A particle moves along line segments from the origin to the points (3, 0, 0), (3, 3, 1), (0, 3, 1), and back to the origin under the influence of the force field
F(x, y, z) = z2i + 5xyj + 4y2k.
I have tried multiple times to solve this problem but the answer is still not right.
To find the work done by the force field, you need to calculate the line integral along the path of the particle. The line integral of a vector field F along a curve C is given by:
∫F • dr = ∫(F • T) ds
Where F is the vector field, dr is the differential displacement along the curve C, T is the unit tangent vector to the curve, and ds is the differential arc length along C.
Let's break down the problem step by step:
1. Determine the parametric equations for the path of the particle.
The path of the particle consists of three line segments: OA, AB, and BO.
OA: (x, y, z) = (t, 0, 0), where 0 ≤ t ≤ 3
AB: (x, y, z) = (3, t, t), where 0 ≤ t ≤ 3
BO: (x, y, z) = (3 - t, 3, 1), where 0 ≤ t ≤ 3
2. Calculate the unit tangent vector T for each segment.
The unit tangent vector T is the normalized derivative of the position vector with respect to the parameter t.
For OA: T = (1, 0, 0)
For AB: T = (0, 1, 1)
For BO: T = (-1, 0, 0)
3. Calculate the line integrals for each segment.
We need to find the dot product of the force field F with the unit tangent vector T and integrate it with respect to the parameter t.
For OA:
∫(F • dr) = ∫[(z^2)i + (5xy)j + (4y^2)k] • (i dt) = ∫t^2 dt = 1/3 t^3 | from 0 to 3 = 1/3(3)^3 - 1/3(0)^3 = 9
For AB:
∫(F • dr) = ∫[(z^2)i + (5xy)j + (4y^2)k] • (j dt) = ∫(5t^2) dt = 5/3 t^3 | from 0 to 3 = 5/3(3)^3 - 5/3(0)^3 = 45
For BO:
∫(F • dr) = ∫[(z^2)i + (5xy)j + (4y^2)k] • (-i dt) = ∫(-t^2) dt = -1/3 t^3 | from 0 to 3 = -1/3(3)^3 - (-1/3(0)^3) = -9
4. Sum up the individual line integrals to find the total work done.
Total work done = ∑(Line integrals) = 9 + 45 + (-9) = 45 units of work.
Therefore, the total work done by the force field along the given path is 45 units of work.