Problem 65 - Calculus Early Transcendentals

Question

An important derivative operation in many applications is called the Laplacian; in Cartesian coordinates, for \(z=f(x, y),\) the Laplacian is \(z_{x x}+z_{y y} .\) Determine the Laplacian in polar coordinates using the following steps. a. Begin with \(z=g(r, \theta)\) and write \(z_{x}\) and \(z_{y}\) in terms of polar coordinates (see Exercise 64). b. Use the Chain Rule to find \(z_{x x}=\frac{\partial}{\partial x}\left(z_{x}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. c. Use the Chain Rule to find \(z_{y y}=\frac{\partial}{\partial y}\left(z_{y}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. d. Combine parts (b) and (c) to show that $$ z_{x x}+z_{y y}=z_{r r}+\frac{1}{r} z_{r}+\frac{1}{r^{2}} z_{\theta \theta}. $$

Step-by-Step Solution

Answer

Answer: The Laplacian in polar coordinates for a function \(z = f(x, y)\) is given by: $$\nabla^2z = z_{rr} + \frac{1}{r}z_r + \frac{1}{r^2}z_{\theta\theta}$$

Step 1: Convert x and y to r and θ

Recall that the conversion from Cartesian coordinates to polar coordinates is given by: $$x = r \cos\theta, \ y = r \sin\theta$$

Step 2: Differentiate z with respect to x and y

Compute the partial derivatives of z: $$z_x = \frac{\partial z}{\partial x} = \frac{\partial g(r,\theta)}{\partial x}, \ z_y = \frac{\partial z}{\partial y} = \frac{\partial g(r,\theta)}{\partial y}$$

Step 3: Express z_x and z_y in polar coordinates

Using the chain rule, rewrite the partial derivatives \(z_x\) and \(z_y\) in terms of polar coordinates: $$z_x = \frac{\partial g}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial g}{\partial \theta} \frac{\partial \theta}{\partial x}, \ z_y = \frac{\partial g}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial g}{\partial \theta} \frac{\partial \theta}{\partial y}$$ Recall that from the conversion formulas, $$\frac{\partial r}{\partial x} = \frac{x}{r} = \frac{\cos\theta}{r}, \ \frac{\partial r}{\partial y} = \frac{y}{r} =\frac{\sin\theta}{r}, \ \frac{\partial \theta}{\partial x} = -\frac{y}{r^2} = -\frac{\sin\theta}{r^2}, \ \frac{\partial \theta}{\partial y} = \frac{x}{r^2} = \frac{\cos\theta}{r^2}$$ Substituting these into the equations for \(z_x\) and \(z_y\), we get: $$z_x = \frac{\partial g}{\partial r} \frac{\cos\theta}{r} - \frac{\partial g}{\partial \theta} \frac{\sin\theta}{r^2}, \ z_y = \frac{\partial g}{\partial r} \frac{\sin\theta}{r} + \frac{\partial g}{\partial \theta} \frac{\cos\theta}{r^2}$$ #b. Finding z_{xx}#

Step 4: Compute z_xx using the chain rule

Differentiate \(z_x\) with respect to x while keeping in mind the chain rule: $$z_{xx} = \frac{\partial}{\partial x}(z_x) = \frac{\partial}{\partial x}\left(\frac{\partial g}{\partial r} \frac{\cos\theta}{r} - \frac{\partial g}{\partial \theta} \frac{\sin\theta}{r^2}\right)$$

Step 5: Expand the derivative of z_xx

Expand the partial derivative \(z_{xx}\) using the product rule and chain rule: $$z_{xx} = \frac{\partial}{\partial x}\frac{\partial g}{\partial r} \frac{\cos\theta}{r} - \frac{\partial}{\partial x}\frac{\partial g}{\partial \theta} \frac{\sin\theta}{r^2}$$ #c. Finding z_{yy}#

Step 6: Compute z_yy using the chain rule

Differentiate \(z_y\) with respect to y while keeping in mind the chain rule: $$z_{yy} = \frac{\partial}{\partial y}(z_y) = \frac{\partial}{\partial y}\left(\frac{\partial g}{\partial r} \frac{\sin\theta}{r} + \frac{\partial g}{\partial \theta} \frac{\cos\theta}{r^2}\right)$$

