The work done by the force F = on a particle traveling along the path C is given
ID: 2892140 • Letter: T
Question
The work done by the force F = on a particle traveling along the path C is given by W = integral_c F.middot dr where C is parameterized by for a lessthanorequalto t lessthanorequalto b. Depending on the nature of the vector field F and the path C, you should apply the appropriate technique to efficiently calculate W. Case I. vector field F is not conservative (ie., F notequalto nabla^vector f for any scalar function f, i.e., PartialdifferentailP/Partialdifferentailx). Compute W using the definition by W = integral^b_a F(x(t), (t)).x'(t), y'(t) > dt. Calculate the work done by the force F = along the path for 0 lesstahnorequalto t lesstahnorequalto 1 from (0, 0) to from (1, 1) and from (1, 1) to (2, 3) along a straight line.Explanation / Answer
C1: from (0,0) to (1,1)
F=<x+y2,y2,x2-y> , r(t)=<t,t2,t3> ,0t1
F(r(t))=<t+t4,t4,t2-t2>=<t+t4,t4,0> , r'(t)=<1,2t,3t2>
C1Fdr=[0 to 1]F(r(t)).r'(t)dt
C1Fdr=[0 to 1]<t+t4,t4,0>.<1,2t,3t2>dt
C1Fdr=[0 to 1][(t+t4)+(t4*2t)+(0*3t2)]dt
C1Fdr=[0 to 1][t+t4+2t5]dt
C1Fdr=[0 to 1][(1/2)t2+(1/5)t5+(1/3)t6]
C1Fdr=[(1/2)12+(1/5)15+(1/3)16]-[(1/2)02+(1/5)05+(1/3)06]
C1Fdr=[(1/2)+(1/5)+(1/3)]-0
C1Fdr=31/30
C2: from (1,1) to (2,3)
F=<x+y2,y2,x2-y> , r(t)=<t,2t-1,0> ,1t2
F(r(t))=<1+(2t-1)2,(2t-1)2,t2-(2t-1)> , r'(t)=<1,2-0,0>
F(r(t))=<1+(2t-1)2,(2t-1)2,t2-2t+1)> , r'(t)=<1,2,0>
C2Fdr=[1 to 2]F(r(t)).r'(t)dt
C2Fdr=[1 to 2]<1+(2t-1)2,(2t-1)2,t2-2t+1)>.<1,2,0>dt
C2Fdr=[1 to 2][1+(2t-1)2+2(2t-1)2+0]dt
C2Fdr=[1 to 2][1+3(2t-1)2]dt
C2Fdr=[1 to 2][t+(1/2)(2t-1)3]
C2Fdr=[2+(1/2)(4-1)3]-[1+(1/2)(2-1)3]
C2Fdr=2+(27/2)-1-(1/2)
C2Fdr=14
work done,W =C1Fdr+C2Fdr
work done,W =(31/30)+14
work done,W =(451/30)
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