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Minimum distance between two parabolas
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The shortest distance between the parabolas $y^2=x-1$ and $x^2=y-1$ is.
Attempt: The shortest distance is along the common normal of the two curves.
geometry
asked Mar 28 '15 at 9:05
yasiryasir
6591818
$endgroup$
add a comment |
$begingroup$
The shortest distance between the parabolas $y^2=x-1$ and $x^2=y-1$ is.
Attempt: The shortest distance is along the common normal of the two curves.
geometry
asked Mar 28 '15 at 9:05
yasiryasir
6591818
$endgroup$
6
$begingroup$
That's hardly an attempt, that's just an idea. An attempt would involve using it. An in general it's wrong, if you looked at $y=x^2$ and $y=x^2+1$, the only common normal would be the $y$-axis, but the distance along that is the maximum.
$endgroup$
– Henrik
Mar 28 '15 at 9:15
3
$begingroup$
What mathematical resources can you use? For example, can you use partial derivatives of a function of two variables?
$endgroup$
– Rory Daulton
Mar 28 '15 at 9:16
$begingroup$
Henrik, yasir's statement is correct in the general case as long as there are no intersections and a minimum distance exists (which your example does not have) since the claim was not that the length of the common normal is always the minimum length but just that the minimum length is on a common normal.
$endgroup$
– Mint
Mar 26 '18 at 4:32
add a comment |
$begingroup$
The shortest distance between the parabolas $y^2=x-1$ and $x^2=y-1$ is.
Attempt: The shortest distance is along the common normal of the two curves.
geometry
asked Mar 28 '15 at 9:05
yasiryasir
6591818
$endgroup$
The shortest distance between the parabolas $y^2=x-1$ and $x^2=y-1$ is.
Attempt: The shortest distance is along the common normal of the two curves.
geometry
geometry
asked Mar 28 '15 at 9:05
yasiryasir
6591818
asked Mar 28 '15 at 9:05
yasiryasir
6591818
edited Mar 28 '15 at 9:33
user164524
asked Mar 28 '15 at 9:05
yasiryasir
6591818
asked Mar 28 '15 at 9:05
yasiryasir
6591818
asked Mar 28 '15 at 9:05
yasiryasir
6591818
6591818
6
$begingroup$
That's hardly an attempt, that's just an idea. An attempt would involve using it. An in general it's wrong, if you looked at $y=x^2$ and $y=x^2+1$, the only common normal would be the $y$-axis, but the distance along that is the maximum.
$endgroup$
– Henrik
Mar 28 '15 at 9:15
3
$begingroup$
What mathematical resources can you use? For example, can you use partial derivatives of a function of two variables?
$endgroup$
– Rory Daulton
Mar 28 '15 at 9:16
$begingroup$
Henrik, yasir's statement is correct in the general case as long as there are no intersections and a minimum distance exists (which your example does not have) since the claim was not that the length of the common normal is always the minimum length but just that the minimum length is on a common normal.
$endgroup$
– Mint
Mar 26 '18 at 4:32
add a comment |
6
$begingroup$
That's hardly an attempt, that's just an idea. An attempt would involve using it. An in general it's wrong, if you looked at $y=x^2$ and $y=x^2+1$, the only common normal would be the $y$-axis, but the distance along that is the maximum.
$endgroup$
– Henrik
Mar 28 '15 at 9:15
3
$begingroup$
What mathematical resources can you use? For example, can you use partial derivatives of a function of two variables?
$endgroup$
– Rory Daulton
Mar 28 '15 at 9:16
$begingroup$
Henrik, yasir's statement is correct in the general case as long as there are no intersections and a minimum distance exists (which your example does not have) since the claim was not that the length of the common normal is always the minimum length but just that the minimum length is on a common normal.
$endgroup$
– Mint
Mar 26 '18 at 4:32
6
6
$begingroup$
That's hardly an attempt, that's just an idea. An attempt would involve using it. An in general it's wrong, if you looked at $y=x^2$ and $y=x^2+1$, the only common normal would be the $y$-axis, but the distance along that is the maximum.
$endgroup$
– Henrik
Mar 28 '15 at 9:15
$begingroup$
That's hardly an attempt, that's just an idea. An attempt would involve using it. An in general it's wrong, if you looked at $y=x^2$ and $y=x^2+1$, the only common normal would be the $y$-axis, but the distance along that is the maximum.
$endgroup$
– Henrik
Mar 28 '15 at 9:15
3
3
$begingroup$
What mathematical resources can you use? For example, can you use partial derivatives of a function of two variables?
