Question

Let \(F(x, y)=x^{2}-y^{2}\). Evaluate a) \(\int_{(0,0)}^{(2,8)} \nabla F \cdot d \mathbf{r}\) on the curve \(y=x^{3}\); b) \(\oint \frac{\partial F}{\partial n} d s\) on the circle \(x^{2}+y^{2}=1\), if \(\mathbf{n}\) is the outer normal and \(\frac{\partial F}{\partial n}=\nabla F \cdot \mathbf{n}\) is the directional derivative of \(F\) in the direction of \(\mathbf{n}\) (Section 2.14).

Short answer

Question: Evaluate the integrals (a) $\int_{(0,0)}^{(2,8)} \nabla F \cdot d\mathbf{r}$ where F(x,y) = x^2 - y^2 and the curve is defined by y = x^3, and (b) the integral of the directional derivative of F in the direction of the unit outer normal along the curve x^2 + y^2 = 1. Answer: (a) The integral value is -60. (b) The integral value is 0.

Step-by-step solution

(Step 1: Compute the gradient of F)

To compute the gradient of F, we need to find the partial derivatives of the function with respect to x and y. So, we have: \(\nabla F = (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}) = (2x, -2y)\) a)

(Step 2A: Determine d\(\mathbf{r}\) on the curve y=x^3)

We are given that the curve is defined by the relation \(y = x^3\). To find d\(\mathbf{r}\), let's differentiate \(y = x^3\) with respect to \(t\). So, we have \(\frac{dy}{dx} = 3x^2\). Then, d\(\mathbf{r} = (dx, dy) = \left( dx, \frac{dy}{dx} dx\right) = (dx, 3x^2 dx)\).

(Step 3A: Evaluate the line integral of ∇F ⋅ d\(\mathbf{r}\))

Now that we have both ∇F and d\(\mathbf{r}\), we can compute the line integral: \(\int_{(0,0)}^{(2,8)} \nabla F \cdot d \mathbf{r} = \int_{(0)}^{(2)} (2x, -2y) \cdot (dx, 3x^2 dx)\) Since \(y = x^3\), the line integral becomes: \(\int_{0}^{2} (2x, -2x^3) \cdot (1, 3x^2) dx = \int_{0}^{2} (2x - 6x^5) dx\) Finally, evaluating the integral: \(\int_{0}^{2} (2x - 6x^5) dx = \left[ x^2 - x^6 \right]_{0}^{2} = (4 - 64) - (0 - 0) = -60\) b)

(Step 2B: Determine n and \(\frac{\partial F}{\partial n}\) along the circle)

The outer normal vector \(\mathbf{n}\) on the circle \(x^2 + y^2 = 1\) can be calculated by computing the gradient of the constraint function: \(\nabla (x^2 + y^2) = (2x, 2y)\) Since we want the outer normal vector, we don't need to normalize this, so we have \(\mathbf{n} = (2x, 2y)\). Now, let's compute the directional derivative of F in the direction of \(\mathbf{n}\): \(\frac{\partial F}{\partial n} = \nabla F \cdot \mathbf{n} = (2x, -2y) \cdot (2x, 2y) = 4x^2 - 4y^2\)

(Step 3B: Evaluate the integral of the directional derivative along the circle)

To evaluate the integral, we will parametrize the circle using polar coordinates: \(x = \cos\theta\) \(y = \sin\theta\) and also we need to find the differential arc length \(ds\): \(ds = \sqrt{(\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2} d\theta\) Differentiating x and y with respect to \(\theta\), we get: \(\frac{dx}{d\theta} = -\sin\theta\) \(\frac{dy}{d\theta} = \cos\theta\) Now, \(ds = \sqrt{(-\sin\theta)^2 + (\cos\theta)^2} d\theta = d\theta\) Therefore, the integral becomes: \(\oint \frac{\partial F}{\partial n} ds = \int_{0}^{2\pi} (4\cos^2\theta - 4\sin^2\theta) d\theta\) Evaluating the integral: \(\int_{0}^{2\pi} (4\cos^2\theta - 4\sin^2\theta) d\theta = 0\) So the answer for part (b) is 0.

Key concepts

Gradient of a Function

In calculus, the gradient of a function is a vector that represents both the direction and the rate of the fastest increase of the function. For a function with two variables, such as in our exercise where the function is given by \( F(x, y) = x^{2} - y^{2} \), the gradient can be calculated by taking the partial derivatives of the function with respect to each variable. Indeed, the solution begins by finding the gradient of \( F \), which is \( abla F = (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}) = (2x, -2y) \).

The geometric interpretation of the gradient is that at any given point, it points in the direction where the function increases most rapidly. The magnitude of the gradient vector represents the rate of increase in that direction. Understanding the gradient is crucial when we evaluate line integrals, because it helps us find the component of a vector field that is tangent to a curve, which is essential for integration along a path in the field.

Directional Derivative

The directional derivative represents the rate at which a function changes at a point in a given direction. In our exercise, we specifically look at the directional derivative of \( F \) in the direction of the outer normal vector \( \mathbf{n} \) on the circle \( x^{2} + y^{2} = 1 \).

The outer normal is crucial because it dictates the direction in which we measure the rate of change of \( F \) along the circle's edge. The step-by-step solution provided calculates this directional derivative as \( \frac{\partial F}{\partial n} = abla F \cdot \mathbf{n} = 4x^{2} - 4y^{2} \) by taking the dot product of the gradient of \( F \) with the normal vector \( \mathbf{n} \).

The concept of the directional derivative is widely used in various fields such as physics, engineering, and economics to describe how a quantity changes not just overall, but in specific directions. It's a fundamental concept in the study of vector fields and fluid flow.

Parametric Equations

Parametric equations define a set of functions which assign a set of quantities as a function of one or more independent variables known as parameters. When dealing with curves, such as the one in our exercise defined by \( y = x^3 \), parametric equations allow us to express the components of vectors or points along the curve in terms of a single parameter.

In the solution provided, parametric equations aren't explicitly mentioned; however, the process involves using the parameter \( x \) to express both \( x \) and \( y \) when calculating \( d\mathbf{r} = (dx, dy) \). Similarly, for evaluating the integral around the circle, polar coordinates, which are a type of parametric description where the parameter is the angle \( \theta \), are used to describe the positions along the circle in terms of \( \cos\theta \) and \( \sin\theta \).

This concept is pivotal in simplifying the integration process of complex shapes by breaking them down into simpler, more manageable parts, much like providing coordinates to navigate a map. Parametric equations are particularly beneficial in physics and engineering for modeling trajectories and motions.