0
$\begingroup$

I'm trying to show that

$$ \frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy = d\left(\tan^{-1}\left(\frac{x}{y}\right)\right) $$

but am having trouble figuring out exactly how to approach the problem. I've tried a few tricks with switching to polar coordinates or trying to make use of right triangles, but have had no luck. Could anyone point me in the right direction here?

$\endgroup$
  • $\begingroup$ You don't need polar or right triangles for this. $\endgroup$ – user_of_math Jul 18 '14 at 13:26
  • $\begingroup$ Do you know the derivative of $\operatorname{tan}^{-1}$? $\endgroup$ – Giulio Bresciani Jul 18 '14 at 13:27
  • $\begingroup$ This is not exactly true, because the LHS is defined in $\mathbb{R}^2-\{(0,0)\}$, while the RHS is defined in $\{(x,y)\in\mathbb{R}^2:y\neq0\}$. $\endgroup$ – enzotib Jul 18 '14 at 14:38
5
$\begingroup$

Hint: $$ \frac{d}{dt}\tan^{-1}(t) = \frac{1}{1+t^2} $$ $$ d f(x,y) = \frac{\partial f(x,y)}{\partial x} dx + \frac{\partial f(x,y)}{\partial y} dy $$

$\endgroup$
2
$\begingroup$

$$\begin{align} \frac{y}{x^2+y^2} \, dx - \frac{x}{x^2+y^2} \, dy &=\frac{1}{(\frac{x}{y})^2+1}\frac{1}{y}dx+ \frac{1}{(\frac{x}{y})^2+1}(-\frac{x}{y^2}) dy\\ &=\frac{1}{(\frac{x}{y})^2+1}(\frac{1}{y}dx+(-\frac{x}{y^2})dy)\\ &=\frac{1}{(\frac{x}{y})^2+1}d(\frac{x}{y})\\ &= d\left(\tan^{-1}\left(\frac{x}{y}\right)\right) \end{align}$$

$\endgroup$
-1
$\begingroup$

Using the Hint from above: \begin{align} d\left(\tan^{-1}\left(\frac{x}{y}\right)\right) &= \frac{\partial \tan^{-1}\left(\frac{x}{y}\right)}{\partial\left(\frac{x}{y}\right)}\frac{\partial\left(\frac{x}{y}\right)}{\partial x} dx + \frac{\partial \tan^{-1}\left(\frac{x}{y}\right)}{\partial\left(\frac{x}{y}\right)}\frac{\partial\left(\frac{x}{y}\right)}{\partial y} dy \\ &= \frac{1}{1+\frac{x^2}{y^2}}\frac{1}{y}dx + \frac{1}{1+\frac{x^2}{y^2}}\left(-\frac{x}{y^2} \right)dy\\ &=\frac{y}{x^2+y^2}dx-\frac{x}{y^2+x^2}dy \end{align}

$\endgroup$

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.