Problem 36 - Calculus with Concepts in Calculus

Question

Let \(z=f(x, y), x=r \cos \theta\), and \(y=r \sin \theta\). a. Show that \(\frac{\partial z}{\partial x}=\frac{\partial z}{\partial r} \cos \theta-\frac{\partial z}{\partial \theta} \frac{\sin \theta}{r}\) and \(\frac{\partial z}{\partial y}=\frac{\partial z}{\partial r} \sin \theta+\frac{\partial z}{\partial \theta} \frac{\cos \theta}{r}\).

Step-by-Step Solution

Answer

We use the chain rule to derive the partial derivatives of \(z\) with respect to \(x\) and \(y\).

Step 1: Chain Rule for Partial Derivatives

To begin, recognize that since we have a function written as \[ z = f(x, y) = f(r \cos \theta, r \sin \theta) \] where \(x\) and \(y\) are replaced by transformations involving \(r\) and \(\theta\), we need to use the chain rule to express partial derivatives of \(z\) with respect to \(x\) and \(y\).

Step 2: Express \( \frac{\partial z}{\partial x} \) Using the Chain Rule

The chain rule states that: \[\frac{\partial z}{\partial x} = \frac{\partial z}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial z}{\partial \theta} \frac{\partial \theta}{\partial x}\]However, since \(x = r \cos \theta\), differentiate both \(r\) and \(\theta\) with respect to \(x\): \[\frac{\partial r}{\partial x} = \cos \theta, \\frac{\partial \theta}{\partial x} = -\frac{\sin \theta}{r}\]Substitute these into the chain rule result:\[\frac{\partial z}{\partial x} = \frac{\partial z}{\partial r} \cos \theta - \frac{\partial z}{\partial \theta} \frac{\sin \theta}{r}\]

Step 3: Express \( \frac{\partial z}{\partial y} \) Using the Chain Rule

Similarly, apply the chain rule for \( \frac{\partial z}{\partial y} \):\[\frac{\partial z}{\partial y} = \frac{\partial z}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial z}{\partial \theta} \frac{\partial \theta}{\partial y}\]Since \(y = r \sin \theta\), we have:\[\frac{\partial r}{\partial y} = \sin \theta, \\frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r}\]Substitute these into the chain rule result:\[\frac{\partial z}{\partial y} = \frac{\partial z}{\partial r} \sin \theta + \frac{\partial z}{\partial \theta} \frac{\cos \theta}{r}\]

Key Concepts

Partial Derivatives

In calculus, partial derivatives are a powerful tool for understanding how functions change. They allow us to examine how a multivariable function, like \(z = f(x, y)\), changes as we tweak one variable at a time, keeping others constant. For example, the partial derivative of \(z\) with respect to \(x\), denoted \(\frac{\partial z}{\partial x}\), tells us how \(z\) changes as \(x\) changes, while leaving \(y\) unchanged.

When dealing with transformations, such as converting Cartesian coordinates \((x, y)\) into polar coordinates \((r, \theta)\), the situation becomes more complex. Here, partial derivatives further demonstrate their utility through the chain rule, helping us connect derivatives in transformed coordinates. This ties in with our exercise, where we leverage partial derivatives to relate changes in \(r\) and \(\theta\) to changes in \(x\) and \(y\).

The application of partial derivatives in transformations can be seen when substituting \(x = r \cos \theta\) and \(y = r \sin \theta\) to express partial derivatives of \(z\) in terms of \(r\) and \(\theta\).

Polar Coordinates

Polar coordinates, represented as \(r\) and \(\theta\), provide an alternative method of pinpointing locations in a plane using distance and angle rather than horizontal and vertical distances. This system is especially advantageous when dealing with circular paths or regions, as calculations often simplify using polar coordinates.

In our problem, we convert from Cartesian to polar coordinates using the relations \(x = r \cos \theta\) and \(y = r \sin \theta\). This change not only demands adjustments in how we calculate derivatives but also highlights the adaptability of mathematical tools like the chain rule. Polar coordinates allow for a more intuitive understanding and solving approach for circular or angular problems.

When functions are expressed in polar coordinates, understanding how small changes in \(r\) and \(\theta\) affect the function becomes critical. It's here that partial derivatives and the chain rule come into play, offering a structured way to venture through such coordinate transformations. This underscores the intertwined nature of coordinate systems and differentiation methods in calculus.

Calculus Problem Solving

Calculus problem solving often involves breaking down complex problems into manageable parts. By applying different techniques, such as the chain rule for derivatives, intricate mathematical transformations become feasible. In typical exercises, this may involve expressing a function in terms of different variables and coordinates, as seen in our original problem.

Understanding the chain rule is crucial. It allows us to compute how one quantity changes in relation to another, even when multiple variables are involved. Specifically, with the transformations \(x = r \cos \theta\) and \(y = r \sin \theta\), we apply the chain rule to express partial derivatives of \(z\) with respect to \(x\) and \(y\). This aids in solving the equations given in the exercise.

The key to solving such calculus problems is a step-by-step approach:

• Identify the variables and their transformations.
• Apply the chain rule for partial derivatives accurately.
• Simplify expressions to reveal the relationship between different derivatives.

Through consistent practice and understanding of these concepts, solving calculus problems becomes a more intuitive and accessible path.