# Show connection of gradient and level curve for function $f(x,y)=2x^2-y^2$ at point A(2,3)?

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:

$\nabla f(2,3) * \vec{r}'(t) = (8,-6)(9\sqrt{2}, 3\sqrt{2}) = 54\sqrt{2} \neq 0$

• Your parametrization is wrong. it should be $x(t) = \dfrac{1}{\sqrt{2}}\tan t$ – dezdichado Aug 1 '17 at 16:23
• @dezdichado I strongly suggest you add your comment as an answer, since it does give $\nabla f(2,3)\cdot \vec{r}'(t)=0$ as required. – projectilemotion Aug 1 '17 at 16:41
• The function $(x,y)\mapsto A(x,y)$ has not been defined. So, what is $A(2,3)$? – Christian Blatter Aug 1 '17 at 17:28
• @ChristianBlatter $A$ is the name of the point. A few lines down, the question says, “At point A...” – amd Aug 1 '17 at 18:58

Your parametrization is wrong, it should be: $$\vec{r}(t) = \Big(\dfrac{1}{\sqrt{2}}\tan t,\, \sec t\Big)$$