Calculating curvature of a contour

up vote
1
down vote

favorite

I have an equation of a scalar field in the form

$$
f(x, y) = x^2 + y^2 + xy + c.
$$

I want to find the curvature of the contour of the curve at $ f_c = f(0.5, 0.5) $.

So I need to calculate the derivative $ fracmathrm dymathrm dx $ and $ fracmathrm d^2ymathrm dx^2 $

I can solve the equation $ f(x, y) = f_c $ and get the derivative of $ f(x, y) $ w.r.t $ x $.

On paper we do,

d/dx (f(x,y))=d/dx(fc)
2*x+2*y*(dy/dx)+y+x(dy/dx)=0
(dy/dx)=-(2*x+y)/(x+2*y)

and further d/dx(dy/dx) for a curvature approximate

how can I rearrange the equation such that I can get the value of dy/dx on mathematica

EDIT:
This is what I tried so far:

Clear["Global`*"]
ftest[x_, y_] = x*x + y*y + x*y + c;
dftest = (D[ftest[x, y], x]/D[ftest[x, y], y]);
d2ftest = D[dftest, x];

Sol :

-((2 x + y)/(x + 2 y)^2) + 2/(x + 2 y)

However this will be wrong as I will be taking the ratio of partials w.r.t. x and y

edited 7 mins ago

Henrik Schumacher

43.5k263129

asked 1 hour ago

Some_guy

New contributor

You could solve your equation for y, then differentiate.
â€“Â b.gatessucks
1 hour ago

Have you tried anything? Rules require you to show a minimal familiarity with Mathematica.
â€“Â Kubaâ™¦
1 hour ago

@b.gatessucks How do I do that.. I used D[Solve[f(x,y)=f(c)] but it returns 0 -> InverseFunction[fs,2,2]^(1,0)[t,0]
â€“Â Some_guy
1 hour ago

Start by solving your ftest for y.
â€“Â b.gatessucks
1 hour ago

1

So your implicit function is something like ` f[x,y]:= x^2 + y^2 + x y - 3/4==0` ?
â€“Â Ulrich Neumann
1 hour ago

Â |Â
show 1 more comment

up vote
1
down vote

favorite

I have an equation of a scalar field in the form

$$
f(x, y) = x^2 + y^2 + xy + c.
$$

I want to find the curvature of the contour of the curve at $ f_c = f(0.5, 0.5) $.

So I need to calculate the derivative $ fracmathrm dymathrm dx $ and $ fracmathrm d^2ymathrm dx^2 $

I can solve the equation $ f(x, y) = f_c $ and get the derivative of $ f(x, y) $ w.r.t $ x $.

On paper we do,

d/dx (f(x,y))=d/dx(fc)
2*x+2*y*(dy/dx)+y+x(dy/dx)=0
(dy/dx)=-(2*x+y)/(x+2*y)

and further d/dx(dy/dx) for a curvature approximate

how can I rearrange the equation such that I can get the value of dy/dx on mathematica

EDIT:
This is what I tried so far:

Clear["Global`*"]
ftest[x_, y_] = x*x + y*y + x*y + c;
dftest = (D[ftest[x, y], x]/D[ftest[x, y], y]);
d2ftest = D[dftest, x];

Sol :

-((2 x + y)/(x + 2 y)^2) + 2/(x + 2 y)

However this will be wrong as I will be taking the ratio of partials w.r.t. x and y

edited 7 mins ago

Henrik Schumacher

43.5k263129

asked 1 hour ago

Some_guy

New contributor

You could solve your equation for y, then differentiate.
â€“Â b.gatessucks
1 hour ago

Have you tried anything? Rules require you to show a minimal familiarity with Mathematica.
â€“Â Kubaâ™¦
1 hour ago

@b.gatessucks How do I do that.. I used D[Solve[f(x,y)=f(c)] but it returns 0 -> InverseFunction[fs,2,2]^(1,0)[t,0]
â€“Â Some_guy
1 hour ago

Start by solving your ftest for y.
â€“Â b.gatessucks
1 hour ago

1

So your implicit function is something like ` f[x,y]:= x^2 + y^2 + x y - 3/4==0` ?
â€“Â Ulrich Neumann
1 hour ago

Â |Â
show 1 more comment

up vote
1
down vote

favorite

I have an equation of a scalar field in the form

$$
f(x, y) = x^2 + y^2 + xy + c.
$$

I want to find the curvature of the contour of the curve at $ f_c = f(0.5, 0.5) $.