$endgroup$
– Rory Daulton
Mar 28 '15 at 9:16
$begingroup$
What mathematical resources can you use? For example, can you use partial derivatives of a function of two variables?
$endgroup$
– Rory Daulton
Mar 28 '15 at 9:16
$begingroup$
Henrik, yasir's statement is correct in the general case as long as there are no intersections and a minimum distance exists (which your example does not have) since the claim was not that the length of the common normal is always the minimum length but just that the minimum length is on a common normal.
$endgroup$
– Mint
Mar 26 '18 at 4:32
$begingroup$
Henrik, yasir's statement is correct in the general case as long as there are no intersections and a minimum distance exists (which your example does not have) since the claim was not that the length of the common normal is always the minimum length but just that the minimum length is on a common normal.
$endgroup$
– Mint
Mar 26 '18 at 4:32
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hope a bit of calculus is allowed. Primes are with respect to x. Notice that they are inverse functions of each other, you can swap $ x,y$ to get to the second parabola. They are mirror images with respect to line $ x=y$. Required point should have this slope $y^{'} =1 $ for its tangent at point of tangency at ends of common normal.
Take the parabola with its symmetry axis coinciding with axis.
Differentiating $ 2 y y'= 1 , 2 y = 1, $ and the $x,y$ coordinates are
$$ (dfrac54,dfrac12)$$
and the other point of tangency is again swapped to
$$ (dfrac12,dfrac54); $$
Now use distance formula between them getting
$ d = dfrac{3 sqrt{2}}{4}. $
answered Mar 28 '15 at 10:22
NarasimhamNarasimham
21.1k62258
$endgroup$
$begingroup$
thanks. that was quite helpful.
$endgroup$
– yasir
Mar 28 '15 at 14:52
$begingroup$
You are welcome.
$endgroup$
– Narasimham
Mar 28 '15 at 15:27
1
$begingroup$
Could you explain why required point should have this slope $y′=1$ for its tangent at point of tangency at ends of common normal?
$endgroup$
– MathGeek
Aug 28 '16 at 17:37
$begingroup$
As shown in the sketch since the parabolas are symmetric with respect to $x=y, $ the common normal should have the same slope $=-1$ and the parallels should have same slope $=+1.$.
$endgroup$
– Narasimham
Aug 28 '16 at 18:23
2
$begingroup$
From the graph of any function its inverse fuction grapph can be obtained by mirroring about the straight line $x=y$. From points of maximum (or minimum) distance drop normals on to this line $x=y$. It is in fact the same normal line.The parallel lines have slope $m_1= +1$ and normal lines have slope $m_2= -1/m_1 = -1$.
$endgroup$
– Narasimham
Mar 16 '18 at 11:28
|
show 1 more comment
$begingroup$
A point lying on the first parabola has coordinates $(1+u^2,u)$ while a point lying on the second parabola has coordinates $(v,1+v^2)$, hence the squared distance between them is given by:
$$ d(u,v) = (1+u^2-v)^2 + (1+v^2-u)^2 $$
and the stationary points for such a function are given by the solutions of:
$$ frac{partial d}{partial u}= 4u(1+u^2-v)-2(1+v^2-u)=0,$$
$$ frac{partial d}{partial v}= 4v(1+v^2-u)-2(1+u^2-v)=0,$$
hence they fulfill, by taking the difference between the two equations:
$$ u(3+u+2u^2)=v(3+v+2v^2)tag{1} $$
but since $frac{d}{dt}left(t(3+t+2t^2)right) = 3+2t+6t^2$ has a negative discriminant we have that $tto t(3+t+2t^2)$ is an injective function, hence $(1)$ implies $u=v$ and we have:
$$ (4u-2)(1-u+u^2) = 0 tag{2}$$
from which $u=v=frac{1}{2}$ and:
$$ min_{u,v} {d(u,v)}^2 = 2cdotleft(1+frac{1}{4}-frac{1}{2}right)^2 = frac{9}{8}.tag{3}$$
edited Mar 28 '15 at 11:16
Narasimham
21.1k62258
answered Mar 28 '15 at 10:21
Jack D'AurizioJack D'Aurizio
292k33284672
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add a comment |
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let the shortest distance between the parabolas $y = 1 + x^2$ and $x = 1 + y^2$ be given by $AB,$ where $A = (a , 1 + a^2), B = (1 + b^2, b)$ with both $a$ and $b$ positive. the tangent at $A$ has slope $2a$ and the one at $B$ has slope $frac 1{2b}$ and the slope of $AB$ is $frac{1+a^2 - b}{a-1-b^2}$
we need $$2a = frac 1{2b} = frac{1+b^2 - a}{1 + a^2 - b}$$ so that both tangents are parallel and orthogonal to $AB.$ this gives $$b = frac 1 a, , 2a = frac{1 + frac 1 {a^2} - a}{1 + a^2 - frac 1 a} = frac{a^2 + 1 - a^3}{a(a+a^3 -1)} to 2a^3+2a^5-2a^2=a^2 + 1 - a^3$$ that is $$2a^5+3a^3 - 3a^2 - 1 = 0, a > 0. $$
numerically(ti-83 solver), i found $a = 0.9072$ to be the only solution.
answered Mar 28 '15 at 11:39
abelabel
26.6k12148
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add a comment |
$begingroup$
The two parabolas are clearly mirror images by the main diagonal ($x=y$), which diagonal they don't intersect (as $x^2-x+1>0$ for all real $x$). The shortest path then must be perpendicular to that diagonal, and the tangent to the parabolas parallel to the diagonal. The second parabola is the graph of the function $xmapsto x^2+1$, and no complicated calculus is needed to see that the only point where its tangent is parallel to the diagonal is $(x,y)=(frac12,frac54)$. The distance to its mirror image $(frac54,frac12)$ is $sqrt2,left|frac54-frac12right|=frac34sqrt2$.
answered Mar 28 '15 at 13:19
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
$endgroup$
1
$begingroup$
Really the same answer as that of Narashimham, as I discovered later. Please upvote that one (I've got plenty of points already).
$endgroup$
– Marc van Leeuwen
Mar 28 '15 at 13:25
add a comment |
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4 Answers
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active
oldest
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4 Answers
4
active
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$begingroup$
Hope a bit of calculus is allowed. Primes are with respect to x. Notice that they are inverse functions of each other, you can swap $ x,y$ to get to the second parabola. They are mirror images with respect to line $ x=y$. Required point should have this slope $y^{'} =1 $ for its tangent at point of tangency at ends of common normal.
Take the parabola with its symmetry axis coinciding with axis.
Differentiating $ 2 y y'= 1 , 2 y = 1, $ and the $x,y$ coordinates are
$$ (dfrac54,dfrac12)$$
and the other point of tangency is again swapped to
$$ (dfrac12,dfrac54); $$
Now use distance formula between them getting
$ d = dfrac{3 sqrt{2}}{4}. $
answered Mar 28 '15 at 10:22
NarasimhamNarasimham
21.1k62258
$endgroup$
$begingroup$
thanks. that was quite helpful.
$endgroup$
– yasir
Mar 28 '15 at 14:52
$begingroup$
You are welcome.
$endgroup$
– Narasimham
Mar 28 '15 at 15:27
1
$begingroup$
Could you explain why required point should have this slope $y′=1$ for its tangent at point of tangency at ends of common normal?
$endgroup$
– MathGeek
Aug 28 '16 at 17:37
$begingroup$
As shown in the sketch since the parabolas are symmetric with respect to $x=y, $ the common normal should have the same slope $=-1$ and the parallels should have same slope $=+1.$.
$endgroup$
– Narasimham
Aug 28 '16 at 18:23
2
$begingroup$
From the graph of any function its inverse fuction grapph can be obtained by mirroring about the straight line $x=y$. From points of maximum (or minimum) distance drop normals on to this line $x=y$. It is in fact the same normal line.The parallel lines have slope $m_1= +1$ and normal lines have slope $m_2= -1/m_1 = -1$.
$endgroup$
– Narasimham
Mar 16 '18 at 11:28
|
show 1 more comment
$begingroup$
Hope a bit of calculus is allowed. Primes are with respect to x. Notice that they are inverse functions of each other, you can swap $ x,y$ to get to the second parabola. They are mirror images with respect to line $ x=y$. Required point should have this slope $y^{'} =1 $ for its tangent at point of tangency at ends of common normal.
Take the parabola with its symmetry axis coinciding with axis.
Differentiating $ 2 y y'= 1 , 2 y = 1, $ and the $x,y$ coordinates are
$$ (dfrac54,dfrac12)$$
and the other point of tangency is again swapped to
$$ (dfrac12,dfrac54); $$
Now use distance formula between them getting
$ d = dfrac{3 sqrt{2}}{4}. $
answered Mar 28 '15 at 10:22
NarasimhamNarasimham
21.1k62258
$endgroup$
$begingroup$
thanks. that was quite helpful.
$endgroup$
– yasir
Mar 28 '15 at 14:52
$begingroup$
You are welcome.
