How to draw or describe Level curves of $xln (y^2-x)$












0












$begingroup$


How to find the level curves of $$f(x,y)=xln (y^2-x)$$



If I had $f(x,y)=ln (y^2-x)$, the level curves would have the equation $x=y^2-e^k$, which is a parabola with horizontal axis of symmetry. But now when I have $x$ that multiples on, I get stuck.






calculus







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  • $begingroup$
    What do you mean with "$x$ that multiplies on"? When you fix $k$, you have a graph. Therefore, for a fixed $k$, $x=y^2 - e^k$ is a level curve. I don't see any problem here.
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:00












  • $begingroup$
    desmos.com/calculator/njpygyntxe
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:01










  • $begingroup$
    @MattiP. OP is talking about $xln(y^2-x)$
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:02










  • $begingroup$
    @ShubhamJohri ahaa, that wasn't very clear in the question definition.
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:03










  • $begingroup$
    Why not just solve $f(x,y)=k$ for $y$ (up to $pm$) and plot the curve?
    $endgroup$
    – MPW
    Dec 18 '18 at 9:04
















0












$begingroup$


How to find the level curves of $$f(x,y)=xln (y^2-x)$$



If I had $f(x,y)=ln (y^2-x)$, the level curves would have the equation $x=y^2-e^k$, which is a parabola with horizontal axis of symmetry. But now when I have $x$ that multiples on, I get stuck.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean with "$x$ that multiplies on"? When you fix $k$, you have a graph. Therefore, for a fixed $k$, $x=y^2 - e^k$ is a level curve. I don't see any problem here.
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:00












  • $begingroup$
    desmos.com/calculator/njpygyntxe
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:01










  • $begingroup$
    @MattiP. OP is talking about $xln(y^2-x)$
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:02










  • $begingroup$
    @ShubhamJohri ahaa, that wasn't very clear in the question definition.
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:03










  • $begingroup$
    Why not just solve $f(x,y)=k$ for $y$ (up to $pm$) and plot the curve?
    $endgroup$
    – MPW
    Dec 18 '18 at 9:04














0












0








0





$begingroup$


How to find the level curves of $$f(x,y)=xln (y^2-x)$$



If I had $f(x,y)=ln (y^2-x)$, the level curves would have the equation $x=y^2-e^k$, which is a parabola with horizontal axis of symmetry. But now when I have $x$ that multiples on, I get stuck.






calculus







share|cite|improve this question











$endgroup$




How to find the level curves of $$f(x,y)=xln (y^2-x)$$



If I had $f(x,y)=ln (y^2-x)$, the level curves would have the equation $x=y^2-e^k$, which is a parabola with horizontal axis of symmetry. But now when I have $x$ that multiples on, I get stuck.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 9:11









Shubham Johri

5,412818




5,412818










asked Dec 18 '18 at 8:55









user575062user575062

62




62












  • $begingroup$
    What do you mean with "$x$ that multiplies on"? When you fix $k$, you have a graph. Therefore, for a fixed $k$, $x=y^2 - e^k$ is a level curve. I don't see any problem here.
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:00












  • $begingroup$
    desmos.com/calculator/njpygyntxe
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:01










  • $begingroup$
    @MattiP. OP is talking about $xln(y^2-x)$
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:02










  • $begingroup$
    @ShubhamJohri ahaa, that wasn't very clear in the question definition.
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:03










  • $begingroup$
    Why not just solve $f(x,y)=k$ for $y$ (up to $pm$) and plot the curve?
    $endgroup$
    – MPW
    Dec 18 '18 at 9:04


















  • $begingroup$
    What do you mean with "$x$ that multiplies on"? When you fix $k$, you have a graph. Therefore, for a fixed $k$, $x=y^2 - e^k$ is a level curve. I don't see any problem here.
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:00












  • $begingroup$
    desmos.com/calculator/njpygyntxe
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:01










  • $begingroup$
    @MattiP. OP is talking about $xln(y^2-x)$
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:02










  • $begingroup$
    @ShubhamJohri ahaa, that wasn't very clear in the question definition.
    $endgroup$
    – Matti P.
    Dec 18 '18 at 9:03










  • $begingroup$
    Why not just solve $f(x,y)=k$ for $y$ (up to $pm$) and plot the curve?
    $endgroup$
    – MPW
    Dec 18 '18 at 9:04
















$begingroup$
What do you mean with "$x$ that multiplies on"? When you fix $k$, you have a graph. Therefore, for a fixed $k$, $x=y^2 - e^k$ is a level curve. I don't see any problem here.
$endgroup$
– Matti P.
Dec 18 '18 at 9:00






$begingroup$
What do you mean with "$x$ that multiplies on"? When you fix $k$, you have a graph. Therefore, for a fixed $k$, $x=y^2 - e^k$ is a level curve. I don't see any problem here.
$endgroup$
– Matti P.
Dec 18 '18 at 9:00














$begingroup$
desmos.com/calculator/njpygyntxe
$endgroup$
– Matti P.
Dec 18 '18 at 9:01




$begingroup$
desmos.com/calculator/njpygyntxe
$endgroup$
– Matti P.
Dec 18 '18 at 9:01












$begingroup$
@MattiP. OP is talking about $xln(y^2-x)$
$endgroup$
– Shubham Johri
Dec 18 '18 at 9:02




$begingroup$
@MattiP. OP is talking about $xln(y^2-x)$
$endgroup$
– Shubham Johri
Dec 18 '18 at 9:02












$begingroup$
@ShubhamJohri ahaa, that wasn't very clear in the question definition.
$endgroup$
– Matti P.
Dec 18 '18 at 9:03




$begingroup$
@ShubhamJohri ahaa, that wasn't very clear in the question definition.
$endgroup$
– Matti P.
Dec 18 '18 at 9:03












$begingroup$
Why not just solve $f(x,y)=k$ for $y$ (up to $pm$) and plot the curve?
$endgroup$
– MPW
Dec 18 '18 at 9:04




$begingroup$
Why not just solve $f(x,y)=k$ for $y$ (up to $pm$) and plot the curve?
$endgroup$
– MPW
Dec 18 '18 at 9:04










1 Answer
1






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oldest

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0












$begingroup$

The level curves have the equation $xln(y^2-x)=kinBbb R$. The point $(0,y)$ lies on the level curve only for $k=0$. For $kne0,xne0$.



