{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 1 16 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 } 1 1 0 0 8 4 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 40 "The work integral in clas sical mechanics" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "Let us demonstrate the path independence of the work int egral for a conservative curl-free force field," }}{PARA 0 "" 0 "" {TEXT -1 44 "such as the gravitational field of a sphere." }}{PARA 0 " " 0 "" {TEXT -1 58 "Let us define the potential energy for a particle \+ of mass " }{TEXT 258 1 "m" }{TEXT -1 46 " in the field of the star (or planet) of mass " }{TEXT 257 1 "M" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "V:=-G*M*m/r;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "F:=-grad(V,[x,y,z]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "T hat didn't work!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "F:=-gra d(V,[r,theta,phi],coords=spherical);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalm(F[1]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 181 "The - sign on front of the grad throws Maple off, because the grad pr ocedure returns a vector, and not a list. The fix is to put the minus \+ sign with the scalar potential function V." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 43 "F:=grad(-V,[r,theta,phi],coords=spherical);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "F[1];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "OK, Maple has its peculiar features! It didn't like \+ the '-' sign before the gradient." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "curl(F,[r,theta,phi],coords=spherical);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "The force field is curl-free by construct ion: it has been derived from a scalar potential." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 132 "For plotting the potenti al energy we need to switch to Cartesian coordinates. We choose the sh ow the potential at a fixed z=1 plane:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "PL1:=plot3d(subs(r=sqrt(x^2+y^2+1),G=1,M=1,m=1,V),x=- 5..5,y=-5..5,axes=boxed): plots[display](PL1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Let us choose a straight-line path:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "r_i:=[2,2,1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "r_f:=[-2,0,1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "r_t:=r_i+(r_f-r_i)*t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "evalm(r_t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 120 " Note: here t is a parameter, not necessarily time. Nevertheless, we ca ll the derivative of the parameter representation " }{TEXT 264 1 "v" } {TEXT -1 3 "_t:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "v_t:=dif f(r_t,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "Now the work integra l is reduced to a parameter integral using the dotproduct of " }{TEXT 263 1 "F" }{TEXT -1 6 " with " }{TEXT 262 1 "v" }{TEXT -1 28 "_t, but \+ also realizing that " }{TEXT 261 1 "F" }{TEXT -1 38 " needs to be eval uated along the path," }}{PARA 0 "" 0 "" {TEXT -1 9 "i.e., as " } {TEXT 260 1 "F" }{TEXT -1 1 "(" }{TEXT 259 1 "r" }{TEXT -1 7 "_t) !!! " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 228 "Also : we have defined the parameter representation in Cartesian coordinate s, so we should use these coordinates for the dot product calculation, which means that the force has to be translated into Cartesians. How \+ do we do this?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 46 "F:=grad(-subs(r=sqrt(x^2+y^2+z^2),V),[x,y,z]); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "We start with the common mist ake made by many beginners:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "Work:=int(dotprod(F,v_t),t=0..1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "This is a bad answer for several reasons:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "1) the work has to c ome out as a number (we have specified initial and final points)." }} {PARA 0 "" 0 "" {TEXT -1 57 "2) we have not evaluated the force along \+ the chosen path." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "r_t:=ev alm(r_t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "Work:=int(dotp rod(subs(x=r_t[1],y=r_t[2],z=r_t[3],F),v_t),t=0..1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "This is frustrating: we do apparently the right thing, yet we get the same wrong answer!!!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "How do we avoid this pitfall?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "1) look at subexpressions, and then realize that something hasn't worked." }} {PARA 0 "" 0 "" {TEXT -1 77 "Substitutions into vectors and matrices i n Maple require the use of the op():" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "dotprod(subs(x=r_t[1],y=r_t[2],z=r_t[3],op(F)),v_t); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "This is the correct integrand : it no longer depends on " }{TEXT 268 1 "x" }{TEXT -1 2 ", " }{TEXT 267 1 "y" }{TEXT -1 2 ", " }{TEXT 266 1 "z" }{TEXT -1 24 " just on th e parameter " }{TEXT 265 1 "t" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 20 "Now do the integral:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 70 "Work:=int(dotprod(subs(x=r_t[1],y=r_t[2],z=r_t[3],op( F)),v_t),t=0..1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "An alternati ve to using op() is to convert the vector produced by grad into a list ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "FL:=convert(F,list);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "Work:=int(dotprod(subs(x= r_t[1],y=r_t[2],z=r_t[3],FL),v_t),t=0..1);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 9 "evalf(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "Th e work integral tells us by how much the kinetic energy has changed be tween " }{TEXT 271 1 "r" }{TEXT -1 7 "_i and " }{TEXT 270 1 "r" } {TEXT -1 45 "_f due to the work done by the force on mass " }{TEXT 269 1 "m" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 68 "Now let us co mpare this result with the potential energy difference:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "r_i;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "We need a function to calculate the magnitude of the posi tion vector:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "v_mag:=v->s qrt(add(v[i]^2,i=1..3));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "v_mag(r_i),v_mag(r_f);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 " subs(r=v_mag(r_i),V)-subs(r=v_mag(r_f),V);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 180 "We recognize that the difference between the potential e nergy at the initial and final points amounts to the same as the work \+ done on the particle, or the change in kinetic energy." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "Work-%;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 16 "with(plottools);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "l := line(r_i, r_f, color=red, linestyle=2, thickness =3): plots[display](l,PL1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "We want to indicate with the height the amount of potential energy, ther efore:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "r_iV:=r_i: r_iV[3 ]:=subs(r=v_mag(r_i),G=1,M=1,m=1,V): r_iV;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 59 "r_fV:=r_f: r_fV[3]:=subs(r=v_mag(r_f),G=1,M=1,m=1,V ): r_fV;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "l := line(r_iV, r_fV, color=red, linestyle=2, thickness=3): plots[display](l,PL1);" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 159 "As the particle traverses the p otential well it picks up a lot of kinetic energy on the way down, in \+ order to give up a good fraction as it climbs up the hill." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "We can verify thi s by calculating the work done at some intermediate point." }}{PARA 0 "" 0 "" {TEXT -1 43 "First recall the total kinetic energy gain:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalf(Work);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "Now break the path up into two halves, ju st by looking at what happens between " }{TEXT 275 1 "t" }{TEXT -1 7 " =0 and " }{TEXT 274 1 "t" }{TEXT -1 23 "=0.5, and then between " } {TEXT 273 1 "t" }{TEXT -1 9 "=0.5 and " }{TEXT 272 1 "t" }{TEXT -1 3 " =1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "Work1:=int(dotprod(s ubs(x=r_t[1],y=r_t[2],z=r_t[3],op(F)),v_t),t=0..1/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "Work2:=int(dotprod(subs(x=r_t[1],y= r_t[2],z=r_t[3],op(F)),v_t),t=1/2..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "evalf([Work1,Work2,Work1+Work2]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 323 "Notice the large amount of work done on the part icle on the first 'half', then the particle works against the potentia l well, but still has a net gain of kinetic energy at the final point. It is, however, at a lower value of potential energy at this point, i .e., it gained kinetic energy at the expense of potential energy." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 177 " Now consider a non-conservative force. It will have no direct physical meaning, i.e., it is a made-up problem to demonstrate what happens wh en a force field is not conservative." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "F:=[x^2+2*y-z,y+z-x,z^3+x*2];" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "curl(F,[x,y,z]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "This force field is not curl-free, which means that it cannot be derived from a scalar potential. We choose the same path as before." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "r_i:=[2 ,2,1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "r_f:=[-2,0,1];" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "r_t:=evalm(r_i+(r_f-r_i)*t );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "v_t:=diff(r_t,t);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "Another one of these Maple-quirks . We defined " }{TEXT 19 3 "r_t" }{TEXT -1 36 " as a combination of li sts, but the " }{TEXT 19 5 "evalm" }{TEXT -1 21 " did something funny. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "whattype(r_t);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "The evalm turned " }{TEXT 19 3 "r_ t" }{TEXT -1 15 " into a symbol." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "v_t:=map(diff,r_t,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "If we had avoided the " }{TEXT 19 5 "evalm" }{TEXT -1 139 ", the simple differentiation would have worked because t appeared as a factor of one of the lists (as carried out in the previous secti on)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "Work1:=int(dotprod( subs(x=r_t[1],y=r_t[2],z=r_t[3],F),v_t),t=0..1/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "Work2:=int(dotprod(subs(x=r_t[1],y=r_t[2],z =r_t[3],F),v_t),t=1/2..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "Work:=int(dotprod(subs(x=r_t[1],y=r_t[2],z=r_t[3],F),v_t),t=0..1); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "evalf([Work1,Work2,Work 1+Work2,Work]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 229 "Of course, th e work integral is additive. What we want to demonstrate is that if we connect the two endpoints in some different way, then a different res ult will be found for the work done on the particle as it moves from A to B." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 "Let us choose as an alternative to the direct connection from A t o B a path that goes via the origin [0,0,1] in the plane." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "r_1: =[0,0,1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "r_tp1:=evalm(r _i+(r_1-r_i)*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "v_tp1:= map(diff,r_tp1,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "r_tp2 :=evalm(r_1+(r_f-r_1)*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "v_tp2:=map(diff,r_tp2,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 " Now we are ready to calculate the work integral." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Work1:=int(dotprod(subs(x=r_tp1[1],y=r_tp1[2] ,z=r_tp1[3],F),v_tp1),t=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Work2:=int(dotprod(subs(x=r_tp2[1],y=r_tp2[2],z=r_tp2[3],F),v_ tp2),t=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "evalf([Wor k1+Work2,Work]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 109 "The amount w ork done on the particle as it moves through the force field along a d ifferent path is different!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "What can we do to visualize the force field, an d how it affects the particle?" }}{PARA 0 "" 0 "" {TEXT -1 36 "First d efine the force field in the " }{TEXT 276 1 "z" }{TEXT -1 90 "=1 plane and ignore the z-component of the force, as we are not moving the par ticle along " }{TEXT 277 1 "z" }{TEXT -1 51 " (no work is done by this component of the force!)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Fzeq1:=subs(z=1,F);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 " F2d:=[Fzeq1[1],Fzeq1[2]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "PL1:=fi eldplot(F2d,x=-2..2,y=-2..2,arrows=thick,color=blue,axes=boxed,grid=[1 2,12]): display(PL1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "wi th(plottools):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "l1 := lin e([r_i[1],r_i[2]], [r_f[1], r_f[2]], color=red, linestyle=2, thickness =3): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "l2 := line([r_i[1] ,r_i[2]], [r_1[1], r_1[2]], color=green, linestyle=2, thickness=3): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "l3 := line([r_1[1],r_1[2] ], [r_f[1], r_f[2]], color=green, linestyle=2, thickness=3): " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "display(l1,l2,l3,PL1);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 210 "The amount of work done on the pa rticle as it moves through this field is different, because in differe nt regions of space the the force vectors are allowed to have differen t lengths and different orientations." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 118 "For a force field which is not curl- free, the travel over a closed contour (loop) allows one to gain (or l ose) energy." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "The graph also shows visually why we refer to the field as not \+ being curl-free." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "evalf([ Work1,Work2,Work1+Work2,Work]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 169 "Note that the amount of work done by the force on the particle is negative. We start the particle at [2,2] (top right-hand corner), and it moves against the force field." }}{PARA 0 "" 0 "" {TEXT -1 278 "Th e fact that it arrives at location [0,-2] with a negative amount of wo rk (it worked against the force field) means that its kinetic energy w ill be less than it was at A. It will be less at B than at A by a diff erent amount depending on whether the red or green path is chosen." }} {PARA 0 "" 0 "" {TEXT -1 24 "An interesting question:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 278 9 "Exercise:" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 114 "Calculate the wor k done on the particle along a path such that it gains energy to go fr om A=[2,2,1] to B=[-2,0,1]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 279 8 "Solution" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "r_i,r_f;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "r_1:=[2,-2,1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "r_tp1:=evalm(r_i+(r_1-r_i)*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "r_2:=[-2,-2,1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "v_tp1:=map(diff,r_tp1,t);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 30 "r_tp2:=evalm(r_1+(r_2-r_1)*t);" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 25 "v_tp2:=map(diff,r_tp2,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "r_tp3:=evalm(r_2+(r_f-r_2)*t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "v_tp3:=map(diff,r_tp3,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Now we are ready to calculate the work integral." }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Work1:=int(dotprod(subs(x=r _tp1[1],y=r_tp1[2],z=r_tp1[3],F),v_tp1),t=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Work2:=int(dotprod(subs(x=r_tp2[1],y=r_tp2[2] ,z=r_tp2[3],F),v_tp2),t=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Work3:=int(dotprod(subs(x=r_tp3[1],y=r_tp3[2],z=r_tp3[3],F),v_ tp3),t=0..1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalf(Work 1+Work2+Work3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 5 } {VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }