Why is the minimum distance between two conics along their common normal?
Throughout my highschool mathematics for coordinate and as well as 3-D geometry.Many All formulae regarding the minimum distance between two curves are derived using the fact that the minimum distance is along their common normal.
Intuitively it's very obvious you draw a parabola, a circle whatever. The shortest distance is always along the common normal but algebraically or using calculus what would be a more mathematical proof?
calculus
asked Nov 30 '18 at 12:01
Avnish KabajAvnish Kabaj
174111
add a comment |
Throughout my highschool mathematics for coordinate and as well as 3-D geometry.Many All formulae regarding the minimum distance between two curves are derived using the fact that the minimum distance is along their common normal.
Intuitively it's very obvious you draw a parabola, a circle whatever. The shortest distance is always along the common normal but algebraically or using calculus what would be a more mathematical proof?
calculus
asked Nov 30 '18 at 12:01
Avnish KabajAvnish Kabaj
174111
The first order variation is $2mathbf{x}_{Ato B}cdotmathrm{d}mathbf{x}_{Ato B}$.
– user10354138
Nov 30 '18 at 12:09
add a comment |
Throughout my highschool mathematics for coordinate and as well as 3-D geometry.Many All formulae regarding the minimum distance between two curves are derived using the fact that the minimum distance is along their common normal.
Intuitively it's very obvious you draw a parabola, a circle whatever. The shortest distance is always along the common normal but algebraically or using calculus what would be a more mathematical proof?
calculus
asked Nov 30 '18 at 12:01
Avnish KabajAvnish Kabaj
174111
Throughout my highschool mathematics for coordinate and as well as 3-D geometry.Many All formulae regarding the minimum distance between two curves are derived using the fact that the minimum distance is along their common normal.
Intuitively it's very obvious you draw a parabola, a circle whatever. The shortest distance is always along the common normal but algebraically or using calculus what would be a more mathematical proof?
calculus
calculus
asked Nov 30 '18 at 12:01
Avnish KabajAvnish Kabaj
174111
asked Nov 30 '18 at 12:01
Avnish KabajAvnish Kabaj
174111
asked Nov 30 '18 at 12:01
Avnish KabajAvnish Kabaj
174111
asked Nov 30 '18 at 12:01
Avnish KabajAvnish Kabaj
174111
asked Nov 30 '18 at 12:01
Avnish KabajAvnish Kabaj
174111
174111
The first order variation is $2mathbf{x}_{Ato B}cdotmathrm{d}mathbf{x}_{Ato B}$.
– user10354138
Nov 30 '18 at 12:09
add a comment |
The first order variation is $2mathbf{x}_{Ato B}cdotmathrm{d}mathbf{x}_{Ato B}$.
– user10354138
Nov 30 '18 at 12:09
The first order variation is $2mathbf{x}_{Ato B}cdotmathrm{d}mathbf{x}_{Ato B}$.
– user10354138
Nov 30 '18 at 12:09
The first order variation is $2mathbf{x}_{Ato B}cdotmathrm{d}mathbf{x}_{Ato B}$.
– user10354138
Nov 30 '18 at 12:09
add a comment |
1 Answer
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This is about two curves $smapsto{bf f}(s)$ and $tmapsto{bf g}(t)$ in ${mathbb R}^n$ and necessary conditions for a local minimum (or maximum) of the distance between two (non-colliding) points on these two curves. The objective function is
$$Phi(s,t):=|{bf f}(s)-{bf g}(t)|^2=bigl( {bf f}(s)-{bf g}(t)bigr)cdotbigl({bf f}(s)-{bf g}(t)bigr) .$$
Assume that for the parameter values $s_0$ and $t_0$ we have ${bf d}:={bf f}(s_0)-{bf g}(t_0)ne{bf 0}$, and that $Phi$ assumes a local minimum at $(s_0,t_0)$. Then
$$Phi_s(s_0,t_0)=2{bf f}'(s_0)cdot({bf f}(s_0)-{bf g}(t_0)bigr)=0quad wedgequadPhi_t(s_0,t_0)=-2{bf g}'(t_0)cdotbigl({bf f}(s_0)-{bf g}(t_0)bigr)=0 .$$
But this is saying that ${bf d}perp{bf f}'(s_0)$ and at the same time ${bf d}perp{bf g}'(t_0)$.
answered Nov 30 '18 at 16:32
Christian BlatterChristian Blatter
172k7113326
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1 Answer
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1 Answer
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oldest
votes
This is about two curves $smapsto{bf f}(s)$ and $tmapsto{bf g}(t)$ in ${mathbb R}^n$ and necessary conditions for a local minimum (or maximum) of the distance between two (non-colliding) points on these two curves. The objective function is
$$Phi(s,t):=|{bf f}(s)-{bf g}(t)|^2=bigl( {bf f}(s)-{bf g}(t)bigr)cdotbigl({bf f}(s)-{bf g}(t)bigr) .$$
Assume that for the parameter values $s_0$ and $t_0$ we have ${bf d}:={bf f}(s_0)-{bf g}(t_0)ne{bf 0}$, and that $Phi$ assumes a local minimum at $(s_0,t_0)$. Then
$$Phi_s(s_0,t_0)=2{bf f}'(s_0)cdot({bf f}(s_0)-{bf g}(t_0)bigr)=0quad wedgequadPhi_t(s_0,t_0)=-2{bf g}'(t_0)cdotbigl({bf f}(s_0)-{bf g}(t_0)bigr)=0 .$$
But this is saying that ${bf d}perp{bf f}'(s_0)$ and at the same time ${bf d}perp{bf g}'(t_0)$.
