Problem 209 - Advanced Calculus

Question

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.

Step-by-Step Solution

Answer

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 \(\int_{\mathrm{C}} \mathrm{F}^{-}\cdot \mathrm{d} \mathrm{C}^{-} = -\frac{2}{3}\).

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 $\int_{\mathrm{C}} \mathrm{F}^{-}\cdot \mathrm{d} \mathrm{C}^{-} = -\frac{2}{3}$.

Key Concepts

Line Integrals

In vector calculus, line integrals are a key concept for assessing how a vector field influences a path.
It's a process that calculates the sum of values that a function takes over a curve. This concept extends the idea of integration from one-dimensional intervals to more complex paths or curves.
For a given vector field, the line integral computes the total effect that the field has over a specific curve. Here is how it works in essence:

• You have a curve, defined by a path, and a vector field through which this curve passes.
• The line integral evaluates this field along the path of the curve.
• It's like "adding up" the influence of the field at every point along the path.

When calculating a line integral of a vector field \(\vec{F}\) over a curve \(C\), you follow the specific directions of the path to evaluate the integral. It’s represented as \(\int_{C} \vec{F} \cdot d\vec{r}\). This allows capturing how much the vector field pushes or flows along the direction of the path.

Vector Fields

A vector field assigns a vector to every point in a space. Imagine it like a map with arrows at each point, indicating direction and magnitude like a wind map showing wind speed and direction at every location.
In mathematical terms, a vector field over a plane can be written as \(\vec{F}(x, y)\).
In the context of the exercise, an example of a vector field used is \(\vec{F}(x, y) = (x^2, xy)\).

• The vector field represents forces, velocities, or other vector quantities at every point in a defined space.
• Evaluating a vector field is how we interpret the behavior of these quantities over specific regions or along certain paths.
• In line integrals, you consider how this field applies force or influence as you traverse the path.

The application can be diverse, from calculating work done by a force field on a particle moving along a path to evaluating electromagnetic fields. Understanding the vector field is crucial for exploring its various integrals and uses.

Parametrization

Parametrization transforms a curve into a form that is easier to work with using parameters, usually represented by \(t\).
This technique involves turning a path or curve into a function of some parameter. This makes calculations of integrals more straightforward.
For a line from point \((x_0, y_0)\) to \((x_1, y_1)\), a common parametrization is \(\vec{r}(t) = (x_0 + t(x_1 - x_0), y_0 + t(y_1 - y_0))\).

• It simplifies evaluating line integrals, as curves are represented using one real parameter.
• By using parametrization, one changes the problem into manageable steps that follow the path more systematically.
• Parametrization aids in calculations by providing clear boundary points and the direction of integration.

In both parts of the exercise, parametrization is used, where the original path is \(\vec{r}(t) = (t,t)\) and the reverse path is expressed as \(\vec{r}^-(t) = (1-t, 1-t)\). These calculations transform abstract curves into mathematical formulas.

Reverse Path

The reverse path of a curve means tracing the same path but in the opposite direction. This concept is essential in vector calculus because it alters the orientation of the integral in question.
Mathematically, for a curve \(C\) defined from \(t = a\) to \(t = b\) by \(\vec{r}(t)\), its reverse path \(C^-\) is represented as \(\vec{r}^-(t) = \vec{r}(a+b-t)\).

• This change results in an opposite sign for the integrals involving the path.
• The change in the direction changes how the vector field interacts with the path, inverting the effects.
• This is because the derivative of the reverse path involves a negative sign, reflected in the integral.

Within the problem context, proving \(\int_{C^-} \vec{F} = -\int_{C} \vec{F}\) involved showcasing this sign change by deriving the reverse parametrization. The reverse path application in part b required computing the line integral using this derived negative path adjustment to reflect this change in direction and total calculation.