Function $f$ has two variables ($y$ and $z$). Show that $\displaystyle \nabla f = {\partial f \over \partial y}\ \hat j \ + \ {\partial f \over \partial z} \ \hat k$ transforms as a vector under rotations. Hint : use $${\partial f\over \partial \overline y} = {\partial f\over \partial y} { \partial y\over\partial \overline y} + {\partial f\over \partial z} { \partial z\over\partial \overline y} \tag{+} $$ and analogous of this for $\displaystyle {\partial f\over \partial \overline z}$
I think I have done this correctly but please check my solution.
I know $\overline y = y\cos \theta + z \sin \theta$ and $\overline z = -y\sin \theta + z\cos \theta$
I solved for $y$ and $z$ as a function of $\overline y$ and $\overline z$.
I got $y = \overline y \cos \theta - \overline z \sin \theta$ and $z = \overline y \sin \theta + \overline z \cos \theta$
Now, as per the hint I calculated
$${ \partial y\over\partial \overline y} = \cos \theta\ \ \& \ \ { \partial z\over\partial \overline y} = \sin\theta$$
From this and $(+)$ I got, $${\partial f\over \partial \overline y} = {\partial f\over \partial y} \cos \theta + {\partial f\over \partial z} \sin\theta $$
Similarly I got,
$${\partial f\over \partial \overline z} = {\partial f\over \partial z} \cos \theta - { \partial y\over\partial \overline z}\sin\theta$$
So now I think transformed gradient would be
$$\overline\nabla f =\left({\partial f\over \partial y} \cos \theta + {\partial f\over \partial z} \sin\theta,{\partial f\over \partial z} \cos \theta - { \partial y\over\partial \overline z}\sin\theta\right) $$
Which is similar to a vector. Hence done.
- I can't seems to visualise this question. Please provide some intuitive insight on this question ?
- Where does we get $(+)$ ?
- Is this correct ?