Show that the families of hyberbolae $x^2-y^2=a$, $xy=b$ create an orthogonal grid: what am I doing wrong?












0












$begingroup$



Show that the families of hyberbolae $x^2-y^2=a$, $xy=b$ create an
orthogonal grid.




My idea is to find the intersection points of the hyperbolae, and then take a look at the $y_1' cdot y_2'$ at those points. It should be $-1$.



From
$begin{cases}
|y|=sqrt{x^2-a} \
y=frac{b}{x}
end{cases}$
we get $sqrt{x^2-a}=pm frac{b}{x}$.



My derivatives are $y_1'=frac{1}{sqrt{x^2-a}}$, $y_2'= frac{b}{x^2}$.



$y_1'cdot y_2' = pm frac{x}{b} cdot frac{b}{x^2} = pm frac{1}{x} neq -1$.



I will be grateful for a hint or a full solution, but what most concerns me is where did my reasoning go astray.



Thank you.






calculus derivatives







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I believe your $y'_1$ is incorrect.
    $endgroup$
    – Rahul
    Nov 29 '18 at 11:49






  • 1




    $begingroup$
    Derivative of a composed function for $y_1$
    $endgroup$
    – user376343
    Nov 29 '18 at 12:08










  • $begingroup$
    @Rahul Thanks, I see now! $y_1'=frac{x}{sqrt{x^2-a}}$, and $y_1' cdot y_2' = pm 1$, but what do I do about $pm$?
    $endgroup$
    – fragileradius
    Nov 29 '18 at 12:18










  • $begingroup$
    Implicit differentiation gives $2x-2y_1y_1'=0$, that is $y_1'=x/y_1$. Now $y_2'=-b/x^2$. Multiply both and use that in the intersection points we have $y_1=y_2$.
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 12:52






  • 1




    $begingroup$
    Check out this answer I made to a similar question! Hope I helped! math.stackexchange.com/questions/3027827/…
    $endgroup$
    – Ricardo B.
    Dec 6 '18 at 1:57
















0












$begingroup$



Show that the families of hyberbolae $x^2-y^2=a$, $xy=b$ create an
orthogonal grid.




My idea is to find the intersection points of the hyperbolae, and then take a look at the $y_1' cdot y_2'$ at those points. It should be $-1$.



From
$begin{cases}
|y|=sqrt{x^2-a} \
y=frac{b}{x}
end{cases}$
we get $sqrt{x^2-a}=pm frac{b}{x}$.



My derivatives are $y_1'=frac{1}{sqrt{x^2-a}}$, $y_2'= frac{b}{x^2}$.



$y_1'cdot y_2' = pm frac{x}{b} cdot frac{b}{x^2} = pm frac{1}{x} neq -1$.



I will be grateful for a hint or a full solution, but what most concerns me is where did my reasoning go astray.



Thank you.






calculus derivatives







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    I believe your $y'_1$ is incorrect.
    $endgroup$
    – Rahul
    Nov 29 '18 at 11:49






  • 1




    $begingroup$
    Derivative of a composed function for $y_1$
    $endgroup$
    – user376343
    Nov 29 '18 at 12:08










  • $begingroup$
    @Rahul Thanks, I see now! $y_1'=frac{x}{sqrt{x^2-a}}$, and $y_1' cdot y_2' = pm 1$, but what do I do about $pm$?
    $endgroup$
    – fragileradius
    Nov 29 '18 at 12:18










  • $begingroup$
    Implicit differentiation gives $2x-2y_1y_1'=0$, that is $y_1'=x/y_1$. Now $y_2'=-b/x^2$. Multiply both and use that in the intersection points we have $y_1=y_2$.
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 12:52






  • 1




    $begingroup$
    Check out this answer I made to a similar question! Hope I helped! math.stackexchange.com/questions/3027827/…
    $endgroup$
    – Ricardo B.
    Dec 6 '18 at 1:57














0












0








0





$begingroup$



Show that the families of hyberbolae $x^2-y^2=a$, $xy=b$ create an
orthogonal grid.