So I need to calculate the derivative $ fracmathrm dymathrm dx $ and $ fracmathrm d^2ymathrm dx^2 $

I can solve the equation $ f(x, y) = f_c $ and get the derivative of $ f(x, y) $ w.r.t $ x $.

On paper we do,

d/dx (f(x,y))=d/dx(fc)
2*x+2*y*(dy/dx)+y+x(dy/dx)=0
(dy/dx)=-(2*x+y)/(x+2*y)

and further d/dx(dy/dx) for a curvature approximate

how can I rearrange the equation such that I can get the value of dy/dx on mathematica

EDIT:
This is what I tried so far:

Clear["Global`*"]
ftest[x_, y_] = x*x + y*y + x*y + c;
dftest = (D[ftest[x, y], x]/D[ftest[x, y], y]);
d2ftest = D[dftest, x];

Sol :

-((2 x + y)/(x + 2 y)^2) + 2/(x + 2 y)

However this will be wrong as I will be taking the ratio of partials w.r.t. x and y

edited 7 mins ago

Henrik Schumacher

43.5k263129

asked 1 hour ago

Some_guy

New contributor

I have an equation of a scalar field in the form

$$
f(x, y) = x^2 + y^2 + xy + c.
$$

I want to find the curvature of the contour of the curve at $ f_c = f(0.5, 0.5) $.

So I need to calculate the derivative $ fracmathrm dymathrm dx $ and $ fracmathrm d^2ymathrm dx^2 $

I can solve the equation $ f(x, y) = f_c $ and get the derivative of $ f(x, y) $ w.r.t $ x $.

On paper we do,

d/dx (f(x,y))=d/dx(fc)
2*x+2*y*(dy/dx)+y+x(dy/dx)=0
(dy/dx)=-(2*x+y)/(x+2*y)

and further d/dx(dy/dx) for a curvature approximate

how can I rearrange the equation such that I can get the value of dy/dx on mathematica

EDIT:
This is what I tried so far:

Clear["Global`*"]
ftest[x_, y_] = x*x + y*y + x*y + c;
dftest = (D[ftest[x, y], x]/D[ftest[x, y], y]);
d2ftest = D[dftest, x];

Sol :

-((2 x + y)/(x + 2 y)^2) + 2/(x + 2 y)

However this will be wrong as I will be taking the ratio of partials w.r.t. x and y

calculus-and-analysis geometry differential-geometry

edited 7 mins ago

Henrik Schumacher

43.5k263129

asked 1 hour ago

Some_guy

New contributor

edited 7 mins ago

Henrik Schumacher

43.5k263129

asked 1 hour ago

Some_guy

New contributor

edited 7 mins ago

Henrik Schumacher

43.5k263129

edited 7 mins ago

Henrik Schumacher

43.5k263129

edited 7 mins ago

Henrik Schumacher

43.5k263129

asked 1 hour ago

Some_guy

New contributor

asked 1 hour ago

Some_guy

asked 1 hour ago

Some_guy

New contributor

Some_guy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

You could solve your equation for y, then differentiate.
â€“Â b.gatessucks
1 hour ago

Have you tried anything? Rules require you to show a minimal familiarity with Mathematica.
â€“Â Kubaâ™¦
1 hour ago

@b.gatessucks How do I do that.. I used D[Solve[f(x,y)=f(c)] but it returns 0 -> InverseFunction[fs,2,2]^(1,0)[t,0]
â€“Â Some_guy
1 hour ago

Start by solving your ftest for y.
â€“Â b.gatessucks
1 hour ago

1

So your implicit function is something like ` f[x,y]:= x^2 + y^2 + x y - 3/4==0` ?
â€“Â Ulrich Neumann
1 hour ago

Â |Â
show 1 more comment

You could solve your equation for y, then differentiate.
â€“Â b.gatessucks
1 hour ago

Have you tried anything? Rules require you to show a minimal familiarity with Mathematica.
â€“Â Kubaâ™¦
1 hour ago

@b.gatessucks How do I do that.. I used D[Solve[f(x,y)=f(c)] but it returns 0 -> InverseFunction[fs,2,2]^(1,0)[t,0]
â€“Â Some_guy
1 hour ago

Start by solving your ftest for y.
â€“Â b.gatessucks
1 hour ago

1

So your implicit function is something like ` f[x,y]:= x^2 + y^2 + x y - 3/4==0` ?
â€“Â Ulrich Neumann
1 hour ago