$endgroup$
– Narasimham
Mar 28 '15 at 15:27
1
$begingroup$
Could you explain why required point should have this slope $y′=1$ for its tangent at point of tangency at ends of common normal?
$endgroup$
– MathGeek
Aug 28 '16 at 17:37
$begingroup$
As shown in the sketch since the parabolas are symmetric with respect to $x=y, $ the common normal should have the same slope $=-1$ and the parallels should have same slope $=+1.$.
$endgroup$
– Narasimham
Aug 28 '16 at 18:23
2
$begingroup$
From the graph of any function its inverse fuction grapph can be obtained by mirroring about the straight line $x=y$. From points of maximum (or minimum) distance drop normals on to this line $x=y$. It is in fact the same normal line.The parallel lines have slope $m_1= +1$ and normal lines have slope $m_2= -1/m_1 = -1$.
$endgroup$
– Narasimham
Mar 16 '18 at 11:28
|
show 1 more comment
$begingroup$
Hope a bit of calculus is allowed. Primes are with respect to x. Notice that they are inverse functions of each other, you can swap $ x,y$ to get to the second parabola. They are mirror images with respect to line $ x=y$. Required point should have this slope $y^{'} =1 $ for its tangent at point of tangency at ends of common normal.
Take the parabola with its symmetry axis coinciding with axis.
Differentiating $ 2 y y'= 1 , 2 y = 1, $ and the $x,y$ coordinates are
$$ (dfrac54,dfrac12)$$
and the other point of tangency is again swapped to
$$ (dfrac12,dfrac54); $$
Now use distance formula between them getting
$ d = dfrac{3 sqrt{2}}{4}. $
answered Mar 28 '15 at 10:22
NarasimhamNarasimham
21.1k62258
$endgroup$
Hope a bit of calculus is allowed. Primes are with respect to x. Notice that they are inverse functions of each other, you can swap $ x,y$ to get to the second parabola. They are mirror images with respect to line $ x=y$. Required point should have this slope $y^{'} =1 $ for its tangent at point of tangency at ends of common normal.
Take the parabola with its symmetry axis coinciding with axis.
Differentiating $ 2 y y'= 1 , 2 y = 1, $ and the $x,y$ coordinates are
$$ (dfrac54,dfrac12)$$
and the other point of tangency is again swapped to
$$ (dfrac12,dfrac54); $$
Now use distance formula between them getting
$ d = dfrac{3 sqrt{2}}{4}. $
answered Mar 28 '15 at 10:22
NarasimhamNarasimham
21.1k62258
edited Jan 31 at 16:20
answered Mar 28 '15 at 10:22
NarasimhamNarasimham
21.1k62258
answered Mar 28 '15 at 10:22
NarasimhamNarasimham
21.1k62258
answered Mar 28 '15 at 10:22
NarasimhamNarasimham
21.1k62258
21.1k62258
$begingroup$
thanks. that was quite helpful.
$endgroup$
– yasir
Mar 28 '15 at 14:52
$begingroup$
You are welcome.
$endgroup$
– Narasimham
Mar 28 '15 at 15:27
1
$begingroup$
Could you explain why required point should have this slope $y′=1$ for its tangent at point of tangency at ends of common normal?
$endgroup$
– MathGeek
Aug 28 '16 at 17:37
$begingroup$
As shown in the sketch since the parabolas are symmetric with respect to $x=y, $ the common normal should have the same slope $=-1$ and the parallels should have same slope $=+1.$.
$endgroup$
– Narasimham
Aug 28 '16 at 18:23
2
$begingroup$
From the graph of any function its inverse fuction grapph can be obtained by mirroring about the straight line $x=y$. From points of maximum (or minimum) distance drop normals on to this line $x=y$. It is in fact the same normal line.The parallel lines have slope $m_1= +1$ and normal lines have slope $m_2= -1/m_1 = -1$.
$endgroup$
– Narasimham
Mar 16 '18 at 11:28
|
show 1 more comment
$begingroup$
thanks. that was quite helpful.
$endgroup$
– yasir
Mar 28 '15 at 14:52
$begingroup$
You are welcome.
$endgroup$
– Narasimham
Mar 28 '15 at 15:27
1
$begingroup$
Could you explain why required point should have this slope $y′=1$ for its tangent at point of tangency at ends of common normal?
$endgroup$
– MathGeek
Aug 28 '16 at 17:37
$begingroup$
As shown in the sketch since the parabolas are symmetric with respect to $x=y, $ the common normal should have the same slope $=-1$ and the parallels should have same slope $=+1.$.
