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Problem 209

# a) Let $$F^{\rightarrow}$$ be a vector field on an open set $$V$$ and $$C$$ a curve in V defined on the interval $$[a, b]$$. Prove $\int_{(C)-} F^{\rightarrow}=-\int_{C} F^{-}$$, where$$C^{-}$ is the reverse path of the curve $$C$$. b) Then evaluate $\int_{\mathrm{C}} \mathrm{F}^{-} \cdot \mathrm{d} \mathrm{C}^{-}$$where$$\mathrm{F}^{-}(\mathrm{x}, \mathrm{y})=\left(\mathrm{x}^{2}, \mathrm{xy}\right)$ along the line segment from the point $$(1,1)$$ to $$(0,0)$$ using the reverse path.

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The short version of the answer based on the provided step-by-step solution would be: We first proved the relationship between the line integrals of curve $$C$$ and its reverse path $$C^-$$ by parametrizing both paths and comparing the results. Then, we found the parametric representation of the reverse path of the line segment from point (1,1) to point (0,0) as $$\vec{r}^-(t) = (1-t,1-t)$$ and evaluated the line integral of the given vector field $$\vec{F}(x, y) = (x^2, xy)$$ along this path. The result of the integral is $$\int_{\mathrm{C}} \mathrm{F}^{-}\cdot \mathrm{d} \mathrm{C}^{-} = -\frac{2}{3}$$.
See the step by step solution

## Step 1: In order to prove the relation between integrals of the curve C and its reverse path, we need to understand what a reverse path is. Given a curve C defined by a parametric representation $$\vec{r}(t)$$ on the interval $$[a,b]$$, the reverse path, denoted by $$C^-$$, is the same curve traced in the opposite direction. Mathematically, this can be represented as $$\vec{r}^-(t) = \vec{r}(a+b-t)$$, for $$t \in [a, b]$$. #Step 2: Prove the relation between integrals#

To prove the relationship between the integrals of curve C and its reverse path $$C^-$$, we will first write down the line integral for both paths. For $$C$$, the line integral is: $\int_{C} \vec{F} \cdot d\vec{r} = \int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt}\,dt.$ For $$C^-$$, the line integral is: $\int_{(C^-)} \vec{F} \cdot d\vec{r}^- = \int_{a}^{b} \vec{F}(\vec{r}^-(t)) \cdot \frac{d\vec{r}^-}{dt}\,dt.$ Since $$\vec{r}^-(t) = \vec{r}(a+b-t)$$, we can substitute this into the second integral, and calculate the derivative of $$\vec{r}^-$$ with respect to t. Now let's substitute $$\vec{r}^-$$ with $$\vec{r}(a+b-t)$$: $\int_{(C^-)} \vec{F} \cdot d\vec{r}^- = \int_{a}^{b} \vec{F}(\vec{r}(a+b-t)) \cdot \frac{d\vec{r}(a+b-t)}{dt}\,dt.$ Then, using the chain rule, we get $$\frac{d\vec{r}(a+b-t)}{dt} = -\frac{d\vec{r}(t)}{dt}$$: $\int_{(C^-)} \vec{F} \cdot d\vec{r}^- = -\int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt}\,dt.$ This shows that $$\int_{(C)-} \vec{F} = -\int_{C} \vec{F}$$, which is what we wanted to prove. #Step 3: Parametrizing the reverse path for part b#

## Step 2: For part b, we have to compute the line integral of the vector field $$\vec{F}(x, y) = (x^2, xy)$$ along the reverse path of the line segment from point (1,1) to point (0,0). We can parametrize the original path from (0,0) to (1,1) as $$\vec{r}(t) = (t,t)$$ for $$0 \le t \le 1$$. To find the reverse path, we simply adjust the parameter so that the endpoints are reversed: $$\vec{r}^-(t) = (1-t,1-t)$$, where $$0 \le t \le 1$$. #Step 4: Computing the line integral for part b#

Using the vector field $$\vec{F}(x, y) = (x^2, xy)$$ and the reverse path $$\vec{r}^-(t) = (1-t,1-t)$$, we can calculate the line integral: First, find the derivative of the reverse path with respect to t: $$\frac{d\vec{r}^-}{dt} = (-1,-1)$$. Then, evaluate the vector field along the reverse path: $$\vec{F}(\vec{r}^-(t)) = ((1-t)^2, (1-t)^2)$$. Now we can write down the integral formula: $\int_{C^-} \vec{F} \cdot d\vec{r}^- = \int_{0}^{1} ((1-t)^2, (1-t)^2) \cdot (-1, -1)\,dt.$ Compute the dot product: $\int_{0}^{1} (-(1-t)^2 - (1-t)^2)\,dt.$ Combine the terms: $\int_{0}^{1} (-2(1-t)^2)\,dt.$ Finally, integrate: $-2 \int_{0}^{1} (1-2t+t^2)\,dt = -2\left[t - t^2 + \frac{t^3}{3}\right]_0^1 = -2\left(1 - 1 + \frac{1}{3}\right) = -\frac{2}{3}.$ Therefore, the result $\int_{\mathrm{C}} \mathrm{F}^{-}\cdot \mathrm{d} \mathrm{C}^{-} = -\frac{2}{3}$.

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