IDNLearn.com: Where your questions are met with thoughtful and precise answers. Get accurate and comprehensive answers to your questions from our community of knowledgeable professionals.
Parameterize each line segment from [tex](x_0,y_0,z_0)[/tex] to [tex](x_1,y_1,z_1)[/tex] by
[tex]\vec r(t) = (1-t) (x_0\,\vec\imath + y_0\,\vec\jmath + z_0\,\vec k) + t (x_1\,\vec\imath + y_1\,\vec\jmath + z_1\,\vec k[/tex]
with [tex]0\le t\le1[/tex]. The work done by [tex]\vec F[/tex] on the particle along each segment is given the line integral of [tex]\vec F[/tex] with respect to that segment,
[tex]\displaystyle \int_{C_i} \vec F \cdot d\vec r = \int_0^1 \vec F(\vec r_i(t)) \cdot \dfrac{d\vec r_i(t)}{dt} \, dt[/tex]
• (3, 0, 0) to (3, 5, 1)
[tex]\vec r_1(t) = 3\,\vec\imath + 5t\,\vec\jmath + t\,\vec k[/tex]
[tex]W_1 = \displaystyle \int_0^1 \left(t^2\,\vec\imath + 75t\,\vec\jmath + 50t^2\,\vec k\right) \cdot \left(5\,\vec\jmath + \vec k\right) \, dt \\\\ ~~~~~~~~ = \int_0^1 (375t + 50t^2) \, dt = \frac{1225}6[/tex]
• (3, 5, 1) to (0, 5, 1)
[tex]\vec r_2(t) = 3(1-t)\,\vec\imath + 5(1-t)\,\vec\jmath + \vec k[/tex]
[tex]W_2 = \displaystyle \int_0^1 \left(\vec\imath + 75(1-t)\,\vec\jmath + 50 \,\vec k\right) \cdot \left(-3\,\vec\imath - 5\,\vec\jmath\right) \, dt \\\\ ~~~~~~~~ = -3 \int_0^1 \,dt = -3[/tex]
• (0, 5, 1) to (0, 0, 0)
[tex]\vec r_3(t) = 5(1-t)\,\vec\jmath + (1-t)\,\vec k[/tex]
[tex]W_3 = \displaystyle \int_0^1 \left((1-t)^2\,\vec\imath + 50(1-t)^2\,\vec k\right) \cdot \left(-5\,\vec\jmath - \vec k\right) \, dt \\\\ ~~~~~~~~ = \int_0^1 (-50 + 100t - 50t^2) \, dt = -\frac{50}3[/tex]
Then the total work done by [tex]\vec F[/tex] on the particle is
[tex]W = W_1 + W_2 + W_3 = \boxed{\dfrac{369}2}[/tex]