Problem 50 - Calculus Early Transcendentals

Question

If \(V(x, y)\) is the electric potential at a point \((x, y)\) in the xy-plane, then the level curves of \(V\) are called equipotential curves because at all points on such a curve the electric potential is the same. Sketch some equipotential curves if \(V(x, y)=c / \sqrt{r^{2}-x^{2}-y^{2}},\) where \(c\) is a positive constant.

Step-by-Step Solution

Answer

Equipotential curves are concentric circles centered at the origin.

Step 1: Understand the Function

The function given is the electric potential function: \(V(x, y) = \frac{c}{\sqrt{r^2 - x^2 - y^2}}\), where \(c\) is a constant. We need to determine where this function is constant, which is the essence of an equipotential curve: it represents points where \(V\) is the same.

Step 2: Set the Function Equal to a Constant

To find the equipotential curves, we set \(V(x, y) = k\), where \(k\) is some constant. Thus, \(\frac{c}{\sqrt{r^2 - x^2 - y^2}} = k\). Rearrange this equation to solve for the expression under the square root.

Step 3: Solve for Equipotential Equation

Start solving the equation \(\frac{c}{\sqrt{r^2 - x^2 - y^2}} = k\). Multiply both sides by \(\sqrt{r^2 - x^2 - y^2}\) to get \(c = k \sqrt{r^2 - x^2 - y^2}\). Divide both sides by \(k\), then square both sides to eliminate the square root: \((\frac{c}{k})^2 = r^2 - x^2 - y^2\). Rearrange to find \(x^2 + y^2 = r^2 - (\frac{c}{k})^2\).

Step 4: Identify the Shape and Domain

The equation \(x^2 + y^2 = r^2 - (\frac{c}{k})^2\) describes a circle centered at the origin with radius \(\sqrt{r^2 - (\frac{c}{k})^2}\). Note that \(r^2 > (\frac{c}{k})^2\) must be satisfied for the radius to be a real positive number.

Step 5: Sketch the Equipotential Curves

Draw circles centered at the origin in the xy-plane. For different values of \(k\) that satisfy \(r^2 > (\frac{c}{k})^2\), we get different circle sizes, representing various equipotential curves. Each circle corresponds to a distinct potential level.

Key Concepts

Electric Potential

Electric potential is a fundamental concept in physics that describes the potential energy per unit charge at a specific point in a field due to a source charge. Think of it as an invisible force field that dictates how a charged particle will behave when placed in the field. The unit of electric potential is the volt.In the equation given, the potential function is defined as \(V(x, y) = \frac{c}{\sqrt{r^2 - x^2 - y^2}}\). This represents the potential at a point \((x, y)\) in the plane. The constant \(c\) determines the strength of the potential, while \(r\) represents a parameter that defines a constraint or boundary in the context.Understanding electric potential helps to predict how charges interact with each other. It's crucial when dealing with static charges, as it indicates potential energy changes and helps determine forces.

Level Curves

Level curves, or contour lines, are curves along which a function of two variables has a constant value. In the context of electric potential, these curves are particularly known as equipotential curves. Every point on an equipotential curve has the same electric potential value.In our equation, by solving \(V(x, y) = k\), where \(k\) is a constant, we determine the equipotential curves. These curves impart a visualization of how the potential varies across the plane.Equipotential curves offer a practical way to represent fields in graphical form. They allow us to easily understand areas of equal potential in a circuit or field. Such curves are beneficial for simplifying complex calculations by providing a clear representation of potential energy distribution.

Circular Geometry

Circular geometry is the study of shapes and figures based on circles. The equation \(x^2 + y^2 = r^2 - \left(\frac{c}{k}\right)^2\) is derived while identifying the equipotential curves in the exercise. This equation describes a circle centered at the origin with a specific radius.The radius of these circles, \(\sqrt{r^2 - \left(\frac{c}{k}\right)^2}\), varies depending on the value of \(k\). These circles help us visualize the equipotential lines in the plane as concentric circles centered around the origin. This reveals how the electric potential dissipates or varies as one moves away from the center.Using this simple circular framework in mathematical problems enhances our ability to solve spatial distribution issues and understand the influence of geometry on function behavior in physics.

Mathematical Problem Solving

Mathematical problem solving involves applying math concepts to solve problems effectively. In this exercise, the process involved setting up the problem, rearranging the equation, and finding solutions that give insight into the geometric nature of the solution.To determine equipotential curves, one must:- Translate the function into an equation involving a constant (set \(V(x, y) = k\)).- Rearrange and manipulate the equation to isolate desired variables or expressions.- Understand the resulting geometry—here, recognizing it as a circle on the two-dimensional plane.By carefully navigating through these steps, students enhance their problem-solving skills, allowing them to tackle more complex physics and mathematics problems. The ability to transform equations into geometric understandings underpins much of mathematical physics and helps bridge conceptual links between theory and application.