Problem 52 - Calculus Early Transcendentals

Question

\(49-54\) Assume that all the given functions have continuous second-order partial derivatives. If \(z=f(x, y),\) where \(x=r \cos \theta\) and \(y=r \sin \theta,\) find (a) \(\partial z / \partial r,\) (b) \(\partial z / \partial \theta,\) and \((\mathrm{c}) \partial^{2} z / \partial r \partial \theta\)

Step-by-Step Solution

Answer

(a) \( \partial z / \partial r = \cos \theta \partial z / \partial x + \sin \theta \partial z / \partial y \); (b) \( \partial z / \partial \theta = -r \sin \theta \partial z / \partial x + r \cos \theta \partial z / \partial y \); (c) \( \partial^2 z / \partial r \partial \theta = -\sin \theta \partial z / \partial x + \cos \theta \partial^2 z / \partial x \partial \theta + \cos \theta \partial z / \partial y + \sin \theta \partial^2 z / \partial y \partial \theta \).

Step 1: Understanding the Relationship between Variables

We know from the problem statement that we have: \(x = r \cos \theta\) and \(y = r \sin \theta\). These are polar coordinates transformed into Cartesian coordinates.

Step 2: Finding \( \partial z / \partial r \)

Using the chain rule, we need to express \(\partial z / \partial r\) in terms of \(x\) and \(y\). Since \(z=f(x, y)\), by the chain rule: \[ \frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial r} \]Calculate: \(\frac{\partial x}{\partial r} = \cos \theta\) and \(\frac{\partial y}{\partial r} = \sin \theta\).Thus, \[\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \cos \theta + \frac{\partial z}{\partial y} \sin \theta\]

Step 3: Finding \( \partial z / \partial \theta \)

Again, we use the chain rule with respect to \(\theta\) as follows:\[ \frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial \theta} \]Calculate:\(\frac{\partial x}{\partial \theta} = -r \sin \theta\) and \(\frac{\partial y}{\partial \theta} = r \cos \theta\).Then,\[\frac{\partial z}{\partial \theta} = -r \sin \theta \frac{\partial z}{\partial x} + r \cos \theta \frac{\partial z}{\partial y}\]

Step 4: Finding \( \partial^2 z / \partial r \partial \theta \)

We need to compute the mixed partial derivative:\[ \frac{\partial^2 z}{\partial r \partial \theta} = \frac{\partial}{\partial \theta} \left( \frac{\partial z}{\partial r} \right) \]Using the expression for \(\frac{\partial z}{\partial r}\) as derived earlier:\[ \frac{\partial^2 z}{\partial r \partial \theta} = \frac{\partial}{\partial \theta} \left( \cos \theta \frac{\partial z}{\partial x} + \sin \theta \frac{\partial z}{\partial y} \right) \]This involves:1. Differentiating \(\cos \theta \frac{\partial z}{\partial x}\) with respect to \(\theta\): \[-\sin \theta \frac{\partial z}{\partial x} + \cos \theta \frac{\partial^2 z}{\partial x \partial \theta}\]2. Differentiating \(\sin \theta \frac{\partial z}{\partial y}\) with respect to \(\theta\): \[\cos \theta \frac{\partial z}{\partial y} + \sin \theta \frac{\partial^2 z}{\partial y \partial \theta}\]Thus,\[ \frac{\partial^2 z}{\partial r \partial \theta} = -\sin \theta \frac{\partial z}{\partial x} + \cos \theta \frac{\partial^2 z}{\partial x \partial \theta} + \cos \theta \frac{\partial z}{\partial y} + \sin \theta \frac{\partial^2 z}{\partial y \partial \theta} \]

Step 5: Final Step: Summary of Results

(a) \( \frac{\partial z}{\partial r} = \cos \theta \frac{\partial z}{\partial x} + \sin \theta \frac{\partial z}{\partial y} \)(b) \( \frac{\partial z}{\partial \theta} = -r \sin \theta \frac{\partial z}{\partial x} + r \cos \theta \frac{\partial z}{\partial y} \)(c) \( \frac{\partial^2 z}{\partial r \partial \theta} = -\sin \theta \frac{\partial z}{\partial x} + \cos \theta \frac{\partial^2 z}{\partial x \partial \theta} + \cos \theta \frac{\partial z}{\partial y} + \sin \theta \frac{\partial^2 z}{\partial y \partial \theta} \)

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus that helps us find the derivative of a composite function. Think of it as a formula that tells you how to "chain" together derivatives of functions that are composed of one another.

