For $f(x,y)=\ln(x-y)$, sketch the level curves for the values $k=-1,0,1$

If we set $f(x,y)=k$, then for the values $k=-1,0,1:$

$$ y= \begin{cases} x-e^{-1}\\ x-1\\ x-e^{} \end{cases} $$

Graphing these gives:

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  • 1
    $\begingroup$ For me it is correct. $\endgroup$ – user376343 May 30 at 9:04

$$ \ln(x-y)=k $$

$$ x-y=e^{k} $$

$$ y=x-e^{k} $$

Therefore, the level curves are straight lines with the slope of $1$ and $y$-intercept $-e^k$

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