- #1

tompenny

- 15

- 3

- 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:)

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