Formula for tangent plane to surface given by parametrization
$begingroup$
I am aware of how to find an equation of the tangent place to a surface that is given as the graph of a function $z = g(x,y)$. Here one finds a normal vector by essentially taking the partial derivatives.
My question is if there is a formula that can be used when the surface is given by a general parametrization $vec{r}(u,v)$. I would assume that there are still some partial derivatives and maybe a cross product somewhere, but I am not quite seeing it.
(I am just asking out of curiosity.)
calculus partial-derivative tangent-spaces
asked Dec 4 '18 at 19:15
John DoeJohn Doe
27721346
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add a comment |
$begingroup$
I am aware of how to find an equation of the tangent place to a surface that is given as the graph of a function $z = g(x,y)$. Here one finds a normal vector by essentially taking the partial derivatives.
My question is if there is a formula that can be used when the surface is given by a general parametrization $vec{r}(u,v)$. I would assume that there are still some partial derivatives and maybe a cross product somewhere, but I am not quite seeing it.
(I am just asking out of curiosity.)
calculus partial-derivative tangent-spaces
asked Dec 4 '18 at 19:15
John DoeJohn Doe
27721346
$endgroup$
add a comment |
$begingroup$
I am aware of how to find an equation of the tangent place to a surface that is given as the graph of a function $z = g(x,y)$. Here one finds a normal vector by essentially taking the partial derivatives.
My question is if there is a formula that can be used when the surface is given by a general parametrization $vec{r}(u,v)$. I would assume that there are still some partial derivatives and maybe a cross product somewhere, but I am not quite seeing it.
(I am just asking out of curiosity.)
calculus partial-derivative tangent-spaces
asked Dec 4 '18 at 19:15
John DoeJohn Doe
27721346
$endgroup$
I am aware of how to find an equation of the tangent place to a surface that is given as the graph of a function $z = g(x,y)$. Here one finds a normal vector by essentially taking the partial derivatives.
My question is if there is a formula that can be used when the surface is given by a general parametrization $vec{r}(u,v)$. I would assume that there are still some partial derivatives and maybe a cross product somewhere, but I am not quite seeing it.
(I am just asking out of curiosity.)
calculus partial-derivative tangent-spaces
calculus partial-derivative tangent-spaces
asked Dec 4 '18 at 19:15
John DoeJohn Doe
27721346
asked Dec 4 '18 at 19:15
John DoeJohn Doe
27721346
asked Dec 4 '18 at 19:15
John DoeJohn Doe
27721346
asked Dec 4 '18 at 19:15
John DoeJohn Doe
27721346
asked Dec 4 '18 at 19:15
John DoeJohn Doe
27721346
27721346
add a comment |
add a comment |
1 Answer
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You know that your plane is parallel to $vec r_u = partialvec r/partial u$ and $vec r_v = partialvec r/partial v$, and also passes through point $vec r(u,v)$. Can you write down the equation from those hints?
answered Dec 4 '18 at 19:19
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
So I would guess that normal vector is $vec{r}_utimes vec{r}_v$ and then I just pick a point.
$endgroup$
– John Doe
Dec 4 '18 at 19:21
$begingroup$
Yep. The equation can be written as a triple product $(vec R - vec r, vec r_u times vec r_v)=0$, where $vec R$ is a point of a plane.
$endgroup$
– Vasily Mitch
Dec 4 '18 at 19:28
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
You know that your plane is parallel to $vec r_u = partialvec r/partial u$ and $vec r_v = partialvec r/partial v$, and also passes through point $vec r(u,v)$. Can you write down the equation from those hints?
answered Dec 4 '18 at 19:19
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
So I would guess that normal vector is $vec{r}_utimes vec{r}_v$ and then I just pick a point.
$endgroup$
– John Doe
Dec 4 '18 at 19:21
$begingroup$
Yep. The equation can be written as a triple product $(vec R - vec r, vec r_u times vec r_v)=0$, where $vec R$ is a point of a plane.
$endgroup$
– Vasily Mitch
Dec 4 '18 at 19:28
add a comment |
$begingroup$
You know that your plane is parallel to $vec r_u = partialvec r/partial u$ and $vec r_v = partialvec r/partial v$, and also passes through point $vec r(u,v)$. Can you write down the equation from those hints?
answered Dec 4 '18 at 19:19
Vasily MitchVasily Mitch
2,3141311
$endgroup$
$begingroup$
So I would guess that normal vector is $vec{r}_utimes vec{r}_v$ and then I just pick a point.
$endgroup$
– John Doe
Dec 4 '18 at 19:21
$begingroup$
Yep. The equation can be written as a triple product $(vec R - vec r, vec r_u times vec r_v)=0$, where $vec R$ is a point of a plane.
$endgroup$
– Vasily Mitch
Dec 4 '18 at 19:28
add a comment |
$begingroup$
You know that your plane is parallel to $vec r_u = partialvec r/partial u$ and $vec r_v = partialvec r/partial v$, and also passes through point $vec r(u,v)$. Can you write down the equation from those hints?
answered Dec 4 '18 at 19:19
Vasily MitchVasily Mitch
2,3141311
$endgroup$
You know that your plane is parallel to $vec r_u = partialvec r/partial u$ and $vec r_v = partialvec r/partial v$, and also passes through point $vec r(u,v)$. Can you write down the equation from those hints?
answered Dec 4 '18 at 19:19
Vasily MitchVasily Mitch
2,3141311
answered Dec 4 '18 at 19:19
Vasily MitchVasily Mitch
2,3141311
answered Dec 4 '18 at 19:19
Vasily MitchVasily Mitch
2,3141311
answered Dec 4 '18 at 19:19
Vasily MitchVasily Mitch
2,3141311
2,3141311
$begingroup$
So I would guess that normal vector is $vec{r}_utimes vec{r}_v$ and then I just pick a point.
$endgroup$
– John Doe
Dec 4 '18 at 19:21
$begingroup$
Yep. The equation can be written as a triple product $(vec R - vec r, vec r_u times vec r_v)=0$, where $vec R$ is a point of a plane.
$endgroup$
– Vasily Mitch
Dec 4 '18 at 19:28
add a comment |
$begingroup$
So I would guess that normal vector is $vec{r}_utimes vec{r}_v$ and then I just pick a point.
$endgroup$
– John Doe
Dec 4 '18 at 19:21
$begingroup$
Yep. The equation can be written as a triple product $(vec R - vec r, vec r_u times vec r_v)=0$, where $vec R$ is a point of a plane.
$endgroup$
– Vasily Mitch
Dec 4 '18 at 19:28
$begingroup$
So I would guess that normal vector is $vec{r}_utimes vec{r}_v$ and then I just pick a point.
$endgroup$
– John Doe
Dec 4 '18 at 19:21
$begingroup$
So I would guess that normal vector is $vec{r}_utimes vec{r}_v$ and then I just pick a point.
$endgroup$
– John Doe
Dec 4 '18 at 19:21
$begingroup$
Yep. The equation can be written as a triple product $(vec R - vec r, vec r_u times vec r_v)=0$, where $vec R$ is a point of a plane.
$endgroup$
– Vasily Mitch
Dec 4 '18 at 19:28
$begingroup$
Yep. The equation can be written as a triple product $(vec R - vec r, vec r_u times vec r_v)=0$, where $vec R$ is a point of a plane.
$endgroup$
– Vasily Mitch
Dec 4 '18 at 19:28
add a comment |
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