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







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












$begingroup$


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






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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044920%2fhow-to-draw-or-describe-level-curves-of-x-ln-y2-x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044920%2fhow-to-draw-or-describe-level-curves-of-x-ln-y2-x%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
Read more

In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

Image
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
Read more

How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

Image
1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
Read more