Question

If \(x=r \cos \theta\) and \(y=r \sin \theta,\) find \((\partial y / \partial \theta)_{r}\) and \((\partial y / \partial \theta)_{x} .\) Also find \((\partial \theta / \partial y)_{x}\) in two ways (by climinating \(r\) from the given equations and then differentiating, or by taking differentials in both equations and then eliminating \(d r\) ). When are \(\partial y / \partial \theta\) and \(\partial \theta / \partial y\) reciprocals?

Short answer

\(\partial y / \partial \theta\) and \(\partial \theta / \partial y\) are reciprocals when their product equals 1. The specific partial derivatives are \( (\partial y / \partial \theta)_{r} = r \cos \theta \), \((\partial y / \partial \theta)_{x}= x \sec^2 \theta \) and \((\partial \theta / \partial y)_{x} = -\frac{x}{x^2 + y^2} \).

Step-by-step solution

Determine \(\partial y / \partial \theta\) at constant \(r\)

Substitute the expression for \(y\): \(y = r \sin \theta\). Differentiate with respect to \(\theta\) while keeping \(r\) constant:\[\left(\frac{\partial y}{\partial \theta}\right)_{r} = r \cos \theta.\]

Determine \(\partial y / \partial \theta\) at constant \(x\)

First, note that \(x = r \cos \theta\). Re-express \(r\) in terms of \(x\) and \(\theta\) from this equation. Thus, \(r = \frac{x}{\cos \theta}\). Substitute this back into the expression for \(y\): \[y = \frac{x \sin \theta}{\cos \theta} = x \tan \theta.\]Now differentiate with respect to \(\theta\) while keeping \(x\) constant:\[\left(\frac{\partial y}{\partial \theta}\right)_{x} = x \sec^2 \theta.\]

Determine \(\partial \theta / \partial y\) by eliminating \(r\)

From \(y = r \sin \theta\), express \(r = \frac{y}{\sin \theta}\). Substitute this into \(x = r \cos \theta\): \[ x = \frac{y \cos \theta}{\sin \theta} = y \cot \theta.\]Therefore: \[ \theta = \cot^{-1}\left(\frac{x}{y}\right). \]Differentiate with respect to \(y\) while keeping \(x\) constant:\[\left(\frac{\partial \theta}{\partial y}\right)_{x} = \frac{-1}{y^2 + \left(\frac{x}{y}\right)^2} \cdot \frac{x}{y^2}.\]This simplifies to:\[\left(\frac{\partial \theta}{\partial y}\right)_{x} = -\frac{x}{x^2 + y^2}.\]

Determine \(\partial \theta / \partial y\) using differentials

Take the differentials of both given equations:\[dx = d(r \cos \theta)= \cos \theta \, dr - r \sin \theta \, d\theta,\]\[dy = d(r \sin \theta)= \sin \theta \, dr + r \cos \theta \, d\theta.\]Eliminate \(dr\) by dividing \(dy\) by \(dx\): \[\frac{dy}{dx} = \frac{\sin \theta \, dr + r \cos \theta \, d\theta}{\cos \theta \, dr - r \sin \theta \, d\theta}.\]Rearrange to solve for \(d\theta\):\[\left(\frac{\partial \theta}{\partial y}\right)_{x} = \frac{dy}{r \cos \theta} = -\frac{x}{x^2 + y^2}.\]

Identify the reciprocal relation

The reciprocals condition is satisfied when:\[\left(\frac{\partial y}{\partial \theta}\right) \cdot \left(\frac{\partial \theta}{\partial y}\right) = 1.\]Substitute \(\left(\frac{\partial y}{\partial \theta}\right)_{x} = x \sec^2 \theta\) and \(\left(\frac{\partial \theta}{\partial y}\right)_{x} = -\frac{x}{x^2 + y^2}\) into the equation when these partial derivatives coincide.

Key concepts

Partial Derivatives

Partial derivatives are a foundational concept in multivariable calculus. They measure the rate of change of a function with respect to one of its variables while keeping the other variables constant. For example, given a function of two variables, say, \(f(x, y)\), the partial derivative with respect to \(x\) is represented as \(\partial f / \partial x\) and tells you how \(f\) changes as \(x\) changes, with \(y\) kept fixed. Similarly, the partial derivative with respect to \(y\) is denoted as \(\partial f / \partial y\).
In the context of the exercise, we calculated partial derivatives \( (\partial y / \partial \theta)_r \) and \( (\partial y / \partial \theta)_x \), where subscripts indicate the variable held constant. When converting between coordinate systems or working with parameters, identifying these constants is crucial. This ensures we correctly apply the differentiation rules for each scenario.
A simple example: if \(z = x y\), then \(\partial z / \partial x = y\) and \(\partial z / \partial y = x\). By understanding these basics, you can tackle more complex problems like the one in the exercise.

Polar Coordinates

Polar coordinates provide a way to describe a point in a plane using the distance from a reference point (called the radius, \(r\)) and an angle from a reference direction (called the angle, \(\theta\)). This is different from Cartesian coordinates \((x, y)\), which use horizontal and vertical distances from a reference point.
Converting between these systems is straightforward using the equations:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

For our exercise, we differentiated these equations to understand how changes in \(\theta\) affect \(y\), both with \(r\) and \(x\) held constant.
For instance, keeping \(r\) constant and differentiating \(y = r \sin \theta\) with respect to \(\theta\), we get \((\partial y / \partial \theta)_r = r \cos \theta\). This shows how \(y\) changes with \(\theta\) when distance from the origin (\(r\)) doesn't change. Conversely, keeping \(x\) constant required more steps. Rewriting \(r\) in terms of \(x\), we differentiated again to find \((\partial y / \partial \theta)_x = x \sec^2 \theta\). These conversions are crucial in many applied fields like physics and engineering.

Differentiation

Differentiation is the process of finding the derivative of a function, which measures how a function changes as its input changes. In multivariable calculus, we often deal with functions of multiple variables. The process becomes partial differentiation, allowing us to see how the function changes with respect to each variable independently.
In the exercise, we applied differentiation rules to find partial derivatives in polar coordinates. We began by expressing \(y\) and \(r\) in terms of \(\theta\). Using differentiation, we then found how \(y\) changes as you vary \(\theta\).
Furthermore, differentiating implicitly, we found reciprocals of partial derivatives. We demonstrated this by finding \((\partial \theta / \partial y)_x\) from two methods: eliminating \(r\) and using differentials. Both methods led to the same result, validating our approach.
Understanding differentiation and its techniques form the backbone for many advanced mathematical concepts and real-world applications. Remember, whether it’s problem-solving or theoretical exploration, a firm grasp of differentiation ensures accurate and insightful results.