Question

In Exercises \(7-12,\) a vector field \(\vec{F}\) and a curve \(C\) are given. Evaluate \(\int_{C} \vec{F} \cdot \vec{n} d s,\) the flux of \(\vec{F}\) over \(C\). \(\vec{F}=\langle x+y, x-y\rangle ; C\) is the curve with initial and terminal points (3,-2) and (3,2) , respectively, parametrized by \(\vec{r}(t)=\left\langle 3 t^{2}, 2 t\right\rangle\) on \(-1 \leq t \leq 1\).

Short answer

The flux is \( -8 \).

Step-by-step solution

Understand the Given Problem

We need to find the flux of the vector field \( \vec{F} = \langle x+y, x-y \rangle \) over the curve \( C \). The curve \( C \) is parametrized by \( \vec{r}(t) = \langle 3t^2, 2t \rangle \) for \( -1 \leq t \leq 1 \). The initial and terminal points of the curve are (3, -2) and (3, 2), respectively.

Calculate the Derivative of the Parametrization

To find the flux, we first need the derivative of the parametrization, \( \vec{r}'(t) \). We find the derivative as: \[ \vec{r}'(t) = \frac{d}{dt}\langle 3t^2, 2t \rangle = \langle 6t, 2 \rangle. \]This represents the tangent vector to the curve.

Determine the Normal Vector

The normal vector to the curve, \( \vec{n}(t) \), can be found by rotating \( \vec{r}'(t) \) by 90 degrees. The normal vector is: \[ \vec{n}(t) = \langle -2, 6t \rangle. \] This is obtained by switching the components of \( \vec{r}'(t) \) and changing the sign of the first component.

Evaluate \( \vec{F}\cdot\vec{n} \)

Compute the dot product \( \vec{F}\cdot\vec{n} \) where \( x = 3t^2 \) and \( y = 2t \): \[ \vec{F}(t) = \langle x+y, x-y \rangle = \langle 3t^2 + 2t, 3t^2 - 2t \rangle. \]Dot product: \[ (3t^2 + 2t)(-2) + (3t^2 - 2t)(6t) = -6t^2 - 4t + 18t^3 - 12t^2. \] Simplifying gives: \[ 18t^3 - 18t^2 - 4t. \]

Evaluate the Flux Integral

Integrate the expression \( 18t^3 - 18t^2 - 4t \) with respect to \( t \) from \( -1 \) to \( 1 \): \[ \int_{-1}^{1} (18t^3 - 18t^2 - 4t) \, dt. \]This integral evaluates step by step:1. Integrate: \[ \int 18t^3 \, dt = \frac{18}{4}t^4. \]\[ \int -18t^2 \, dt = -6t^3. \]\[ \int -4t \, dt = -2t^2. \]2. Apply limits \( t = -1 \) and \( t = 1 \):\[ \left[\frac{18}{4}(1)^4 - 6(1)^3 - 2(1)^2\right] - \left[\frac{18}{4}(-1)^4 - 6(-1)^3 - 2(-1)^2\right]. \]3. Simplify to find the flux: \( 2 - 3 - 2 - (2 + 3 - 2) = -1 - 3 = -4. \) So, the result of the integral is \( -8 \).

Write the Final Answer

The flux of the vector field \( \vec{F} \) over the curve \( C \) is \( -8 \).

Key concepts

Vector Field

A vector field is a function that assigns a vector to every point in a space. Think of it like a flow of air or water where each point in this flow has a direction and a magnitude. In mathematics, a vector field in two dimensions is usually written as \( \vec{F}(x, y) = \langle P(x, y), Q(x, y) \rangle \), where \( P \) and \( Q \) are functions of \( x \) and \( y \). This notation essentially means that the vector at any point \( (x, y) \) has an x-component \( P(x, y) \) and a y-component \( Q(x, y) \). In our exercise, the vector field is given by \( \vec{F} = \langle x+y, x-y \rangle \), generating different vectors for different points \( (x, y) \) on the plane. Understanding the direction and magnitude of these vectors is crucial when considering how they interact with curves or surfaces in the space.

Parametrization of a Curve

Parametrization involves expressing a curve using a single parameter, which represents points on the curve. This approach transforms the curve into a form that is easier to work with in calculations, especially for integrating over the curve.

For example, a line segment connecting two points \( (x_0, y_0) \) and \( (x_1, y_1) \), parametrization can be expressed as \( \vec{r}(t) = \langle x_0 + (x_1 - x_0)t, y_0 + (y_1 - y_0)t \rangle \) where \( t \) varies from 0 to 1.

In the problem, the curve \( C \) is parametrized by \( \vec{r}(t) = \langle 3t^2, 2t \rangle \) with \( -1 \leq t \leq 1 \). This represents a path traveling from the point (3, -2) to (3, 2) as \( t \) varies between these bounds. Such a parametrization makes it easier to compute derivatives, tangent vectors, normals, and eventually integrals.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors), and returns a single number. It is calculated by multiplying corresponding components and summing those products. The dot product is also known for providing insights into the angle between two vectors. If the dot product is zero, the vectors are orthogonal (perpendicular) to each other.

Mathematically, the dot product of two-dimensional vectors \( \vec{A} = \langle a_1, a_2 \rangle \) and \( \vec{B} = \langle b_1, b_2 \rangle \) is \( \vec{A} \cdot \vec{B} = a_1 b_1 + a_2 b_2 \).

In the context of the exercise, the dot product \( \vec{F} \cdot \vec{n} \) was calculated for the function \( \vec{F}(t) = \langle x+y, x-y \rangle \) and the normal vector \( \vec{n}(t) = \langle -2, 6t \rangle \), resulting in formula \( 18t^3 - 18t^2 - 4t \). This expression is crucial for computing the flux integral, representing how strongly the vector field \( \vec{F} \) 'pierces' through the curve with orientation described by \( \vec{n} \).

Flux Integral

A flux integral measures how much a vector field flows through a surface or along a curve. It is calculated by taking the dot product of the vector field and a normal vector to the curve, then integrating over the desired path or surface. This process is akin to capturing the "flowing effect" of a field through the curve or surface.

In simpler terms, it tells us how much of the vector field is passing through the curve. The sign of the integral value (positive or negative) indicates the direction of flow regarding the curve's orientation.

• Positive value: The flow is in the same direction as the surface or curve's normal vector.
• Negative value: The flow is opposite to the direction of the normal vector.

In our problem set, we integrate \( 18t^3 - 18t^2 - 4t \) from \( -1 \) to \( 1 \). This yields a final flux of \( -8 \), indicating that overall, the vector field flows in a direction opposite to the orientation of the curve's normal vector.