Find the work done by the force field f(x,y,z)=3xi+3yj+7kf(x,y,z)=3xi+3yj+7k on a particle that moves along the helix r(t)=3cos(t)i+3sin(t)j+3tk,0≤t≤2π

1 Answer

  • Call the path [tex]\mathcal C[/tex]. Then the work done by [tex]\mathbf f(x,y,z)[/tex] along [tex]\mathcal C[/tex] is given by the line integral,[tex]\displaystyle\int_{\mathcal C}\mathbf f(x,y,z)\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt[/tex]
    Swapping the [tex]\mathbf{ijk}[/tex] notation out for ordered component notation, I'll write
    [tex]\mathbf r(t)=(x(t),y(t),z(t))=(3\cos t,3\sin t,3t)[/tex]
    so that
    [tex]\dfrac{\mathrm d\mathbf r}{\mathrm dt}=(-3\sin t,3\cos t,3)[/tex]
    The line integral reduces to[tex]\displaystyle\int_0^1(9\cos t,9\sin t,7)\cdot(-3\sin t,3\cos t,3)\,\mathrm dt[/tex][tex]=\displaystyle\int_0^1(-27\cos t\sin t+27\sin t\cos t+21)\,\mathrm dt[/tex][tex]=\displaystyle21\int_0^1\mathrm dt=21[/tex]