measured value is within 14 of the 7correct7 value we can draw the points and the plane to see how well they t gt display plot3dsubs sol axbyc x1 3 y1 4 colorgrey stylewireframe pointplot3dop data symbolcross colorred axesboxed 2 hunting for a local extreme value of fxy This procedure takes a function fof two variables7 a starting point ado7 a step length h7 a maximum number of steps H7 and a stopping distance er It returns a list abfab 7a2lo2fa2lo27 fab lt fa2b2 lt The method is to compute the gradient of f at ab and move along it a step of length h to a new point a b fa b is compared with fab and if it is smaller than falo ez7 then we cut steplength in half and start again This is repeated until fa b gt fab or the steplength is less than er it is added to the list V V V V indets xquot2yquot2 symbol 967 y hunt procfnabhner local ipgdaibiajbjhif if not typefnfunction then f else f o fn fi p abevalffab i 1 aizaz bizbz while iltn1 do hih gdevalfsubsxaiybidifffxyxdifffXyyl gd 1gd1quot2gd2quot2gd ajaihigd1 bjbihigd2 while fajbjltfaibi and higter do hi5hi ajaihigd1 bjbihigd2 d unapplyfn opindets fn o if fajbjgtfaibi then aizajz bibj p paibifaibi V fi i od matrix p end i1z Here we test hunt on the function 10 XAQ yA2 gt hunt10xquot2yquot21111000003 8 8875000000 9652777778 9932777778 9950339762 1 07500000000 04166666667 701833333333 01575757577 700407415498 9996680252 000719783910 9999896382 70001281810890 9999996714 1 07500000000 04166666667 701833333333 01575757577 700407415498 000719783910 70001281810890 Showhunt draws the graph of the track of the hunt for a maximum on the graph of f It has the same inputs as hunt7 except for the parameter w which controls the size of the window used to show the track V V V V V w untprocfabhn1erw local lstxrngyrngn lsthuntfabhn1er nlinalgrowdimlst xrngaseqlsti1i1n xrngminopxrngmaxopxrng xrngxrng1wxrng2xrng1xrng2wxrng2xrng1 yrngbseqlsti2i1n yrngminopyrngmaxopyrng yrngyrng1wyrng2yrng1yrng2wyrng2yrng1 plotsdisplayplot3dfxyxxrngyyrng plotspointplot3dseqlsti1lsti2lsti3i11 symbolcirclecolorblue plotspointplot3dseqlsti1lsti2lsti3i1n1 symbolcrosscolorblue plotspointplot3dseqlsti1lsti2lsti3inn symbolcirclecolorred axesboxedstylepatchcontour end showhunt10xquot2yquot21152000015 Play around with hunt and showhunt gt plot3d4xy1xquot4yquot4x22y22view222216 gt showhunt4xy1xquot42yquot4 1 1 2 10 034 nding critical points algebraically and applying the 2nd deriva tive test Here is a function to study gt fxquot4yquot44xy 1 f z4y474xy1 Calculating the rst partial derivatives of f gt fx difffx fydifffy fx 4x3 7 4y fy 4y3 7 4x Using solve to nd the critical points We see that there are only three The other two are complex pairs gt evalf solvefxfy xy m 0 y 0 m 1 y 1 m 71 y 71 m 11 y 711 x 07071067812 07071067812 I y 707071067812 070710678121 You can also use fsolve to nd the critical points gt slfsolvefxfyxyx2 2y2 2 51m 0 y 0 gt s2fsolvefxfy xy x2 2 y2 2 avoidsl 52 z 71000000000 y 71000000000 gt 3zfsolvefxfy xy x2 2 y2 2 avoidsl s2 55 z 1000000000 y 1000000000 Use the second derivative test on the 3 critical points gt fxxdifffxx fyydifffyy fxydifffxy fII 12 m2 fyy 12142 fry 74 gt discrcimzfxxfyyfxyquot2 discrcz39m 144 m2 y2 7 16 We see that at 00 the discriminant is lt0 so there is no local extreme At 11 and 1 1 however the discriminant is gt0 and fxx gt0 so the function has local maxima at 11 and 1 1 4 maximizing the volume of a box with a half top and a quarter front A box is to be made from 10 square feet of metal so that its top is only half there and its front is 34 missing Find the maximum volume of the box Maximimize V xyz subject to 10 32 xy 2yz 74 xz rst solve the constraint for one of the variables and reduce V to a function of 2 variables gt constraint110 32 xy 2yz 74 xz 3 7 constraint 10 my2yz zz 4 gt sol solveconstraintz 501727203my 8y7m gt V xysol V72my7203zy 8 y 7 m Find critical points gt Vx diffVx VydiffVy 72y7203zy7 6my2 14my7203xy V I 8y7m 8y7m 8y7m2 V 7 2m7203xy 6m2y 16my7203my y39i 8y7m 8y7m 8y7m2 These are ugly Lets simplify gt szsimplifyVx VI 7232 7160 48mg 21x2 39 8 y 7 96V gt VysimplifyVy 4x2 770 12y2 21mg 8 y 7 902 By inspection7 we see that Vx and Vy will 0 when X 0 y However these values are not in the domain of V So we next look for the zeros of the other two factors of the top of Vx and Vy gt eqns 16048xy21xquot20 7012yquot221xy0 Vy7 eqns 716048my 21m2 O7 77012y2 21mg 0 We can draw these