1
$\begingroup$

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 $

$\endgroup$
  • 3
    $\begingroup$ Your parametrization is wrong. it should be $x(t) = \dfrac{1}{\sqrt{2}}\tan t$ $\endgroup$ – dezdichado Aug 1 '17 at 16:23
  • $\begingroup$ @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. $\endgroup$ – projectilemotion Aug 1 '17 at 16:41
  • $\begingroup$ The function $(x,y)\mapsto A(x,y)$ has not been defined. So, what is $A(2,3)$? $\endgroup$ – Christian Blatter Aug 1 '17 at 17:28
  • $\begingroup$ @ChristianBlatter $A$ is the name of the point. A few lines down, the question says, “At point A...” $\endgroup$ – amd Aug 1 '17 at 18:58
2
$\begingroup$

Making my comment an answer, so this question no longer counts as unanswered.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.