$endgroup$
– Narasimham
Aug 28 '16 at 18:23
2
$begingroup$
From the graph of any function its inverse fuction grapph can be obtained by mirroring about the straight line $x=y$. From points of maximum (or minimum) distance drop normals on to this line $x=y$. It is in fact the same normal line.The parallel lines have slope $m_1= +1$ and normal lines have slope $m_2= -1/m_1 = -1$.
$endgroup$
– Narasimham
Mar 16 '18 at 11:28
$begingroup$
thanks. that was quite helpful.
$endgroup$
– yasir
Mar 28 '15 at 14:52
$begingroup$
thanks. that was quite helpful.
$endgroup$
– yasir
Mar 28 '15 at 14:52
$begingroup$
You are welcome.
$endgroup$
– Narasimham
Mar 28 '15 at 15:27
$begingroup$
You are welcome.
$endgroup$
– Narasimham
Mar 28 '15 at 15:27
1
1
$begingroup$
Could you explain why required point should have this slope $y′=1$ for its tangent at point of tangency at ends of common normal?
$endgroup$
– MathGeek
Aug 28 '16 at 17:37
$begingroup$
Could you explain why required point should have this slope $y′=1$ for its tangent at point of tangency at ends of common normal?
$endgroup$
– MathGeek
Aug 28 '16 at 17:37
$begingroup$
As shown in the sketch since the parabolas are symmetric with respect to $x=y, $ the common normal should have the same slope $=-1$ and the parallels should have same slope $=+1.$.
$endgroup$
– Narasimham
Aug 28 '16 at 18:23
$begingroup$
As shown in the sketch since the parabolas are symmetric with respect to $x=y, $ the common normal should have the same slope $=-1$ and the parallels should have same slope $=+1.$.
$endgroup$
– Narasimham
Aug 28 '16 at 18:23
2
2
$begingroup$
From the graph of any function its inverse fuction grapph can be obtained by mirroring about the straight line $x=y$. From points of maximum (or minimum) distance drop normals on to this line $x=y$. It is in fact the same normal line.The parallel lines have slope $m_1= +1$ and normal lines have slope $m_2= -1/m_1 = -1$.
$endgroup$
– Narasimham
Mar 16 '18 at 11:28
$begingroup$
From the graph of any function its inverse fuction grapph can be obtained by mirroring about the straight line $x=y$. From points of maximum (or minimum) distance drop normals on to this line $x=y$. It is in fact the same normal line.The parallel lines have slope $m_1= +1$ and normal lines have slope $m_2= -1/m_1 = -1$.
$endgroup$
– Narasimham
Mar 16 '18 at 11:28
|
show 1 more comment
$begingroup$
A point lying on the first parabola has coordinates $(1+u^2,u)$ while a point lying on the second parabola has coordinates $(v,1+v^2)$, hence the squared distance between them is given by:
$$ d(u,v) = (1+u^2-v)^2 + (1+v^2-u)^2 $$
and the stationary points for such a function are given by the solutions of:
$$ frac{partial d}{partial u}= 4u(1+u^2-v)-2(1+v^2-u)=0,$$
$$ frac{partial d}{partial v}= 4v(1+v^2-u)-2(1+u^2-v)=0,$$
hence they fulfill, by taking the difference between the two equations:
$$ u(3+u+2u^2)=v(3+v+2v^2)tag{1} $$
but since $frac{d}{dt}left(t(3+t+2t^2)right) = 3+2t+6t^2$ has a negative discriminant we have that $tto t(3+t+2t^2)$ is an injective function, hence $(1)$ implies $u=v$ and we have:
$$ (4u-2)(1-u+u^2) = 0 tag{2}$$
from which $u=v=frac{1}{2}$ and:
$$ min_{u,v} {d(u,v)}^2 = 2cdotleft(1+frac{1}{4}-frac{1}{2}right)^2 = frac{9}{8}.tag{3}$$
edited Mar 28 '15 at 11:16
Narasimham
21.1k62258
answered Mar 28 '15 at 10:21
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
A point lying on the first parabola has coordinates $(1+u^2,u)$ while a point lying on the second parabola has coordinates $(v,1+v^2)$, hence the squared distance between them is given by:
$$ d(u,v) = (1+u^2-v)^2 + (1+v^2-u)^2 $$
and the stationary points for such a function are given by the solutions of:
$$ frac{partial d}{partial u}= 4u(1+u^2-v)-2(1+v^2-u)=0,$$
