Problem 23 - The Calculus with Analytic Geometry

Question

The electric potential is \(V\) volts at any point \((x, y)\) in the \(x y\) plane and \(V=e^{-2 x} \cos 2 y .\) Distance is measured in feet. (a) Find the rate of change of the potential at the point \(\left(0, \frac{1}{4} \pi\right)\) in the direction of the unit vector \(\cos \frac{1}{6} \pi \mathrm{i}+\sin \frac{1}{6} \pi \mathrm{j}\). (b) Find the direction and magnitude of the greatest rate of change of \(V\) at \(\left(0, \frac{1}{4} \pi\right) .\)

Step-by-Step Solution

Answer

(a) -1.(b) Direction: (0, -2). Magnitude: 2.

Step 1: Write the given function

The electric potential function is given by:\[ V = e^{-2x} \, \cos 2y \]

Step 2: Find the gradient of V

The gradient \( abla V \) is the vector of partial derivatives of \( V \):\[ abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right) \]First, calculate the partial derivative with respect to \( x \):\[ \frac{\partial V}{\partial x} = -2e^{-2x} \, \cos 2y \]Next, calculate the partial derivative with respect to \( y \):\[ \frac{\partial V}{\partial y} = -2e^{-2x} \, \sin 2y \]

Step 3: Evaluate the gradient at the given point

Evaluate \( abla V \) at \( \left(0, \frac{1}{4} \pi\right) \):\[ abla V \bigg|_{\left(0, \frac{1}{4} \pi\right)} = \left( -2e^{0} \, \cos \left( \frac{1}{2} \pi \right), -2e^{0} \, \sin \left( \frac{1}{2} \pi \right) \right) = (0, -2) \]

Step 4: Find the rate of change in the given direction

The unit vector in the direction given is:\[ \mathbf{u} = \cos \frac{1}{6} \pi \mathbf{i} + \sin \frac{1}{6} \pi \mathbf{j} \]Calculate \( \cos \frac{1}{6} \pi = \frac{\sqrt{3}}{2} \) and \( \sin \frac{1}{6} \pi = \frac{1}{2} \):\[ \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \]The rate of change of \( V \) in this direction is:\[ abla V \cdot \mathbf{u} = (0, -2) \cdot \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) = 0 \cdot \frac{\sqrt{3}}{2} + (-2) \cdot \frac{1}{2} = -1 \]

Step 5: Determine the direction of the steepest ascent

The direction of the greatest rate of change of \( V \) is the same as the direction of the gradient vector \( abla V \):\[ abla V = (0, -2) \]

Step 6: Find the magnitude of the greatest rate of change

The magnitude of the gradient vector \( abla V \) gives the magnitude of the greatest rate of change:\[ \| abla V \| = \sqrt{0^2 + (-2)^2} = 2 \]

Key Concepts

Gradient Calculation

The gradient of a function represents its rate of change in all directions. For a multivariable function, the gradient is a vector of partial derivatives. Given the electric potential function as \[ V = e^{-2x} \, \cos 2y \], we start by finding its gradient.
The gradient \( abla V \) is expressed as \[ abla V = \left( \frac{\partial V}{\partial x}, \frac{\partial V}{\partial y} \right) \].
We calculate each partial derivative:
\( \frac{\partial V}{\partial x} = -2e^{-2x} \, \cos 2y \) and
\( \frac{\partial V}{\partial y} = -2e^{-2x} \, \sin 2y \).
At the point \( \left(0, \frac{\pi}{4} \right) \), this gives \( \abla V|_{\left(0, \frac{\pi}{4}\right)} = (0, -2) \).
The gradient tells us how the function changes at any point.

Directional Derivative

The directional derivative measures how a function changes as we move in a specific direction.
To find it, we project the gradient onto a unit vector in the desired direction.
Using the unit vector \( \mathbf{u} = \cos \frac{1}{6} \pi \mathbf{i} + \sin \frac{1}{6} \pi \mathbf{j} \), we calculate the rate of change:
First, find the components: \( \cos \frac{1}{6} \pi = \frac{\sqrt{3}}{2} \) and \( \sin \frac{1}{6} \pi = \frac{1}{2} \).
The unit vector becomes \( \mathbf{u} = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) \).
The rate of change in this direction is \( \abla V \cdot \mathbf{u} = (0, -2) \cdot \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right) = -1 \).
This tells us how the electric potential changes as we move in the specified direction.

Rate of Change in Multivariable Functions

Understanding how functions change with multiple variables is crucial.
The rate of change in a multivariable function at any point is given by its gradient.
If we consider the electric potential \( V = e^{-2x} \, \cos 2y \), the gradient at the point \( \left(0, \frac{\pi}{4} \right) \) is \( (0, -2) \).
This vector represents the rates of change in the \( x \) and \( y \) directions.
Overall, it tells us how rapidly the potential is increasing or decreasing based on our direction.

Steepest Ascent

The direction of the steepest ascent is given by the direction of the gradient vector.
For our electric potential function \( V \), the gradient at \( \left(0, \frac{\pi}{4}\right) \) is \( (0, -2) \).
Therefore, the direction of the greatest increase (or steepest ascent) is along this gradient vector.
If we move in this direction, the potential will increase the fastest.
This concept is widely used in optimization problems and in understanding how changes propagate through a system.

Magnitude of Gradient

The magnitude of the gradient tells us the largest rate of change at a point.
Mathematically, it is the length of the gradient vector.
For our function at the point \( \left(0, \frac{\pi}{4} \right) \), the gradient is \( (0, -2) \).
The magnitude is calculated as: \[ \| \abla V \| = \sqrt{0^2 + (-2)^2} = 2 \].
This means the maximum rate at which the electric potential is changing, at this point, is 2 volts per foot.
The magnitude of the gradient is an essential measure in physics and engineering.