two curves in the plane using in1plicitplot gt implicitploteqns x0 10 y0 10 We want the point of intersection of the two curves gt solve eqns X y 8 x 7 7 RootOf6 22 35 y RootOf6 22 35 x RootOf2 22 7 35 label L8 y RootOfQ 22 7 35 label L8 Here again this is ugly Lets evalf it convert to decimal approximations gt solxy evalf solve eqns x y 2 solxy z 1593638146 y 1394433377 The rst solution is complex and out of domain The second one is clearly the one we want So we declare that gt Maxvolume subs solxy V Maxwolume 2656063576 Around 2656 cubic feet is the best we can do here What is the height 2 of the best box gt subs solxy sol 1195228609 Note this box doesn t have a square base or side More extreme values gt withplots execute this line first 5 Discriminant 0 examples for the second derivative test We know that if the discriminant D fm fW 7 fwz lt 0 at a critical point m0 yo7 then there is not a local max or local min at m0 yo This is because the signs of fm and fW are different so fis concave at m0 yo on one of the lines mg g and m yo and concave down on the other line Also7 we know that if D is positive7 then fz07 yo is a max or min value depending on whether fmz0 yo is negative or positive respectively This is because of the second derivative test for functions of 1 variable Now we need examples to show that when D 07 then fz07 yo could be a max value or a min value or neither max nor min Example 1 D0 at the critical point 007 and f00 is a local minimum value fm7 y m2 7 2mg y2 Note that fm fW 7 fwz 2 2 7 22 07 fz7 y m 7 y2 is a parabolic cylinder with min imum value 0 everywhere on the line y 71 V display pointplot3d O O O symbolcircle colorblue spacecurve t t 0 t1 1 colorred plot3dxquot22xyyquot2 x1 1 y1 1 axesboxedorientation 7357 V Example 2 D0 at 00 but f00 is a local min Just take the negative of example 1 Example 3 D 0 at 00 but f00 is neither a local max nor a local min there isn t a quadratic example of this But if we take example and add xA3 to it7 this turns 00 into a 7hanging valley7 and does not alter the value of D display pointplot3d O O O symbolcircle colorblue spacecurve t t tquot3 t1 1 colorred plot3dxquot22xyyquot2xquot3 x1 1 y1 1 axesboxed orientation 73 57 V 6 nding extreme values of a continuous function on a closed and bounded set in the plane Here is an 7algorithm7 for nding the extreme values of a continuous function f on a closed and bounded domain D 1 Identify the boundary of D and the interior of D The interior of D is the set of points a7b in D for which there is a positive 6 so that if x7y is within 6 of alo7 then x7y lies in D The boundary of D is the points of D that are not in the interior of D Those points have the property that there are points arbirarily close to them which are not in D We assume that fis differentiable on the interior of D 2 Find the critical points of f in the interior of D7 and tabulate the function there 3 Find the extreme values of f on the boundary of D The boundary of D is usually a curve of some sort So one way to optimize fon the boundary would be to parameterize the boundary with a function rt and then optimize the composition f r Note 1 r might be piecewise de ned If the boundary is complicated7 you might have to break it into pieces and parameterize the pieces Note 2 Sometimes it is useful to draw a few gradient vectors for fto see how fis increasing This might help narrow down the search for maximum and minimum values So if all the gradient vectors are pointing into the interior away from boundary at a certain place7 probably there is not a maximum value of f on the boundary there7 because the gradient vectors point in the direction of greatest increasing f values Examples 1 Optimize ie nd all extreme values of fxy my 1 2 on D the triangular region with vertices 027 07 27 and 17 2 The boundary of D consists of the three segments connecting the vertices and the interior is all points inside the triangle One checks that fm 0 fy only at 00 which is not in the interior of D Thus the maximum value of f on D must lie on the boundary of D We can nd the extreme values on each edge separately on the edge 02 to 07 2 fxy 0 on the edge 02 to 17 27 y 2 7 4x so we want to optimize fz7 y fz7 2 7 4m z 2 7 4m 2 2 z for 0 g x g 1 the derivative 2 7 6 z 0 when x 137 so the extreme values on this interval must be among the values f02 O7 f1323 13 and f1 2 1 on the edge 07 2 to 17 27 