Find the orthogonal family of curves to the level lines of $f(x,y)=xy-1$.

My attemp:

The level lines are $$\left\lbrace(x,y)\in\mathbb R^2:xy-1=K\quad\text{for some constant }K\right\rbrace\text.$$

Differentiating both sides: $$\mathfrak F:y+xy'=0\Rightarrow\mathfrak F^\perp:y-\frac x{y'}=0\Rightarrow\mathfrak F^\perp:\frac{y^2}2-\frac{x^2}2=c\Rightarrow\boxed{\mathfrak F^\perp:y^2-x^2=C,\;C\in\mathbb R}\text.$$

Is that correct?



Yes, your answer is correct.

You can also verify your result by graphing those curves and see the level curves are orthogonal to each other at every point of intersection.

  • $\begingroup$ Thanks! I have plotted some functions and they seem to be orthogonal in the points I chose. Now I think, there is any problem if $K=C=-1$? Should we distinguish cases? $\endgroup$ – manooooh Aug 28 '18 at 16:26
  • 1
    $\begingroup$ $K=C=-1$ will give you the coordinate axes and the bisectors and they are also acceptable. They intersect at the origin and are perpendicular. $\endgroup$ – Mohammad Riazi-Kermani Aug 28 '18 at 16:29

Yes it is correct indeed we have

  • $xy-1=K \implies ydx+xdy=0 \quad m_1=\frac{dy}{dx}=-\frac{y}{x}$

  • $\frac{y^2}2-\frac{x^2}2=c \implies ydy-xdx=0 \quad m_2=\frac{dy}{dx}=\frac{x}{y}$

and $$m_1\cdot m_2=-\frac{y}{x}\cdot \frac{x}{y}=-1$$


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