Problem 149 - Calculus 3

Question

Show that \(z=e^{x} \sin y\) satisfies the equation \(\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0\)

Step-by-Step Solution

Answer

The function satisfies the equation as the sum of the second partial derivatives is zero.

Step 1: Compute the First Partial Derivative with Respect to x

Start with the function given: \( z = e^{x} \sin y \). Compute the first partial derivative with respect to \( x \). This means we differentiate \( z \) while treating \( y \) as a constant: \[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(e^{x} \sin y) = e^{x} \sin y. \]

Step 2: Compute the Second Partial Derivative with Respect to x

Now differentiate \( \frac{\partial z}{\partial x} = e^{x} \sin y \) again with respect to \( x \): \[ \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial}{\partial x}(e^{x} \sin y) = e^{x} \sin y. \]

Step 3: Compute the First Partial Derivative with Respect to y

Use the original function \( z = e^{x} \sin y \) to compute the first partial derivative with respect to \( y \). Differentiate with respect to \( y \): \[ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(e^{x} \sin y) = e^{x} \cos y. \]

Step 4: Compute the Second Partial Derivative with Respect to y

Differentiate \( \frac{\partial z}{\partial y} = e^{x} \cos y \) with respect to \( y \): \[ \frac{\partial^{2} z}{\partial y^{2}} = \frac{\partial}{\partial y}(e^{x} \cos y) = -e^{x} \sin y. \]

Step 5: Verify the Partial Differential Equation

Sum the second partial derivatives and verify if they satisfy the equation: \[ \frac{\partial^{2} z}{\partial x^{2}} + \frac{\partial^{2} z}{\partial y^{2}} = e^{x} \sin y + (-e^{x} \sin y) = 0. \] Thus, the equation is satisfied.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus that involve differentiating a multivariable function with respect to one variable at a time.
When working with partial derivatives, it's critical to treat all other variables as constants during differentiation. This approach is different from ordinary derivatives where functions have only one variable.When you compute the partial derivative of a function, say \( z = f(x, y) \), with respect to \( x \), notation like \( \frac{\partial z}{\partial x} \) is used. Here, you differentiate treating \( y \) as constant. Let's consider an example: for \( z = e^{x} \sin y \), the partial derivative with respect to \( x \) is:

• Keep \( \sin y \) constant, differentiate \( e^{x} \), resulting in \( e^{x} \sin y \).

Similarly, find the partial derivative with respect to \( y \) by treating \( x \) as constant, resulting in \( e^{x} \cos y \). Partial derivatives are useful in understanding how a function's value changes as one variable changes while others remain fixed.

Mathematical Proof

A mathematical proof is a logical process where we verify the truth of a mathematical statement. In our exercise, we are asked to prove that the function \( z = e^{x} \sin y \) satisfies a specific partial differential equation.First, calculate the second partial derivatives required by the equation: the second derivative with respect to \( x \) and that with respect to \( y \).
This is achieved by further differentiating the respective first partial derivatives.Once the second derivatives are computed:

• For \( x \): \( \frac{\partial^{2} z}{\partial x^{2}} = e^{x} \sin y \).
• For \( y \): \( \frac{\partial^{2} z}{\partial y^{2}} = -e^{x} \sin y \).

Add these two derivatives and observe the sum is zero, thus satisfying the given equation. This step completes the proof, affirming \( z \) solves the differential equation.

Step-by-Step Solution

Step-by-step solutions are detailed guides that walk you through mathematical processes. Let's break down our original exercise for clarity.First, compute the partial derivatives. Start with the given function \( z = e^{x} \sin y \):- **Step 1**: Find \( \frac{\partial z}{\partial x} \) resulting in \( e^{x} \sin y \).- **Step 2**: Again with respect to \( x \), find \( \frac{\partial^{2} z}{\partial x^{2}} = e^{x} \sin y \).Then switch to the \( y \) variable:- **Step 3**: Compute \( \frac{\partial z}{\partial y} = e^{x} \cos y \).- **Step 4**: Differentiate again for \( \frac{\partial^{2} z}{\partial y^{2}} = -e^{x} \sin y \).Finally, verify the equation by adding these second derivatives:

• \( e^{x} \sin y + (-e^{x} \sin y) = 0 \).

This approach ensures each step is logical and transparent, guiding through the derivation and verification smoothly.