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gtfitzpatrick
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cos\theta = 1 when \theta = 0
so that gives r = x, the initial condition?
so that gives r = x, the initial condition?
Solution for Expressing Derivatives in Terms of u
- Thread starter gtfitzpatrick
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SUMMARY
The discussion focuses on expressing the partial derivatives of the function \( w(r, \theta) = u(r \cos \theta, r \sin \theta) \) in terms of the derivatives of \( u \). Participants derived the equations \( \frac{\partial w}{\partial r} = \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta \) and \( \frac{\partial w}{\partial \theta} = \frac{\partial u}{\partial x} r \sin \theta - \frac{\partial u}{\partial y} r \cos \theta \). The conversation also emphasized the need to rewrite the initial conditions and the PDE \( y \frac{\partial u}{\partial x} - x \frac{\partial u}{\partial y} = 1 \) in polar coordinates. The final solution was confirmed as \( w(r, \theta) = \theta + f(r) \), where the initial condition \( u(x, 0) = 0 \) was applied to determine \( f(r) \).
PREREQUISITES- Understanding of partial derivatives and their applications in multivariable calculus.
- Familiarity with polar coordinates and their relationship to Cartesian coordinates.
- Knowledge of solving partial differential equations (PDEs).
- Experience with initial conditions in mathematical problems.
- Study the method of characteristics for solving first-order PDEs.
- Learn about the transformation of coordinates from Cartesian to polar systems.
- Explore the implications of initial conditions on the solutions of PDEs.
- Investigate the properties of functions defined in polar coordinates.
Students and professionals in mathematics, particularly those focusing on differential equations, multivariable calculus, and mathematical physics. This discussion is especially beneficial for anyone tackling problems involving polar coordinates and partial derivatives.
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