When functions are defined in terms of other functions, direct differentiation becomes challenging. For this exercise, we have the function \(z = f(x, y)\), where \(x\) and \(y\) themselves depend on \(r\) and \(\theta\).

To differentiate \(z\) with respect to \(r\) or \(\theta\), we use the chain rule, which allows us to express \(\partial z/\partial r\) and \(\partial z/\partial \theta\) using the derivatives of \(z\) with respect to \(x\) and \(y\).

- For \(\partial z/\partial r\), use: \[\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial r}\]
- For \(\partial z/\partial \theta\), use: \[\frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial \theta}\]

The rule helps deconstruct the problem into manageable pieces, by focusing on simpler partial derivatives, and combining them to find the desired derivative.

Mixed Partial Derivatives

Mixed partial derivatives are partial derivatives taken with respect to different variables. In this exercise, mixed partial derivatives involve \(r\) and \(\theta\).

Imagine them as performing one partial derivative, and then performing another, each with respect to a different variable. Mixed derivatives can provide insight into how changes in these variables simultaneously affect the function.

Here, the task was to find \(\partial^2 z / \partial r \partial \theta\). First, you differentiate \(z\) with respect to \(r\) to find \(\partial z/\partial r\), and then differentiate this result with respect to \(\theta\).

To achieve this, we follow the components of chain derivation:- Start by obtaining \(\partial z / \partial r\) as derived using the chain rule.- Then, differentiate \(\partial z / \partial r\) with respect to \(\theta\).

This emphasizes the idea that order matters in partial differentiation, but under the conditions of continuous second derivatives, the mixed partial derivatives \(\partial^2 z / \partial r \partial \theta\) and \(\partial^2 z / \partial \theta \partial r\) are equal.

Polar Coordinates

Polar coordinates are a system of coordinates where each point on a plane is determined by a distance from a reference point and an angle from a reference direction.

- The reference point (analogous to the origin in Cartesian coordinates) is called the pole.- The reference direction, usually a line parallel to the positive x-axis, defines how angles are measured.

The coordinates \((r, \theta)\) relate to Cartesian coordinates \((x,y)\) through:- \(x = r \cos \theta\)- \(y = r \sin \theta\)

These transformations are crucial for problems involving rotational symmetry or periodicity. Converting between polar and Cartesian representations allows us to use familiar calculus techniques. In this problem, polar coordinates facilitate differentiating \(z = f(x, y)\) in terms of \(r\) and \(\theta\).

Using polar coordinates simplifies expressing complex relationships in mathematics, especially when dealing with circles and angles.

Second-order Derivatives

Second-order derivatives provide information on how a first-order derivative is changing.

In this context, second-order partial derivatives give us a deeper look into the curvature or concavity of functions in multiple dimensions. They reveal how the rate of change itself is varying with respect to the variables involved.

For a function \(z = f(x, y)\), the second-order derivatives include:

• \(\partial^2 z / \partial x^2\) – the rate of change of the first derivative \(\partial z / \partial x\) with respect to \(x\)
• \(\partial^2 z / \partial y^2\) – the rate of change of the first derivative \(\partial z / \partial y\) with respect to \(y\)
• Mixed derivative \(\partial^2 z / \partial x \partial y\) – rate of change of \(\partial z / \partial x\) with respect to \(y\) or vice versa



In applications such as physics and engineering, second-order derivatives can describe things like acceleration, which is the derivative of velocity (a first-order derivative of position).

For our exercise, having continuous second-order derivatives implies smoothness and the reliability of mixed partial derivatives being equal (Clairaut's theorem). This property is often used to simplify calculations and prove mathematical results.