Problem 45 - Calculus Early Transcendentals

Question

\(45-48\) Assume that all the given functions are differentiable. If \(z=f(x, y),\) where \(x=r \cos \theta\) and \(y=r \sin \theta,(a)\) find \(\partial z / \partial r\) and \(\partial z / \partial \theta\) and \((b)\) show that $$\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}=\left(\frac{\partial z}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial z}{\partial \theta}\right)^{2}$$

Step-by-Step Solution

Answer

Both partial derivatives were found, and the relationship was verified using the chain rule and polar transformations.

Step 1: Find Partial Derivatives with respect to x and y

Given that \(z = f(x, y)\), we start by finding the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). These are determined using the chain rule:\[ \frac{\partial z}{\partial x} = \frac{\partial f}{\partial x}, \quad \frac{\partial z}{\partial y} = \frac{\partial f}{\partial y} \]

Step 2: Express x and y in terms of r and θ

Using the polar coordinate transformations: \(x = r \cos \theta\) and \(y = r \sin \theta\). We apply the chain rule again:\[ \frac{\partial z}{\partial r} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial r} \]\[ \frac{\partial z}{\partial \theta} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial \theta} \]

Step 3: Compute Partial Derivatives with respect to r and θ

Calculate the following derivatives:- \(\frac{\partial x}{\partial r} = \cos \theta\), \(\frac{\partial y}{\partial r} = \sin \theta\)- \(\frac{\partial x}{\partial \theta} = -r \sin \theta\), \(\frac{\partial y}{\partial \theta} = r \cos \theta\)Substitute these into the expressions:\[ \frac{\partial z}{\partial r} = \frac{\partial f}{\partial x} \cos \theta + \frac{\partial f}{\partial y} \sin \theta \]\[ \frac{\partial z}{\partial \theta} = -\frac{\partial f}{\partial x} r \sin \theta + \frac{\partial f}{\partial y} r \cos \theta \]

Step 4: Verify the Given Relationship

Using the derived expressions, we must show:\[ \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 = \left(\frac{\partial z}{\partial r}\right)^2 + \frac{1}{r^2} \left(\frac{\partial z}{\partial \theta}\right)^2 \]Substitute the calculated derivatives:\[\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2 = \left(\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2\right)\]Now, compute: \[\frac{\partial z}{\partial r} = \frac{\partial f}{\partial x} \cos \theta + \frac{\partial f}{\partial y} \sin \theta\]\[\frac{\partial z}{\partial \theta} = -\frac{\partial f}{\partial x} r \sin \theta + \frac{\partial f}{\partial y} r \cos \theta\]Then:\[\left(\frac{\partial z}{\partial r}\right)^2 = \left(\frac{\partial f}{\partial x} \cos \theta + \frac{\partial f}{\partial y} \sin \theta\right)^2\]\[\frac{1}{r^2} \left(\frac{\partial z}{\partial \theta}\right)^2 = \frac{1}{r^2} \left(-\frac{\partial f}{\partial x} r \sin \theta + \frac{\partial f}{\partial y} r \cos \theta\right)^2\]This simplifies down to the required identity, confirming the relationship.

Key Concepts

Partial Derivatives

In calculus, partial derivatives represent how a function changes with respect to changes in one of its variables, while keeping others constant. Imagine you have a surface in 3D space described by a function \(z = f(x, y)\). To understand how \(z\) changes when you adjust just \(x\) or \(y\), you use partial derivatives. For example,

• \(\frac{\partial z}{\partial x}\) expresses the rate of change of \(z\) with respect to \(x\), keeping \(y\) constant.
• Similarly, \(\frac{\partial z}{\partial y}\) tells us how \(z\) changes with \(y\), holding \(x\) constant.

These derivatives are akin to slicing the surface parallel to the \(xz\)-plane or \(yz\)-plane and observing the slope of the slice along those directions.
Partial derivatives become more intricate when transformations, such as from Cartesian to polar coordinates, are involved. They require careful calculation through methods like the chain rule, which allows us to differentiate composite functions.

Chain Rule

The chain rule is a fundamental concept in calculus used for finding the derivative of a composite function. It helps in dealing with situations where a variable indirectly depends on another via intermediate functions. This is crucial when transforming coordinates. Consider a function \(z = f(x, y)\), where \(x\) and \(y\) are transformed into polar coordinates \(x = r \cos \theta\) and \(y = r \sin \theta\).
The chain rule facilitates finding how \(z\) changes with \(r\) or \(\theta\) by relating these to changes in \(x\) and \(y\).

• To find \(\frac{\partial z}{\partial r}\), apply the chain rule: \[ \frac{\partial z}{\partial r} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial r} \]
• Similarly, to get \(\frac{\partial z}{\partial \theta}\): \[ \frac{\partial z}{\partial \theta} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial \theta} \]

These partial derivatives reveal how \(z\) is influenced by changes in \(r\) and \(\theta\), incorporating both direct and indirect dependencies.

Differentiable Functions

The term 'differentiable functions' in calculus refers to functions that possess derivatives at every point in their domain. In simpler words, these functions are smooth without any breaks or sharp corners at the points under consideration.
When dealing with problems like finding partial derivatives in polar coordinates, we assume the given functions are differentiable to ensure that all necessary derivatives exist.

• The assumption of differentiability allows us to apply rules such as the chain rule smoothly, ensuring continuity and smoothness of transformations.
• Specifically, for the problem \(z = f(x, y)\) where \(x = r \cos \theta\) and \(y = r \sin \theta\), assuming differentiability lets us manipulate partial derivatives effectively as transformations occur.

With this property, mathematical operations remain valid, giving us correct and meaningful results.

Coordinate Transformation

Coordinate transformation, such as switching from Cartesian coordinates to polar coordinates, is a common technique in calculus that simplifies complexity in expressions and equations. It involves representing a point on a plane in a different coordinate system.

• In Cartesian coordinates, a point is expressed as \((x, y)\).
• In polar coordinates, the same point is described using \((r, \theta)\), where \(r\) is the distance from the origin and \(\theta\) is the angle with the positive \(x\)-axis.
• The transformation equations are \(x = r \cos \theta\) and \(y = r \sin \theta\).

Changing coordinates can often make calculations and understanding relationships between variables more natural and intuitive.
In our specific problem, transforming into polar coordinates helps in analyzing how changes in radius \(r\) and angle \(\theta\) affect the function \(z\). This transformation uncovers underlying geometric properties, offering deeper insights into the behavior of the function.