$$ frac{partial d}{partial v}= 4v(1+v^2-u)-2(1+u^2-v)=0,$$
hence they fulfill, by taking the difference between the two equations:
$$ u(3+u+2u^2)=v(3+v+2v^2)tag{1} $$
but since $frac{d}{dt}left(t(3+t+2t^2)right) = 3+2t+6t^2$ has a negative discriminant we have that $tto t(3+t+2t^2)$ is an injective function, hence $(1)$ implies $u=v$ and we have:
$$ (4u-2)(1-u+u^2) = 0 tag{2}$$
from which $u=v=frac{1}{2}$ and:
$$ min_{u,v} {d(u,v)}^2 = 2cdotleft(1+frac{1}{4}-frac{1}{2}right)^2 = frac{9}{8}.tag{3}$$
edited Mar 28 '15 at 11:16
Narasimham
21.1k62258
answered Mar 28 '15 at 10:21
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
add a comment |
$begingroup$
A point lying on the first parabola has coordinates $(1+u^2,u)$ while a point lying on the second parabola has coordinates $(v,1+v^2)$, hence the squared distance between them is given by:
$$ d(u,v) = (1+u^2-v)^2 + (1+v^2-u)^2 $$
and the stationary points for such a function are given by the solutions of:
$$ frac{partial d}{partial u}= 4u(1+u^2-v)-2(1+v^2-u)=0,$$
$$ frac{partial d}{partial v}= 4v(1+v^2-u)-2(1+u^2-v)=0,$$
hence they fulfill, by taking the difference between the two equations:
$$ u(3+u+2u^2)=v(3+v+2v^2)tag{1} $$
but since $frac{d}{dt}left(t(3+t+2t^2)right) = 3+2t+6t^2$ has a negative discriminant we have that $tto t(3+t+2t^2)$ is an injective function, hence $(1)$ implies $u=v$ and we have:
$$ (4u-2)(1-u+u^2) = 0 tag{2}$$
from which $u=v=frac{1}{2}$ and:
$$ min_{u,v} {d(u,v)}^2 = 2cdotleft(1+frac{1}{4}-frac{1}{2}right)^2 = frac{9}{8}.tag{3}$$
edited Mar 28 '15 at 11:16
Narasimham
21.1k62258
answered Mar 28 '15 at 10:21
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
A point lying on the first parabola has coordinates $(1+u^2,u)$ while a point lying on the second parabola has coordinates $(v,1+v^2)$, hence the squared distance between them is given by:
$$ d(u,v) = (1+u^2-v)^2 + (1+v^2-u)^2 $$
and the stationary points for such a function are given by the solutions of:
$$ frac{partial d}{partial u}= 4u(1+u^2-v)-2(1+v^2-u)=0,$$
$$ frac{partial d}{partial v}= 4v(1+v^2-u)-2(1+u^2-v)=0,$$
hence they fulfill, by taking the difference between the two equations:
$$ u(3+u+2u^2)=v(3+v+2v^2)tag{1} $$
but since $frac{d}{dt}left(t(3+t+2t^2)right) = 3+2t+6t^2$ has a negative discriminant we have that $tto t(3+t+2t^2)$ is an injective function, hence $(1)$ implies $u=v$ and we have:
$$ (4u-2)(1-u+u^2) = 0 tag{2}$$
from which $u=v=frac{1}{2}$ and:
$$ min_{u,v} {d(u,v)}^2 = 2cdotleft(1+frac{1}{4}-frac{1}{2}right)^2 = frac{9}{8}.tag{3}$$
edited Mar 28 '15 at 11:16
Narasimham
21.1k62258
answered Mar 28 '15 at 10:21
Jack D'AurizioJack D'Aurizio
292k33284672
edited Mar 28 '15 at 11:16
Narasimham
21.1k62258
edited Mar 28 '15 at 11:16
Narasimham
21.1k62258
edited Mar 28 '15 at 11:16
Narasimham
21.1k62258
21.1k62258
answered Mar 28 '15 at 10:21
Jack D'AurizioJack D'Aurizio
292k33284672
answered Mar 28 '15 at 10:21
Jack D'AurizioJack D'Aurizio
292k33284672
answered Mar 28 '15 at 10:21
Jack D'AurizioJack D'Aurizio
292k33284672
292k33284672
add a comment |
add a comment |
$begingroup$
let the shortest distance between the parabolas $y = 1 + x^2$ and $x = 1 + y^2$ be given by $AB,$ where $A = (a , 1 + a^2), B = (1 + b^2, b)$ with both $a$ and $b$ positive. the tangent at $A$ has slope $2a$ and the one at $B$ has slope $frac 1{2b}$ and the slope of $AB$ is $frac{1+a^2 - b}{a-1-b^2}$
we need $$2a = frac 1{2b} = frac{1+b^2 - a}{1 + a^2 - b}$$ so that both tangents are parallel and orthogonal to $AB.$ this gives $$b = frac 1 a, , 2a = frac{1 + frac 1 {a^2} - a}{1 + a^2 - frac 1 a} = frac{a^2 + 1 - a^3}{a(a+a^3 -1)} to 2a^3+2a^5-2a^2=a^2 + 1 - a^3$$ that is $$2a^5+3a^3 - 3a^2 - 1 = 0, a > 0. $$
numerically(ti-83 solver), i found $a = 0.9072$ to be the only solution.
