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Chapter 8: Problem 25

Consider a system of equations such as ( \(1 a, b\) ). Transform variables by defining \(x=r \cos \theta, \quad y=r \sin \theta .\) (a) Show that $$ r^{2}=x^{2}+y^{2}, \quad \theta=\arctan \frac{y}{x} . $$ (b) Show that $$ \frac{1}{2} \frac{d\left(r^{2}\right)}{d t}=x \frac{d x}{d t}+y \frac{d y}{d t} $$ (c) Verify that $$ r^{2} \frac{d \theta}{d t}=x \frac{d y}{d t}-y \frac{d x}{d t} $$

Short Answer

Expert verified
\( r^2 = x^2 + y^2 \), \( \theta = \arctan \left( \frac{y}{x} \right) \), \( \frac{1}{2} \frac{d(r^2)}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt} \), and \(r^2 \frac{d \theta}{dt} = x \frac{dy}{dt} - y \frac{dx}{dt}\).

Step by step solution

01

Transform the Variables

Use the definitions provided for the variable transformations: \(x = r \, \cos \theta \) and \( y = r \, \sin \theta \).
02

Show that \( r^2 = x^2 + y^2 \)

Square both equations from Step 1:\[ x^2 = (r \, \cos \theta)^2 = r^2 \, \cos^2 \theta \]\[ y^2 = (r \, \sin \theta)^2 = r^2 \, \sin^2 \theta \]Add these two equations: \(x^2 + y^2 = r^2 \, \cos^2 \theta + r^2 \, \sin^2 \theta \)Factor out \( r^2 \) from the right side: \( x^2 + y^2 = r^2 (\cos^2 \theta + \sin^2 \theta) \)Since \( \cos^2 \theta + \sin^2 \theta = 1 \): \( x^2 + y^2 = r^2 \), showing that \( r^2 = x^2 + y^2 \).
03

Show that \( \theta = \arctan \frac{y}{x} \)

Divide \(y = r \, \sin \theta \) by \(x = r \, \cos \theta \):\[ \frac{y}{x} = \frac{r \, \sin \theta}{r \, \cos \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta \]Thus, \(\theta = \arctan \left( \frac{y}{x} \right) \), showing that \( \theta = \arctan \left( \frac{y}{x} \right) \).
04

Show \( \frac{1}{2} \frac{d(r^2)}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt} \)

Start by differentiating \( r^2 = x^2 + y^2 \) implicitly with respect to time \(t\):\[ \frac{d}{dt}(r^2) = \frac{d}{dt}(x^2 + y^2) \]Using the chain rule:\[ 2r \frac{dr}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \]Divide through by 2:\[ r \frac{dr}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt} \]Since \( r^2 = x^2 + y^2 \), we have:\[ \frac{1}{2} \frac{d(r^2)}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt} \]
05

Verify \( r^2 \frac{d \theta}{dt} = x \frac{dy}{dt} - y \frac{dx}{dt} \)

Differentiate \( \theta = \arctan \left( \frac{y}{x} \right) \) implicitly with respect to time \( t \):\[ \frac{d \theta}{dt} = \frac{1}{1 + (\frac{y}{x})^2} \frac{d}{dt} \left( \frac{y}{x} \right) \]Using the quotient rule on \( \frac{d}{dt} \left( \frac{y}{x} \right) \):\[ \frac{d}{dt} \left( \frac{y}{x} \right) = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2} \]Thus,\[ \frac{d \theta}{dt} = \frac{1}{1 + (\frac{y}{x})^2} \cdot \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2} = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2 + y^2} \]Recall from Step 2: \( x^2 + y^2 = r^2 \). Therefore, we have:\[ \frac{d \theta}{dt} = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{r^2} \]Multiplying both sides by \( r^2 \), we get:\[ r^2 \frac{d \theta}{dt} = x \frac{dy}{dt} - y \frac{dx}{dt} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variable Transformation
In this context, variable transformation involves converting Cartesian coordinates \((x, y)\) into polar coordinates \((r, \theta)\). In polar coordinates, we represent a point's location based on its distance from the origin (\(r\)) and the angle (\(\theta\)) it forms with the positive x-axis. The following transformations are key:
  • \(x = r \cos \theta\)
  • \(y = r \sin \theta\)

By expressing \((x, y)\) in this way, we simplify the equations and expressions in some mathematical problems, making them easier to solve.
Implicit Differentiation
Implicit differentiation finds the derivative of a variable where it's not easily isolated. When dealing with relations like \((x^2 + y^2 = r^2)\), we differentiate both sides with respect to time \(t\), treating some variables as functions of \(t\).
For example, differentiating \((r^2)\) gives: \(\frac{d}{dt}(r^2) = \frac{d}{dt}(x^2 + y^2)\). By using the chain rule:
  • \(2r \frac{dr}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}\)

Implicit differentiation is crucial in this exercise to handle complex chain rules seamlessly.
Quotient Rule
The quotient rule helps differentiate a function that is the ratio of two differentiable functions. If \(u\) and \(v\) are functions of \(t\), the rule is:
  • \(\frac{d}{dt} \frac{u}{v} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}\)

In our exercise, consider \( \theta = \arctan \frac{y}{x}\): differentiating \( \frac{y}{x} \) with respect to \( t \):
  • \(\frac{d}{dt}(\frac{y}{x}) = \frac{x \frac{dy}{dt} - y \frac{dx}{dt}}{x^2}\)

Quotient rule allows handling complex fractions accurately in differentiation problems.
Polar Coordinates
Polar coordinates offer a different way to describe points in a plane using \( r \) and \( \theta \). Coordinates such as \( (r, \theta) \) simplify many problems, especially those involving symmetry. By the equations:
  • \( x = r \cos \theta \)
  • \(y = r \sin \theta \)
we can convert back and forth between coordinate systems as needed.
For example, showing \( r^2 = x^2 + y^2 \) stems from converting polar coordinates to Cartesian and simplifying.
Mathematical Proofs
Mathematical proofs involve logical steps to demonstrate the truth of a statement.
For instance, proving \( r^2 = x^2 + y^2 \) involves:
  • Square \(( x = r \cos \theta \) and \( y = r \sin \theta \)
  • Add the equations
  • Simplify using trigonometric identities like \( \cos^2 \theta + \sin^2 \theta = 1 \)

Each proof step is clear and logical, ensuring no gaps in the reasoning, making the concepts easier to understand with practice.

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