Mixing
Finding Tangent Plane
Clash Royale CLAN TAG#URR8PPP
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I was given the following theorems:
let $M$ be a $k$ dimension manifold defined by $M=ain mathbbR^n:F(x)=0$ when $Fin C^1$ and Rank$(D_f(x))$ is maximal for all $ain M$ so $T_a(M)=vin mathbbR^n:D_F(x)v=0$
Let $M$ be a $k$ dimension manifold with atlas $A$ so for all $ain M$
$T_a(M)=t_1fracpartial phipartial x_1(x)+...t_kfracpartial phipartial x_k(x)$ where $t_1,t_2,...,t_kin mathbbR$, $phiin A$ such that $phi(x)=a$
Now I am trying to find the tangent plane of using those theorems:
a. $S=(x,y): x^2+y^2=R^2$ at $(fracRsqrt2,-fracRsqrt2)$
b. $z=x^2+y^2+5$ at $(1,1,7)$
a by 1: First we set $F=x^2+y^2$ now we take $D_F=beginpmatrix
fracpartial Fpartial x & fracpartial Fpartial y
endpmatrix=beginpmatrix
2x & 2y \
endpmatrix$
(we have to check that that the Rank of $D_F$ is maximal for all $ain M$? what if it is not?)
For any $(a_1,a_2)=ain x^2+y^2$ the Rank of $(2a_1 , 2a_2)$ will be at most $1$ as else we had $S=(0,0)$
now we take the dot product $beginpmatrix
2x & 2y
endpmatrix beginpmatrix
fracRsqrt2 \
-fracRsqrt2
endpmatrix=0iff=Rsqrt2x-Rsqrt2y=0iff Rsqrt2(x-y)=0$ so $x=y$?
b by 1: $F=z-x^2-y^2$ now $D_F=beginpmatrix
fracpartial Fpartial x & fracpartial Fpartial y & fracpartial Fpartial z
endpmatrix=beginpmatrix
-2x & -2y & 1 \
endpmatrix$ So $beginpmatrix
-2x & -2y & 1 \
endpmatrixbeginpmatrix
1 \
1 \
7
endpmatrix=0iff -2x-2y+7=0iff 2x+2y=7$
Is it correct? How can I use method $2$ to find the tangent plane?
real-analysis multivariable-calculus
asked Jul 30 at 9:56
newhere
742310
add a comment |Â
up vote
0
down vote
favorite
I was given the following theorems:
let $M$ be a $k$ dimension manifold defined by $M=ain mathbbR^n:F(x)=0$ when $Fin C^1$ and Rank$(D_f(x))$ is maximal for all $ain M$ so $T_a(M)=vin mathbbR^n:D_F(x)v=0$
Let $M$ be a $k$ dimension manifold with atlas $A$ so for all $ain M$
$T_a(M)=t_1fracpartial phipartial x_1(x)+...t_kfracpartial phipartial x_k(x)$ where $t_1,t_2,...,t_kin mathbbR$, $phiin A$ such that $phi(x)=a$
Now I am trying to find the tangent plane of using those theorems:
a. $S=(x,y): x^2+y^2=R^2$ at $(fracRsqrt2,-fracRsqrt2)$
b. $z=x^2+y^2+5$ at $(1,1,7)$
a by 1: First we set $F=x^2+y^2$ now we take $D_F=beginpmatrix
fracpartial Fpartial x & fracpartial Fpartial y
endpmatrix=beginpmatrix
2x & 2y \
endpmatrix$
(we have to check that that the Rank of $D_F$ is maximal for all $ain M$? what if it is not?)
For any $(a_1,a_2)=ain x^2+y^2$ the Rank of $(2a_1 , 2a_2)$ will be at most $1$ as else we had $S=(0,0)$
now we take the dot product $beginpmatrix
2x & 2y
endpmatrix beginpmatrix
fracRsqrt2 \
-fracRsqrt2
endpmatrix=0iff=Rsqrt2x-Rsqrt2y=0iff Rsqrt2(x-y)=0$ so $x=y$?
b by 1: $F=z-x^2-y^2$ now $D_F=beginpmatrix
fracpartial Fpartial x & fracpartial Fpartial y & fracpartial Fpartial z
endpmatrix=beginpmatrix
-2x & -2y & 1 \
endpmatrix$ So $beginpmatrix
-2x & -2y & 1 \
endpmatrixbeginpmatrix
1 \
1 \
7
endpmatrix=0iff -2x-2y+7=0iff 2x+2y=7$
Is it correct? How can I use method $2$ to find the tangent plane?