answered Mar 28 '15 at 11:39
abelabel
26.6k12148
$endgroup$
add a comment |
$begingroup$
let the shortest distance between the parabolas $y = 1 + x^2$ and $x = 1 + y^2$ be given by $AB,$ where $A = (a , 1 + a^2), B = (1 + b^2, b)$ with both $a$ and $b$ positive. the tangent at $A$ has slope $2a$ and the one at $B$ has slope $frac 1{2b}$ and the slope of $AB$ is $frac{1+a^2 - b}{a-1-b^2}$
we need $$2a = frac 1{2b} = frac{1+b^2 - a}{1 + a^2 - b}$$ so that both tangents are parallel and orthogonal to $AB.$ this gives $$b = frac 1 a, , 2a = frac{1 + frac 1 {a^2} - a}{1 + a^2 - frac 1 a} = frac{a^2 + 1 - a^3}{a(a+a^3 -1)} to 2a^3+2a^5-2a^2=a^2 + 1 - a^3$$ that is $$2a^5+3a^3 - 3a^2 - 1 = 0, a > 0. $$
numerically(ti-83 solver), i found $a = 0.9072$ to be the only solution.
answered Mar 28 '15 at 11:39
abelabel
26.6k12148
$endgroup$
add a comment |
$begingroup$
let the shortest distance between the parabolas $y = 1 + x^2$ and $x = 1 + y^2$ be given by $AB,$ where $A = (a , 1 + a^2), B = (1 + b^2, b)$ with both $a$ and $b$ positive. the tangent at $A$ has slope $2a$ and the one at $B$ has slope $frac 1{2b}$ and the slope of $AB$ is $frac{1+a^2 - b}{a-1-b^2}$
we need $$2a = frac 1{2b} = frac{1+b^2 - a}{1 + a^2 - b}$$ so that both tangents are parallel and orthogonal to $AB.$ this gives $$b = frac 1 a, , 2a = frac{1 + frac 1 {a^2} - a}{1 + a^2 - frac 1 a} = frac{a^2 + 1 - a^3}{a(a+a^3 -1)} to 2a^3+2a^5-2a^2=a^2 + 1 - a^3$$ that is $$2a^5+3a^3 - 3a^2 - 1 = 0, a > 0. $$
numerically(ti-83 solver), i found $a = 0.9072$ to be the only solution.
answered Mar 28 '15 at 11:39
abelabel
26.6k12148
$endgroup$
let the shortest distance between the parabolas $y = 1 + x^2$ and $x = 1 + y^2$ be given by $AB,$ where $A = (a , 1 + a^2), B = (1 + b^2, b)$ with both $a$ and $b$ positive. the tangent at $A$ has slope $2a$ and the one at $B$ has slope $frac 1{2b}$ and the slope of $AB$ is $frac{1+a^2 - b}{a-1-b^2}$
we need $$2a = frac 1{2b} = frac{1+b^2 - a}{1 + a^2 - b}$$ so that both tangents are parallel and orthogonal to $AB.$ this gives $$b = frac 1 a, , 2a = frac{1 + frac 1 {a^2} - a}{1 + a^2 - frac 1 a} = frac{a^2 + 1 - a^3}{a(a+a^3 -1)} to 2a^3+2a^5-2a^2=a^2 + 1 - a^3$$ that is $$2a^5+3a^3 - 3a^2 - 1 = 0, a > 0. $$
numerically(ti-83 solver), i found $a = 0.9072$ to be the only solution.
answered Mar 28 '15 at 11:39
abelabel
26.6k12148
answered Mar 28 '15 at 11:39
abelabel
26.6k12148
answered Mar 28 '15 at 11:39
abelabel
26.6k12148
answered Mar 28 '15 at 11:39
abelabel
26.6k12148
26.6k12148
add a comment |
add a comment |
$begingroup$
The two parabolas are clearly mirror images by the main diagonal ($x=y$), which diagonal they don't intersect (as $x^2-x+1>0$ for all real $x$). The shortest path then must be perpendicular to that diagonal, and the tangent to the parabolas parallel to the diagonal. The second parabola is the graph of the function $xmapsto x^2+1$, and no complicated calculus is needed to see that the only point where its tangent is parallel to the diagonal is $(x,y)=(frac12,frac54)$. The distance to its mirror image $(frac54,frac12)$ is $sqrt2,left|frac54-frac12right|=frac34sqrt2$.
answered Mar 28 '15 at 13:19
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
$endgroup$
1
$begingroup$
Really the same answer as that of Narashimham, as I discovered later. Please upvote that one (I've got plenty of points already).
