Final Answer:
The work done by the field F(x, y, z) = (e^x, e^y, xyz) when the object moves along the path r = (t^2, t, t/4) for 0 <= t <= 1 is approximately 4.57 units.
Step-by-step explanation:
To calculate the work done by the field along the given path, we use the line integral formula ∫ F ⋅ dr, where F(x, y, z) is the field and dr is the differential displacement vector along the path. Given F(x, y, z) = (e^x, e^y, xyz) and r = (t^2, t, t/4), we need to find the dot product and integrate it along the path.
The differential displacement vector dr = (dx, dy, dz) is calculated by taking the derivatives of the parameterized path equation r = (t^2, t, t/4) with respect to t. So, dr = (2t, 1, 1/4) dt.
Next, compute F ⋅ dr, which is the dot product of F and dr: F ⋅ dr = (e^(t^2) * 2t) + (e^t) + ((t^3)/4).
Then, set up the integral ∫ F ⋅ dr from t = 0 to t = 1 and evaluate the integral: ∫ F ⋅ dr = ∫[0 to 1] ((e^(t^2) * 2t) + (e^t) + ((t^3)/4)) dt ≈ 4.57 units.
This result represents the approximate work done by the field along the specified path. Integrating the dot product of the field and the differential displacement vector along the path provides the overall work done by the field.