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**renjer****Member**- Registered: 2006-04-29
- Posts: 50

Now here's a problem I have, it's often that I know the function of f but in this particular question, I do not know what f is at all. Is it okay if I let f be a function of my choice and prove it from there?

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 18,117

renjer,

It is given w=f(x,y), x=rcosθ and y=rsinθ

and you are asked to show

You are right, w is given as a function of x and y, but the function is not defined

Truth alone triumphs - Mundaka Upanishad.

Character is who you are when no one is looking.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

IT IS A GOOD QUESTION!!!

It's a question about coords, combination of derivatives, and vector calculus altogether.

It's about expressing the original combination of derivatives in another co-ords, usually orthogonal curvy ones. For example, polar co-ords are curvy, and dr and dθ at any given point are orthogonal to each other.

It's usually explained in a vector calculus book or a calculus book for physicians. Untill now, I haven't got a satisfactory explaination for this kind of transfer in any books.

However, I do find a particular solution for your coords and your case.

Define

After substition,

According to Chain Rule

Thus you can compute the right of your equation and use cos²θ+sin²θ=1 to prove it equates the left.

*Last edited by George,Y (2006-05-12 23:39:22)*

**X'(y-Xβ)=0**

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**renjer****Member**- Registered: 2006-04-29
- Posts: 50

I remember the chain rule telling me this:

So I'm quite unsure about your method from the chain rule onwards. Can you please explain further?

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 18,117

renjer,

In this link, at the bottom of the page, you would find the chain rule for partial derivatives, which I think, George has referred.

Truth alone triumphs - Mundaka Upanishad.

Character is who you are when no one is looking.

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