Transforming to 2d polar coordinates, dtheta/dx != 1/(dx/dtheta)

[galgal]

Hi, I'm puzzled by the following, and wondering what am I missing:
x = r cos(theta)
y = r sin(theta)

r = sqrt(x^2+y^2)
up to +-pi: theta = arctan(y/x)

we get:
dx/dtheta = -r sin(theta)
dtheta/dx = 1/(1+y^2/x^2) * (-y/x^2) = -y/(x^2 + y^2) = -sin(theta)/r

if we use the inverse function derivative (for a constant r) we get a contradiction, since 1/(r sin(theta) != sin(theta)/r

What has happened here?

Thanks a lot!

 

[HallsofIvy]

I'm not sure I understand your question. In general, \(\displaystyle \frac{1}{\frac{\partial y}{\partial x}}\) is NOT equal to \(\displaystyle \frac{\partial y}{\partial x}\). You seem to think it should be. Why?

 

[Dr.Peterson]

Hi, I'm puzzled by the following, and wondering what am I missing:
x = r cos(theta)
y = r sin(theta)

r = sqrt(x^2+y^2)
up to +-pi: theta = arctan(y/x)

we get:
dx/dtheta = -r sin(theta)
dtheta/dx = 1/(1+y^2/x^2) * (-y/x^2) = -y/(x^2 + y^2) = -sin(theta)/r

if we use the inverse function derivative (for a constant r) we get a contradiction, since 1/(r sin(theta) != sin(theta)/r

What has happened here?

The quick answer is that when you differentiated with respect to theta, you assumed r is constant; but when you differentiated with respect to x, you assumed y is constant. Those are different situations.

How much do you know about partial derivatives?

 

[galgal]

The quick answer is that when you differentiated with respect to theta, you assumed r is constant; but when you differentiated with respect to x, you assumed y is constant. Those are different situations.

How much do you know about partial derivatives?

Alright! Indeed, when I worked with x, r as independent variables and done these calculations, I also got -1/(r sin(theta). That's, of course, because I assumed again r is constant...
Thanks a lot!

HallsofIvy - I did flip the dy/dx: 1/(dy/dx) = dx/dy (assuming the function is invertible and the derivative exists)..

 

[HallsofIvy]

Alright! Indeed, when I worked with x, r as independent variables and done these calculations, I also got -1/(r sin(theta). That's, of course, because I assumed again r is constant...
Thanks a lot!

HallsofIvy - I did flip the dy/dx: 1/(dy/dx) = dx/dy (assuming the function is invertible and the derivative exists)..

Here, you are dealing with ordinary derivatives of y with respect to the single variable, x. Your original question was about partial derivatives. Those are not the same thing!