Question

In Exercises 33-38, find the work done by the force field \(\mathbf{F}\) on a particle moving along the given path. \(\mathbf{F}(x, y)=-x \mathbf{i}-2 y \mathbf{j}\) \(C: y=x^{3}\) from (0,0) to (2,8)

Short answer

The work done by the force field on a particle moving along the given path is -66 units.

Step-by-step solution

Parameterize the Curve

Start by parameterizing the given curve with respect to \(x\). Since \(x\) ranges from 0 to 2, use the letter \(t\) to represent \(x\). So, let \(x=t\) and \(y=t^{3}\). Therefore, the parameterized curve \(C(t)\) is expressed as: \(C(t)=\) where \(t\) is in \([0,2]\).

Compute the Differential of the Parameterized Curve

Compute the differential \(dC(t)\) of the parameterized curve which will be used in the line integral. The differential \(dC(t)\) is simply the derivative of \(C(t)\), and can be obtained as: \(dC(t)=C'(t)dt=<1,3t^{2}>dt\). Thus, we will use \(F(C(t))dC(t)\) to compute the line integral in the next step.

Calculate the Line Integral

Compute the line integral of \(F\) along the curve \(C\), which represents the work done. This is given by the formula: \[W=\int_{C} \mathbf{F} \cdot d \mathbf{r}=\int_{a}^{b} \mathbf{F}(C(t)) \cdot d C(t) d t\]Here, \(\mathbf{a}\) and \(\mathbf{b}\) are the limits of the parameter, which in this case are 0 and 2. Substituting the given force field \(\mathbf{F}(x, y)=-x \mathbf{i}-2 y \mathbf{j}\) in terms of \(t\) we get \(\mathbf{F}(C(t))=-t \mathbf{i}-2 t^{3} \mathbf{j}\). Now,\(F(C(t)) \cdot dC(t)=(-t)(1)-2t^{3}(3t^{2})=-t-6t^{5}\). Putting these pieces together, the work \(W\) becomes \[W=\int_{0}^{2} (-t-6t^{5}) dt\]

Solve the Integral

Evaluate the integral to get the work done. This is a simple power rule problem, which gives \[W= -\frac{1}{2}t^{2}-\frac{6}{6}t^{6}\Bigg|_{0}^{2}=-2-64+0=-66\]