Step 7: Expand the derivative of z_yy

Expand the partial derivative \(z_{yy}\) using the product rule and chain rule: $$z_{yy} = \frac{\partial}{\partial y}\frac{\partial g}{\partial r} \frac{\sin\theta}{r} + \frac{\partial}{\partial y}\frac{\partial g}{\partial \theta} \frac{\cos\theta}{r^2}$$ #d. Finding Laplacian in polar coordinates#

Step 8: Combine z_xx and z_yy

Add \(z_{xx}\) and \(z_{yy}\) and simplify: $$z_{xx} + z_{yy} = z_{rr} + \frac{1}{r}z_r + \frac{1}{r^2}z_{\theta\theta}$$ The Laplacian in polar coordinates for a function \(z = f(x, y)\) is given by: $$\nabla^2z = z_{rr} + \frac{1}{r}z_r + \frac{1}{r^2}z_{\theta\theta}$$

Key Concepts

Partial Derivatives

Partial derivatives are essential tools in calculus to understand how a multivariable function changes when its individual variables change. When we have a function of two variables like \(z = f(x, y)\), the partial derivative with respect to \(x\) (denoted as \(z_x\)) measures how \(z\) changes with small changes in \(x\), keeping \(y\) constant. Similarly, \(z_y\) is the rate of change of \(z\) with respect to \(y\), keeping \(x\) constant.

To find these derivatives, you treat all other variables as constants, just like you would differentiate a single-variable function. This process allows you to analyze the function more deeply, considering how the change in one direction specifically influences the outcome. For instance, in physics and engineering, this is crucial for understanding phenomena like heat distribution or waves.

In our specific problem, once \(z_x\) and \(z_y\) are determined in Cartesian coordinates, they facilitate transformation into polar coordinates, laying the groundwork for further derivative operations.

Chain Rule

The chain rule is a fundamental technique not just in single-variable calculus, but especially in multivariable differentiations where functions depend on other functions. The chain rule provides a method to differentiate composite functions by breaking them down into simpler parts.

For example, consider a function \(z = g(r, \theta)\) derived from \(x = r \cos\theta\) and \(y = r \sin\theta\). Using the chain rule, we calculate partial derivatives like \(z_x\) and \(z_y\) by considering how \(z\) changes through \(r\) and \(\theta\). The rule asserts that

• \(z_x = \frac{\partial g}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial g}{\partial \theta} \frac{\partial \theta}{\partial x}\)
• \(z_y = \frac{\partial g}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial g}{\partial \theta} \frac{\partial \theta}{\partial y}\)

This process involves determining how each intermediate variable—\(r\) and \(\theta\)—depends on \(x\) and \(y\). Applying the chain rule rigorously allows us to express derivatives in terms of the new variables, thereby enabling accurate transformation and further operations in problems involving coordination transformation, like converting Cartesian derivatives to polar derivatives.

Conversion to Polar Coordinates

The conversion between Cartesian and polar coordinates is a common technique in mathematics, especially useful for simplifying problems with circular symmetry. Polar coordinates use the distance from the origin \(r\) and the angle \(\theta\) from the positive x-axis to locate a point in a plane.

The formulas \(x = r \cos\theta\) and \(y = r \sin\theta\) allow conversion from Cartesian \((x, y)\) to polar \((r, \theta)\). This is particularly helpful when dealing with circular or radial problems, where polar coordinates can make the math much more intuitive.

In practice, this conversion is crucial for tasks such as finding the Laplacian in different coordinates like polar. Once the original problem is restated in terms of \(r\) and \(\theta\), partial derivatives and other operations must also be converted, using relationships formed by the chain rule.

Finally, after expressing the problem in polar coordinates, it typically results in expressions that encapsulate radial and angular components separately, as seen in the transformation of the Laplacian. This separation often leads to simplified analysis and solutions for differential equations in physics and engineering.