# Assistance needed with partial derivatives, change on $x^{2}y$ surface with two variables?

$f(x,y) =x^{2}y=12$

$$\begin{cases} \partial_{x}f = 2xy+x^{2}\dot{y} \\ \partial_{y}f = (2x \dot{x}) y + x^{2} \end{cases}$$

now $$\partial_{x}(2,3) =12+4\dot{y}$$ and $$\partial_{y}(2,3) =12\dot{x}+4$$ but what are the terms $\dot{x}$ and $\dot{y}$? I need to calculate the change at point $x=2$ (so putting $x=2$ into $f$, I get point (2,3)). But I am unable make the leap here. Change in $f$ is? Is it a tuple $(\partial_{x}f, \partial_{y}f)$ or what does it mean? Help appreciated.

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I tried to fix your LaTeX. Is this what you intended? –  t.b. Oct 8 '11 at 16:40
if $x=2 \Rightarrow y=\frac{5}{2}$ –  pedja Oct 8 '11 at 16:41
@pedja: good notice, thansk. t.b. yes, that is right. –  hhh Oct 8 '11 at 16:45

I don't know what your $\dot{x}$ and $\dot{y}$ are but I assume from your calculation that they are supposed to be $\partial_x y = \dot{y}$ and $\partial_y x = \dot{x}$. But, these are both zero, assuming $x$ and $y$ are just the standard coordinates in $\mathbf{R}^2$ for example.
So, since $\partial_x y = 0$ and $\partial_y x = 0$ you get $\partial_x f = 2xy$ and $\partial_y f = x^2$. Now you can evaluate at each at the point $(2,3)$.
so $\nabla f = (12, 4)$ in $\mathbb R^{2}$? What does $12/4=3$ mean and what does $(12,4)$ actually mean? Change with respect to $x$ divided by the change with respect to $y$? How do you state the change here? And how would it change if you was not in $\mathbb R^{2}$, when is it btw save to assume $\mathbb R^{2}$? The change with $\dot{x}$ and $\dot{y}$ is to some extent more revealing, why are they not included in $\mathbb R^{2}$? –  hhh Oct 8 '11 at 16:41
..thinking this twice, only if I had a ball (which this is not, I would have one dim so the other dim would be depended on the other). Here no such case so I must have some sense of the surface in making this decision about indepedence. I am unsure with different spaces (not this easy) whether it is as easy to just assume $\dot{x}=0$ and $\dot{y}=0$, can be easy source of err. –  hhh Oct 8 '11 at 17:08