Question #183526

Evaluate LaTeX: \int_cF.dr\:\: where LaTeX: F\left(x,y,z\right)=xzi-yzkF(x,y,z)=xzi−yzk and c is the line segment from (3,0,1) to (-1,2,0)


1
Expert's answer
2021-05-04T13:04:50-0400

Given, F(x,y,z)=xziyzkF(x,y,z)=xz\bold{i}-yz\bold{k} and line segment c from (3,0,1) to (-1,2,0).

Then,

r(t)=(1t)<3,0,1>+t<1,2,0>=<34t,2t,1t>Differentiate with respect to t,drdt=<4,2,1>dr=(4i+2jk)dtNow, F(r(t))=(34t)(1t)i2t(1t)k=(3t27t+3)i+(2t22t)kTherefore,F.dr=[(3t27t+3)i+(2t22t)k].[4i+2jk]dt=014(3t27t+3)(2t22t)dt=0114t2+30t12dt=[143t3+15t212t]01=53r(t)=(1-t)<3,0,1>+t<-1,2,0>=<3-4t,2t,1-t>\newline \text{Differentiate with respect to t,} \frac{dr}{dt}=<-4,2,-1>\newline dr=(-4i+2j-k)dt\newline \text{Now, }F(r(t))=(3-4t)(1-t)i-2t(1-t)k\newline \hspace{2.1cm}=(3t^2-7t+3)i+(2t^2-2t)k\newline \text{Therefore,}\newline \intop F.dr=[(3t^2-7t+3)i+(2t^2-2t)k].[−4i+2j−k]dt\newline \hspace{1.1cm}=\int^1_{0}-4(3t^2-7t+3)-(2t^2-2t)dt\newline \hspace{1.1cm}=\int^1_{0}-14t^2+30t-12dt\newline \hspace{1.1cm}=[-\frac{14}{3}t^3+15t^2-12t]^1_{0}\newline \hspace{1.1cm}=\frac{-5}{3}

Thus,the requrired answer is 5/3.


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