Mixing
Tangent space at a point and optimization under constraint with the sub-manifold $x^2+2y^2-z^2=1$
$begingroup$
Given is the following set M containing all points $subset mathbb{R}^3$such that $x^2+2y^2-z^2=1$
a) Prove that M is a two dimensional sub-manifold of $mathbb{R}^3$
b) Calculate the tangential space of M in the point $p=(1,0,0)$
c) Determine the points on M with minimal distance to the origin with respect to the euclidean Norm.
This problem is very similar to this one that I asked a month ago : How to find the set of all points in a submanifold minimizing the distance to a given point?
But I want to be sure that I proceeded correctly and made no mistakes/ forgot nothing. Thanks for your feedback.
a) To prove that this is a sub-manifold, we simply take the gradient, find all points that yield zero, and plug those points into our set. If they aren't included in the set, then this set is a sub-manifold. So $(2x,4y,-2z)=(0,0,0)$ which only yields the point $x=y=z=0$ which is not included in our set because $0neq 1$. The dimension is $2$ because the set takes a point in $mathbb{R}^3$ and outputs a point in $mathbb{R}^1$ so $3-1=2$
b) To find the tangent space in a point, we input our point into the gradient and find its kernel. So $(2,0,0)cdot v =(0,0,0)$. So $a_1=0$ and $ a_2 = b, a_3=c$. So $a_2, a_3$ are free variables. Thus the kernel is the span of $b(0,1,0)+c(0,0,1)$
c) This is an optimization problem with constraint. The function we want to optimize is the euclidean distance squared (We can square it because it does not change the result since it is a "monotonically increasing" function. This allows us to get rid of the square roots to make the algebra nicer.) Our constraint is the set M.
So $f(x,y,z)=(x-0)^2+(y-0)^2+(z-0)^2$ and $g(x,y,z)=x^2+2y^2-z^2-1$.
Using Lagrange multiplier, we get :
$(2x,2y,2z)=lambda (2x,4y,-2z)$. If $lambda =1, y=z=0, x^2=1-2*0^2+0^2=1$ so $x=pm 1$
If $lambda = -1, x=y=0$ then we have no solutions since $-z^2$ never equals $1$
If $lambda = frac{1}{2}, x=z=0, 2y^2=1$ so $y=pm frac{1}{sqrt{2}}$
So our candidates are $(pm 1,0,0), (0,pm frac{1}{sqrt{2}},0)$. Now,given that $(0,pm frac{1}{sqrt{2}},0)$ is closer to the origin than $(pm 1,0,0)$, our set of points is $(0,pm frac{1}{sqrt{2}},0)$
Thanks for your feedback !
real-analysis general-topology analysis multivariable-calculus manifolds
asked Jan 25 at 15:13
PoujhPoujh
6161516
$endgroup$
add a comment |
$begingroup$
Given is the following set M containing all points $subset mathbb{R}^3$such that $x^2+2y^2-z^2=1$
a) Prove that M is a two dimensional sub-manifold of $mathbb{R}^3$
b) Calculate the tangential space of M in the point $p=(1,0,0)$
c) Determine the points on M with minimal distance to the origin with respect to the euclidean Norm.
This problem is very similar to this one that I asked a month ago : How to find the set of all points in a submanifold minimizing the distance to a given point?
But I want to be sure that I proceeded correctly and made no mistakes/ forgot nothing. Thanks for your feedback.
a) To prove that this is a sub-manifold, we simply take the gradient, find all points that yield zero, and plug those points into our set. If they aren't included in the set, then this set is a sub-manifold. So $(2x,4y,-2z)=(0,0,0)$ which only yields the point $x=y=z=0$ which is not included in our set because $0neq 1$. The dimension is $2$ because the set takes a point in $mathbb{R}^3$ and outputs a point in $mathbb{R}^1$ so $3-1=2$
b) To find the tangent space in a point, we input our point into the gradient and find its kernel. So $(2,0,0)cdot v =(0,0,0)$. So $a_1=0$ and $ a_2 = b, a_3=c$. So $a_2, a_3$ are free variables. Thus the kernel is the span of $b(0,1,0)+c(0,0,1)$
c) This is an optimization problem with constraint. The function we want to optimize is the euclidean distance squared (We can square it because it does not change the result since it is a "monotonically increasing" function. This allows us to get rid of the square roots to make the algebra nicer.) Our constraint is the set M.
So $f(x,y,z)=(x-0)^2+(y-0)^2+(z-0)^2$ and $g(x,y,z)=x^2+2y^2-z^2-1$.
Using Lagrange multiplier, we get :
$(2x,2y,2z)=lambda (2x,4y,-2z)$. If $lambda =1, y=z=0, x^2=1-2*0^2+0^2=1$ so $x=pm 1$
If $lambda = -1, x=y=0$ then we have no solutions since $-z^2$ never equals $1$
If $lambda = frac{1}{2}, x=z=0, 2y^2=1$ so $y=pm frac{1}{sqrt{2}}$
So our candidates are $(pm 1,0,0), (0,pm frac{1}{sqrt{2}},0)$. Now,given that $(0,pm frac{1}{sqrt{2}},0)$ is closer to the origin than $(pm 1,0,0)$, our set of points is $(0,pm frac{1}{sqrt{2}},0)$
Thanks for your feedback !
real-analysis general-topology analysis multivariable-calculus manifolds
asked Jan 25 at 15:13
PoujhPoujh
6161516
$endgroup$
add a comment |
$begingroup$
Given is the following set M containing all points $subset mathbb{R}^3$such that $x^2+2y^2-z^2=1$
a) Prove that M is a two dimensional sub-manifold of $mathbb{R}^3$
b) Calculate the tangential space of M in the point $p=(1,0,0)$
c) Determine the points on M with minimal distance to the origin with respect to the euclidean Norm.
