Question

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 1

Answer:

0

Step-by-step explanation:

We want to compute the curve integral (or line integral)

$\bf \int_{C}F$

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 = $\bf C_1+C_2$

where  

$\bf C_1$ =  line segment from (1, 5) to (0, 0)  

$\bf C_2$ = line segment from (0, 0) to (0, 9)

so,

$\bf \int_{C}F=\int_{C_1}F+\int_{C_2}F$

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 $\bf C_1$ can be parameterized as

$\bf r_1(t) = (1-t, 5-5t)$ for 0 ≤ t ≤ 1

and $\bf C_2$ can be parameterized as

$\bf r_2(t) = (0, 9t)$ for 0 ≤ t ≤ 1

The derivatives are

$\bf r_1'(t) = (-1, -5)$

$\bf r_2'(t) = (0, 9)$

and

$\bf \int_{C_1}F=\int_{0}^{1}F(r_1(t))\circ r_1'(t)dt=\int_{0}^{1}(5-5t,t-1)\circ (-1,-5)dt=0$

$\bf \int_{C_2}F=\int_{0}^{1}F(r_2(t))\circ r_2'(t)dt=\int_{0}^{1}(9t,0)\circ (0,-9)dt=0$

In consequence,

$\bf \int_{C}F=0$