- #1
unscientific
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Homework Statement
Change of coordinates from rectangular (x,y) to polar (r,θ). Not sure what's wrong with my working..
What's wrong with my Jacobian of polar coordinates?
- Thread starter unscientific
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SUMMARY
The discussion focuses on the differentiation errors encountered when converting from rectangular coordinates (x, y) to polar coordinates (r, θ). The correct differentiation of y = x tan(θ) is given as ∂y/∂θ = x sec²(θ) + ∂x/∂θ tan(θ), highlighting that x is a function of θ. Additionally, to find ∂x/∂r, the equation r² = x² + y² must be differentiated with respect to r while holding θ constant, leading to the equation 2r = 2x ∂x/∂r + 2y ∂y/∂r. The discussion emphasizes starting from the definitions x = r cos(θ) and y = r sin(θ) for clarity in differentiation.
PREREQUISITES- Understanding of polar coordinates and their relationship to rectangular coordinates
- Knowledge of partial differentiation and its application in multivariable calculus
- Familiarity with trigonometric identities, particularly secant and tangent functions
- Basic algebraic manipulation skills for handling equations involving multiple variables
- Study the derivation of polar coordinates from rectangular coordinates in detail
- Learn about partial derivatives and their applications in multivariable calculus
- Explore trigonometric identities and their relevance in calculus
- Practice problems involving differentiation of functions with multiple variables
Students studying calculus, particularly those focusing on multivariable calculus and coordinate transformations, as well as educators teaching these concepts.
Change of coordinates from rectangular (x,y) to polar (r,θ). Not sure what's wrong with my working..
- #2
pasmith
Science Advisor
Homework Helper
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Your error is that you have differentiated y = x \tan\theta incorrectly. You should get
\frac{\partial y}{\partial \theta} = x\sec^2\theta + \frac{\partial x}{\partial \theta}\tan\theta
because x is also a function of \theta.
Also, to work out \partial x/\partial r, you would need to differentiate
r^2 = x^2 + y^2
with respect to r with \theta held constant, which again gives
2r = 2x\frac{\partial x}{\partial r} + 2y\frac{\partial y}{\partial r}
because y is also a function of r.
If you're trying to find derivatives with respect to r and \theta, it's best to start from x = r\cos\theta, y = r\sin\theta.
\frac{\partial y}{\partial \theta} = x\sec^2\theta + \frac{\partial x}{\partial \theta}\tan\theta
because x is also a function of \theta.
Also, to work out \partial x/\partial r, you would need to differentiate
r^2 = x^2 + y^2
with respect to r with \theta held constant, which again gives
2r = 2x\frac{\partial x}{\partial r} + 2y\frac{\partial y}{\partial r}
because y is also a function of r.
If you're trying to find derivatives with respect to r and \theta, it's best to start from x = r\cos\theta, y = r\sin\theta.
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