Q:

Find a parametrization for the curve「and determine the work done on a particle moving along Γ in R3 through the force field F:R^3--R^3'where F(x,y,z) = (1,-x,z) and (a) Im (Γ) is the line segment from (0,0,0) to (1,2,1) (b) Im (Γ) is the polygonal curve with successive vertices (1,0,0), (0,1,1), and (2,2,2) (c) Im (Γ) is the unit circle in the plane z = 1 with center (0,0,1) beginning and ending at (1,0,1), and starting towards (0,1,1)

Accepted Solution

A:
a. Parameterize [tex]\Gamma[/tex] by[tex]\vec r(t)=(t,2t,t)[/tex]with [tex]0\le t\le1[/tex]. The work done by [tex]\vec F[/tex] along [tex]\Gamma[/tex] is[tex]\displaystyle\int_\Gamma\vec F\cdot\mathrm d\vec r=\int_0^1(1,-t,t)\cdot(1,2,1)\,\mathrm dt=\int_0^1(1-t)\,\mathrm dt=\boxed{\frac12}[/tex]b. Break up [tex]\Gamma[/tex] into each component line segment, denoting them by [tex]\Gamma_1[/tex] and [tex]\Gamma_2[/tex], and parameterize each respectively by[tex]\vec r_1(t)=(1-t,t,t)[/tex] and[tex]\vec r_2(t)=(2t,1+t,1+t)[/tex]both with [tex]0\le t\le1[/tex]. Then the work done by [tex]\vec F[/tex] along each component path is[tex]\displaystyle\int_{\Gamma_1}\vec F\cdot\mathrm d\vec r_1=\int_0^1(1,t-1,t)\cdot(-1,1,1)\,\mathrm dt=\int_0^1(2t-2)\,\mathrm dt=-1[/tex][tex]\displaystyle\int_{\Gamma_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(1,-2t,1+t)\cdot(2,1,1)\,\mathrm dt=\int_0^1(3-t)\,\mathrm dt=\frac52[/tex]giving a total work done of [tex]-1+\dfrac52=\boxed{\dfrac32}[/tex].c. Parameterize [tex]\Gamma[/tex] by[tex]\vec r(t)=(\cos t,\sin t,1)[/tex]with [tex]0\le t\le2\pi[/tex]. Then the work done by [tex]\vec F[/tex] is[tex]\displaystyle\int_\Gamma\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}(1,-\cos t,1)\cdot(-\sin t,\cos t,0)\,\mathrm dr=-\int_0^{2\pi}(\sin t+\cos^2t)\,\mathrm dt=\boxed{-\pi}[/tex]