Problem 308 - Calculus 3

Question

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. \(\mathbf{E} = [2e^{-\frac{\pi}{2}}, 0]\) b. \(\mathbf{E}\) is the steepest descent direction of potential.

Step 1: Find the Gradient of the Potential

To find the electric intensity vector, we first need to calculate the gradient of the potential function. The gradient of a function \(V(x, y)\) is given by \(abla V(x, y) = \left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}\right)\). Here, \(V(x, y) = e^{-2x} \cos(2y)\). Compute the partial derivatives: - \(\frac{\partial V}{\partial x} = \frac{d}{dx}[e^{-2x}] \cos(2y) = -2e^{-2x} \cos(2y)\)- \(\frac{\partial V}{\partial y} = e^{-2x} \cdot [-\sin(2y) \cdot 2] = -2e^{-2x} \sin(2y)\)So, \(abla V(x, y) = (-2e^{-2x}\cos(2y), -2e^{-2x}\sin(2y))\).

Step 2: Compute Electric Intensity Vector at Given Point

The electric intensity vector \(\mathbf{E}\) is given by \(\mathbf{E} = -abla V(x, y)\). Substitute \(x = \frac{\pi}{4}\) and \(y = 0\) into the gradient to find \(\mathbf{E}\) at the point:- \(\mathbf{E} = [2e^{-2\left(\frac{\pi}{4}\right)}\cos(2 \times 0), 2e^{-2\left(\frac{\pi}{4}\right)}\sin(2 \times 0)]\)- Simplify: \(\mathbf{E} = [2e^{-\frac{\pi}{2}}, 0]\) as \(\cos(0) = 1\) and \(\sin(0) = 0\).

Step 3: Demonstrate Maximum Decrease Direction

The electric intensity vector \(\mathbf{E}\) is pointing in the direction of steepest decrease of electric potential. This is because the gradient \(abla V\) indicates the direction of steepest ascent, and \(\mathbf{E} = -abla V\) thus points in the steepest descent direction. This illustrates that at every point in the plane, the potential decreases most rapidly in the direction of \(\mathbf{E}\).

Key Concepts

Electric Potential

The electric potential, often denoted as \(V(x, y)\), is a scalar quantity that represents the electric potential energy per unit charge at a point in space. It is a measure of the work needed to move a positive test charge from a reference point to a specific point inside an electric field without producing any acceleration.
Electric potential is measured in volts and is related to the electric field by the formula \(\mathbf{E} = -abla V(x, y)\), where \(\mathbf{E}\) is the electric field intensity vector, and \(-abla V(x, y)\) is the negative gradient of the potential.
In our exercise, the potential function given is \(V(x, y) = e^{-2x} \cos(2y)\). This function tells us how the potential energy varies with \(x\) and \(y\) in the plane. The task is to find how this potential changes, which we do by calculating its gradient.

Gradient

The gradient of a function is a vector that points in the direction of the greatest rate of increase of the function. It combines partial derivatives with respect to each variable to form a single vector.
For a function \(V(x, y)\), its gradient can be expressed as \(abla V(x, y) = \left(\frac{\partial V}{\partial x}, \frac{\partial V}{\partial y}\right)\). In our problem, the gradient provides the rate and direction of change for the electric potential \(V(x, y)\).
By finding the gradient \(abla V(x, y)\) of the potential function \(V(x, y) = e^{-2x} \cos(2y)\), we obtain a vector indicating where the potential increases most rapidly. The individual components, partial derivatives, show how the function changes in \(x\) and \(y\) directions.

Partial Derivatives

Partial derivatives are crucial in multivariable calculus, representing the rate at which a function changes as one of its variables changes while others are held constant. For the potential function \(V(x, y) = e^{-2x} \cos(2y)\), you find:

• \(\frac{\partial V}{\partial x} = -2e^{-2x} \cos(2y)\): This derivative shows how \(V\) changes as \(x\) varies, with \(y\) unchanged.
• \(\frac{\partial V}{\partial y} = -2e^{-2x} \sin(2y)\): This derivative illustrates how \(V\) changes with variations in \(y\), while \(x\) remains fixed.

The partial derivatives are essential to construct the gradient, as they are the components of this vector. By understanding how each variable influences the function, we can predict and analyze the function's behavior over the plane.

Direction of Steepest Descent

The direction of the steepest descent is crucial when dealing with optimization problems or finding the path of quickest decrease in potential energy. For an electric potential, this means finding the direction in which the potential drops the fastest.
The gradient \(abla V\) indicates the direction of the steepest ascent, i.e., where the potential increases fastest. As the electric field intensity is \(\mathbf{E} = -abla V\), it points in the opposite direction, providing the steepest descent path.
In our context, the electric intensity vector \(\mathbf{E} = [2e^{-\frac{\pi}{2}}, 0]\) at \((\frac{\pi}{4}, 0)\) demonstrates this principle. It highlights the fact that at any chosen point in space, moving opposite to the gradient's direction results in the most rapid decrease of potential. This concept explains why charges move towards lower potential energy areas, aligning naturally with electric field lines.