Question

For the potential function \(\varphi\) and points \(A, B, C,\) and D on the level curve \(\varphi(x, y)=0,\) complete the following steps. a. Find the gradient field \(\mathbf{F}=\nabla \varphi.\) b. Evaluate \(\mathbf{F}\) at the points \(A, B, C,\) and \(D.\) c. Plot the level curve \(\varphi(x, y)=0\) and the vectors \(\mathbf{F}\) at the points \(A\) \(B, C,\) and \(D.\) $$\begin{aligned} &\varphi(x, y)=\frac{32-x^{4}-y^{4}}{32} ; A(2,2), B(-2,2), C(-2,-2) \text { and } D(2,-2) \end{aligned}$$

Short answer

In summary, we computed the gradient field of the function \(\varphi(x, y)\) and evaluated it at points A, B, C, and D. The gradient field components were found to be: $$ \mathbf{F}=\begin{bmatrix}-\frac{x^3}{8}\\-\frac{y^3}{8}\end{bmatrix} $$ The gradient field evaluated at the given points is: $$ \mathbf{F}(A) = \begin{bmatrix}-1\\-1\end{bmatrix}, \mathbf{F}(B) = \begin{bmatrix}1\\-1\end{bmatrix}, \mathbf{F}(C) = \begin{bmatrix}1\\1\end{bmatrix}, \mathbf{F}(D) = \begin{bmatrix}-1\\1\end{bmatrix} $$ To plot the level curve and gradient field, first graph the level curve \(x^4 + y^4 = 32\), which should be an elongated diamond shape. Then, sketch the gradient vectors at points A, B, C, and D. The vectors should point in directions mostly orthogonal to the level curve at those points.

Step-by-step solution

Compute the gradient field \(\mathbf{F}=\nabla \varphi\)#

To compute the gradient field, we need to find the partial derivatives of \(\varphi(x, y)\) with respect to \(x\) and \(y\). $$ \begin{aligned} \frac{\partial \varphi}{\partial x} &= -\frac{4x^3}{32} = -\frac{x^3}{8} \\ \frac{\partial \varphi}{\partial y} &= -\frac{4y^3}{32} = -\frac{y^3}{8} \end{aligned} $$ So the gradient field (also known as the gradient vector) \(\mathbf{F} = \nabla \varphi\) is: $$ \mathbf{F}=\begin{bmatrix}\frac{\partial \varphi}{\partial x}\\\frac{\partial \varphi}{\partial y}\end{bmatrix} = \begin{bmatrix}-\frac{x^3}{8}\\-\frac{y^3}{8}\end{bmatrix} $$ Now we can evaluate the gradient field at the given points A, B, C, and D.

Evaluate the gradient field at the given points#

To evaluate \(\mathbf{F}\) at the points A, B, C, and D, substitute their coordinates into the gradient field components: $$ \mathbf{F}(A) = \begin{bmatrix}-\frac{2^3}{8}\\-\frac{2^3}{8}\end{bmatrix} = \begin{bmatrix}-1\\-1\end{bmatrix} $$ $$ \mathbf{F}(B) = \begin{bmatrix}-\frac{(-2)^3}{8}\\-\frac{2^3}{8}\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix} $$ $$ \mathbf{F}(C) = \begin{bmatrix}-\frac{(-2)^3}{8}\\-\frac{(-2)^3}{8}\end{bmatrix} = \begin{bmatrix}1\\1\end{bmatrix} $$ $$ \mathbf{F}(D) = \begin{bmatrix}-\frac{2^3}{8}\\-\frac{(-2)^3}{8}\end{bmatrix} = \begin{bmatrix}-1\\1\end{bmatrix} $$

Plot the level curve and the gradient field#

Unfortunately, as an AI, I cannot physically plot graphs. But I can give you directions on how to graph the level curve \(\varphi(x, y) = 0\) and indicate the gradient field at points A, B, C, and D. 1. Set \(\varphi(x, y)=\frac{32-x^{4}-y^{4}}{32}\) equal to zero and solve for the level curve: $$ \begin{aligned} \frac{32-x^{4}-y^{4}}{32}&=0 \\ 32&=x^4+y^4 \end{aligned} $$ 2. Plot the level curve \(x^4 + y^4 = 32\) on the Cartesian plane. It should be an elongated diamond shape with its vertices being (2,0), (-2,0), (0,2), and (0,-2). 3. Now add the gradient vectors \(\mathbf{F}(A)\), \(\mathbf{F}(B)\), \(\mathbf{F}(C)\), and \(\mathbf{F}(D)\) on the level curve at their respective points (A, B, C, and D). These vectors should point in different directions, mostly orthogonal to the level curve at those points.

