Problem 151 - Calculus 3

Question

Show that \(z=e^{-t} \cos \left(\frac{x}{c}\right)\) satisfies the heat equation \(\frac{\partial z}{\partial t}=-e^{-t} \cos \left(\frac{x}{c}\right)\)

Step-by-Step Solution

Answer

The function \( z = e^{-t} \cos\left(\frac{x}{c}\right) \) satisfies the heat equation as shown by matching the partial derivative with the equation.

Step 1: Identify the Problem

The goal is to show that the function \( z = e^{-t} \cos\left(\frac{x}{c}\right) \) satisfies the given heat equation: \[ \frac{\partial z}{\partial t} = -e^{-t} \cos\left(\frac{x}{c}\right) \] The function involves both \( t \) and \( x \). We will compute the partial derivative with respect to \( t \) to verify if it matches the right-hand side of the equation.

Step 2: Compute \( \frac{\partial z}{\partial t} \)

Differentiate \( z = e^{-t} \cos\left(\frac{x}{c}\right) \) with respect to \( t \). Here, \( \cos\left(\frac{x}{c}\right) \) is treated as a constant since it doesn't depend on \( t \):\[\frac{\partial z}{\partial t} = \frac{\partial}{\partial t}\left(e^{-t} \cos\left(\frac{x}{c}\right)\right) = -e^{-t} \cos\left(\frac{x}{c}\right).\]

Step 3: Compare with the Given Equation

The calculated partial derivative \( \frac{\partial z}{\partial t} = -e^{-t} \cos\left(\frac{x}{c}\right) \) perfectly matches the right-hand side of the given heat equation. This confirms that the function \( z \) indeed satisfies the heat equation.

Key Concepts

Partial Derivatives

Partial derivatives are a way to measure how a function changes as one of the variables is varied, holding all others constant. When dealing with a function of several variables, take the derivative with respect to one variable at a time. In the context of the given function, \( z = e^{-t} \cos\left(\frac{x}{c}\right) \), we differentiate with respect to \( t \), treating \( x \) as a constant.
This means we look at how \( z \) changes only as \( t \) changes, ignoring the variations in \( x \). For the exercise, it's shown that the partial derivative of \( z \) with respect to \( t \) is computed as \( \frac{\partial z}{\partial t} = -e^{-t} \cos\left(\frac{x}{c}\right) \).
Here are some key points about partial derivatives:

• They focus on the rate of change of the function with respect to one variable.
• Other variables are treated as constants.
• Useful for understanding the behavior of multivariable functions.

By understanding partial derivatives, you get a better glimpse of the function's behavior and how it responds to changes in its parameters.

Differential Equations

Differential equations are fundamental in describing systems and phenomena in many areas, like engineering and physics. These equations relate a function to its derivatives, helping model how things change over time or space. The heat equation in this exercise is a type of differential equation used to model the distribution of heat (or variation in temperature) in a given region over time.
For this problem, we consider the heat equation \( \frac{\partial z}{\partial t} = -e^{-t} \cos\left(\frac{x}{c}\right) \). This involves the partial derivative of \( z \) with respect to time \( t \), indicating how heat or temperature changes at any given point \( x \). In solving differential equations, particularly this type:

• We determine if a proposed solution satisfies the equation.
• It involves comparing the derivative with the right side of the equation.

Differential equations form the basis for modeling dynamic systems and predicting future behavior, ensuring that we understand how changes propagate through a system.

Function Verification

Function verification involves checking whether a proposed function is a valid solution to a given equation. This is crucial when testing solutions for differential equations to ensure they meet all necessary conditions.
In this instance, verification means proving that \( z = e^{-t} \cos\left(\frac{x}{c}\right) \) is a solution to the heat equation by calculating its partial derivative and showing it's equal to the specified right-hand side. This consists of:

• Calculating the derivative as the equation prescribes.
• Ensuring the calculated result matches what's given in the differential equation.

It's a validation step that confirms the correctness of a solution.
By doing function verification, you make sure mathematical solutions conform to expected conditions, solidifying the solution's reliability and applicability in its intended context.