fxy fx 2 72z 1 2 for x between 0 and 1 the derivative 2 2x 0 when x 17 so the extreme values on this interval are f0 2 0 and f1 2 1 Combining these efforts7 we see that f1323 13 is the maximum value and f1 2 1 is the minimum value of fxy on the given triangular region We can check our work visually in Maple gt withplots This is a domain restriction function It returns x7y if x7y is in the domain of f7 otherwise it returns 00 so that frxy fxy if x7y is in the domain of f and frxy 0 otherwise gt r procxy if xgt O and y gt 2 and y lt 2 4 then op xy else op 00 fi end r pr0cz7 y ifO z and 7 2 g y and y g 2 74 mthen opz7 yelseop07 0end if end proc gt f xy gtxy xquot2 f 7 z y 7zyz2 is the composition operator in Maple gt h f O I h f r gt h1323 1 3 Here is the graph of f7 with the absolute minimum and maximum values shown gt displaypointplot3d 1 2h1 2 1323h1323 colorredsymbolcircle plot3dh gt 1 1 2 2 1 2 1 view1 35 stylepatchcontour numpoints4000 orie ntation 160 77 axesboxed 2 Optimize fz7 y z y on the closed disk 2 y2 g 1 you do this one Also try to draw a picture to verify your conclusions Visually First change the de nition of r so that rxy X737 if XAQ yA2 lt 17 otherwise rXy 070 gt r procxy if xgt O and y gt 2 and y lt 2 4 then opxy else op00 fi end 7 p1quot0cm7 y ifO z and 7 2 g y and y g 2 7 4 mthen opz7 yelseop07 0end if end proc the de nition of f remains unchanged gt f xy gtxy xquot2 f 7 y Hmyz2 Also7 h will remain unchanged as the composition of f with r gt h f O r h f r Here is the picture you need to modify You will need to modify the location of the extreme points on the graph colorredsymbolcircle gt displaypointplot3d12h12 1323h1323 plot3dh 1 2 2 1 2 1 view1 1 stylepatchcontour numpoints4000 7 The locus of the centroid of a pie slice Here s a problem Find the centroid of a pie slice as a function of its radius R and its central angle T First look at the extremes of this problem If T 0 then the centroid is going to be the midpoint of the radius If T 2 71397 then the centroid is going to be at the center of the circle If T is held xed and R is allowed to change7 the centroid will move along the midline of the pie slice The interesting question is what happens when R is held xed and T changes from 0 to 2quoti7 what is the locus of the centroid We can work this out if we can determine the centroid as a function of T and R First the area of the slice is gt A intintrr0 R t0 T RZT A 2 Next the moment of the slice about the y axis is gt My intintrcostrr0 R t0 T 1 My 3 sinT R3 So xbar7 the x coordinate of the centroid is MyA gt xbarz unapplyltMyART 2 sinTR b R T I am 7 a 3 T gt Mx intintrsintrr0 R t0 T R3 1 Mr 7 cosTR3 gt ybarz unapplysimplifyMxA RT W R T g Lets check our calculation by drawing the pie slice and its centroid to see if it looks reasonable gt withplottools 0mg arrow circle cone7 cuboid7 curve cutm7 cutout7 cylinder digc7 dodecahedmn7 ellz39ps e7 elliptic4T0 hemisphere hexahedmn ho rnothe ty7 hyperbola7 icossujiedron7 line7 octahedron7 parallelepiped7 pieslice point7 polygon7 project7 rectangle7 reflect7 rotate7 Scale7 sse rnz39torus 7 sphere7 sitella te7 tetrahedron torus7 tramsfor rn7 tramslate7 urml gt pslicez procRT use plotsplottools in displaydiskevalf xbarRT ybarRT 01Rcolormagenta pieslic gt e O O R O T coloryellow scalingconstrained end use end Checking the centroid of the quarter pie gt pslice4Pi2 gt xbarRPi2 3 7139 Note this agrees with our calculation in class on Wednesday We can make a movie showing the locus being traced out by the centroid of the pie slice as the angle of the slice goes from 0 to 360 gt plots display seqpslice 4 i 0001 2Pi36 i0 36 scalingcon strainedaxesnone insequencetrue titlequotLocus of the centroid of an expanding pie slicequot Locus oflhe centroid of an expanding pie slice What are some questions that occur to us as we watch this movie First we notice that when T is small the centroid is not close to the midpoint as we might expect it to be The reason becomes clear when look again at the centroid XbarRTybarRT gt xbarRT ybarRT What happens to xbar as T appoaches 0 Since 2 goes to 1 and 2 goes to 0 we see the centroid approaches the point 2311 0 So there is a jump discontinuity in the locus at T Other questions In all of these consider R to be xed What is the maximum value of ybar What is the minimum value of xbar What is the length of the locus What is area under the locus What