Key concepts

Gradient Vector

The gradient vector is a central concept in calculus, especially when dealing with functions of several variables. It compiles all partial derivatives of a function into a single vector. In essence, the gradient points in the direction of the greatest rate of increase of the function and its magnitude represents the rate of change in that direction.

Mathematically, for a function \(\varphi(x, y)\) the gradient, denoted by \(abla \varphi\), is expressed as \[abla \varphi = \begin{bmatrix} \frac{\partial \varphi}{\partial x} \ \frac{\partial \varphi}{\partial y}\end{bmatrix} = \begin{bmatrix} -\frac{x^3}{8} \ -\frac{y^3}{8}\end{bmatrix}\] where \(\frac{\partial \varphi}{\partial x}\) and \(\frac{\partial \varphi}{\partial y}\) are the partial derivatives with respect to \(x\) and \(y\), respectively.
To understand the gradient vector's significance in the context of the potential function \(\varphi(x, y) = \frac{32 - x^4 - y^4}{32}\), we saw that the gradient field, when evaluated at specific points, provides us with vectors that indicate the slope and direction of steepest ascent from those points.

Partial Derivatives

Partial derivatives are used to measure how a function changes as only one of its variables changes while all other variables are held constant. They are essential in computing the gradient vector and in multivariable analysis in general. The notation \(\frac{\partial}{\partial x} \varphi(x, y)\) exposes how \(\varphi\) changes in the \(x\)-direction at a particular point and similarly for \(y\).

In our exercise, the partial derivatives were \[\frac{\partial \varphi}{\partial x} = -\frac{4x^3}{32}\] and \[\frac{\partial \varphi}{\partial y} = -\frac{4y^3}{32}\].

Understanding these derivatives allowed us to find the gradient vector \(\mathbf{F}\), which in turn provided valuable insights into the behavior of \(\varphi\) at given points. For example, evaluating \(\mathbf{F}\) at point \(A(2,2)\) yielded a gradient vector of \(\mathbf{F}(A) = \begin{bmatrix}-1\-1\end{bmatrix}\), which shows us the direction and steepness of the slope at \(A\) in the negative \(x\) and \(y\) direction.

Level Curve

Level curves, or contour lines, represent a set of points at which a function has the same value—think of them as equal-altitude lines on a geographic map. In a two-dimensional space, these curves can help visualize how a function behaves without graphing the entire 3D surface.

To plot the level curve of the potential function \(\varphi(x, y)\), we set \(\varphi(x, y)\) equal to a constant. In the exercise, the level curve was determined by setting \(\varphi(x, y)=0\). This yielded the equation \(x^4 + y^4 = 32\), a unique curve in Cartesian coordinates.

Plotting this level curve provides insight into the topography of \(\varphi\) and helps to visualize how the function changes in space. On this curve, we can illustrate the gradient vectors at specific points, offering a tangible understanding of how the function rises and falls in various regions of the \(xy\)-plane.

Potential Function

A potential function represents a scalar field from which a gradient field can be derived. It is particularly important in physics as it often signifies potential energy, where the gradient vector field corresponds to the force experienced by a particle in the field.

In calculus, the potential function \(\varphi(x, y)\) has a gradient \(\mathbf{F}=abla\varphi\) associated with it. For the example in our exercise, the potential function was given by \(\varphi(x, y) = \frac{32 - x^4 - y^4}{32}\). By finding the gradient vector field from this potential function, we could analyze vector behavior in multiple points in the field.

The significance of \(\varphi\) lies in the fact that its gradient vector field \(\mathbf{F}\) gives direction and magnitude of the maximum increase of the potential. By evaluating \(\mathbf{F}\) at specific points, we were able to discern directional tendencies and the inherent properties of the scalar field, further enhancing the comprehension of the potential function's nature.