$endgroup$
– Marc van Leeuwen
Mar 28 '15 at 13:25
add a comment |
$begingroup$
The two parabolas are clearly mirror images by the main diagonal ($x=y$), which diagonal they don't intersect (as $x^2-x+1>0$ for all real $x$). The shortest path then must be perpendicular to that diagonal, and the tangent to the parabolas parallel to the diagonal. The second parabola is the graph of the function $xmapsto x^2+1$, and no complicated calculus is needed to see that the only point where its tangent is parallel to the diagonal is $(x,y)=(frac12,frac54)$. The distance to its mirror image $(frac54,frac12)$ is $sqrt2,left|frac54-frac12right|=frac34sqrt2$.
answered Mar 28 '15 at 13:19
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
$endgroup$
1
$begingroup$
Really the same answer as that of Narashimham, as I discovered later. Please upvote that one (I've got plenty of points already).
$endgroup$
– Marc van Leeuwen
Mar 28 '15 at 13:25
add a comment |
$begingroup$
The two parabolas are clearly mirror images by the main diagonal ($x=y$), which diagonal they don't intersect (as $x^2-x+1>0$ for all real $x$). The shortest path then must be perpendicular to that diagonal, and the tangent to the parabolas parallel to the diagonal. The second parabola is the graph of the function $xmapsto x^2+1$, and no complicated calculus is needed to see that the only point where its tangent is parallel to the diagonal is $(x,y)=(frac12,frac54)$. The distance to its mirror image $(frac54,frac12)$ is $sqrt2,left|frac54-frac12right|=frac34sqrt2$.
answered Mar 28 '15 at 13:19
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
$endgroup$
The two parabolas are clearly mirror images by the main diagonal ($x=y$), which diagonal they don't intersect (as $x^2-x+1>0$ for all real $x$). The shortest path then must be perpendicular to that diagonal, and the tangent to the parabolas parallel to the diagonal. The second parabola is the graph of the function $xmapsto x^2+1$, and no complicated calculus is needed to see that the only point where its tangent is parallel to the diagonal is $(x,y)=(frac12,frac54)$. The distance to its mirror image $(frac54,frac12)$ is $sqrt2,left|frac54-frac12right|=frac34sqrt2$.
answered Mar 28 '15 at 13:19
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
answered Mar 28 '15 at 13:19
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
answered Mar 28 '15 at 13:19
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
answered Mar 28 '15 at 13:19
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
88.7k5111230
1
$begingroup$
Really the same answer as that of Narashimham, as I discovered later. Please upvote that one (I've got plenty of points already).
$endgroup$
– Marc van Leeuwen
Mar 28 '15 at 13:25
add a comment |
1
$begingroup$
Really the same answer as that of Narashimham, as I discovered later. Please upvote that one (I've got plenty of points already).
$endgroup$
– Marc van Leeuwen
Mar 28 '15 at 13:25
1
1
$begingroup$
Really the same answer as that of Narashimham, as I discovered later. Please upvote that one (I've got plenty of points already).
$endgroup$
– Marc van Leeuwen
Mar 28 '15 at 13:25
$begingroup$
Really the same answer as that of Narashimham, as I discovered later. Please upvote that one (I've got plenty of points already).
$endgroup$
– Marc van Leeuwen
Mar 28 '15 at 13:25
add a comment |
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That's hardly an attempt, that's just an idea. An attempt would involve using it. An in general it's wrong, if you looked at $y=x^2$ and $y=x^2+1$, the only common normal would be the $y$-axis, but the distance along that is the maximum.
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– Henrik
Mar 28 '15 at 9:15
3
$begingroup$
What mathematical resources can you use? For example, can you use partial derivatives of a function of two variables?
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– Rory Daulton
Mar 28 '15 at 9:16
$begingroup$
Henrik, yasir's statement is correct in the general case as long as there are no intersections and a minimum distance exists (which your example does not have) since the claim was not that the length of the common normal is always the minimum length but just that the minimum length is on a common normal.
$endgroup$
– Mint
Mar 26 '18 at 4:32