Question

Find the work done by the force field \(F\) in moving an object from \(P\) to \(Q\). $$ \mathbf{F}(x, y)=\frac{2 x}{y} \mathbf{i}-\frac{x^{2}}{y^{2}} \mathbf{j} ; P(-3,2), Q(1,4) $$

Short answer

The short answer could be computed as the definite integral obtained in Step 5. However, without the specific computation, the answer is not available

Step-by-step solution

Calculation of Displacement

Calculate the displacement vector \(\delta\) as the difference between the end point and initial point: \( \delta = Q - P = (1-(-3), 4-2) = (4, 2) \)

Definition of Parametric Equations

Assume the object is moving along a straight line. That line can be represented by the parametric equations: \(x(t) = -3 + 4t, y(t) = 2 + 2t\), where \(0\leq t\leq1\). This is based on the fact that, when \(t = 0\), we are at \(P(-3,2)\) and when \(t = 1\), we are at \(Q(1,4)\)

Calculation of the Vector Field along the Path

Substitute \(x(t)\) and \(y(t)\) into \(F\) to get the force field along the path: \(\mathbf{F}(t) = \frac{2 x(t)}{y(t)} \mathbf{i}-\frac{x(t)^{2}}{y(t)^{2}} \mathbf{j}\)

Compute the Dot Product of Force Field and the Differential of the Displacement Vector

The work done is the integral from 0 to 1 of the dot product of force field and the differential of the displacement vector, \(d\delta\), along the path. Since \(dx = 4 dt\) and \(dy = 2 dt\), \(d\delta = (dx, dy) = (4 dt, 2 dt)\). Calculate the dot product\(F(t) \cdot d\delta(t)\) and then integrate from 0 to 1

Final Integration

Take the integral of the product obtained in Step 4 from 0 to 1. The result is the work done by the force