is the centroid of the region under the locus Other questions Take any nice 1 parameter family of regions and ask for the locus of the centroid For example it was asked in class where the centroid of a trapezpoidal region is Making this question very speci c for the right triangle with vertices 00 a0 010 and h between 0 and b let T be the trapezoid with base 00 to a0 and top 0h to a 7 h Calulate the centroid of T as a function of a b and h Use this to investigate the locus of the centroid with a and b xed and h going from 0 up to b Draw pictures to check your work gt restart withplottools withplots 8 The centroid of an ice cream cone spherical box Fix 10 between 0 and 7139 We want to compute the centroid of the solid S O consisting of all points 07 j 0 such that p is between 0 and 17 j is between 0 and 10 and 9 is between 0 and 27139 Here is it s picture for 10 7quot icc phi0 n gtdisplay plot3d1 t0 2Pi p0 phiO coordsspherical stylewireframe colorblue seqline O O O sinphi0 cos 2Piin gt sinphi0 sin2Piin cos phi0 thickness2 colorred i1 n scal ingconstrained axesnormal orientation 80 70 gt icc Pi472 V gt Intxquot3x0 3 3 m3 dm 0 2025000000 gt evalf 70 By inspection7 we can see that the centroid is 00a where a is somewhere between 0 and 17 depending on 0 The volume and moment about the xy plane can best be evaluated as iterated integrals in spherical coordinates gt IntIntInt1 z xS phi 0 yIntIntInt1rhoquot2sinphi r ho0 1 phi0 phiO theta0 2Pi S ldzdmdy AZWOwOlpzsin dpd d0 So the volume of the ice cream cone with cone angle 2 10 and cone slant height 1 is gt Vintintint1rhoquot2sinphi rho0 1 phi0 phiO theta0 2P i 2 2 V g 7 geos gt07r And the moment about the xy plane is gt Intlntlntz z Xs Phi 0 yIntIntIntrhocosphirh0quot25i nphi rho0 1 phi0 phi 0 theta0 2Pi sOZd2d yAh otlpgwswsmw This evaluates to Mxy int int int rhocos phirhoquot2sinphi rho0 1 phi0 phiO t heta0 2Pi 1 Mxy g 7 Zcos gt02 7139 So we can calculate the z coordinate of the centroid zbar as gt zbarz unapplyMxyVphi0 1 27 Z cos gt02 7r zbar gt0 a 2 2 Y 7 g cos 0 7139 And for example when the cone angle is 90 degrees zbar is gt simplifyzbar Pi4 evalf zbar Pi4 3 7 06401650420 8 72 V We can add the centroid to the picture and check to see if it looks reasonable gt picturez phiO gtdisplay pointplot3d O O evalf zbar phiO symbolcircle color black thickness3 icc phiO 20 scalingconstrained orientation 116 80 picture gt0 a plots idisplay nlots ipointplot d O 0 evalfzbar gt0 symbol plottools icircle color black thickness 3 icc gt0 20 scaling constrained orientation 116 80 picturePi4 V This seems eminately reasonable Now lets make a movie of the locus of the centroid of the spherical box as 10 goes from 0 to Pi gt displayseqpictureiPi20 i1 20 scalingconstrainedinsequence true As in the case of the centroid of the pie slice we see that as 10 approaches 7139 the centroid moves to the center of the ball At the other extreme when 10 is getting closer to O the ice cream cone is approaching a segment with centroid at the middle of the segment but the centroid is approaching a number closer to 1 than 0 In the case of the pie slice this number was 23 the limit of the centroid of the triangle approximation to the pie slice In this case the number is gt Limit zbar t t0 rightlimit zbar t t0 right Eilcost27r 3 lim 2 2 taO iECOSlttgtW 4 Question Why that number Perhaps if you solve the problem below you could come up with an explanation Problem Calculate the centroid of the cone of slant height 1 and base radius r 9 drawing the intersection of perpendicular cylinders 91 Two cylinders The goal is to draw a picture of the intersection of the two cylinders of radius 1 centered on the x axis and y axis respectively7 and calculate the volume gt withplots gt withplottools We can draw a cylinder of color clr and radius r about the line y tanthetax in the xy plane using tubeplot plots tubeplot cylint r theta clr gtdisplaytubeplot tcostheta tsintheta 0 t1 1 numpoints40 tubepoints36 radiusr scalingconstrained colorclr axe gt sboxed The intersection of the cylinders of radius 1 about the x axis and y axis can be obtained by solving the equations XAQ 2A2 1 and yA2z121 simultaneously What we get is two ellipses X y7 XA2ZA2 1 and X y7 XA2ZA2 1 These can be parameterized with xsint7 sint7 z cost t O2Pi and xsint7 y sint7 z cost7 t 02Pi We can use spacecurve to draw these gt 1nte VV displayspacecurve sint sint cost t0 