Given the force field F, find the work required to move an object on the given oriented curve. F = (y, - x) on the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9) The amount of work required is ____ (Simplify your answer.)

Answer:0Step-by-step explanation:We want to compute the curve integral (or line integral) [tex] \bf \int_{C}F[/tex] where the force field F is defined by F(x,y) = (y, -x) and C is the path consisting of the line segment from (1, 5) to (0, 0) followed by the line segment from (0, 0) to (0, 9). We can write  C = [tex] \bf C_1+C_2[/tex] where  [tex] \bf C_1[/tex] =  line segment from (1, 5) to (0, 0)  [tex] \bf C_2[/tex] = line segment from (0, 0) to (0, 9) so, [tex] \bf \int_{C}F=\int_{C_1}F+\int_{C_2}F[/tex] Given 2 points P, Q in the plane, we can parameterize the line segment joining P and Q with r(t) = tQ + (1-t)P for 0 ≤ t ≤ 1 Hence [tex] \bf C_1[/tex] can be parameterized as [tex] \bf r_1(t) = (1-t, 5-5t)[/tex] for 0 ≤ t ≤ 1 and [tex] \bf C_2[/tex] can be parameterized as [tex] \bf r_2(t) = (0, 9t)[/tex] for 0 ≤ t ≤ 1 The derivatives are [tex] \bf r_1'(t) = (-1, -5)[/tex] [tex] \bf r_2'(t) = (0, 9)[/tex] and [tex] \bf \int_{C_1}F=\int_{0}^{1}F(r_1(t))\circ r_1'(t)dt=\int_{0}^{1}(5-5t,t-1)\circ (-1,-5)dt=0[/tex] [tex] \bf \int_{C_2}F=\int_{0}^{1}F(r_2(t))\circ r_2'(t)dt=\int_{0}^{1}(9t,0)\circ (0,-9)dt=0[/tex] In consequence, [tex] \bf \int_{C}F=0[/tex]