As title is saying I need to show connection of gradient and level curve for function
$f(x,y)=2x^2-y^2$
at point
$A(2,3)$.
First I find gradient of function:
$\nabla f(x,y) = 4x-2y $
At point A:
$\nabla f(2,3) = (8,-6)$
Then for $f(x,y) = f(2,3)$:
$2x^2-y^2=-1$
We see that this equation of hyperbola. So I am having problem with parameterization of this equation. Because this is vertical hyperbola:
$ x=b \tan(t) $
$ y=a \sec(t) $
I have tried to do it this way. Vector $\vec{r}$ is equal to:
$ \vec{r} = (x(t), y(t)) = (\sqrt{2} \tan(t), \sec(t)) $
So how to get tangent vector from r? Is this right way?
$ \vec{r}'(t) = (\sqrt{2}sec^2(t), sec(t)tan(t)) $
We know that $\vec{r}(t) = (2,3)$
$ (\sqrt{2} \tan(t), \sec(t)) = (2,3) $
So we get $\tan(t) = \sqrt{2} $ and $\sec(t) = 3$
Then $ \vec{r}'(t) = (9\sqrt{2}, 3\sqrt{2}) $
As theorem states:
$ \nabla f(2,3) * \vec{r}'(t) = 0 $
But I get:
$ \nabla f(2,3) * \vec{r}'(t) = (8,-6)(9\sqrt{2}, 3\sqrt{2}) = 54\sqrt{2} \neq 0 $
Making my comment an answer, so this question no longer counts as unanswered.
Your parametrization is wrong, it should be: $$\vec{r}(t) = \Big(\dfrac{1}{\sqrt{2}}\tan t,\, \sec t\Big)$$