Problem 35 - Calculus Early Transcendentals

Question

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) along the curve \(C\) $$ \begin{array}{l}{\mathbf{F}(x, y)=\left(x^{2}+y^{2}\right)^{-3 / 2}(x \mathbf{i}+y \mathbf{j})} \\ {C: \mathbf{r}(t)=e^{t} \sin t \mathbf{i}+e^{t} \cos t \mathbf{j} \quad(0 \leq t \leq 1)}\end{array} $$

Step-by-Step Solution

Answer

The line integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \) evaluates to 0.

Step 1: Rewrite the Vector Field

The given vector field is \( \mathbf{F}(x, y) = \left(x^{2}+y^{2}\right)^{-3 / 2}(x \mathbf{i}+y \mathbf{j}) \). Express it using elements \( x \) and \( y \) derived from the parameterization of the curve \( C \): \( x = e^t \sin t \) and \( y = e^t \cos t \).

Step 2: Find the Differential Length Element

Differentiate the parameterization \( \mathbf{r}(t) = e^{t} \sin t \mathbf{i} + e^{t} \cos t \mathbf{j} \) to find \( d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt \). This gives:\[ \frac{d\mathbf{r}}{dt} = (e^t \cos t + e^t \sin t) \mathbf{i} + (e^t \cos t - e^t \sin t) \mathbf{j} \].

Step 3: Dot Product of \( \mathbf{F}(\mathbf{r}(t)) \) and \( d\mathbf{r} \)

Substitute the parametric expressions into \( \mathbf{F} \) to express it as a function of \( t \). Then compute the dot product of \( \mathbf{F}(e^t \sin t, e^t \cos t) \) and \( \frac{d\mathbf{r}}{dt} \). Simplify the expression: \[ \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} = (e^t \sin t \cos t + e^{2t}) \cdot (e^t\cos t + e^t \sin t) \].

Step 4: Set Up the Integral

Integrate the dot product over the interval \( t = 0 \) to \( t = 1 \). The integral becomes: \[ \int_0^1 \frac{e^{2t}}{(e^{2t})^{3/2}} \cdot \left( (e^t \cos t + e^t \sin t) \right) \, dt \].

Step 5: Compute the Integral

Simplify the expression inside the integral and find the antiderivative. Substitute the limits to find the value of the integral. Since provided field is of the form of a potential field, if accurate taken, integral returns 0:\( \int_0^1 \mathbf{F}(t) \cdot \frac{d\mathbf{r}}{dt} \ dt = 0 \).

Step 6: Final Evaluation

Since the integral represents the work done by a central potential field along a closed curve, and the curve is within the context provided of being centered around, the resultant work is zero.

Key Concepts

Vector Fields

Vector fields are mathematical constructs that assign a vector to every point within a certain space. You can think of them like a gentle breeze or gravity pulling down—there's a direction and intensity at every spot. In our exercise, the vector field \[ \mathbf{F}(x, y) = \left(x^{2}+y^{2}\right)^{-3 / 2}(x \mathbf{i}+y \mathbf{j}) \] assigns vectors based on the coordinates \((x, y)\).

• Direction: The direction is given by the vector components \(x \mathbf{i} + y \mathbf{j}\).
• Magnitude: The magnitude depends on the term \((x^{2}+y^{2})^{-3/2}\) which diminishes with distance from the origin.

This field represents an inverse-square law often seen in physics, like gravitational or electric fields. Understanding how these vector fields operate is essential, especially as they show how force or velocity might act in a spatial region.

Parametric Curves

Parametric curves are really useful in defining the paths within vector fields. These curves describe a path in terms of a parameter, usually denoted as \(t\), representing time or another variable. In this exercise, the path is given by:\[ \mathbf{r}(t) = e^{t} \sin t \mathbf{i} + e^{t} \cos t \mathbf{j} \]

• Parameterization: This form expresses both x and y coordinates as functions of \(t\).
• Curve Description: Different values of \(t\) trace out points along the curve \(C\).

This parameterization allows us to translate problems from two-dimensional space into a single parameter, making complex integrations or calculations simpler and more manageable. We find derivatives to compute changes over time, just as in step-by-step solutions.

Potential Fields

Potential fields are a special kind of vector field with the property that they can be expressed as the gradient of some scalar potential function. This means that the work done by the field around any closed path is zero, which inherently simplifies many problems.
For example, gravitational and electrostatic forces are known to be potential fields, as they are conservative fields:

• Conservative Nature: The line integral of a conservative field depends only on the endpoints of the path, not the path itself.
• Zero Work Example: In a closed loop, like in the given exercise, if the field is conservative, the total work done is zero.

Recognizing potential fields is key in physics and engineering as it assures that energy is conserved, making calculations and predictions reliable. The exercise's conclusion hinges on this very principle, assuring us that despite the path's complexity, the work computed turns out to be zero.