Question

find the work done by the force field f on a particle moving along the given path. f(x, y) = x2i − xyj c: x = cos3 t, y = sin3 t from (1, 0) to (0, 1)

Answer 1

The work done by a force field on a particle along a path is calculated using a line integral. The process involves substituting the parametrized path into the force field, finding the differential displacements dx and dy, and evaluating the integral over the given interval.

Step-by-step explanation:

To find the work done by the force field ℝ on a particle moving along a given path, we need to compute the line integral of the force along the path. The force field is given as f(x, y) = x2i − xyj, and the path is parametrized by c: x = cos3t, y = sin3t. To compute the work done, we utilize the line integral formula:

W = ∫C F ⋅ dr

where W is the work done, F is the force field, and dr is the differential displacement along the path. Since the force field can be expressed as F(x, y) = Fxi + Fyj, and the path is parametrized by x(t) and y(t), we substitute these expressions into the line integral:

W = ∫ Fxdx + Fydy

By substituting x(t) and y(t) into F(x, y) and calculating dx and dy in terms of dt, we obtain:

dx = -3sin2t cos(t) dt and dy = 3cos2t sin(t) dt

Then, we evaluate the integral over the interval from t = 0 to t = π/2, which corresponds to the path from (1, 0) to (0, 1).

The computation involves integrating the product of the force field components and the parametric derivatives of x and y with respect to t:

W = ∫ Fxdx + Fydy = ∫ (x2)(-3sin2t cos(t)) dt + (x(-y))(3cos2t sin(t)) dt

By solving this integral, we find the total work done by the force field on the particle.

Answer 2

The work done by the force field
`\( \mathbf{F}(x, y) = x^2\mathbf{i} - xy\mathbf{j} \)` on a particle moving along the path
`\( x = \cos^3 t, \, y = \sin^3 t \) from \((1, 0)\)`to
`\((0, 1)\)`is
`\( (1)/(4) \)` joules.

Explanation:

To find the work done, we use the line integral of the force field
`\( \mathbf{F} \)`along the given path. The line integral is given by:

`\[ W = \int_C \mathbf{F} \cdot d\mathbf{r} \]`

Here,
`\( \mathbf{F}(x, y) = x^2\mathbf{i} - xy\mathbf{j} \)`and the path is parameterized as
`\( x = \cos^3 t, \, y = \sin^3 t \)`. The limits of integration are from
`\( t = 0 \) to \( t = (\pi)/(2) \)` to cover the path from
`\((1, 0)\) to \((0, 1)\)`.

The differential displacement
`\( d\mathbf{r} \)`is given by
`\( dx\mathbf{i} + dy\mathbf{j} \),` which, in terms of
`\( t \)`, becomes
`\( -3\cos^2 t \sin t dt\mathbf{i} + 3\sin^2 t \cos t dt\mathbf{j} \)`.

Substitute these into the line integral formula and evaluate:

`\[ W = \int_0^{(\pi)/(2)} \left( \cos^6 t - \cos^3 t \sin^4 t \right) dt \]`

Solving this integral yields
`\( (1)/(4) \)`joules. This indicates that the work done by the force field on the particle as it moves along the specified path is
`\( (1)/(4) \)` joules.