Problem 52 Find the indicated higher-order ... [FREE SOLUTION]

Chapter 3: Problem 52

Find the indicated higher-order partial derivatives. Show that \(z=e^{x} \sin y\) satisfies the equation \(\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0\)

Short Answer

Expert verified

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

Step by step solution

Compute the First Partial Derivative with respect to x

First, let's find the first-order partial derivative of \(z\) with respect to \(x\). Since \(z = e^x \sin y\), treat \(\sin y\) as a constant and differentiate \(e^x\) to get: \ \(\frac{\partial z}{\partial x} = e^x \sin y\).

Compute the Second Partial Derivative with respect to x

Now, differentiate the result from Step 1 with respect to \(x\) again to compute \(\frac{\partial^2 z}{\partial x^2}\). Since the derivative of \(e^x\) is still \(e^x\), we have: \ \(\frac{\partial^2 z}{\partial x^2} = e^x \sin y\).

Compute the First Partial Derivative with respect to y

Differentiate \(z = e^x \sin y\) with respect to \(y\). Here, \(e^x\) is treated as a constant, and the derivative of \(\sin y\) is \(\cos y\). Thus, \ \(\frac{\partial z}{\partial y} = e^x \cos y\).

Compute the Second Partial Derivative with respect to y

Differentiate the result from Step 3 with respect to \(y\) again. The derivative of \(\cos y\) is \(-\sin y\), so \ \(\frac{\partial^2 z}{\partial y^2} = -e^x \sin y\).

Verify the Given Equation is Satisfied

Now, we need to check if the equation \(\frac{\partial^{2}z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0\) holds true. Substitute the second derivatives from Steps 2 and 4: \ \(e^x \sin y - e^x \sin y = 0\). \ As \(0 = 0\), the equation is satisfied.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Differentiation

Partial differentiation involves finding the derivative of a function with multiple variables with respect to one of these variables, while treating the others as constants. This method allows us to understand how a function changes as just one variable changes, holding others constant.

For instance, consider the function given as: \( z = e^x \sin y \). Here, by treating either \( x \) or \( y \) as constant, we can unveil specific partial derivatives.

This technique is particularly useful when analyzing functions in multivariable calculus as it allows a deeper look at the rate of change in various directions, providing insights into the function's local and global behavior.

Second-Order Derivatives

Second-order derivatives are the derivatives of the derivatives of a function. In multivariable calculus, these are higher-order partial derivatives, telling us about the curvature or concavity of a function with respect to each variable.

For the provided problem, after determining the first partial derivatives, we take it a step further by differentiating again. We find:

The second-order partial derivative with respect to \(x\): \( \frac{\partial^2 z}{\partial x^2} = e^x \sin y \)
The second-order partial derivative with respect to \(y\): \( \frac{\partial^2 z}{\partial y^2} = -e^x \sin y \)

Taken together, these second-order derivatives can be used to evaluate the concavity of the surface described by \(z\), as well as contribute to solving partial differential equations like the given one.

Partial Derivative with Respect to x

To compute the partial derivative with respect to \(x\) for the function \( z = e^x \sin y \), we treat \( \sin y \) as a constant.

Differentiating \(e^x\) where only \(x\) changes gives us: \( \frac{\partial z}{\partial x} = e^x \sin y \).
From this result, differentiate with respect to \(x\) again to get: \( \frac{\partial^2 z}{\partial x^2} = e^x \sin y \).

This simplification comes from the exponential function's unique property; its derivative is the function itself. These steps underline the systematic approach in handling multivariable functions for deeper analysis.

Partial Derivative with Respect to y

When calculating the partial derivative with respect to \(y\), we need to consider \( e^x \) as constant and focus on \( \sin y \).

Differentiate \( \sin y\) to get: \( \frac{\partial z}{\partial y} = e^x \cos y \).
Then, differentiate \( e^x \cos y \) again with respect to \(y\). The derivative of \( \cos y \) is \(-\sin y\), so we have: \( \frac{\partial^2 z}{\partial y^2} = -e^x \sin y \).

These calculations are crucial when solving partial differential equations, like confirming the given relation: \( \frac{\partial^{2}z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0\). This approach reveals the functional relationship representation and behavior.

91影视

Find the indicated higher-order partial derivatives. Show that \(z=e^{x} \sin y\) satisfies the equation \(\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partial y^{2}}=0\)

Short Answer

Step by step solution

Compute the First Partial Derivative with respect to x

Compute the Second Partial Derivative with respect to x

Compute the First Partial Derivative with respect to y

Compute the Second Partial Derivative with respect to y

Verify the Given Equation is Satisfied

Key Concepts

Partial Differentiation

Second-Order Derivatives

Partial Derivative with Respect to x

Partial Derivative with Respect to y

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