Problem 26 - The Calculus with Analytic Geometry

Question

If \(f\) and \(g\) are differentiable functions of \(x\) and \(y\) and \(u=f(x, y)\) and \(v=g(x, y)\), such that \(\partial u / \partial x=\partial v / \partial y\) and \(\partial u / \partial y=\) \(-\partial v / \partial x\), then if \(x=r \cos \theta\) and \(y=r \sin \theta\), show that $$ \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \text { and } \quad \frac{\partial v}{\partial r}=-\frac{1}{r} \frac{\partial u}{\partial \theta} $$

Step-by-Step Solution

Answer

In polar coordinates, the given conditions lead to: \( \frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta} \) and \( \frac{\partial v}{\partial r} = -\frac{1}{r} \frac{\partial u}{\partial \theta} \).

Step 1: Express partial derivatives in polar coordinates

Given the relationships, we switch to polar coordinates where the transformations are: \( x = r \cos \theta \) and \( y = r \sin \theta \). We need to find the partial derivatives of \(u\) and \(v\) in terms of \(r\) and \(\theta\).

Step 2: Apply Chain Rule

Use the chain rule to express the partial derivatives: \[ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\frac{dx}{dr} + \frac{\partial u}{\partial y}\frac{dy}{dr} \] and \[ \frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x}\frac{dx}{d\theta} + \frac{\partial u}{\partial y}\frac{dy}{d\theta} \]

Step 3: Compute derivatives of x and y

For \(x\) and \(y\) in terms of \(r\) and \(\theta\): \[ \frac{dx}{dr} = \cos \theta, \, \frac{dx}{d\theta} = -r \sin \theta \] and \[ \frac{dy}{dr} = \sin \theta, \, \frac{dy}{d\theta} = r \cos \theta \]

Step 4: Substitute and simplify

Substitute the results into the chain rule expressions: \[ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta \] and \[ \frac{\partial u}{\partial \theta} = -r \frac{\partial u}{\partial x} \sin \theta + r \frac{\partial u}{\partial y} \cos \theta \]

Step 5: Use given conditions

Given \( \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \) and \( \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x} \), substitute these into our expressions: \[ \frac{\partial u}{\partial r} = \frac{\partial v}{\partial y} \cos \theta - \frac{\partial v}{\partial x} \sin \theta \] \[ \frac{\partial u}{\partial \theta} = r \left( \frac{\partial v}{\partial y} \cos \theta - \frac{\partial v}{\partial x} \sin \theta \right) \]

Step 6: Rearrange and identify

Thus, \( \frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial u}{\partial \theta} \) can be rewritten as \[ \frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta} \]

Step 7: Repeat for v

Repeat similar steps for \(v\) to show: \[ \frac{\partial v}{\partial r} = -\frac{1}{r} \frac{\partial u}{\partial \theta} \]

Key Concepts

Differentiable Functions

Differentiable functions are functions that have derivatives at each point in their domains. This means you can determine the slope of the function at any given point.
A function's differentiability implies continuity, but not vice versa. That is, if a function is differentiable, it is also continuous.
This concept is crucial in calculus and analysis because it allows us to study and compute rates of change.
In our exercise, we assume that the functions \(f\) and \(g\) are differentiable with respect to \(x\) and \(y\). This is necessary to use the properties and rules of differentiation, such as the chain rule.

Chain Rule

The chain rule is a fundamental technique in calculus used for differentiating compositions of functions.
The chain rule allows us to differentiate a function with respect to an intermediate variable.
It states that if you have functions \(u\) and \(v\) that are composed of \(x\) and \(y\), you can find the derivative of \(u\) and \(v\) with respect to \(x\) and \(y\) by considering the intermediate step.
In our current problem, we use the chain rule to find partial derivatives of \(u\) and \(v\) with respect to polar coordinates \(r\) and \( \theta \).
This involves expressions such as:
\[ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \frac{dx}{dr} + \frac{\partial u}{\partial y} \frac{dy}{dr} \]
and
\[ \frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x} \frac{dx}{d \theta} + \frac{\partial u}{\partial y} \frac{dy}{d \theta} \].
This approach breaks down the differentiation process into simpler steps, making it more manageable.

Coordinate Transformation

Coordinate transformations involve changing from one coordinate system to another, providing a new perspective for solving problems.
Here, we transform from Cartesian coordinates \( x \) and \( y \) to polar coordinates \( r \) and \( \theta \).
Polar coordinates are useful when dealing with circular or rotational symmetry, as they simplify the equations and derivatives.
For our exercise, the transformations are:
\[ x = r \cos \theta \] and \[ y = r \sin \theta \].
These equations convert Cartesian coordinates into polar coordinates.
Such transformations allow us to express partial derivatives in terms of \(r\) and \(\theta\), making calculations easier and more intuitive in certain contexts.

Polar Coordinates

Polar coordinates \((r, \theta)\) offer an alternative to the Cartesian system, especially for problems involving circles or angles.
In this system, \(r\) represents the radial distance from the origin, while \(\theta\) is the angle from the positive x-axis.
Converting functions to polar coordinates often simplifies differentiation and integration, especially when working with circular symmetry.
In our problem, the partial derivatives of \(u\) and \(v\) are shown in polar coordinates:
\[ \frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta} \] and \[ \frac{\partial v}{\partial r} = - \frac{1}{r} \frac{\partial u}{\partial \theta} \].
These results make it evident how the functions change with respect to \(r\) and \( \theta \), illustrating the compact and elegant nature of polar coordinates in solving such problems.