Solving Polar Coordinates in System of Equations

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SUMMARY

This discussion focuses on solving a system of equations in polar coordinates, specifically examining the transformation from Cartesian coordinates (x, y) to polar coordinates (r, θ). The equations x' = f(x, y) and y' = g(x, y) are analyzed, leading to the expression r' = f(r cos θ, r sin θ) cos θ + g(r cos θ, r sin θ) sin θ. The derivation of the sine and cosine components is clarified through the relationships x = r cos θ and y = r sin θ, demonstrating how these trigonometric functions arise in the context of polar coordinate transformations.

PREREQUISITES
  • Understanding of polar coordinates and their relationship to Cartesian coordinates
  • Familiarity with vector functions and their derivatives
  • Knowledge of basic calculus, particularly differentiation
  • Experience with systems of equations in mathematical modeling
NEXT STEPS
  • Study the derivation of polar coordinates from Cartesian coordinates in detail
  • Learn about the Jacobian transformation in multivariable calculus
  • Explore applications of polar coordinates in physics and engineering problems
  • Investigate numerical methods for solving systems of equations in polar coordinates
USEFUL FOR

Mathematicians, physics students, engineers, and anyone interested in understanding the application of polar coordinates in solving systems of equations.

onie mti
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given that

x'=f(x,y)
y'=g(x,y)
iff the vector function (r, θ) is a sloution of the system
r'=f(rcosθ,rsinθ)cosθ +g(rcosθ,rsinθ)sinθ

am trying to show that this is true but i just don't get where the si
nθ and cosθ come from, how do i get to that
 
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Are x and y functions of some other variable, say t?
 
onie mti said:
given that

x'=f(x,y)
y'=g(x,y)
iff the vector function (r, θ) is a sloution of the system
r'=f(rcosθ,rsinθ)cosθ +g(rcosθ,rsinθ)sinθ

am trying to show that this is true but i just don't get where the si
nθ and cosθ come from, how do i get to that

Operating in polar coordinates You have $\displaystyle x= r\ \cos \theta$ and $\displaystyle y= r\ \sin \theta$, so that is...

$\displaystyle r^{\ '} = \frac{d}{d t} \sqrt{x^{2} + y^{2}} = \frac{1}{2}\ \frac{2\ x\ x^{\ '} + 2\ y\ y^{\ '}}{\sqrt{x^{2} + y^{2}}} = f(r\ \cos \theta, r\ \sin \theta)\ \cos \theta + g(r\ \cos \theta, r\ \sin \theta)\ \sin \theta $

Kind regards

$\chi$ $\sigma$
 

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