give answer in 2 step with explanation at the end of each step and final answer at the end:Consider the curve given by the parametrization:
r(t) = (vV2tet, et) fort (0,00).
Now, Suppose that you are

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:Consider the curve given by the parametrization:
r(t) = (vV2tet, et) fort (0,00).
Now, Suppose that you are at the point (0, 1, 1) and subjected to a force given by:
F(z,y,2) = | —2—, — ~~, ——=——| (it's very similar to gravity)
(EHP) (Re) Ee)
How much work will you do traveling along the curve from (0,1,1) to (v2, ee!) 2
Make sure to provide a clear, complete, and detailed solution. If you use any theorems, state which ones you are
using clearly and also show why those theorems are appropriate to this problem.

VIPSolutions · Accepted Answer

Let's solve the problem in **2 steps** as requested, with explanations at each step and the final answer at the end.

---

### **Step 1: Set up the Work Integral**

The work done by a force **F** along a curve **C** is given by the line integral:

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

Here, the curve is parametrized by $ \mathbf{r}(t) = (\sqrt{2}t, e^t, e^{-t}) $, for $ t \in [0, 1] $ (since at $ t=0 $ we get $ (0,1,1) $ and at $ t=1 $ we get $ (\sqrt{2}, e, e^{-1}) $).

The force field is:
$$
\mathbf{F}(x, y, z) = \left( \frac{-x}{(x^2+y^2+z^2)^{3/2}}, \frac{-y}{(x^2+y^2+z^2)^{3/2}}, \frac{-z}{(x^2+y^2+z^2)^{3/2}} \right)
$$

We need to compute:
$$
W = \int_{t=0}^{t=1} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt
$$

**Explanation:**  
This step sets up the line integral for work done by a vector field along a curve. This is appropriate because the definition of work in a vector field is the dot product of the force and the displacement along the path.

---

### **Step 2: Substitute, Simplify, and Evaluate**

First, compute $ \mathbf{r}'(t) $:
$$
\mathbf{r}(t) = (\sqrt{2}t, e^t, e^{-t})
$$
$$
\mathbf{r}'(t) = (\sqrt{2}, e^t, -e^{-t})
$$

Now, substitute $ \mathbf{r}(t) $ into $ \mathbf{F} $:

- $ x = \sqrt{2}t $
- $ y = e^t $
- $ z = e^{-t} $
- $ x^2 + y^2 + z^2 = 2t^2 + e^{2t} + e^{-2t} $

So,
$$
\mathbf{F}(\mathbf{r}(t)) = \left( \frac{-\sqrt{2}t}{(2t^2+e^{2t}+e^{-2t})^{3/2}}, \frac{-e^t}{(2t^2+e^{2t}+e^{-2t})^{3/2}}, \frac{-e^{-t}}{(2t^2+e^{2t}+e^{-2t})^{3/2}} \right)
$$

The dot product:
$$
\mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = 
\frac{-\sqrt{2}t}{A^{3/2}} \cdot \sqrt{2}
+ \frac{-e^t}{A^{3/2}} \cdot e^t
+ \frac{-e^{-t}}{A^{3/2}} \cdot (-e^{-t})
$$
where $ A = 2t^2 + e^{2t} + e^{-2t} $.

$$
= \frac{-2t}{A^{3/2}} + \frac{-(e^{2t})}{A^{3/2}} + \frac{(e^{-2t})}{A^{3/2}}
$$
$$
= \frac{-2t - e^{2t} + e^{-2t}}{A^{3/2}}
$$

So the work integral becomes:
$$
W = \int_{0}^{1} \frac{-2t - e^{2t} + e^{-2t}}{(2t^2 + e^{2t} + e^{-2t})^{3/2}} dt
$$

---

### **Final Answer:**

$$
\boxed{
W = \int_{0}^{1} \frac{-2t - e^{2t} + e^{-2t}}{(2t^2 + e^{2t} + e^{-2t})^{3/2}} \, dt
}
$$

This is the exact value of the work done. If needed, you can evaluate this integral numerically.

---

**Summary of Theorems Used:**
- **Line Integral of a Vector Field:** Used because work done by a force along a path is computed by the line integral $ \int_C \mathbf{F} \cdot d\mathbf{r} $.
- **Parameterization:** Used to convert the line integral into an integral over $ t $ by substituting the path and its derivative.

Let me know if you'd like the numerical value as well!

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