- #31
gtfitzpatrick
- 372
- 0
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
- Start date
-
- Tags
- Derivatives Terms
Click For Summary
Homework Help Overview
The discussion revolves around expressing derivatives of a function \( w(r, \theta) \) in terms of another function \( u(x, y) \) within the context of a partial differential equation (PDE). The original poster seeks to compute the partial derivatives \( \frac{\partial w}{\partial r} \) and \( \frac{\partial w}{\partial \theta} \) based on the relationships between the variables in polar coordinates.
Discussion Character
- Mixed
Approaches and Questions Raised
- Participants explore the relationships between the derivatives of \( w \) and \( u \) using the chain rule, with attempts to express these derivatives in terms of \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \). There are questions about the correctness of the derivatives and how to substitute variables appropriately. Some participants suggest simplifying the expressions using relationships between \( x, y \) and \( r, \theta \).
Discussion Status
The discussion is ongoing, with participants providing guidance on computing derivatives and expressing them in terms of the original variables. There is a recognition of the need to rewrite the PDE in polar coordinates, and some participants are exploring the implications of initial conditions on the solution.
Contextual Notes
Participants note the constraints of the problem, including the initial condition \( u(x, 0) = 0 \) and the range \( 0 < x < \infty \). There is also a discussion about the implications of these conditions in the context of polar coordinates.
Similar threads
- · Replies 3 ·
- · Replies 18 ·
- · Replies 12 ·