Finding a function in x,y from function in polar coordinates

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SUMMARY

The discussion focuses on converting a function defined in polar coordinates, v(r, θ) = 9 + 18cos(2θ) - 9sin(4θ), into Cartesian coordinates, u(x, y). The correct transformation is u(x, y) = 9 + 18cos(2arctan(y/x)) - 9sin(4arctan(y/x)). Further simplification is possible by applying double-angle formulas for sine and cosine, allowing for substitution of θ with arctan(y/x) to yield algebraic expressions.

PREREQUISITES
  • Understanding of polar and Cartesian coordinate systems
  • Familiarity with trigonometric identities, specifically double-angle formulas
  • Knowledge of the arctangent function and its properties
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study double-angle formulas for sine and cosine
  • Learn about the properties of the arctangent function
  • Practice converting functions between polar and Cartesian coordinates
  • Explore algebraic simplification techniques for trigonometric expressions
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Students studying calculus, particularly those focusing on polar coordinates and their applications in Cartesian systems, as well as educators teaching these concepts.

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Homework Statement


v is in polar coordinates and i want to fin u(x,y) knowing that v(r,theta)=u(rcos(theta),rsin(theta))
therefore, u(x,y)=v(sqrt(x^2+y^2), arctan(y/x))
v(r,theta) = 9+18cos(2(theta))-9sin(4(theta))
question: what is u(x,y)?

Homework Equations





The Attempt at a Solution



u(x,y)=9+18cos(2arctan(y/x))-9sin(4arctan(y/x))

Is this correct and can i simplify this more?
Thank you.
 
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It's correct, but you can simplify a lot more. I would use double-angle formulas to express [itex]\sin(4\theta)[/itex] and [itex]\cos(2\theta)[/itex] in terms of [itex]\sin(\theta)[/itex] and [itex]\cos(\theta)[/itex]. Then I could substitute [itex]\theta=\tan^{-1}(y/x)[/itex] and work the resulting expressions into algebraic expressions.
 

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