As title is saying I need to show connection of gradient and level curve for function


at point


First I find gradient of function:

$\nabla f(x,y) = 4x-2y $

At point A:

$\nabla f(2,3) = (8,-6)$

Then for $f(x,y) = f(2,3)$:


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 $

  • 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

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

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


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