2Pi thickness3color black gt inter2 displayspacecurve sint sint cost t0 2Pi thickness3color black Now we can draw roughly the intersection of the cylinders gt displayinter inter2 cylint1 Pi2blue cylint1 0 red labels xy z stylewireframe orientation 60 50 If you look at this from the top7 you can see that the cross sections by planes parallel with the xy plane with the intersection are squares We can draw those squares with sq below Then frame de ned underneath shows a bunch of the cross sections gt sq t clr gtdisplay polygon sint sint cos 13 sint sint cost sint sint cost sint sint cos 13 colorclr gt framez ngtdisplayseqsqiPi36 black i1 n inter inter2 labels xyz s tylewireframe orientation 66 50 i n39 367 style wirefmme7 orientation 667 50 gt picdisplayframe36 axes boxed frame n gt plots 7displayseqsq black7 i 1n7 inter interg labels m y z gt pic a W WW 3 pw wm d 1mamps s v s a w amp w w s s A a v a From this picture we can see that the cross section by a plane at height h between 1 and 1 is a square of side 2x17 hz Hence the volume of the intersection of the two cylinders is 1 16 4 7 4 h dh 3 gt Int41hquot2 h1 1 int41hquot2 h1 1 1 474712001E 1 3 Further questions What is the surface area of the intersection also what is the total length of the edges of the intersection 92 Three Cylinders Now lets draw the intersection of the 3 mutually perpendicular cylinders of radius 1 about the x7 y7 and z axes The intersection of the cylinder about the z axis with the other two cylinders in succession give 4 more ellipses which are parameterized and drawn below gt displayspacecurvesintcostsintt02Pithickness3color black inter4 displayspacecurvesintcostsintt02Pithickness3color red interS displayspacecurvecostsintsintt02Pithickness3color blue inter6 displayspacecurvecostsintsintt02Pithickness3color brown V V V gt displayinter5 inter6 inter4 inter inter inter2 labels x y z scali ngconstrained orientation 4O 60 axesboxed We can see that the surface of the intersection consists of 12 congruent quadrilateral pieces of a cylinder They are grouped into 3 groups of 4 pieces of each of the intersecting cylinders Here is a color coded picture of the surface gt V V V V V f proct options operator if evalftltevalfPi4 then sintsintcost else costsintcost fi end g proct options operator if evalftltevalfPi4 then sintsintcost else costsintcost fi end surfclI gtdisplayseqlinefiPi236giPi236colorclr i136seqpointplot3dfiPi236giPi236 symbolcirclecolorblacki036 bluesurfrotatedisplayseqlinefiPi236giPi236color bluei1 3600 Pi2 picdisplayseqrotatebluesurfOiPi20i03seqrotaterefle ctsurfgreen000001100iPi200 i03 seqrotaterotatesurfred0 Pi2000iPi2i03 scalingconstrainedlabelsxyzaxesboxedorientation7050titl equotIntersection of 3 cylindersquot pic Intersection of3 c 39ders Further examination shows there are 6 vertices of order 4 2 for each pair of intersecting cylinders and 8 vertices of order 37 representing the 23 8 points of intersection of the 3 cylinders There are also 24 edges It makes for ball you wouldn t want to play soccer with What about the volume Well7 by the symmetry it will 16V where V is the part trapped above the region 0 lt X lt y and 2 y2 g 1 in the xy plane and the cylinder 2 22 1 This can be m 1 V 17m2 written as the sum of two iterated integrals 1 7 2 dy dz 1 1 7 2 dy dz 0 0 0 gt V1intsqrt1xquot2xx01sqrt2 4 2 2 V 7r 73 m gt V2intsqrt1xquot2sqrt1xquot2 x1sqrt2 1 2 5 2 V2 7 T So the total volume of the intersection is gt 16V1V2evalf 16V1V2 4V2 2 32 202 2 2F 7 m 7 4686291496 7 3 Further questions Calculate the surface area of the intersection calculate the total length of the edges of the surface Problem Find a formula for the volume of a truncated prism TP with base a triangle with sides a and b and included angle t and heights h1h2 and h3 Solution For starters we might guess that the formula is the area of the triangle times the average of the heights V absm t h h g h3 But we can develop a formula by setting up a coordinate system expressing the volume as the integral over the triangle of the height function and then hoping that the integral can be evaluated as an iterated integral to reveal a 7nice7 formula Let s choose the origin 000 to be at the vertex of the included angle t Then label the height over the vertex h1 and give it its coordinates 00h1 The a leg we ll put along the positive x axis with endpoint a00 the b leg then will have coordinates