Question

Given the force field F, find the work required to move an object on the given oriented curve.

F = on the path consisting of the line segment from (-1,0) to (0,8) followed by the line segment from (0,8) to (2,8)

Answer 1

67/4

Explanation:

Let C, D be the line segments that join (-1,0) with (0,8) and (0,8) with (2,8) respectively.

The work required by the force field F(x,y)=(x,y) is the curve integral

`\large \int_(C+D)^{} F=\int_(C)^{}F+\int_(D)^{}F`

the line segments that joins (-1,0) with (0,8) can be parameterized as

r(t) = t(0,8)+(1-t)(-1,0)=(t-1,8t) with 0≤ t≤ 1

So

`\large \int_(C)^{}F=\int_(0)^(1)F(r(t))\bullet r'(t)dt=\int_(0)^(1)(t-1,8t)\bullet (1,8)dt=\int_(0)^(1)(t-1+64t)dt=\\\\=\int_(0)^(1)(65t-1)dt=65\int_(0)^(1)tdt-\int_(0)^(1)dt=(65)/(2)-1=(63)/(2)`

the line segments that joins (0,8) with (2,8) can be parameterized as

r(t) = t(2,8)+(1-t)(0,8)=(2t,8) with 0≤ t≤ 1

hence

`\large \int_(D)^{}F=\int_(0)^(1)F(r(t))\bullet r'(t)dt=\int_(0)^(1)(2t,8)\bullet (2,0)dt=\int_(0)^(1)4tdt=4\int_(0)^(1)tdt=2`

and we have the work required by F to move a particle alon the path is

`\large \int_(C)^{}F+\int_(D)^{}F=(63)/(2)+2=(67)/(4)`