Let $F_1,F_2:\mathbb{R^2} \to \mathbb{R}$ be the functions $ F_1(x_1,x_2)={\frac{-x_2}{x_1^2+x_2^2}}$ and $ F_2(x_1,x_2)={\frac{x_1}{x_1^2+x_2^2}}$. Which of the following is correct?

  1. ${\frac{\partial F_1}{\partial x_2}}$ =${\frac{\partial F_2}{\partial x_1}}$

  2. There exists a function $f: \mathbb{R^2}\setminus\{(0,0)\}\to \mathbb{R}$ such that ${\frac{\partial f}{\partial x_1}}=F_1$ and ${\frac{\partial f}{\partial x_2}}=F_2$

  3. There exists no function $f:\mathbb{R^2}\setminus\{(0,0)\}\to \mathbb{R}$ such that ${\frac{\partial f}{\partial x_1}}=F_1$ and ${\frac{\partial f}{\partial x_2}}=F_2$

  4. There exists a function $f:D\to \mathbb{R}$ where $D$ is the open disc of radius $1$ centred at $(2,0)$, which satisfies ${\frac{\partial f}{\partial x_1}}=F_1$ and ${\frac{\partial f}{\partial x_2}}=F_2$ on $D$.

I got first option.

2 $\to$ suppose such function exist then ${\frac{\partial f}{\partial x_1}}=F_1$ gives $f=-\tan^{-1} \left(\frac{x_1}{x_2}\right) +V(x_2)$ where $V(x_2)$ is some function of $x_2$ and ${\frac{\partial f}{\partial x_2}}=F_2$ gives $V'(x_2)=0$ hence $V(x_2)=C$, a constant .

hence such function $f=-\tan^{-1}\left(\frac{x_1}{x_2}\right) +C$ exists.

3 $\to$ not true

how to conclude for 4

is my explanation for 3 is perfect?

  • $\begingroup$ Isn't writing all along $\;x_1,\,x_2\;$ way more cumbersome than writing simply $\;x,\,y\;$ ? Anyway, about 2-3: if what you said is true, then the line integral of the given vector field on the unit circle would be zero, right? But it isn't... $\endgroup$
    – DonAntonio
    May 15, 2017 at 11:46

1 Answer 1


Your answers for $2$ and $3$ are incorrect, since $-\tan^{-1}(x_1/x_2)$ is not defined on all of $\mathbb{R}^2\backslash \{0\}$. Even if you were to define the function at the undefined points, it would not be differentiable. Indeed, there is no such function, since $$\oint_{S^1} F\cdot dr = 2\pi$$ where $F = F_1\, dx+F_2 \, dy$, and where the integral is taken counterclockwise. Since the integral of this vector field around a closed loop is nonzero, $F$ is not conservative, so it's not equal to the gradient of any scalar function. This is true even tough the $(F_1)_y-(F_2)_x=0$, because the domain of $F$ is not simply connected.

For $4$, such a function does exist, because $F$ is defined on all of $D$, and $D$ is simply connected. In fact, $-\tan^{-1}(x_1/x_2)$ does the trick.

  • $\begingroup$ $-tan^{-1}\frac{x_1}{x_2}$ is not defined at which points $\endgroup$ May 16, 2017 at 4:20
  • $\begingroup$ Points where $x_2=0$. $\endgroup$
    – florence
    May 16, 2017 at 5:09
  • $\begingroup$ I am not getting $tan^{-1}\infty=\frac{\pi}{2},...$ $\endgroup$ May 17, 2017 at 3:45
  • $\begingroup$ If you define the function to be $\pi/2$ at those points, then the function will not be continuous. From one direction it will approach $pi/2$, and from the other it will approach $-\pi/2$. Either way, the fact that the integral around a closed loop is nonzero proves that $F$ is not conservative. $\endgroup$
    – florence
    May 17, 2017 at 6:03

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