bcostbsint0 The height h2 and h3 sit over these two points at a0h2 and bcostbsinth3 Here is what the base looks like gt withplots withplottools display polygon 0 O 2 O 3 2 coloryellow textplot 4 5 1 quotx a bsintbcosta yquot textplot 2 1 quotx costsint yquot textplot 02 quot 00 quot textplot 22 quot 210 quot textplot 3 1 2 2 quot bcos t bsint quot scalingconstrainedxtickmarks l ytickmarks V bcost bsmt x a bsmtbcost 395 y The way we have set this up it is best to think of the base triangle T as a type 2 region As y sin b cost 7 a The height function fxy can be written fxy z h h i m wy where w is obtained by substituting in the point z h3 x bcost y bsint and solving for w Now we will use Maple to do the evaluation First get the height as a function of x and y goes from 0 to bsint x goes from cott y to a gt z h1h2ax wy h z h wy a gt zsubssolvesubsxbcost ybsintzh3 w z hgm hl a h2bcost 7 hi 01 Z h a bsinta 22 Now perform the evaluation gt VIntIntz xcostsinty abcostabsinty y0 bs int b cost 7 a y bsinlttgt at 581110 he m M a he bcost 7 hi a y V 3970 COS75 y M T 7 bsint a dm dy sint gt Vintintzxcostsinty abcostabsinty y0 bs int h b cost 7 12 cost2 7 1 1 2 sint2 7 sint2 7 3 2 a bcost 7 a cost h h2bcost7h3 i 7i i a agtlt bs1nt s1nt b3 sinlttgt3 1 bsint a 2 bcost 7 a 7 cost h b cost 7 a 7 h a h bcost 7 hffa 2 h bsint sint bsint bsint i 1 i sint2 h a b s1nt 5 fr a b s1nt This is not the formula I want to remember Let s simplify it gt VsimplifyV 1 V gabsint h 2 h 1 hi gt VunapplyVabth1 h2h3 1 Vabth1h2h37 gabsint h 2h h Much much better We can rewrite this formula as V absm t 2 h h g h area of base times 7a weighted sum of the heights This is close to our guess7 but we have to double the weight of the height over the vertex in the calculation Here is a procedure to draw the truncated prism with its volume on the front face gt drawitprocabth1h2h3 local RPQABCv use plotsplottools in gt R 000 P a00 Q bcostbsint0 gt A00h1 Ba0h2 C39Q00h3 vevalfVabth1h2h34 gt displaytextplot3d a22h12v polygon RPQ colorgrey poly gon RPBA coloryellow polygon PQCB colorturquoise polygon Q RAC colorqnagenta polygon ABC colorgrey scalingc 23 V onstrained end use end gt displaydrawit23Pi6423 translatedrawit152Pi4224 30 O orientation 40 57 axesboxed labels x y evalnz Problem Use this same idea to develop a formula for the volume of a truncated rectangular solid with base rectangle a by lo7 and three given heights h17 h27 and h3 10 Line integral problems Problem Let C be the curve sitting on the surface of the graph of smooth function z fxy over the arc from 00 to 127 and let gxyz be a nice smooth density function de ned on C Represent the mass and center of mass of C in terms of line integrals Evaluate these integrals for fXy 2 a and gxyz 2 Draw a picture showing the surface f and the curve C together with its center of mass A solution The mass ofC is M gz7 y 2ds The moment about the xy plane is me 2gm7 y zds The moments about the gther two planes are de ned in the same way 0 We can parameterize C by rt Xyz t2tft2t So we can evaluate these 4 line integrals like so gt M Int2tquot2sqrt1416tquot2tO1 1 M 2t2516t2dt 0 gt Mint2tquot2sqrt1416tquot2 t0 1 5W7 5 1 37 m 5 1 m M 5m a i nw dn w TA 1n T 7r M 2517952442 gt MxyInt2tquot2 quot2sqrt1416tquot2 t0 1 1 MI 4t4x5 16t2dt 0 gt Mxyint2tquot2quot2sqrt1416tquot2 t0 1 gt M evalf M 5 MI 71 H 7711 m71334 1n6 2 M 25 gi6ln21n5 1280 8 T 12288 2048 W Mxy 3222893594 gt Mxyzevalf Mxy gt zbar MxyM zbar 1279966031 gt MyzInt2tquot2tsqrt1416tquot2 t0 1 1 Myz 2t35 16t2dt 0 gt Myzint2tquot2tsqrt1416tquot2 t0 1 133 x x3 5 5 7 200 7 24 Myz 7T gt Myzzevalf Myz Myra 1962863960 gt xbarMyzM Ibar 07795476703 Note that since y 2quot7 sz 2Myz and so ybar 2Xbar gt sz2Myz MIZ 3925727920 gt ybarszM ybar 1559095341 Now to draw the picture gt use plots in display pointplot3d xbar ybarzbar symbolcircle colorred put this in last gt spacecuIve t2t2tquot2 t0 1 colorbluethickness3 then added this plot3dxquot2xy2x0 1y0 2coloryellow drew this first gt scalingconstrained these plot options were added to make the picture more legible axesboxed gt labelsxyz orientation 147 72 gt end use mh2a xm ahTh mqth3 afnsntla y If this seems a little high7 recall that the density function is the z coordinate and so the center of mass is pulled up signi cantly toward the upper end of the wire Problem for you Recalculate the center of mass if the density of the Wire is constant say 1 