 #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:)
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:)
Last edited by a moderator: