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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)

1 Answer

4 votes

a. Parameterize
\Gamma by


\vec r(t)=(t,2t,t)

with
0\le t\le1. The work done by
\vec F along
\Gamma is


\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}

b. Break up
\Gamma into each component line segment, denoting them by
\Gamma_1 and
\Gamma_2, and parameterize each respectively by


  • \vec r_1(t)=(1-t,t,t) and

  • \vec r_2(t)=(2t,1+t,1+t)

both with
0\le t\le1. Then the work done by
\vec F along each component path is


\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


\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

giving a total work done of
-1+\frac52=\boxed{\frac32}.

c. Parameterize
\Gamma by


\vec r(t)=(\cos t,\sin t,1)

with
0\le t\le2\pi. Then the work done by
\vec F is


\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}

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