Redraw the wire 11 Drawing vector elds Given a vector eld F ltP7Qgt we want to plot enough of its values to be able to sketch in or at least to visualize the streamlines of F the curves which are tangent to the eld value at each point where they are Maple has a nice word eldplot in the plots package to draw these values for you gt withplots 7107 J I I 1 I r animate7 animate d animatecume arrow 39 y 7 I r r T rnal7 conformalix d7 comfomplo l7 contou nlot d7 coonlplot7 coonlplotix d7 confo J 39r eldplo l7 f 1 7 gmdplot7 gmdplotix d7 quot zlz39siplay7 J my 1 i r 1 z z i 1 i z m 7 i r r r r r WWWL r intem ectplo l7 quot r 7 quot r m l quot U quot l lls tplot7 lz39sstplo l ix zl7 loglogplo l7 logplo l7 rnalrz39xplot7 multiple7 odeplot7 pareto7 plotco rnpare7 pomtplot7 polntplotix d7 polmjnlot7 polygonplot7 polygonplot d7 polyhezlm5uppo fezl7 polyhedmplo l7 Tootlocus 7 s e rnz39logplot7 Setcolom 7 ssetoptlons 7 5675010ifz39ons ix d7 spacecume Sparse rnalrz39xplot7 surfdata7 textplo l7 textplo lix d7 tubeplot you can get examples of how to use the word and the 3d version eldplot3d by looking at the bottom of the helpsheet that pops when you execute gt fieldplot I just copied the examples below from that helpsheet and pasted them into this worksheet 1 11 Examples gt withplots fieldplot xxquot2yquot24 quot 12 yxquot2yquot24 quot 12 x2 2 y2 2 fieldplot ysinxy10 x1O 10 y1O 10 arrowsSL M colorx The following two examples gives the same results as above gt xygt xxquot2yquot24quot12 g xy gt yxquot2yquot24quot12 fieldplot fg2222 f xy gt y g xy gt sinx ylO fieldplot fg10101010arrowsSLIM An inverse square law without eldstrength adjustment gt fieldplot x xquot2yquot2 quot 32 y xquot2yquot2 quot 32 x11y11 Where we note that only the arrows very close to 00 are Visible Now using eldstrengthlog fieldplot XXquot2yquot2quot32 yxquot2yquot2quot32l gt x1 1 y1 1 fieldstrengthlog Which makes the direction of the arrows much more Visible Alternatively a pure direction plot can be produced gt fieldplot xxquot2yquot2 quot32 yxquot2yquot2 quot 32 x1 1 y1 1 fieldstrengthfixed R 7 TN A radial eld in polar coordinates gt fieldplot I O r0 1 t0 Pi2 coordspolar A eld in polar coordinates that will not draw the arrows at the origin the eld direction is unde ned there gt fieldplot O 1 r0 1 t0 Pi2 coordspolar 12 Greens theorem Green s theorem states that if F ltPQgt has continuous partial derivatives on an open set containing the positively oriented simple closed curve C and the region D it bounds then the line integral of the the tangential component of F with repect to arc length around C is the double integral over D of the difference Q 7 P The left hand side of Green s theorem has a nice meaning If the integral is positive then on average the tangential component of F is positive so if we take the view that F is a velocity ow for a uid that says that the net ow of the uid on the boundary curve C is counterclockwise If we divide by the length of the curve we would get average signed speed that a point on the curve is rolled around the curve by the velocity eld If the curve is a small circle of radius r then Green s theorem says that this average angular velocity is approximately one half the above difference of partial derivatives For this reason the the difference is a measure of the rotational tendency of the velocity eld at each point and as such is called the curl of F Note The curl of a 3 dimensional velocity eld F ltP Q Rgt is the vector Del X F lt gym 7 Q P 7 R g Q 7 Pgt If F is 2 dimensional R 0 we can call the 3rd component the curl of F Stokes theorem is a generalization of Greens theorem to the situation in space where C is a space curve and D is a surface bounded by C It says that the line integral of the tangential component of F around C is equal to the surface integral of the normal component of curl F over D If instead of integrating the tangential component of the velocity eld F around the curve C we integrate the normal component of F we get another line integral which can also be equated to a double integral by Green s theorem The outward normal n to the curve C is obtained by rotating the tangent vector ltdXdt dydtgtir ti 90 degrees clockwise to get n ltdydt dXdtgtir ti The line integral of the normal component of F ltPQgt then becomes 0 Poly 7 C Q dm By Greens theorem this is equal to the the double integral P Q 1 1A This is called the normal