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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?

Thanks!

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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.

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  • $\begingroup$ 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
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    $\begingroup$ $K=C=-1$ 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
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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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