To find the work done by the force $\vec{F}$ along the given path, we can use the line integral of the force along the path.

Step 1: Express the Force and Path

Given: $\vec{F} = xy\hat{i} - y^2\hat{j}$ Path: $x = 2t^3, \quad y = t^2$

The work done by the force along the path is given by: $W = \int_C \vec{F} \cdot d\vec{r}$ where $d\vec{r} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j}$.

Step 2: Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Given $x = 2t^3$ and $y = t^2$, we calculate: $\frac{dx}{dt} = \frac{d}{dt}(2t^3) = 6t^2$ $\frac{dy}{dt} = \frac{d}{dt}(t^2) = 2t$

So, $d\vec{r} = 6t^2\hat{i} + 2t\hat{j} \, dt$

Step 3: Substitute into the Work Integral

Substitute $\vec{F}$ and $d\vec{r}$ into the work integral: $W = \int_0^1 \left[ (xy)\hat{i} - y^2\hat{j} \right] \cdot \left( 6t^2\hat{i} + 2t\hat{j} \right) dt$

Step 4: Compute the Dot Product

Calculate the dot product: $\vec{F} \cdot d\vec{r} = (xy)(6t^2) + (-y^2)(2t)$ Substitute $x = 2t^3$ and $y = t^2$ into the expression: $= \left( (2t^3)(t^2) \right)(6t^2) + \left( -(t^4) \right)(2t)$ Simplify the expression: $= \left( 2t^5 \right)(6t^2) - 2t^5$ $= 12t^7 - 2t^5$

Step 5: Integrate from $t = 0$ to $t = 1$

Now, integrate this expression with respect to $t$: $W = \int_0^1 (12t^7 - 2t^5) dt$ $W = \left[ \frac{12t^8}{8} - \frac{2t^6}{6} \right]_0^1$ $W = \left[ \frac{12}{8}t^8 - \frac{2}{6}t^6 \right]_0^1$ $W = \left[ \frac{3}{2}t^8 - \frac{1}{3}t^6 \right]_0^1$ $W = \left( \frac{3}{2} - \frac{1}{3} \right) - (0)$ $W = \frac{9}{6} - \frac{2}{6} = \frac{7}{6}$

Final Answer:

The work done by the force $\vec{F}$ along the path is $\frac{7}{6}$ units.

Would you like further details or have any questions?

Here are some related questions you might find useful:

1. How does the work done change if the path is altered?
2. What if the force vector had an additional component, say $\hat{k}$?
3. How is work computed if the force field is conservative?
4. Can you calculate the work done along a different parametric path?
5. What are some real-life examples where such work integrals are applicable?

Tip: Always check the limits of integration carefully, especially when dealing with parametric equations.