For $k,xne0$, you can isolate $x,y$ as under:



$displaystyle xln(y^2-x)=kimplies y^2=x+e^{frac kx} (k,xne0)$



When $k=0$, you get the level curves $x=0ne y,y^2=x+1$ in the $xy$ plane.



The level curves are $begin{cases}displaystyle y^2=x+e^frac zx,&zne0\x=0ne y,y^2=x+1, &z=0end{cases}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't think that you should divide by $x$ or exclude 0 since it is inside the domain of the function. Or can you explain why you exclude 0?
    $endgroup$
    – user575062
    Dec 18 '18 at 9:09












  • $begingroup$
    @user575062 Please see the modified answer
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:22











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

The level curves have the equation $xln(y^2-x)=kinBbb R$. The point $(0,y)$ lies on the level curve only for $k=0$. For $kne0,xne0$.



For $k,xne0$, you can isolate $x,y$ as under:



$displaystyle xln(y^2-x)=kimplies y^2=x+e^{frac kx} (k,xne0)$



When $k=0$, you get the level curves $x=0ne y,y^2=x+1$ in the $xy$ plane.



The level curves are $begin{cases}displaystyle y^2=x+e^frac zx,&zne0\x=0ne y,y^2=x+1, &z=0end{cases}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't think that you should divide by $x$ or exclude 0 since it is inside the domain of the function. Or can you explain why you exclude 0?
    $endgroup$
    – user575062
    Dec 18 '18 at 9:09












  • $begingroup$
    @user575062 Please see the modified answer
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:22
















0












$begingroup$

The level curves have the equation $xln(y^2-x)=kinBbb R$. The point $(0,y)$ lies on the level curve only for $k=0$. For $kne0,xne0$.



For $k,xne0$, you can isolate $x,y$ as under:



$displaystyle xln(y^2-x)=kimplies y^2=x+e^{frac kx} (k,xne0)$



When $k=0$, you get the level curves $x=0ne y,y^2=x+1$ in the $xy$ plane.



The level curves are $begin{cases}displaystyle y^2=x+e^frac zx,&zne0\x=0ne y,y^2=x+1, &z=0end{cases}$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I don't think that you should divide by $x$ or exclude 0 since it is inside the domain of the function. Or can you explain why you exclude 0?
    $endgroup$
    – user575062
    Dec 18 '18 at 9:09












  • $begingroup$
    @user575062 Please see the modified answer
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:22














0












0








0





$begingroup$

The level curves have the equation $xln(y^2-x)=kinBbb R$. The point $(0,y)$ lies on the level curve only for $k=0$. For $kne0,xne0$.



For $k,xne0$, you can isolate $x,y$ as under:



$displaystyle xln(y^2-x)=kimplies y^2=x+e^{frac kx} (k,xne0)$



When $k=0$, you get the level curves $x=0ne y,y^2=x+1$ in the $xy$ plane.



The level curves are $begin{cases}displaystyle y^2=x+e^frac zx,&zne0\x=0ne y,y^2=x+1, &z=0end{cases}$






share|cite|improve this answer











$endgroup$



The level curves have the equation $xln(y^2-x)=kinBbb R$. The point $(0,y)$ lies on the level curve only for $k=0$. For $kne0,xne0$.



For $k,xne0$, you can isolate $x,y$ as under:



$displaystyle xln(y^2-x)=kimplies y^2=x+e^{frac kx} (k,xne0)$



When $k=0$, you get the level curves $x=0ne y,y^2=x+1$ in the $xy$ plane.



The level curves are $begin{cases}displaystyle y^2=x+e^frac zx,&zne0\x=0ne y,y^2=x+1, &z=0end{cases}$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 9:22

























answered Dec 18 '18 at 9:01









Shubham JohriShubham Johri

5,412818




5,412818












  • $begingroup$
    I don't think that you should divide by $x$ or exclude 0 since it is inside the domain of the function. Or can you explain why you exclude 0?
    $endgroup$
    – user575062
    Dec 18 '18 at 9:09












  • $begingroup$
    @user575062 Please see the modified answer
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:22


















  • $begingroup$
    I don't think that you should divide by $x$ or exclude 0 since it is inside the domain of the function. Or can you explain why you exclude 0?
    $endgroup$
    – user575062
    Dec 18 '18 at 9:09












  • $begingroup$
    @user575062 Please see the modified answer
    $endgroup$
    – Shubham Johri
    Dec 18 '18 at 9:22
















$begingroup$
I don't think that you should divide by $x$ or exclude 0 since it is inside the domain of the function. Or can you explain why you exclude 0?
$endgroup$
– user575062
Dec 18 '18 at 9:09






$begingroup$
I don't think that you should divide by $x$ or exclude 0 since it is inside the domain of the function. Or can you explain why you exclude 0?
$endgroup$
– user575062
Dec 18 '18 at 9:09














$begingroup$
@user575062 Please see the modified answer
$endgroup$
– Shubham Johri
Dec 18 '18 at 9:22




$begingroup$
@user575062 Please see the modified answer
$endgroup$
– Shubham Johri
Dec 18 '18 at 9:22


















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