Re: Please help with PDE in polar coordinates.

Or should I write this as differentiate respect to x only in my example.

$\displaystyle r^2=x^2+y^2\;\Rightarrow\;\frac{dr}{dx}=\frac {x}{r}+\frac{y}{r}\frac{dy}{dx}$

$\displaystyle x=r\cos\theta\;\Rightarrow dx=dr\cos\theta\;\Rightarrow\;\frac{dr}{dx}=\frac{ 1}{\cos\theta}=\frac{r}{x}$

$\displaystyle \Rightarrow\;\frac{r}{x}=\frac {x}{r}+\frac{y}{r}\frac{dy}{dx}$

$\displaystyle \Rightarrow\;\frac{dy}{dx}=\left(\frac{r}{x}-\frac{x}{r}\right)\frac{r}{y}$

$\displaystyle y=r\sin\theta$ and $\displaystyle x=r\cos\theta$

$\displaystyle \Rightarrow\;\frac{dy}{dx}=\left(\frac{r}{x}-\frac{x}{r}\right)\frac{r}{y} = \frac{r^2}{r^2 \cos\theta \sin\theta}-\frac{\cos\theta}{\sin\theta}=\tan\theta$

Re: Please help with PDE in polar coordinates.

If you are using implicit differentiation, that is different from partial differentiation, and you need a bit more:

$\displaystyle \begin{align*}& x = r\cos \theta \\ \Rightarrow & 1 = \dfrac{dr}{dx}\cos \theta - r \sin \theta \dfrac{d\theta}{dx} \\ \Rightarrow & \dfrac{dr}{dx} = \dfrac{1}{\cos\theta}+r\tan \theta \dfrac{d\theta}{dx}\end{align*}$

And you already have the other formula:

$\displaystyle \dfrac{dr}{dx} = \dfrac{x}{r} + \dfrac{y}{r}\dfrac{dy}{dx}$

Now, you can set these equal to each other, but you won't get $\displaystyle \tan \theta$.

Edit: But $\displaystyle y = r\sin \theta = r\sin\theta \dfrac{\cos\theta}{\cos\theta} = r\cos\theta \tan\theta = x\tan \theta$.

Differentiating we get $\displaystyle \dfrac{dy}{dx} = \tan \theta + x\sec^2\theta \dfrac{d\theta}{dx}$

Solving for $\displaystyle \dfrac{d\theta}{dx}$, we get $\displaystyle \dfrac{d\theta}{dx} = \dfrac{\cos^2\theta}{x} \dfrac{dy}{dx} - \dfrac{\sin\theta \cos \theta}{x}$

Now, you can plug that in:

$\displaystyle \begin{align*}\dfrac{dr}{dx} & = \dfrac{1}{\cos\theta}+r\tan \theta \dfrac{d\theta}{dx} \\ & = \dfrac{1}{\cos\theta}+r\tan \theta\left(\dfrac{\cos^2\theta}{x} \dfrac{dy}{dx} - \dfrac{\sin\theta \cos \theta}{x}\right)\end{align*}$

Now, setting those equal, you can solve for $\displaystyle \dfrac{dy}{dx}$.

Edit 2: But, setting those equal just yields the identity $\displaystyle \cos^2\theta + \sin^2\theta = 1$.

Re: Please help with PDE in polar coordinates.

Thanks, I am not trying to use implicit differentiation. I just try to right it out in the way it make sense.

I know in partial differentiation, I only use one of either x or y as a variable in rectangle coordinates, and use either r or $\displaystyle \theta$ in polar coordinates. Then treat the other variable as constant. That's the reason:

$\displaystyle \frac{\partial{r}}{\partial {x}}=\frac{x}{r}+\frac{y}{r}\frac{\partial{y}}{ \partial{x} }=\frac{x}{r}\;$ where $\displaystyle \;\frac{\partial{y}}{\partial {x}}= 0\;$ as y is treated as constant.

How should I write it if I want to treat $\displaystyle \;\frac{\partial{y}}{\partial {x}}= \tan\theta\;$?

Again, thanks for your help.

Re: Please help with PDE in polar coordinates.

Sorry, I just looked up Implicit differentiation which I forgot, ignore my last post.

But again, from my last example, it is clear $\displaystyle \frac{dr}{dx}=\frac{r}{x}=\frac{1}{\cos\theta}$ for polar cordinates.

