To calculate the work done by the force $F(x, y, z) = 2x \, \mathbf{i} + 3y \, \mathbf{j} - z \, \mathbf{k}$ along the path $r(t) = t \, \mathbf{i} + t^2 \, \mathbf{j} + t^3 \, \mathbf{k}$ from $t = 0$ to $t = 1$, we use the line integral of the force field along the given path. The work done $W$ is given by:

$W = \int_C \mathbf{F} \cdot d\mathbf{r}$

where $C$ is the path described by $\mathbf{r}(t)$, and $d\mathbf{r}$ represents the differential displacement vector along this path.

Step 1: Parameterize the Force Field $\mathbf{F}$ along the Path

For $r(t) = t \, \mathbf{i} + t^2 \, \mathbf{j} + t^3 \, \mathbf{k}$, we have: $x = t, \quad y = t^2, \quad z = t^3$

Substitute these into $\mathbf{F}(x, y, z)$: $\mathbf{F}(t) = 2x \, \mathbf{i} + 3y \, \mathbf{j} - z \, \mathbf{k} = 2t \, \mathbf{i} + 3t^2 \, \mathbf{j} - t^3 \, \mathbf{k}$

Step 2: Find $d\mathbf{r}$

Differentiate $\mathbf{r}(t) = t \, \mathbf{i} + t^2 \, \mathbf{j} + t^3 \, \mathbf{k}$ with respect to $t$: $\frac{d\mathbf{r}}{dt} = \frac{d}{dt}(t \, \mathbf{i} + t^2 \, \mathbf{j} + t^3 \, \mathbf{k}) = \mathbf{i} + 2t \, \mathbf{j} + 3t^2 \, \mathbf{k}$

So, $d\mathbf{r} = (\mathbf{i} + 2t \, \mathbf{j} + 3t^2 \, \mathbf{k}) \, dt$.

Step 3: Compute $\mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt}$

Now we calculate the dot product $\mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt}$: $\mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = (2t \, \mathbf{i} + 3t^2 \, \mathbf{j} - t^3 \, \mathbf{k}) \cdot (\mathbf{i} + 2t \, \mathbf{j} + 3t^2 \, \mathbf{k})$

Calculate each component of the dot product:

• For $\mathbf{i}$-components: $2t \cdot 1 = 2t$
• For $\mathbf{j}$-components: $3t^2 \cdot 2t = 6t^3$
• For $\mathbf{k}$-components: $-t^3 \cdot 3t^2 = -3t^5$

Thus, $\mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = 2t + 6t^3 - 3t^5$

Step 4: Integrate from $t = 0$ to $t = 1$

Now, integrate $\mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt}$ with respect to $t$ from 0 to 1: $W = \int_0^1 (2t + 6t^3 - 3t^5) \, dt$

Evaluate each term separately:

1. $\int_0^1 2t \, dt = \left[ t^2 \right]_0^1 = 1$
2. $\int_0^1 6t^3 \, dt = \left[ \frac{6t^4}{4} \right]_0^1 = \frac{3}{2}$
3. $\int_0^1 -3t^5 \, dt = \left[ -\frac{3t^6}{6} \right]_0^1 = -\frac{1}{2}$

Adding these results: $W = 1 + \frac{3}{2} - \frac{1}{2} = 2$

Final Answer

The work done by the force along the path is: $W = 2$

Would you like a more detailed explanation on any part of this solution?

---

Here are 5 related questions to deepen your understanding:

1. How does the line integral change if the force field has a different functional form?
2. What is the physical interpretation of work in this context?
3. How would the solution differ if the path $r(t)$ had a different parameterization?
4. Can the work done be negative, and under what conditions?
5. How would you interpret the result if the path were closed?

Tip: Always check that the parameterized path and force field components match before computing line integrals to ensure accuracy.