Higher order derivatives using the chain rule

[tompenny]

Homework Statement

I need to show that higher order derivatives is applicable on a rotation in the plane

Relevant Equations

[u v] = [ cos θ -sin θ * [x y]
sin θ cos θ]

∂2f∂x2+∂2f∂y2=∂2fu2+∂2fv2

Mentor note: Fixed the LaTeX in the following
I have the following statement:

\begin{cases} u=x \cos \theta - y\sin \theta \\ v=x\sin \theta + y\cos \theta \end{cases}

I wan't to calculate:

$$\dfrac{\partial^2}{\partial x^2}$$

My solution for $\dfrac{\partial^2}{\partial x^2}$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v}$
$\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} f = \left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right) f = \cos \theta \dfrac{\partial f}{\partial u} - \sin \theta\dfrac{\partial f}{\partial v}$.Proceeding on to the second derivative I get:$\dfrac{\partial^2}{\partial x^2} = \left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right)\left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right) = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} - 2\sin \theta \cos \theta \dfrac{\partial^2}{\partial u\partial v} + \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$.Similarly I get $\dfrac{\partial^2}{\partial y^2} = \left( \sin \theta \dfrac{\partial}{\partial u} \cos \theta\dfrac{\partial}{\partial v} \right)\left( \sin\theta \dfrac{\partial}{\partial u} \cos \theta\dfrac{\partial}{\partial v} \right) = \sin^2 \theta \dfrac{\partial^2}{\partial u^2} - 2\sin \theta \cos \theta \dfrac{\partial^2}{\partial u\partial v} + \cos^2 \theta \dfrac{\partial^2}{\partial v^2}$.

But how do I calculate $\dfrac{\partial^2}{\partial u^2}$ , $\dfrac{\partial^2}{\partial v^2}$ ?

Any tips would be greatly appreciated:)

 

[PeroK]

You maybe need to try Latex:

https://www.physicsforums.com/help/latexhelp/

 

[epenguin]

You maybe need to try Latex:

https://www.physicsforums.com/help/latexhelp/

I think he is trying, but should first try replacing $ with $$, then we'll see.

 

[PeroK]

Homework Statement:: I need to show that higher order derivatives is applicable on a rotation in the plane
Relevant Equations:: [u v] = [ cos θ -sin θ * [x y]
sin θ cos θ]

I have the following statement:

$$\begin{cases} u=x \cos \theta - y\sin \theta \\ v=x\sin \theta + y\cos \theta \end{cases}$$

My solution for $\dfrac{\partial^2}{\partial x^2}$

$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v}$$

$$\dfrac{\partial f}{\partial x} = \dfrac{\partial}{\partial x} f = \left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right) f = \cos \theta \dfrac{\partial f}{\partial u} - \sin \theta\dfrac{\partial f}{\partial v}$$

Proceeding on to the second derivative I get:

$$\dfrac{\partial^2}{\partial x^2} = \left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right)\left( \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v} \right) = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} - 2\sin \theta \cos \theta \dfrac{\partial^2}{\partial u\partial v} + \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$$.Similarly I get $$\dfrac{\partial^2}{\partial y^2} = \left( \sin \theta \dfrac{\partial}{\partial u} \cos \theta\dfrac{\partial}{\partial v} \right)\left( \sin\theta \dfrac{\partial}{\partial u} \cos \theta\dfrac{\partial}{\partial v} \right) = \sin^2 \theta \dfrac{\partial^2}{\partial u^2} - 2\sin \theta \cos \theta \dfrac{\partial^2}{\partial u\partial v} + \cos^2 \theta \dfrac{\partial^2}{\partial v^2}$$.

But how do I calculate $$\dfrac{\partial^2}{\partial u^2}$$ , $$\dfrac{\partial^2}{\partial v^2}$$ ?

Any tips would be greatly appreciated:)

You don't have to calculate others. You can simplify what you have.

However, you have a mistake in the first line. This is wrong:

$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v}$$

 

[tompenny]

You don't have to calculate others. You can simplify what you have.

However, you have a mistake in the first line. This is wrong:

$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} - \sin \theta\dfrac{\partial}{\partial v}$$

Thank you so much for helping me.. should the first line be:
$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} + \sin \theta\dfrac{\partial}{\partial v}$$ ??

 

[Mark44]

@tompenny, this site uses MathJax, which accepts some LaTeX stuff. The delimiters are pairs of $$ characters (standalone LaTeX) or ## (inline).

 

[PeroK]

Thank you so much for helping me.. should the first line be:
$$\dfrac{\partial}{\partial x} = \dfrac{\partial u}{\partial x}\dfrac{\partial}{\partial u} + \dfrac{\partial v}{\partial x}\dfrac{\partial}{\partial v} = \cos \theta \dfrac{\partial}{\partial u} + \sin \theta\dfrac{\partial}{\partial v}$$ ??

Yes. What about $$\dfrac{\partial^2}{\partial x^2}$$.

Hint: write everything out carefully.

 

[tompenny]

$$\dfrac{\partial^2}{\partial x^2} = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} + 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$$

Is that correct? :)

Can you give me any tips on how to calculate $$\dfrac{\partial^2}{\partial u^2} $$

MAny thanks for all your help:)

 

[PeroK]

$$\dfrac{\partial^2}{\partial x^2} = \cos^2 \theta \dfrac{\partial^2}{\partial u^2} + 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \sin^2 \theta \dfrac{\partial^2}{\partial v^2}$$

Is that correct? :)

Can you give me any tips on how to calculate $$\dfrac{\partial^2}{\partial u^2} $$

MAny thanks for all your help:)

Why not try $$\dfrac{\partial^2}{\partial y^2} $$

 

[tompenny]

then I get:
$$\dfrac{\partial^2}{\partial y^2} = \sin^2 \theta \dfrac{\partial^2}{\partial u^2} - 2sin \theta\cos \theta\dfrac{\partial}{\partial u\partial v}+ \cos^2 \theta \dfrac{\partial^2}{\partial v^2}$$

Can I somewhat use this to calculate $$\dfrac{\partial^2}{\partial u^2}$$ ?

Thank you:)

 

[PeroK]

What about adding those two together?

 

[tompenny]

Hahaha.. I finally got it! the mixed partial cancels and if I use the pythagorean trigonometric identity on what's left I get the answer!

I can't explain how much you've made my day! <3