0
$\begingroup$

Let $O$ be an open subset of $\mathbb R^2$ and suppose the function $f:O\to\mathbb R$ is continuous at the point $(x_0,y_0)$ in $O$.

Define tangent plane as $\phi(x,y)=a+b(x-x_0)+c(y-y_0)$ where $a,b,c$ are real numbers, which has the property that

$$\lim_{(x,y)\to(x_0,y_0)}\frac{f(x,y)-\phi(x,y)}{\sqrt{(x-x_0)^2+(y-y_0)^2}}=0$$

clearly that $\phi $ is a first order approximation of $f$.

I know that if $f$ is continuously differentiable, then the tangent plane at the point $(x_0,y_0)$ is defined by $$\phi(x,y)=f(x_0,y_0)+\frac{\partial f}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)$$

However I don't know how to prove the converse theorem:

If $\phi$ is a tangent plane of a function $f:O\to\mathbb R$ at the point $(x_0,y_0,f(x_0,y_0))$ (which at the point is continuous?), then $f$ has first order partial derivatives at $(x_0,y_0)$ and $a=f(x_0,y_0)$, $b=\frac{\partial f}{\partial x}(x_0,y_0)$, $c=\frac{\partial f}{\partial y}(x_0,y_0)$ such that the tangent plane is unique. Also $f$ has directional derivatives in all direction at the point $(x_0,y_0)$

The bracket (which at the point is continuous?) means that I don't know whether it is a necessary condition for the theorem to hold, since I do not know whether tangent plane exist implies continuous at the point.

| cite | improve this question | | | | |
$\endgroup$
0
$\begingroup$

If you can draw a tangent plane to a multivaraite function $f$ at $(x_0,y_0)$, then this function is surely differentiable at $(x_0,y_0)$. If $f$ is differentiable at $(x_0,y_0)$, then it is also continuous at that point.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Sorry that I am not familiarize with differentiability and can you give me an analytically or brief proof please? $\endgroup$ – Y.H. Chan Jul 23 '14 at 14:36
0
$\begingroup$

You can find all the directional derivatives by the inner product of the vector $\left( \frac{\partial f(x_0,y_0)}{\partial x}, \frac{\partial f(x_0,y_0)}{\partial y} \right)$ and the direction vector $(v_1,v_2)$ (please normalize the direction vector first).

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ But how can you ensure that the derivatives exist? And why derivatives exist implies all directional derivatives exist? (I mean in the case that the derivatives is not continuous) $\endgroup$ – Y.H. Chan Jul 23 '14 at 14:34

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.