find the total work done in moving a parcticle in a force field given by F = 3x * y deg - 52 deg + 10xk along the x = 1 + t ^ 2 y = 24 ^ 2 z = f ^ 3 from 1/x = 1 + f = 2

To find the total work done in moving a particle in a given force field along a specified path, we need to use the line integral of the force field along the path.

Given:
\[
\mathbf{F} = 3x \mathbf{i} + y \mathbf{j} + 10x \mathbf{k}
\]
where \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) are the unit vectors along the \(x\)-, \(y\)-, and \(z\)-axes, respectively.

The path is defined by the parametric equations:
\[
x = 1 + t^2, \quad y = 24^2, \quad z = t^3
\]
where \(t\) is the parameter that varies.

The limits for \(t\) are derived from the condition \( \frac{1}{x} = 1 + t = 2 \). We need to solve this equation to find the limits of \(t\).

### Step 1: Determine the Limits of Integration
From the equation \( \frac{1}{x} = 2 \), solve for \(x\):

\[
x = \frac{1}{2}
\]
But from the parametric equation \( x = 1 + t^2 \), set this equal to \( \frac{1}{2} \):

\[
1 + t^2 = \frac{1}{2}
\]

This equation has no real solution since \( t^2 \geq 0 \), and \( 1 + t^2 \) is always greater than or equal to 1. This suggests there's an issue with the setup or conditions of the problem. Let's address this.

### Clarification and Redefinition
Let's assume instead that the path is simply defined for a certain range of \(t\). We would then define the work done as the integral over this range, and you'd specify this range separately.

Assuming \( t \) varies from \( t = 0 \) to \( t = 1 \) as an example, we compute the line integral.

### Step 2: Parametrize the Force Field
Substitute the parametric equations into the force field:

\[
\mathbf{F}(t) = 3(1 + t^2) \mathbf{i} + (24^2) \mathbf{j} + 10(1 + t^2) \mathbf{k}
\]

### Step 3: Calculate the Work Done
The work done \( W \) is given by:
\[
W = \int_{C} \mathbf{F} \cdot d\mathbf{r}
\]
where \( d\mathbf{r} \) is the differential displacement vector along the curve.

Substitute \( \mathbf{r}(t) = (1 + t^2) \mathbf{i} + 24^2 \mathbf{j} + t^3 \mathbf{k} \).

The derivative \( d\mathbf{r}/dt \) gives:
\[
\frac{d\mathbf{r}}{dt} = 2t \mathbf{i} + 0 \mathbf{j} + 3t^2 \mathbf{k}
\]

So, the dot product \( \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \) is:
\[
\mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = \left[ 3(1 + t^2) \right] \cdot 2t + \left[ 24^2 \right] \cdot 0 + \left[ 10(1 + t^2) \right] \cdot 3t^2
\]

Simplifying:
\[
W = \int_{0}^{1} \left[ 6t(1 + t^2) + 30t^2(1 + t^2) \right] dt
\]

Expand and combine like terms:
\[
W = \int_{0}^{1} \left[ 6t + 6t^3 + 30t^2 + 30t^4 \right] dt
\]

### Step 4: Evaluate the Integral
Evaluate each term separately:
\[
W = \left[ 3t^2 + \frac{6t^4}{4} + 10t^3 + \frac{30t^5}{5} \right]_{0}^{1}
\]

Substituting the limits:
\[
W = \left[ 3(1)^2 + \frac{6(1)^4}{4} + 10(1)^3 + \frac{30(1)^5}{5} \right] - 0
\]

This simplifies to:
\[
W = 3 + 1.5 + 10 + 6 = 20.5 \text{ units of work}
\]

So, the total work done is **20.5 units**.

### Additional Questions
1. What would happen if we changed the path in the integral to another form?
2. How does the force field affect the work done if the path changes direction?
3. How can we generalize this to three-dimensional curves?
4. What if the force field is not conservative? How would this affect the calculation?
5. How does the choice of parameter \( t \) affect the integral?

**Tip:** When dealing with line integrals, always verify the parameter limits and the curve's definition to avoid contradictions.

Learn how to calculate the total work done by a force field along a given parametric path. Detailed step-by-step solution using line integrals and vector field theory. Suitable for college-level students.

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