Given the force →F = xyî-(y^2)ĵ. Find the work done by the path given by x = 2t^3and y=t^2 from t=0 to t=1

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.

Learn how to calculate the work done by the force F = xyî - y^2ĵ along the path x = 2t^3 and y = t^2 using line integrals and parametric equations. Detailed step-by-step solution included. Suitable for college-level students studying vector calculus and line integrals.

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