real-analysis multivariable-calculus
asked Jul 30 at 9:56
newhere
742310
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was given the following theorems:
let $M$ be a $k$ dimension manifold defined by $M=ain mathbbR^n:F(x)=0$ when $Fin C^1$ and Rank$(D_f(x))$ is maximal for all $ain M$ so $T_a(M)=vin mathbbR^n:D_F(x)v=0$
Let $M$ be a $k$ dimension manifold with atlas $A$ so for all $ain M$
$T_a(M)=t_1fracpartial phipartial x_1(x)+...t_kfracpartial phipartial x_k(x)$ where $t_1,t_2,...,t_kin mathbbR$, $phiin A$ such that $phi(x)=a$
Now I am trying to find the tangent plane of using those theorems:
a. $S=(x,y): x^2+y^2=R^2$ at $(fracRsqrt2,-fracRsqrt2)$
b. $z=x^2+y^2+5$ at $(1,1,7)$
a by 1: First we set $F=x^2+y^2$ now we take $D_F=beginpmatrix
fracpartial Fpartial x & fracpartial Fpartial y
endpmatrix=beginpmatrix
2x & 2y \
endpmatrix$
(we have to check that that the Rank of $D_F$ is maximal for all $ain M$? what if it is not?)
For any $(a_1,a_2)=ain x^2+y^2$ the Rank of $(2a_1 , 2a_2)$ will be at most $1$ as else we had $S=(0,0)$
now we take the dot product $beginpmatrix
2x & 2y
endpmatrix beginpmatrix
fracRsqrt2 \
-fracRsqrt2
endpmatrix=0iff=Rsqrt2x-Rsqrt2y=0iff Rsqrt2(x-y)=0$ so $x=y$?
b by 1: $F=z-x^2-y^2$ now $D_F=beginpmatrix
fracpartial Fpartial x & fracpartial Fpartial y & fracpartial Fpartial z
endpmatrix=beginpmatrix
-2x & -2y & 1 \
endpmatrix$ So $beginpmatrix
-2x & -2y & 1 \
endpmatrixbeginpmatrix
1 \
1 \
7
endpmatrix=0iff -2x-2y+7=0iff 2x+2y=7$
Is it correct? How can I use method $2$ to find the tangent plane?
real-analysis multivariable-calculus
asked Jul 30 at 9:56
newhere
742310
I was given the following theorems:
let $M$ be a $k$ dimension manifold defined by $M=ain mathbbR^n:F(x)=0$ when $Fin C^1$ and Rank$(D_f(x))$ is maximal for all $ain M$ so $T_a(M)=vin mathbbR^n:D_F(x)v=0$
Let $M$ be a $k$ dimension manifold with atlas $A$ so for all $ain M$
$T_a(M)=t_1fracpartial phipartial x_1(x)+...t_kfracpartial phipartial x_k(x)$ where $t_1,t_2,...,t_kin mathbbR$, $phiin A$ such that $phi(x)=a$
Now I am trying to find the tangent plane of using those theorems:
a. $S=(x,y): x^2+y^2=R^2$ at $(fracRsqrt2,-fracRsqrt2)$
b. $z=x^2+y^2+5$ at $(1,1,7)$
a by 1: First we set $F=x^2+y^2$ now we take $D_F=beginpmatrix
fracpartial Fpartial x & fracpartial Fpartial y
endpmatrix=beginpmatrix
2x & 2y \
endpmatrix$
(we have to check that that the Rank of $D_F$ is maximal for all $ain M$? what if it is not?)
For any $(a_1,a_2)=ain x^2+y^2$ the Rank of $(2a_1 , 2a_2)$ will be at most $1$ as else we had $S=(0,0)$
now we take the dot product $beginpmatrix
2x & 2y
endpmatrix beginpmatrix
fracRsqrt2 \
-fracRsqrt2
endpmatrix=0iff=Rsqrt2x-Rsqrt2y=0iff Rsqrt2(x-y)=0$ so $x=y$?
b by 1: $F=z-x^2-y^2$ now $D_F=beginpmatrix
fracpartial Fpartial x & fracpartial Fpartial y & fracpartial Fpartial z
endpmatrix=beginpmatrix
-2x & -2y & 1 \
endpmatrix$ So $beginpmatrix
-2x & -2y & 1 \
endpmatrixbeginpmatrix
1 \
1 \
7
endpmatrix=0iff -2x-2y+7=0iff 2x+2y=7$
Is it correct? How can I use method $2$ to find the tangent plane?
real-analysis multivariable-calculus
asked Jul 30 at 9:56
newhere
742310
edited Jul 30 at 13:01
asked Jul 30 at 9:56
newhere
742310
asked Jul 30 at 9:56
newhere
742310
asked Jul 30 at 9:56
newhere
742310
742310
add a comment |Â
add a comment |Â
1 Answer
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If the smooth manifold is given by $vecF(X) = vec0$, then the tangent space to the manifold at point $A$ is the null space (or kernel) of $D_XF(A)$.
In your case, $X = beginbmatrixx \ y \ z endbmatrix$ ,
$A=beginbmatrix1 \ 1 \ 7 endbmatrix$,
$F(X) = z-x^2-y^2$, and the manifold is given by $F(X)=0$.
$D_XF(A) = [-2*1, -2*1, 1*7] = [-2,-2,7]$.
The tangent space at $A$ is the solution $vecdotX = beginbmatrixdotx \ doty \ dotz endbmatrix$ to $[-2,-2,7]vecdotX = 0$
To get the points on the tangent plane at A, add $veca$ to the vectors $vecdotX$ belonging to tangent space.
Hope that helps!
answered Jul 30 at 11:50
Suhan Shetty
835
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If the smooth manifold is given by $vecF(X) = vec0$, then the tangent space to the manifold at point $A$ is the null space (or kernel) of $D_XF(A)$.
In your case, $X = beginbmatrixx \ y \ z endbmatrix$ ,
$A=beginbmatrix1 \ 1 \ 7 endbmatrix$,
$F(X) = z-x^2-y^2$, and the manifold is given by $F(X)=0$.
$D_XF(A) = [-2*1, -2*1, 1*7] = [-2,-2,7]$.
The tangent space at $A$ is the solution $vecdotX = beginbmatrixdotx \ doty \ dotz endbmatrix$ to $[-2,-2,7]vecdotX = 0$
To get the points on the tangent plane at A, add $veca$ to the vectors $vecdotX$ belonging to tangent space.
Hope that helps!
answered Jul 30 at 11:50
Suhan Shetty
835
add a comment |Â
up vote
0
down vote
If the smooth manifold is given by $vecF(X) = vec0$, then the tangent space to the manifold at point $A$ is the null space (or kernel) of $D_XF(A)$.
In your case, $X = beginbmatrixx \ y \ z endbmatrix$ ,
$A=beginbmatrix1 \ 1 \ 7 endbmatrix$,
$F(X) = z-x^2-y^2$, and the manifold is given by $F(X)=0$.
$D_XF(A) = [-2*1, -2*1, 1*7] = [-2,-2,7]$.
The tangent space at $A$ is the solution $vecdotX = beginbmatrixdotx \ doty \ dotz endbmatrix$ to $[-2,-2,7]vecdotX = 0$
To get the points on the tangent plane at A, add $veca$ to the vectors $vecdotX$ belonging to tangent space.
Hope that helps!
answered Jul 30 at 11:50
Suhan Shetty
835
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If the smooth manifold is given by $vecF(X) = vec0$, then the tangent space to the manifold at point $A$ is the null space (or kernel) of $D_XF(A)$.
In your case, $X = beginbmatrixx \ y \ z endbmatrix$ ,
$A=beginbmatrix1 \ 1 \ 7 endbmatrix$,
$F(X) = z-x^2-y^2$, and the manifold is given by $F(X)=0$.
$D_XF(A) = [-2*1, -2*1, 1*7] = [-2,-2,7]$.
The tangent space at $A$ is the solution $vecdotX = beginbmatrixdotx \ doty \ dotz endbmatrix$ to $[-2,-2,7]vecdotX = 0$
To get the points on the tangent plane at A, add $veca$ to the vectors $vecdotX$ belonging to tangent space.
Hope that helps!
answered Jul 30 at 11:50
Suhan Shetty
835
If the smooth manifold is given by $vecF(X) = vec0$, then the tangent space to the manifold at point $A$ is the null space (or kernel) of $D_XF(A)$.
In your case, $X = beginbmatrixx \ y \ z endbmatrix$ ,
$A=beginbmatrix1 \ 1 \ 7 endbmatrix$,
$F(X) = z-x^2-y^2$, and the manifold is given by $F(X)=0$.
$D_XF(A) = [-2*1, -2*1, 1*7] = [-2,-2,7]$.
The tangent space at $A$ is the solution $vecdotX = beginbmatrixdotx \ doty \ dotz endbmatrix$ to $[-2,-2,7]vecdotX = 0$
To get the points on the tangent plane at A, add $veca$ to the vectors $vecdotX$ belonging to tangent space.
Hope that helps!
answered Jul 30 at 11:50
Suhan Shetty
835
answered Jul 30 at 11:50
Suhan Shetty
835
answered Jul 30 at 11:50
Suhan Shetty
835
answered Jul 30 at 11:50
Suhan Shetty
835
835
add a comment |Â
add a comment |Â
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