My idea is to find the intersection points of the hyperbolae, and then take a look at the $y_1' cdot y_2'$ at those points. It should be $-1$.



From
$begin{cases}
|y|=sqrt{x^2-a} \
y=frac{b}{x}
end{cases}$
we get $sqrt{x^2-a}=pm frac{b}{x}$.



My derivatives are $y_1'=frac{1}{sqrt{x^2-a}}$, $y_2'= frac{b}{x^2}$.



$y_1'cdot y_2' = pm frac{x}{b} cdot frac{b}{x^2} = pm frac{1}{x} neq -1$.



I will be grateful for a hint or a full solution, but what most concerns me is where did my reasoning go astray.



Thank you.






calculus derivatives







share|cite|improve this question









$endgroup$





Show that the families of hyberbolae $x^2-y^2=a$, $xy=b$ create an
orthogonal grid.




My idea is to find the intersection points of the hyperbolae, and then take a look at the $y_1' cdot y_2'$ at those points. It should be $-1$.



From
$begin{cases}
|y|=sqrt{x^2-a} \
y=frac{b}{x}
end{cases}$
we get $sqrt{x^2-a}=pm frac{b}{x}$.



My derivatives are $y_1'=frac{1}{sqrt{x^2-a}}$, $y_2'= frac{b}{x^2}$.



$y_1'cdot y_2' = pm frac{x}{b} cdot frac{b}{x^2} = pm frac{1}{x} neq -1$.



I will be grateful for a hint or a full solution, but what most concerns me is where did my reasoning go astray.



Thank you.






calculus derivatives




calculus derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 11:40









fragileradiusfragileradius

297114




297114








  • 2




    $begingroup$
    I believe your $y'_1$ is incorrect.
    $endgroup$
    – Rahul
    Nov 29 '18 at 11:49






  • 1




    $begingroup$
    Derivative of a composed function for $y_1$
    $endgroup$
    – user376343
    Nov 29 '18 at 12:08










  • $begingroup$
    @Rahul Thanks, I see now! $y_1'=frac{x}{sqrt{x^2-a}}$, and $y_1' cdot y_2' = pm 1$, but what do I do about $pm$?
    $endgroup$
    – fragileradius
    Nov 29 '18 at 12:18










  • $begingroup$
    Implicit differentiation gives $2x-2y_1y_1'=0$, that is $y_1'=x/y_1$. Now $y_2'=-b/x^2$. Multiply both and use that in the intersection points we have $y_1=y_2$.
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 12:52






  • 1




    $begingroup$
    Check out this answer I made to a similar question! Hope I helped! math.stackexchange.com/questions/3027827/…
    $endgroup$
    – Ricardo B.
    Dec 6 '18 at 1:57














  • 2




    $begingroup$
    I believe your $y'_1$ is incorrect.
    $endgroup$
    – Rahul
    Nov 29 '18 at 11:49






  • 1




    $begingroup$
    Derivative of a composed function for $y_1$
    $endgroup$
    – user376343
    Nov 29 '18 at 12:08










  • $begingroup$
    @Rahul Thanks, I see now! $y_1'=frac{x}{sqrt{x^2-a}}$, and $y_1' cdot y_2' = pm 1$, but what do I do about $pm$?
    $endgroup$
    – fragileradius
    Nov 29 '18 at 12:18










  • $begingroup$
    Implicit differentiation gives $2x-2y_1y_1'=0$, that is $y_1'=x/y_1$. Now $y_2'=-b/x^2$. Multiply both and use that in the intersection points we have $y_1=y_2$.
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 12:52






  • 1




    $begingroup$
    Check out this answer I made to a similar question! Hope I helped! math.stackexchange.com/questions/3027827/…
    $endgroup$
    – Ricardo B.
    Dec 6 '18 at 1:57








2




2




$begingroup$
I believe your $y'_1$ is incorrect.
$endgroup$
– Rahul
Nov 29 '18 at 11:49