form of Greens theorem The right hand side of this equation measures the transport of iud across the curve C positive means uid is owing out on average negative means uid is owing in The integrand of the right hand integral is called the divergence of the velocity eld F More generally the divergence of a vector eld F ltPQRgt is de ned as div F del dot F P Q R The Divergence theorem is a generalization of the normal form of Green s theorem to the situation in space where you have a surface R bounding a soliid S in space where there is a nice velocity eld F de ned on an open set containing S and R It says that the surface integral of the outward normal component of F over R is equal to the triple integral of the divergence of F over the solid S 121 procedure 1 to investigate velocity elds Tangential form of Greens the orem This procedure takes a vector eld F1 a list of two expressions in X and y a radius R and a center P and draws the circle C of radius R centered at P Then it draws the the red graph of the 31 tangential component of F over the curve and shades vertically the red fence trapped between the graph and the circle C This enables you to estimate the net circulation of uid around G Then it also draws the graph of the curl of F over a square containg D and draw the blue projection of C up onto that graph together with a blue fence This enables you to visually estimate the net curl of F over D Green s theorem says that these two quantities are equal numerically gt restartwithplots greenpiclprocRPF1 Rradius of D Pcenter of D F1yx velocity field local xrngyrngtrngrpFFdrQPrwrkRdcrl use plottools in FunapplyF11F12Oxy xrngP111RP111R yrngP211RP211Rtrng02Pi IIP1Rcost P2 Rsint 0 rpz diffr1tdiffr2t0 Fdr Fr1r21rp1Fr1r22rp2 QP diffFxy2x diffFxy1y wrkzevalfIntFdrttrng6 evalfIntIntsubsxP1RdcosthetayP2Rdsintheta QPRd Rd0 R theta0 2Pi 6 display plot3dQPxxrngyyrngstylewireframecolorblue fieldplot3dFxyxxrngyyrngz001arrowsSLIM seqlineP1Rcosioptrng236P2Rsinioptrng236O P1Rcosioptrng236P2Rsinioptrng236substio ptrng236Fdrcolorredi036 rotatedisplayseqlineP1Rcosioptrng236P2Rsiniop trng236OP1Rcosioptrng236P2Rsinioptrng2 36substioptrng236subsxr1yr2 QP Colorblue i0 36 Pi36 P 1 P2 0 P 1 P2 1 spacecurver1r2Fdrttrngcolorred spacecurver1r2subsxr1yr2 QPttrngthickness1colorblue spacecurverttrngcolorblackaxesboxedlabelsxyztitlecatquot net circulation red area quotconvertwrkstringquot and net curlblue volume quotconvertcrlstring end use V V V a H V V V V V V gt end V greenpic1111sinxyy 122 This procedure takes a vector eld F1 a list of two expressions in X and y7 a radius R7 and a center P and draws the circle C of radius R centered at P Then it draws the the magenta graph of the normal component of F over the curve and shades vertically the magenta fence trapped between the graph and the circle C This enables you to visually estimate the net out ow of uid thru G Then it also draws the graph of the div of F over a square containg D and draw the navy projection of C up onto that graph together with a navy fence of the solid This enables you to visually estimate the net divergencel of F over D Green s theorem says that these two quantities net circulation red area 784177 an et curlblue volume 784177 procedure 2 to investigate velocity elds Normal form of Greens theorem are equal numerically V gt V V V V V V V withplots greenpicQ procRPF1 Rradius of D Pcenter of D F1yx velocity field local xrngyrngtrngrpFFdnPQrprpwkRddv use plottools in FunapplyF11lF12lOlXyI xrngP111RP111R yrngP211RP211Rtrng02Pi rP1RcostP2Rsint0 rpz diffr1tdiffr2t0 Fdn Fr1lr21lIP2 FI1I22IP11 PQ diffFxy1xdiffFxy2y prpwkzevalfIntFdnttrng6 dvzevalfIntIntsubsxP1RdcosthetayP2Rdsintheta PQRd Rd0 R theta0 2Pi 6 display plot3dPQxxrngyyrngstylewireframecolornavy fieldplot3dFxyxxrngyyrngz001arrowsSLIM seqlineP1Rcosioptrng236P2Rsinioptrng236O P1Rcosioptrng236P2Rsinioptrng236substio ptrng236Fdncolormagentai036 rotatedisplayseqlineP1Rcosioptrng236P2Rsiniop trng236OP1Rcosioptrng236P2Rsinioptrng2 36substioptrng236subsxr1yr2 V V V V PQgtgt colormaroon i0 36 Pi36 P1 P2 0 P1 P2 111 spacecurver1r2Fdnttrngcolormagenta spacecurver1r2subsxr1yr2 PQttrngthickness1colornavy spacecurverttrngcolorblackaxesboxedlabelsxyztitlecatquot net outflow magenta area quotconvertprpwkstringquot and net divergencenavy volume quotconvertdvstring end use end greenpic2111sinxyy


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