You could solve your equation for y, then differentiate.
â€“Â b.gatessucks
1 hour ago

Have you tried anything? Rules require you to show a minimal familiarity with Mathematica.
â€“Â Kubaâ™¦
1 hour ago

@b.gatessucks How do I do that.. I used D[Solve[f(x,y)=f(c)] but it returns 0 -> InverseFunction[fs,2,2]^(1,0)[t,0]
â€“Â Some_guy
1 hour ago

Start by solving your ftest for y.
â€“Â b.gatessucks
1 hour ago

So your implicit function is something like ` f[x,y]:= x^2 + y^2 + x y - 3/4==0` ?
â€“Â Ulrich Neumann
1 hour ago

Â |Â
show 1 more comment

2 Answers
2

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up vote
2
down vote

Clear[f]
f[x_, y_] := x^2 + y[x]^2 + x y[x] + c == fc

Solve[D[f[x, y], x, 1], y'[x]][[1, 1]] // FullSimplify
Solve[D[f[x, y], x, 2], y''[x]][[1, 1]] /. % // FullSimplify

y'[x] -> -(2 x + y[x])/(x + 2 y[x])
y''[x] -> -6 (x^2 + y[x] (x + y[x])/(x + 2 y[x])^3

answered 1 hour ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

3,2341927

add a commentÂ |Â

up vote
1
down vote

IIRC, the second fundamental form $I!I$ of a levelset $M = varPhi^-1(0)$ of a mapping $varPhi colon mathbbR^n to mathbbR^m$, $m<n$ at point $x in mathbbR^n$ is given (up to sign that I use to mix up) by

$$ I!I(x) = pm DvarPhi(x)^dagger , D^2varPhi(x),$$

where $DvarPhi(x)^dagger$ denotes the Moore-Penrose pseudoinverse of the Jacobian $DvarPhi(x)$ of $varPhi$. Here, I assume that $DvarPhi(x)$ is surjective (but a similar formula can be derived if $DvarPhi$ has constant rank in a neighborhood of $x$).

More precisely, we have

$$ I!I(x)(u,v) = pm DvarPhi(x)^dagger , D^2varPhi(x)(u,v) quad textfor all $u,,v in T_xM = operatornameker(DvarPhi(x))$$$

as $I!I(x)(u,v)$ is not really meaningful for non-tangent vectors $u, , v not in T_xM$.

This can be obtained as follows:

ÃŽÂ¦ = x, y [Function] x^2 + y^2 + x y + c;
DÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 1];
DDÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 2];
II = Transpose[LinearSolve[DÃŽÂ¦. Transpose [DÃŽÂ¦], DÃŽÂ¦]].DDÃŽÂ¦

In this 2-dimensional example, the curvature $kappa$ of the level set at point $(x,y)$ should be (up to sign) computable as

ÃŽÂ½ = DÃŽÂ¦[[1]]/Sqrt[Total[DÃŽÂ¦^2, 2]];
ÃÂ„ = RotationMatrix[Pi/2].ÃŽÂ½
ÃŽÂº = ÃŽÂ½.(II.ÃÂ„).ÃÂ„ // Simplify

(6 (x^2 + x y + y^2))/(5 x^2 + 8 x y + 5 y^2)^(3/2)

Here, ÃŽÂ½ is the normal to the levelset (point upwards with respect to the function ÃŽÂ¦) and ÃÂ„ is the tangent obtained by counterclockwise 90-degree-rotation of ÃŽÂ½.

edited 8 mins ago

answered 14 mins ago

Henrik Schumacher

43.5k263129

add a commentÂ |Â

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2 Answers
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2 Answers
2

active

oldest

votes

up vote
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Clear[f]
f[x_, y_] := x^2 + y[x]^2 + x y[x] + c == fc