$begingroup$
I believe your $y'_1$ is incorrect.
$endgroup$
– Rahul
Nov 29 '18 at 11:49




1




1




$begingroup$
Derivative of a composed function for $y_1$
$endgroup$
– user376343
Nov 29 '18 at 12:08




$begingroup$
Derivative of a composed function for $y_1$
$endgroup$
– user376343
Nov 29 '18 at 12:08












$begingroup$
@Rahul Thanks, I see now! $y_1'=frac{x}{sqrt{x^2-a}}$, and $y_1' cdot y_2' = pm 1$, but what do I do about $pm$?
$endgroup$
– fragileradius
Nov 29 '18 at 12:18




$begingroup$
@Rahul Thanks, I see now! $y_1'=frac{x}{sqrt{x^2-a}}$, and $y_1' cdot y_2' = pm 1$, but what do I do about $pm$?
$endgroup$
– fragileradius
Nov 29 '18 at 12:18












$begingroup$
Implicit differentiation gives $2x-2y_1y_1'=0$, that is $y_1'=x/y_1$. Now $y_2'=-b/x^2$. Multiply both and use that in the intersection points we have $y_1=y_2$.
$endgroup$
– Michael Hoppe
Nov 29 '18 at 12:52




$begingroup$
Implicit differentiation gives $2x-2y_1y_1'=0$, that is $y_1'=x/y_1$. Now $y_2'=-b/x^2$. Multiply both and use that in the intersection points we have $y_1=y_2$.
$endgroup$
– Michael Hoppe
Nov 29 '18 at 12:52




1




1




$begingroup$
Check out this answer I made to a similar question! Hope I helped! math.stackexchange.com/questions/3027827/…
$endgroup$
– Ricardo B.
Dec 6 '18 at 1:57




$begingroup$
Check out this answer I made to a similar question! Hope I helped! math.stackexchange.com/questions/3027827/…
$endgroup$
– Ricardo B.
Dec 6 '18 at 1:57










2 Answers
2






active

oldest

votes


















2












$begingroup$

Your curves are level lines of the functions
$$F(x,y):=x^2-y^2,qquad G(x,y):=xy .$$
Given a point ${bf z}_0=(x_0,y_0)ne{bf 0}$ the level line of $F$ through ${bf z}_0$ is orthogonal to $nabla F({bf z}_0)=(2x_0,-2y_0)$, and the level line of $G$ through ${bf z}_0$ is orthogonal to $nabla G({bf z}_0)=(y_0,x_0)$. As $nabla F({bf z}_0)cdotnabla F({bf z}_0)=0$ these two gradients are orthogonal to each other, and so are the tangents to the two level lines.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now this is cool!
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 19:20



















0












$begingroup$

These are the real and imaginary parts of $zmapsto z^2$. Then use the Cauchy-Riemann equations.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Which incidentally shows you how to generate many other examples.
    $endgroup$
    – Richard Martin
    Nov 29 '18 at 18:01











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%2f3018519%2fshow-that-the-families-of-hyberbolae-x2-y2-a-xy-b-create-an-orthogonal-gr%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Your curves are level lines of the functions
$$F(x,y):=x^2-y^2,qquad G(x,y):=xy .$$
Given a point ${bf z}_0=(x_0,y_0)ne{bf 0}$ the level line of $F$ through ${bf z}_0$ is orthogonal to $nabla F({bf z}_0)=(2x_0,-2y_0)$, and the level line of $G$ through ${bf z}_0$ is orthogonal to $nabla G({bf z}_0)=(y_0,x_0)$. As $nabla F({bf z}_0)cdotnabla F({bf z}_0)=0$ these two gradients are orthogonal to each other, and so are the tangents to the two level lines.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now this is cool!
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 19:20
