Re: Please help with PDE in polar coordinates.

Quote:

Originally Posted by

**Alan0354** Thanks, I am not trying to use implicit differentiation. I just try to right it out in the way it make sense.

I know in partial differentiation, I only use one of either x or y as a variable in rectangle coordinates, and use either r or $\displaystyle \theta$ in polar coordinates. Then treat the other variable as constant. That's the reason:

$\displaystyle \frac{\partial{r}}{\partial {x}}=\frac{x}{r}+\frac{y}{r}\frac{\partial{y}}{ \partial{x} }=\frac{x}{r}\;$ where $\displaystyle \;\frac{\partial{y}}{\partial {x}}= 0\;$ as y is treated as constant.

How should I write it if I want to treat $\displaystyle \;\frac{\partial{y}}{\partial {x}}= \tan\theta\;$?

Again, thanks for your help.

You need to treat $\displaystyle \theta$ as a constant if you want $\displaystyle \dfrac{\partial y}{\partial x} = \tan\theta$.

Re: Please help with PDE in polar coordinates.

Quote:

Originally Posted by

**SlipEternal** You need to treat $\displaystyle \theta$ as a constant if you want $\displaystyle \dfrac{\partial y}{\partial x} = \tan\theta$.

I understand this, as it should be.

But, so far, if you use partial differentiation, you get $\displaystyle \frac{\partial r}{\partial x}=\frac {x}{r}$, you use implicit differentiation, $\displaystyle \frac{\partial r}{\partial x}\neq\frac {1}{\cos\theta}$. So what formulas can be applied in my example? How do I get to $\displaystyle \frac{\partial r}{\partial x}=\frac {r}{x}$

Re: Please help with PDE in polar coordinates.

I rechecked your posts. I don't understand what example you mean. There was no example given. This whole thread seems to be devoted to trying to find simple formulas for the derivatives of the change of coordinates formulas when you hold certain variables constant.

Re: Please help with PDE in polar coordinates.

Quote:

Originally Posted by

**SlipEternal** I rechecked your posts. I don't understand what example you mean. There was no example given. This whole thread seems to be devoted to trying to find simple formulas for the derivatives of the change of coordinates formulas when you hold certain variables constant.

I am referring to post #7 where I did not fix the theta.

Again, thanks for all your help. You are the only one that help me this much so far out of the 3 forums I posted.

Re: Please help with PDE in polar coordinates.

Well, if you fix $\displaystyle \theta$, then $\displaystyle \dfrac{\partial y}{\partial x} = \tan \theta$. If $\displaystyle \theta$ is 60 degrees (as you suggested), then $\displaystyle \dfrac{\partial y}{\partial x} = \tan(60^\circ) = \sqrt{3}$, just as I calculated in post #4.

If you don't fix any of the variables, then the partial derivatives may not be zero.

Re: Please help with PDE in polar coordinates.

Quote:

Originally Posted by

**SlipEternal** Well, if you fix $\displaystyle \theta$, then $\displaystyle \dfrac{\partial y}{\partial x} = \tan \theta$. If $\displaystyle \theta$ is 60 degrees (as you suggested), then $\displaystyle \dfrac{\partial y}{\partial x} = \tan(60^\circ) = \sqrt{3}$, just as I calculated in post #4.

If you don't fix any of the variables, then the partial derivatives may not be zero.

Yes, but is there a differential formula that I can point to that get me the answer without all the "if"? I am just looking for a calculus formula way to solve this.

Thanks

Re: Please help with PDE in polar coordinates.

Quote:

Originally Posted by

**Alan0354** Yes, but is there a differential formula that I can point to that get me the answer without all the "if"? I am just looking for a calculus formula way to solve this.

Thanks

I don't know what you are looking for then. The "if"s are necessary. Whatever differential formula you are pointing to either looks at a special case (when some variable is fixed), or it is more complex than what you want to hear. But, I can't help you force the mathematics to do what you want. The mathematics works just as I showed you. Either the partial derivatives are fixed or they are not. If they are fixed, you get the formula you wrote in the original post. If they are not fixed, then you don't get the formula you wrote in your original post. So, whatever answer you are looking for may not exist.

Re: Please help with PDE in polar coordinates.

Thanks for all your help.