Solve[D[f[x, y], x, 1], y'[x]][[1, 1]] // FullSimplify
Solve[D[f[x, y], x, 2], y''[x]][[1, 1]] /. % // FullSimplify

y'[x] -> -(2 x + y[x])/(x + 2 y[x])
y''[x] -> -6 (x^2 + y[x] (x + y[x])/(x + 2 y[x])^3

answered 1 hour ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

3,2341927

add a commentÂ |Â

up vote
2
down vote

Clear[f]
f[x_, y_] := x^2 + y[x]^2 + x y[x] + c == fc

Solve[D[f[x, y], x, 1], y'[x]][[1, 1]] // FullSimplify
Solve[D[f[x, y], x, 2], y''[x]][[1, 1]] /. % // FullSimplify

y'[x] -> -(2 x + y[x])/(x + 2 y[x])
y''[x] -> -6 (x^2 + y[x] (x + y[x])/(x + 2 y[x])^3

answered 1 hour ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

3,2341927

add a commentÂ |Â

up vote
2
down vote

Clear[f]
f[x_, y_] := x^2 + y[x]^2 + x y[x] + c == fc

Solve[D[f[x, y], x, 1], y'[x]][[1, 1]] // FullSimplify
Solve[D[f[x, y], x, 2], y''[x]][[1, 1]] /. % // FullSimplify

y'[x] -> -(2 x + y[x])/(x + 2 y[x])
y''[x] -> -6 (x^2 + y[x] (x + y[x])/(x + 2 y[x])^3

answered 1 hour ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

3,2341927

Clear[f]
f[x_, y_] := x^2 + y[x]^2 + x y[x] + c == fc

Solve[D[f[x, y], x, 1], y'[x]][[1, 1]] // FullSimplify
Solve[D[f[x, y], x, 2], y''[x]][[1, 1]] /. % // FullSimplify

y'[x] -> -(2 x + y[x])/(x + 2 y[x])
y''[x] -> -6 (x^2 + y[x] (x + y[x])/(x + 2 y[x])^3

answered 1 hour ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

3,2341927

answered 1 hour ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

3,2341927

answered 1 hour ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

3,2341927

answered 1 hour ago

ÃŽÂ‘ÃŽÂ»ÃŽÂÃŽÂ¾ÃŽÂ±ÃŽÂ½ÃŽÂ´ÃÂÃŽÂ¿ÃÂ‚ ÃŽÂ–ÃŽÂµÃŽÂ³ÃŽÂ³

3,2341927

add a commentÂ |Â

up vote
1
down vote

$$ I!I(x) = pm DvarPhi(x)^dagger , D^2varPhi(x),$$

More precisely, we have

$$ I!I(x)(u,v) = pm DvarPhi(x)^dagger , D^2varPhi(x)(u,v) quad textfor all $u,,v in T_xM = operatornameker(DvarPhi(x))$$$

as $I!I(x)(u,v)$ is not really meaningful for non-tangent vectors $u, , v not in T_xM$.

This can be obtained as follows:

ÃŽÂ¦ = x, y [Function] x^2 + y^2 + x y + c;
DÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 1];
DDÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 2];
II = Transpose[LinearSolve[DÃŽÂ¦. Transpose [DÃŽÂ¦], DÃŽÂ¦]].DDÃŽÂ¦

In this 2-dimensional example, the curvature $kappa$ of the level set at point $(x,y)$ should be (up to sign) computable as

ÃŽÂ½ = DÃŽÂ¦[[1]]/Sqrt[Total[DÃŽÂ¦^2, 2]];
ÃÂ„ = RotationMatrix[Pi/2].ÃŽÂ½
ÃŽÂº = ÃŽÂ½.(II.ÃÂ„).ÃÂ„ // Simplify

(6 (x^2 + x y + y^2))/(5 x^2 + 8 x y + 5 y^2)^(3/2)

Here, ÃŽÂ½ is the normal to the levelset (point upwards with respect to the function ÃŽÂ¦) and ÃÂ„ is the tangent obtained by counterclockwise 90-degree-rotation of ÃŽÂ½.

edited 8 mins ago

answered 14 mins ago

Henrik Schumacher

43.5k263129

add a commentÂ |Â

up vote
1
down vote

$$ I!I(x) = pm DvarPhi(x)^dagger , D^2varPhi(x),$$

More precisely, we have

$$ I!I(x)(u,v) = pm DvarPhi(x)^dagger , D^2varPhi(x)(u,v) quad textfor all $u,,v in T_xM = operatornameker(DvarPhi(x))$$$

as $I!I(x)(u,v)$ is not really meaningful for non-tangent vectors $u, , v not in T_xM$.