answered Nov 30 '18 at 16:32
Christian BlatterChristian Blatter
172k7113326
add a comment |
This is about two curves $smapsto{bf f}(s)$ and $tmapsto{bf g}(t)$ in ${mathbb R}^n$ and necessary conditions for a local minimum (or maximum) of the distance between two (non-colliding) points on these two curves. The objective function is
$$Phi(s,t):=|{bf f}(s)-{bf g}(t)|^2=bigl( {bf f}(s)-{bf g}(t)bigr)cdotbigl({bf f}(s)-{bf g}(t)bigr) .$$
Assume that for the parameter values $s_0$ and $t_0$ we have ${bf d}:={bf f}(s_0)-{bf g}(t_0)ne{bf 0}$, and that $Phi$ assumes a local minimum at $(s_0,t_0)$. Then
$$Phi_s(s_0,t_0)=2{bf f}'(s_0)cdot({bf f}(s_0)-{bf g}(t_0)bigr)=0quad wedgequadPhi_t(s_0,t_0)=-2{bf g}'(t_0)cdotbigl({bf f}(s_0)-{bf g}(t_0)bigr)=0 .$$
But this is saying that ${bf d}perp{bf f}'(s_0)$ and at the same time ${bf d}perp{bf g}'(t_0)$.
answered Nov 30 '18 at 16:32
Christian BlatterChristian Blatter
172k7113326
add a comment |
This is about two curves $smapsto{bf f}(s)$ and $tmapsto{bf g}(t)$ in ${mathbb R}^n$ and necessary conditions for a local minimum (or maximum) of the distance between two (non-colliding) points on these two curves. The objective function is
$$Phi(s,t):=|{bf f}(s)-{bf g}(t)|^2=bigl( {bf f}(s)-{bf g}(t)bigr)cdotbigl({bf f}(s)-{bf g}(t)bigr) .$$
Assume that for the parameter values $s_0$ and $t_0$ we have ${bf d}:={bf f}(s_0)-{bf g}(t_0)ne{bf 0}$, and that $Phi$ assumes a local minimum at $(s_0,t_0)$. Then
$$Phi_s(s_0,t_0)=2{bf f}'(s_0)cdot({bf f}(s_0)-{bf g}(t_0)bigr)=0quad wedgequadPhi_t(s_0,t_0)=-2{bf g}'(t_0)cdotbigl({bf f}(s_0)-{bf g}(t_0)bigr)=0 .$$
But this is saying that ${bf d}perp{bf f}'(s_0)$ and at the same time ${bf d}perp{bf g}'(t_0)$.
answered Nov 30 '18 at 16:32
Christian BlatterChristian Blatter
172k7113326
This is about two curves $smapsto{bf f}(s)$ and $tmapsto{bf g}(t)$ in ${mathbb R}^n$ and necessary conditions for a local minimum (or maximum) of the distance between two (non-colliding) points on these two curves. The objective function is
$$Phi(s,t):=|{bf f}(s)-{bf g}(t)|^2=bigl( {bf f}(s)-{bf g}(t)bigr)cdotbigl({bf f}(s)-{bf g}(t)bigr) .$$
Assume that for the parameter values $s_0$ and $t_0$ we have ${bf d}:={bf f}(s_0)-{bf g}(t_0)ne{bf 0}$, and that $Phi$ assumes a local minimum at $(s_0,t_0)$. Then
$$Phi_s(s_0,t_0)=2{bf f}'(s_0)cdot({bf f}(s_0)-{bf g}(t_0)bigr)=0quad wedgequadPhi_t(s_0,t_0)=-2{bf g}'(t_0)cdotbigl({bf f}(s_0)-{bf g}(t_0)bigr)=0 .$$
But this is saying that ${bf d}perp{bf f}'(s_0)$ and at the same time ${bf d}perp{bf g}'(t_0)$.
answered Nov 30 '18 at 16:32
Christian BlatterChristian Blatter
172k7113326
answered Nov 30 '18 at 16:32
Christian BlatterChristian Blatter
172k7113326
answered Nov 30 '18 at 16:32
Christian BlatterChristian Blatter
172k7113326
answered Nov 30 '18 at 16:32
Christian BlatterChristian Blatter
172k7113326
172k7113326
add a comment |
add a comment |
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The first order variation is $2mathbf{x}_{Ato B}cdotmathrm{d}mathbf{x}_{Ato B}$.
– user10354138
Nov 30 '18 at 12:09