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











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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











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2 Answers
2






active

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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


















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