Problem 25 - The Calculus with Analytic Geometry

Question

If \(f\) is a differentiable function of \(x\) and \(y\) and \(u=f(x, y), x=r \cos \theta\), and \(y=r \sin \theta\), show that $$ \begin{aligned} &\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r} \cos \theta-\frac{\partial u}{\partial \theta} \frac{\sin \theta}{r} \\ &\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r} \sin \theta+\frac{\partial u}{\partial \theta} \frac{\cos \theta}{r} \end{aligned} $$

Step-by-Step Solution

Answer

Use the chain rule for partial derivatives and simplify: \(\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cos \theta - \frac{\partial u}{\partial \theta} \frac{\sin \theta}{r} \), \(\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \sin \theta + \frac{\partial u}{\partial \theta} \frac{\cos \theta}{r} \).

Step 1: Recall the Chain Rule for Multivariable Functions

The function in question is a composite function, meaning we can use the chain rule for partial derivatives. Recall that for a function of multiple variables, the total derivative can be expressed as the sum of its partial derivatives with respect to each variable times the derivative of those variables with respect to the given variable.

Step 2: Express Partial Derivatives Using the Chain Rule

Write the expression for the partial derivatives of the composite function:\[ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial x} \]\[ \frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial y} \]

Step 3: Find Partial Derivatives of Intermediate Variables

Identify the intermediate variables from the transformation equations: \(x = r \cos \theta\) and \(y = r \sin \theta\). Compute the partial derivatives of these intermediate variables:\[ \frac{\partial r}{\partial x} = \cos \theta, \frac{\partial \theta}{\partial x} = -\frac{\sin \theta}{r} \]\[ \frac{\partial r}{\partial y} = \sin \theta, \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r} \]

Step 4: Substitute Partial Derivatives Back into the Chain Rule

Substitute the partial derivatives found in Step 3 back into the chain rule expressions from Step 2:\[ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cos \theta + \frac{\partial u}{\partial \theta} \left(-\frac{\sin \theta}{r}\right) \]\[ \frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \sin \theta + \frac{\partial u}{\partial \theta} \left(\frac{\cos \theta}{r}\right) \]

Step 5: Simplify the Expressions

Combine and simplify the terms to match the desired format. The final expressions should be:\[ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cos \theta - \frac{\partial u}{\partial \theta} \frac{\sin \theta}{r} \]\[ \frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \sin \theta + \frac{\partial u}{\partial \theta} \frac{\cos \theta}{r} \]

Key Concepts

Chain Rule

The Chain Rule is essential for differentiating composite functions, especially in multivariable calculus. When a function depends on several variables, and each of these variables is a function of another variable, we use the chain rule. For instance, if we have a function \( u = f(x, y) \), and both \( x \) and \( y \) are functions of \( r \) and \( \theta \), we can find the partial derivatives of \( u \) with respect to \( x \) and \( y \) by applying the chain rule. This rule allows us to express the derivative of \( u \) in terms of the intermediate variables \( r \) and \( \theta \), making it easier to work with more complex functions.

Partial Derivatives

Partial derivatives measure how a function changes with respect to one of its variables while keeping other variables constant. If \( u = f(x, y) \), the partial derivative of \( u \) with respect to \( x \) (denoted as \( \frac{\partial u}{\partial x} \)) tells us how \( u \) changes as \( x \) changes, holding \( y \) constant. Similarly, \( \frac{\partial u}{\partial y} \) measures the change in \( u \) as \( y \) changes, holding \( x \) constant. In our exercise, by applying the chain rule, we express these partial derivatives in terms of \( r \) and \( \theta \). This is crucial for transforming coordinates and working with functions in polar coordinates.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point in the plane is represented by a distance from a reference point and an angle from a reference direction. The transformation equations \( x = r \cos \theta \) and \( y = r \sin \theta \) convert polar coordinates (\( r, \theta \)) to Cartesian coordinates (\( x, y \)). In this context, we need to find the partial derivatives of \( u \) with respect to \( x \) and \( y \). By understanding the relationship between Cartesian and polar coordinates, we use the chain rule effectively to transform these coordinates, making complex differentials more manageable.

Composite Functions

A composite function is a function that depends on another function. In our scenario, \( u = f(x, y) \) is a composite function because \( x \) and \( y \) themselves are functions of \( r \) and \( \theta \). To differentiate such functions, we use the chain rule. The partial derivatives of \( u \) with respect to \( r \) and \( \theta \) are substituted back using the expressions we derive from the transformation equations. This process breaks down complicated derivatives into simpler parts, making it easier to handle situations where functions depend on multiple variables.

Transformation Equations

Transformation equations are formulas that relate different coordinate systems, allowing us to switch between them effectively. In our exercise, the equations \( x = r \cos \theta \) and \( y = r \sin \theta \) transform between polar and Cartesian coordinates. By computing the partial derivatives of these equations, we derive expressions like \( \frac{\partial r}{\partial x} = \cos \theta \) and \( \frac{\partial \theta}{\partial x} = -\frac{\sin \theta}{r} \). These values are then used in the chain rule to find the partial derivatives of \( u \) with respect to \( x \) and \( y \). Understanding and using transformation equations is pivotal in various fields, including physics and engineering, where different coordinate systems describe complex objects and phenomena.