Convert Derivative in Polar to Cartesian

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    Derivative Polar
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SUMMARY

The discussion focuses on converting derivatives from cylindrical coordinates (r, θ, z) to Cartesian coordinates (x, y, z). The user seeks to apply the chain rule for this conversion, specifically addressing the derivative dm/dr. The correct approach involves using the relationship between the derivatives, where the partial derivative of r with respect to x is defined as ∂r/∂x = x/√(x² + y²). This method ensures accurate transformation of derivatives between the two coordinate systems.

PREREQUISITES
  • Understanding of cylindrical coordinates (r, θ, z)
  • Familiarity with Cartesian coordinates (x, y, z)
  • Knowledge of partial derivatives and the chain rule
  • Basic calculus concepts related to derivatives
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  • Study the application of the chain rule in multivariable calculus
  • Learn about converting between different coordinate systems in calculus
  • Explore examples of partial derivatives in polar and Cartesian coordinates
  • Investigate the implications of coordinate transformations in physics and engineering
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Students and professionals in mathematics, physics, and engineering who need to understand the conversion of derivatives between cylindrical and Cartesian coordinates.

mysubs
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Hello,

I want to convert from cylindrical (r,a,z) --to--> cartesian (x,y,z). However, I'm a not very confident about my level at this.

Say, I have dm/dr, a derivative in polar, and I want to find the equivalent expression in cartesian. (d: is partial derivative here).

I thought about using chain rule, but [r] depends on both [x] and [y]. Is doing it this way: dr=(xdx+ydy)/(x2+y2)1/2, correct? and If not why?

How should I work with this?
 
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Yes, that's the way.
So

[tex]\frac{\partial m}{\partial x} = \frac{\partial m}{\partial r} \frac{\partial r}{\partial x}[/tex]
where
[tex]\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}}[/tex]
 

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