if a force F=5x² y i +3 x y j displaces a particle in xy plane from (0,0) to (1,5) along a curve Y=4x² find the work done

To find the work done by the force \( \mathbf{F} = 5x^2y \, \mathbf{i} + 3xy \, \mathbf{j} \) as it displaces a particle along the curve \( y = 4x^2 \) from the point \( (0,0) \) to \( (1,5) \), we can use the line integral of the force along the specified path. The work done \( W \) is given by the integral of the force vector along the curve defined by the displacement.

Setting Up the Problem

First, we need to express the curve in terms of a single variable. Since we have \( y = 4x^2 \), we can parameterize the curve using \( x \) as our parameter. The limits for \( x \) will range from 0 to 1, corresponding to the points \( (0,0) \) and \( (1,5) \).

Parameterization of the Curve

Using the parameterization:

• Let \( x = t \)
• Then \( y = 4t^2 \)

Thus, the parameterization of the curve can be written as:

\( \mathbf{r}(t) = (t, 4t^2) \) for \( t \) in the interval \([0, 1]\).

Calculating the Work Done

The work done by the force along the curve is given by the line integral:

\( W = \int_C \mathbf{F} \cdot d\mathbf{r} \)

Where \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt \). First, we need to compute \( d\mathbf{r} \):

Since \( \mathbf{r}(t) = (t, 4t^2) \), we differentiate:

\( \frac{d\mathbf{r}}{dt} = \left( \frac{dx}{dt}, \frac{dy}{dt} \right) = (1, 8t) \)

Thus, \( d\mathbf{r} = (1, 8t) dt \).

Substituting into the Integral

Now, we substitute \( x = t \) and \( y = 4t^2 \) into the force vector:

\( \mathbf{F}(t) = 5t^2(4t^2) \, \mathbf{i} + 3t(4t^2) \, \mathbf{j} = 20t^4 \, \mathbf{i} + 12t^3 \, \mathbf{j} \)

Now, we can compute the dot product \( \mathbf{F} \cdot d\mathbf{r} \):

\( \mathbf{F} \cdot d\mathbf{r} = (20t^4, 12t^3) \cdot (1, 8t) = 20t^4 \cdot 1 + 12t^3 \cdot 8t = 20t^4 + 96t^4 = 116t^4 \)

Evaluating the Integral

The work done is then:

\( W = \int_0^1 116t^4 \, dt \)

Calculating this integral:

\( W = 116 \int_0^1 t^4 \, dt = 116 \left[ \frac{t^5}{5} \right]_0^1 = 116 \cdot \frac{1}{5} = \frac{116}{5} = 23.2 \)

Final Result

Therefore, the work done by the force as the particle moves along the curve from \( (0,0) \) to \( (1,5) \) is:

W = 23.2 units of work.