let us consider I have a function $f(x,y)$, I know a value $f_0=f(x=0,y=0)$ but I don't know how $f$ is done. By the way I know his partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. I want to plot $f$ in a 3D plot respect to $x$ and $y$ in an interval $x_L\leq x\leq x_U$ and $y_L\leq y\leq y_U$.

Which is the relation that I should use to do that? How to evaluate $f$?

I tought about:


which depends on derivatives. Am i right?


After some calculation: $$\frac{\partial f}{\partial x}=a$$ with $a$ scalar. $$\frac{\partial f}{\partial y}=b$$ with $b$ scalar.

And $f_0=c$ with $c$ scalar

  • $\begingroup$ Do you know what the partial derivatives are everywhere, or just at the origin? $\endgroup$ – Arthur Aug 20 '18 at 13:55
  • $\begingroup$ I know them everywhere or at least in the interval of interest. $\endgroup$ – it8 Aug 20 '18 at 14:00

If you know the partial derivatives of the function everywere you can come back to $f(x,y)$ by integration. You just have to be careful when integrating a $2$ variable function with respect to only one variable because the constant of integration will be a function of the variable you didn't integrate on. So $$\int\partial_xf(x,y)dx = f_1(x,y)+g(y)\color{red}{+c_1}\\ \int\partial_yf(x,y)dy = f_2(x,y)+h(x)\color{red}{+c_2}$$ I marked the constants in red because they can be incorporated in the functions $g,h$. This two integral have to be the same because they represent the same function, so, with the value of the function in $(0,0)$ you just have to solve the system $$\begin{cases}f_1(x,y)+g(y) = f_2(x,y)+h(x) \\ f(0,0) = f_0\end{cases}$$


Your case is very easy! If both partial derivatives are just constants, your function has to be linear in both $x$ and $y$. Integrating as I showed earlier you get $$\int a dx = ax + g(y) \\\int b dy = by + h(x)$$ by confronting the two we get that $$f(x,y) = ax +by +f_0$$ because to be equal the two integrated function you have $$g(y) = by \\ h(x) = ax$$ obviously there only remains the constant which is clearly $f_0$. So in the end $$f(x,y) = ax+by+c$$

  • $\begingroup$ I add an edit to the post. Should be useful $\endgroup$ – it8 Aug 20 '18 at 14:25
  • 1
    $\begingroup$ Is it been useful in any way? $\endgroup$ – Davide Morgante Aug 21 '18 at 18:14
  • 1
    $\begingroup$ Yes very useful, thank you. Rating this as best answer of course. $\endgroup$ – it8 Aug 22 '18 at 9:32

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