2












$begingroup$

Your curves are level lines of the functions
$$F(x,y):=x^2-y^2,qquad G(x,y):=xy .$$
Given a point ${bf z}_0=(x_0,y_0)ne{bf 0}$ the level line of $F$ through ${bf z}_0$ is orthogonal to $nabla F({bf z}_0)=(2x_0,-2y_0)$, and the level line of $G$ through ${bf z}_0$ is orthogonal to $nabla G({bf z}_0)=(y_0,x_0)$. As $nabla F({bf z}_0)cdotnabla F({bf z}_0)=0$ these two gradients are orthogonal to each other, and so are the tangents to the two level lines.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now this is cool!
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 19:20














2












2








2





$begingroup$

Your curves are level lines of the functions
$$F(x,y):=x^2-y^2,qquad G(x,y):=xy .$$
Given a point ${bf z}_0=(x_0,y_0)ne{bf 0}$ the level line of $F$ through ${bf z}_0$ is orthogonal to $nabla F({bf z}_0)=(2x_0,-2y_0)$, and the level line of $G$ through ${bf z}_0$ is orthogonal to $nabla G({bf z}_0)=(y_0,x_0)$. As $nabla F({bf z}_0)cdotnabla F({bf z}_0)=0$ these two gradients are orthogonal to each other, and so are the tangents to the two level lines.






share|cite|improve this answer









$endgroup$



Your curves are level lines of the functions
$$F(x,y):=x^2-y^2,qquad G(x,y):=xy .$$
Given a point ${bf z}_0=(x_0,y_0)ne{bf 0}$ the level line of $F$ through ${bf z}_0$ is orthogonal to $nabla F({bf z}_0)=(2x_0,-2y_0)$, and the level line of $G$ through ${bf z}_0$ is orthogonal to $nabla G({bf z}_0)=(y_0,x_0)$. As $nabla F({bf z}_0)cdotnabla F({bf z}_0)=0$ these two gradients are orthogonal to each other, and so are the tangents to the two level lines.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 '18 at 16:58









Christian BlatterChristian Blatter

172k7113326




172k7113326












  • $begingroup$
    Now this is cool!
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 19:20


















  • $begingroup$
    Now this is cool!
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 19:20
















$begingroup$
Now this is cool!
$endgroup$
– Michael Hoppe
Nov 29 '18 at 19:20




$begingroup$
Now this is cool!
$endgroup$
– Michael Hoppe
Nov 29 '18 at 19:20











0












$begingroup$

These are the real and imaginary parts of $zmapsto z^2$. Then use the Cauchy-Riemann equations.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Which incidentally shows you how to generate many other examples.
    $endgroup$
    – Richard Martin
    Nov 29 '18 at 18:01
















0












$begingroup$

These are the real and imaginary parts of $zmapsto z^2$. Then use the Cauchy-Riemann equations.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Which incidentally shows you how to generate many other examples.
    $endgroup$
    – Richard Martin
    Nov 29 '18 at 18:01














0












0








0





$begingroup$

These are the real and imaginary parts of $zmapsto z^2$. Then use the Cauchy-Riemann equations.






share|cite|improve this answer









$endgroup$



These are the real and imaginary parts of $zmapsto z^2$. Then use the Cauchy-Riemann equations.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 29 '18 at 12:02









Richard MartinRichard Martin

1,61118




1,61118












  • $begingroup$
    Which incidentally shows you how to generate many other examples.
    $endgroup$
    – Richard Martin
    Nov 29 '18 at 18:01


















  • $begingroup$
    Which incidentally shows you how to generate many other examples.
    $endgroup$
    – Richard Martin
    Nov 29 '18 at 18:01
















$begingroup$
Which incidentally shows you how to generate many other examples.
$endgroup$
– Richard Martin
Nov 29 '18 at 18:01




$begingroup$
Which incidentally shows you how to generate many other examples.
$endgroup$
– Richard Martin
Nov 29 '18 at 18:01


















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%2f3018519%2fshow-that-the-families-of-hyberbolae-x2-y2-a-xy-b-create-an-orthogonal-gr%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

Berounka

Image
Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
Read more

Sphinx de Gizeh

Image
Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
Read more

Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

Image
3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
Read more