This problem is very similar to this one that I asked a month ago : How to find the set of all points in a submanifold minimizing the distance to a given point?
But I want to be sure that I proceeded correctly and made no mistakes/ forgot nothing. Thanks for your feedback.
a) To prove that this is a sub-manifold, we simply take the gradient, find all points that yield zero, and plug those points into our set. If they aren't included in the set, then this set is a sub-manifold. So $(2x,4y,-2z)=(0,0,0)$ which only yields the point $x=y=z=0$ which is not included in our set because $0neq 1$. The dimension is $2$ because the set takes a point in $mathbb{R}^3$ and outputs a point in $mathbb{R}^1$ so $3-1=2$
b) To find the tangent space in a point, we input our point into the gradient and find its kernel. So $(2,0,0)cdot v =(0,0,0)$. So $a_1=0$ and $ a_2 = b, a_3=c$. So $a_2, a_3$ are free variables. Thus the kernel is the span of $b(0,1,0)+c(0,0,1)$
c) This is an optimization problem with constraint. The function we want to optimize is the euclidean distance squared (We can square it because it does not change the result since it is a "monotonically increasing" function. This allows us to get rid of the square roots to make the algebra nicer.) Our constraint is the set M.
So $f(x,y,z)=(x-0)^2+(y-0)^2+(z-0)^2$ and $g(x,y,z)=x^2+2y^2-z^2-1$.
Using Lagrange multiplier, we get :
$(2x,2y,2z)=lambda (2x,4y,-2z)$. If $lambda =1, y=z=0, x^2=1-2*0^2+0^2=1$ so $x=pm 1$
If $lambda = -1, x=y=0$ then we have no solutions since $-z^2$ never equals $1$
If $lambda = frac{1}{2}, x=z=0, 2y^2=1$ so $y=pm frac{1}{sqrt{2}}$
So our candidates are $(pm 1,0,0), (0,pm frac{1}{sqrt{2}},0)$. Now,given that $(0,pm frac{1}{sqrt{2}},0)$ is closer to the origin than $(pm 1,0,0)$, our set of points is $(0,pm frac{1}{sqrt{2}},0)$
Thanks for your feedback !
real-analysis general-topology analysis multivariable-calculus manifolds
asked Jan 25 at 15:13
PoujhPoujh
6161516
$endgroup$
Given is the following set M containing all points $subset mathbb{R}^3$such that $x^2+2y^2-z^2=1$
a) Prove that M is a two dimensional sub-manifold of $mathbb{R}^3$
b) Calculate the tangential space of M in the point $p=(1,0,0)$
c) Determine the points on M with minimal distance to the origin with respect to the euclidean Norm.
This problem is very similar to this one that I asked a month ago : How to find the set of all points in a submanifold minimizing the distance to a given point?
But I want to be sure that I proceeded correctly and made no mistakes/ forgot nothing. Thanks for your feedback.
a) To prove that this is a sub-manifold, we simply take the gradient, find all points that yield zero, and plug those points into our set. If they aren't included in the set, then this set is a sub-manifold. So $(2x,4y,-2z)=(0,0,0)$ which only yields the point $x=y=z=0$ which is not included in our set because $0neq 1$. The dimension is $2$ because the set takes a point in $mathbb{R}^3$ and outputs a point in $mathbb{R}^1$ so $3-1=2$
b) To find the tangent space in a point, we input our point into the gradient and find its kernel. So $(2,0,0)cdot v =(0,0,0)$. So $a_1=0$ and $ a_2 = b, a_3=c$. So $a_2, a_3$ are free variables. Thus the kernel is the span of $b(0,1,0)+c(0,0,1)$
c) This is an optimization problem with constraint. The function we want to optimize is the euclidean distance squared (We can square it because it does not change the result since it is a "monotonically increasing" function. This allows us to get rid of the square roots to make the algebra nicer.) Our constraint is the set M.
So $f(x,y,z)=(x-0)^2+(y-0)^2+(z-0)^2$ and $g(x,y,z)=x^2+2y^2-z^2-1$.
Using Lagrange multiplier, we get :
$(2x,2y,2z)=lambda (2x,4y,-2z)$. If $lambda =1, y=z=0, x^2=1-2*0^2+0^2=1$ so $x=pm 1$
If $lambda = -1, x=y=0$ then we have no solutions since $-z^2$ never equals $1$
If $lambda = frac{1}{2}, x=z=0, 2y^2=1$ so $y=pm frac{1}{sqrt{2}}$
So our candidates are $(pm 1,0,0), (0,pm frac{1}{sqrt{2}},0)$. Now,given that $(0,pm frac{1}{sqrt{2}},0)$ is closer to the origin than $(pm 1,0,0)$, our set of points is $(0,pm frac{1}{sqrt{2}},0)$
Thanks for your feedback !
real-analysis general-topology analysis multivariable-calculus manifolds
real-analysis general-topology analysis multivariable-calculus manifolds
asked Jan 25 at 15:13
PoujhPoujh
6161516
asked Jan 25 at 15:13
PoujhPoujh
6161516
asked Jan 25 at 15:13
PoujhPoujh
6161516
asked Jan 25 at 15:13
PoujhPoujh
6161516
asked Jan 25 at 15:13
PoujhPoujh
6161516
6161516
add a comment |
add a comment |
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