Problem 308 - Calculus Volume 3

Question

Solve the problem. If the electric potential at a point \((x, y)\) in the \(x y\) -plane is \(V(x, y)=e^{-2 x} \cos (2 y),\) then the electric intensity vector at \((x, y)\) is \(\mathbf{E}=-\nabla V(x, y)\) a. Find the electric intensity vector at \(\left(\frac{\pi}{4}, 0\right)\). b. Show that, at each point in the plane, the electric potential decreases most rapidly in the direction of the vector \(\mathbf{E}\)

Step-by-Step Solution

Answer

a. \(\left(2e^{-\frac{\pi}{2}}, 0\right)\). b. The electric potential decreases most rapidly in the direction of \(\mathbf{E}\).

Step 1: Identify the Function for Electric Intensity Vector

The electric intensity vector \(\mathbf{E}\) is given by \(\mathbf{E} = -abla V(x, y)\), where \(V(x, y) = e^{-2x} \cos(2y)\). This means we need to calculate the gradient of \(V(x, y)\) to find \(\mathbf{E}\).

Step 2: Calculate the Gradient of V(x,y)

To find \(-abla V(x, y)\), we need to compute the partial derivatives.- The partial derivative of \(V(x, y)\) with respect to \(x\) is: \[\frac{\partial V}{\partial x} = -2e^{-2x} \cos(2y)\].- The partial derivative of \(V(x, y)\) with respect to \(y\) is: \[\frac{\partial V}{\partial y} = -2e^{-2x} \sin(2y)\].Thus, \(-abla V(x, y) = \left(2e^{-2x} \cos(2y), 2e^{-2x} \sin(2y)\right)\).

Step 3: Compute the Electric Intensity Vector at Specific Point

Substitute \(\left(\frac{\pi}{4}, 0\right)\) into the electric intensity vector:\[\mathbf{E}\left(\frac{\pi}{4}, 0\right) = \left(2e^{-\frac{\pi}{2}} \cos(0), 2e^{-\frac{\pi}{2}} \sin(0)\right) = \left(2e^{-\frac{\pi}{2}} \cdot 1, 2e^{-\frac{\pi}{2}} \cdot 0\right) = \left(2e^{-\frac{\pi}{2}}, 0\right)\].

Step 4: Explain the Direction of Maximum Decrease

The gradient vector \(abla V(x, y)\) indicates the direction of the steepest increase of the function \(V(x, y)\). Consequently, \(-abla V(x, y)\) indicates the direction of the steepest decrease. Thus, at any given point in the plane, moving in the direction of \(-abla V(x, y)\) guarantees the fastest reduction in the electric potential \(V(x, y)\). Because the electric intensity vector \(\mathbf{E} = -abla V(x, y)\), the electric potential decreases most rapidly in the direction of \(\mathbf{E}\).

Key Concepts

Electric Potential

Electric potential is an important concept in electromagnetism. It allows us to understand how charges interact in a field.
It is essentially the potential energy per unit charge at a specific point in space.
Given the equation for the electric potential, such as \[ V(x, y) = e^{-2x} \cos(2y) \] this represents how potential changes with respect to coordinates \((x, y)\).
In simpler terms, electric potential tells us how much work is needed to move a charge to a specific point from infinity.

• It is usually measured in volts.
• Higher electric potential means more potential energy is available.
• It is scalar and does not have direction, unlike vectors.

Understanding electric potential helps us determine the behavior of charges and how forces act on them in a given field.

Gradient of a Function

The gradient is a vector that describes the slope or direction and rate of change of a scalar field, like our electric potential.
For any multivariable function, the gradient is composed of its partial derivatives.
In our context, finding the gradient of \( V(x, y) \) tells us how electric potential changes at different points.
The gradient can be expressed as:\[abla V(x, y) = \left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}\right)\]This is crucial when describing how rapidly electric potential can increase in different directions. A positive slope indicates how potential grows, while a negative one shows a decrease.
Wherever this gradient vector points, that's the direction of the steepest increase for the potential, but importantly the electric intensity is often the opposite of this vector.

• The gradient helps in recognizing the steepest slope or increase direction.
• It acts like a compass showing where the potential increases most rapidly.

Partial Derivatives

Partial derivatives are a fundamental concept used to analyze how functions change with multiple variables.
In our function \( V(x, y) = e^{-2x} \cos(2y) \), we took partial derivatives to find out how the potential independently varies with respect to \(x\) and \(y\).
For this exercise, obtaining the partial derivatives furnished us with:- \( \frac{\partial V}{\partial x} = -2e^{-2x} \cos(2y) \)- \( \frac{\partial V}{\partial y} = -2e^{-2x} \sin(2y) \)These derivatives form the components of the gradient vector.
Essentially, partial derivatives without changing other variables reveal localized trends, shedding light on their individual impact on the overall function.

• Help decompose changes in multivariable functions.
• Critical for understanding the gradient and directionality in space.

Direction of Steepest Descent

The direction of steepest descent is key for optimizing functions and understanding physical phenomena like electric intensity.
This concept highlights where a function decreases at the fastest rate.
In this problem, we look at \(-abla V(x, y)\), which represents the steepest descent direction of potential.
Since our electric intensity vector \( \mathbf{E} = -abla V(x, y) \), it readily informs us where the electric potential decreases most rapidly.
Moving in this direction ensures that the electric potential value reduces as quickly as possible, aligning force directions accordingly.

• Provides insights into stabilization and equilibrium processes.
• Useful for algorithms that optimize or minimize functions like gradients.
• Descent paths are crucial in fields like physics, engineering, and machine learning.