This can be obtained as follows:

ÃŽÂ¦ = x, y [Function] x^2 + y^2 + x y + c;
DÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 1];
DDÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 2];
II = Transpose[LinearSolve[DÃŽÂ¦. Transpose [DÃŽÂ¦], DÃŽÂ¦]].DDÃŽÂ¦

In this 2-dimensional example, the curvature $kappa$ of the level set at point $(x,y)$ should be (up to sign) computable as

ÃŽÂ½ = DÃŽÂ¦[[1]]/Sqrt[Total[DÃŽÂ¦^2, 2]];
ÃÂ„ = RotationMatrix[Pi/2].ÃŽÂ½
ÃŽÂº = ÃŽÂ½.(II.ÃÂ„).ÃÂ„ // Simplify

(6 (x^2 + x y + y^2))/(5 x^2 + 8 x y + 5 y^2)^(3/2)

Here, ÃŽÂ½ is the normal to the levelset (point upwards with respect to the function ÃŽÂ¦) and ÃÂ„ is the tangent obtained by counterclockwise 90-degree-rotation of ÃŽÂ½.

edited 8 mins ago

answered 14 mins ago

Henrik Schumacher

43.5k263129

add a commentÂ |Â

up vote
1
down vote

$$ I!I(x) = pm DvarPhi(x)^dagger , D^2varPhi(x),$$

More precisely, we have

$$ I!I(x)(u,v) = pm DvarPhi(x)^dagger , D^2varPhi(x)(u,v) quad textfor all $u,,v in T_xM = operatornameker(DvarPhi(x))$$$

as $I!I(x)(u,v)$ is not really meaningful for non-tangent vectors $u, , v not in T_xM$.

This can be obtained as follows:

ÃŽÂ¦ = x, y [Function] x^2 + y^2 + x y + c;
DÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 1];
DDÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 2];
II = Transpose[LinearSolve[DÃŽÂ¦. Transpose [DÃŽÂ¦], DÃŽÂ¦]].DDÃŽÂ¦

In this 2-dimensional example, the curvature $kappa$ of the level set at point $(x,y)$ should be (up to sign) computable as

ÃŽÂ½ = DÃŽÂ¦[[1]]/Sqrt[Total[DÃŽÂ¦^2, 2]];
ÃÂ„ = RotationMatrix[Pi/2].ÃŽÂ½
ÃŽÂº = ÃŽÂ½.(II.ÃÂ„).ÃÂ„ // Simplify

(6 (x^2 + x y + y^2))/(5 x^2 + 8 x y + 5 y^2)^(3/2)

Here, ÃŽÂ½ is the normal to the levelset (point upwards with respect to the function ÃŽÂ¦) and ÃÂ„ is the tangent obtained by counterclockwise 90-degree-rotation of ÃŽÂ½.

edited 8 mins ago

answered 14 mins ago

Henrik Schumacher

43.5k263129

$$ I!I(x) = pm DvarPhi(x)^dagger , D^2varPhi(x),$$

More precisely, we have

$$ I!I(x)(u,v) = pm DvarPhi(x)^dagger , D^2varPhi(x)(u,v) quad textfor all $u,,v in T_xM = operatornameker(DvarPhi(x))$$$

as $I!I(x)(u,v)$ is not really meaningful for non-tangent vectors $u, , v not in T_xM$.

This can be obtained as follows:

ÃŽÂ¦ = x, y [Function] x^2 + y^2 + x y + c;
DÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 1];
DDÃŽÂ¦ = D[ÃŽÂ¦[x, y], x, y, 2];
II = Transpose[LinearSolve[DÃŽÂ¦. Transpose [DÃŽÂ¦], DÃŽÂ¦]].DDÃŽÂ¦

In this 2-dimensional example, the curvature $kappa$ of the level set at point $(x,y)$ should be (up to sign) computable as

ÃŽÂ½ = DÃŽÂ¦[[1]]/Sqrt[Total[DÃŽÂ¦^2, 2]];
ÃÂ„ = RotationMatrix[Pi/2].ÃŽÂ½
ÃŽÂº = ÃŽÂ½.(II.ÃÂ„).ÃÂ„ // Simplify

(6 (x^2 + x y + y^2))/(5 x^2 + 8 x y + 5 y^2)^(3/2)

Here, ÃŽÂ½ is the normal to the levelset (point upwards with respect to the function ÃŽÂ¦) and ÃÂ„ is the tangent obtained by counterclockwise 90-degree-rotation of ÃŽÂ½.

edited 8 mins ago

answered 14 mins ago

Henrik Schumacher

43.5k263129

edited 8 mins ago

answered 14 mins ago

Henrik Schumacher

43.5k263129

answered 14 mins ago

Henrik Schumacher

43.5k263129

answered 14 mins ago

Henrik Schumacher

43.5k263129

add a commentÂ |Â

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