How Do You Find an Equation of the Tangent Plane for a Torus?
I've parameterized the torus given by $z^2 + (r - 2)^2$ = 1 using:
x = (2 + $costheta$)$cosalpha$,
y = (2 + $costheta$)$sinalpha$,
z = $sintheta$.
I'm really stuck on how to find the equation of the tangent plane to the torus.
vector-analysis parametrization
edited Nov 21 '18 at 17:08
Federico
4,709514
asked Nov 21 '18 at 16:48
Jake Thompson
11
add a comment |
I've parameterized the torus given by $z^2 + (r - 2)^2$ = 1 using:
x = (2 + $costheta$)$cosalpha$,
y = (2 + $costheta$)$sinalpha$,
z = $sintheta$.
I'm really stuck on how to find the equation of the tangent plane to the torus.
vector-analysis parametrization
edited Nov 21 '18 at 17:08
Federico
4,709514
asked Nov 21 '18 at 16:48
Jake Thompson
11
add a comment |
I've parameterized the torus given by $z^2 + (r - 2)^2$ = 1 using:
x = (2 + $costheta$)$cosalpha$,
y = (2 + $costheta$)$sinalpha$,
z = $sintheta$.
I'm really stuck on how to find the equation of the tangent plane to the torus.
vector-analysis parametrization
edited Nov 21 '18 at 17:08
Federico
4,709514
asked Nov 21 '18 at 16:48
Jake Thompson
11
I've parameterized the torus given by $z^2 + (r - 2)^2$ = 1 using:
x = (2 + $costheta$)$cosalpha$,
y = (2 + $costheta$)$sinalpha$,
z = $sintheta$.
I'm really stuck on how to find the equation of the tangent plane to the torus.
vector-analysis parametrization
vector-analysis parametrization
edited Nov 21 '18 at 17:08
Federico
4,709514
asked Nov 21 '18 at 16:48
Jake Thompson
11
edited Nov 21 '18 at 17:08
Federico
4,709514
asked Nov 21 '18 at 16:48
Jake Thompson
11
edited Nov 21 '18 at 17:08
Federico
4,709514
edited Nov 21 '18 at 17:08
Federico
4,709514
edited Nov 21 '18 at 17:08
Federico
4,709514
4,709514
asked Nov 21 '18 at 16:48
Jake Thompson
11
asked Nov 21 '18 at 16:48
Jake Thompson
11
asked Nov 21 '18 at 16:48
Jake Thompson
11
11
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Call $phi:mathbb R^2tomathbb R^3:(alpha,theta)mapsto(x,y,z)$ your parametrization. Then the vectors $left{frac{partialphi}{partialalpha},frac{partialphi}{partialtheta}right}$ span the tangent space $T$. If you want to find an equation, take a normal vector $n=frac{partialphi}{partialalpha}timesfrac{partialphi}{partialtheta}$, and your tangent space becomes
$$
T={vinmathbb R^3:ncdot v=0}.
$$
If you are interested in the affine tangent plane, you just translate $T$ to the corresponding point on the torus.
Alternatively, from the implicit definition
$$
f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1=0,
$$
you know that the tangent space is defined by ${v:vcdotnabla f=0}$. Again, you can translate it to
$$
T_{(x,y,z)}={v:[v-(x,y,z)]cdotnabla f=0}
$$
if you are interested in the affine plane.
answered Nov 21 '18 at 17:02
Federico
4,709514
... in just the same way you would do it for the sphere.
– Jean Marie
Nov 21 '18 at 17:05
add a comment |
If you're only looking for the equation of the tangent plane of the torus
$$ z^2+left(sqrt{x^2+y^2}-2right)^2 = 1, $$
then you can consider the function
$$ f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1, $$
so that your torus is the preimage $f^{-1}({0}),$ and so that
the gradient $nabla f(x_0,y_0,z_0)$ is a normal vector to the torus at the point $(x_0, y_0, z_0)$, which will immediately give you the equation you are after. Can you fill in the specific details on your own?
answered Nov 21 '18 at 17:02
MisterRiemann
5,7841624
A little more complex than parametric representation, easier to tackle (see solution by @Federico) in particular for locating special planes like bi-tangent Villarceau planes)
– Jean Marie
Nov 21 '18 at 17:09
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Call $phi:mathbb R^2tomathbb R^3:(alpha,theta)mapsto(x,y,z)$ your parametrization. Then the vectors $left{frac{partialphi}{partialalpha},frac{partialphi}{partialtheta}right}$ span the tangent space $T$. If you want to find an equation, take a normal vector $n=frac{partialphi}{partialalpha}timesfrac{partialphi}{partialtheta}$, and your tangent space becomes
$$
T={vinmathbb R^3:ncdot v=0}.
$$
If you are interested in the affine tangent plane, you just translate $T$ to the corresponding point on the torus.
Alternatively, from the implicit definition
$$
f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1=0,
$$
you know that the tangent space is defined by ${v:vcdotnabla f=0}$. Again, you can translate it to
$$
T_{(x,y,z)}={v:[v-(x,y,z)]cdotnabla f=0}
$$
if you are interested in the affine plane.
answered Nov 21 '18 at 17:02
Federico
4,709514
... in just the same way you would do it for the sphere.
– Jean Marie
Nov 21 '18 at 17:05
add a comment |
Call $phi:mathbb R^2tomathbb R^3:(alpha,theta)mapsto(x,y,z)$ your parametrization. Then the vectors $left{frac{partialphi}{partialalpha},frac{partialphi}{partialtheta}right}$ span the tangent space $T$. If you want to find an equation, take a normal vector $n=frac{partialphi}{partialalpha}timesfrac{partialphi}{partialtheta}$, and your tangent space becomes
$$
T={vinmathbb R^3:ncdot v=0}.
$$
If you are interested in the affine tangent plane, you just translate $T$ to the corresponding point on the torus.
Alternatively, from the implicit definition
$$
f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1=0,
$$
you know that the tangent space is defined by ${v:vcdotnabla f=0}$. Again, you can translate it to
$$
T_{(x,y,z)}={v:[v-(x,y,z)]cdotnabla f=0}
$$
if you are interested in the affine plane.
answered Nov 21 '18 at 17:02
Federico
4,709514
... in just the same way you would do it for the sphere.
– Jean Marie
Nov 21 '18 at 17:05
add a comment |
Call $phi:mathbb R^2tomathbb R^3:(alpha,theta)mapsto(x,y,z)$ your parametrization. Then the vectors $left{frac{partialphi}{partialalpha},frac{partialphi}{partialtheta}right}$ span the tangent space $T$. If you want to find an equation, take a normal vector $n=frac{partialphi}{partialalpha}timesfrac{partialphi}{partialtheta}$, and your tangent space becomes
$$
T={vinmathbb R^3:ncdot v=0}.
$$
If you are interested in the affine tangent plane, you just translate $T$ to the corresponding point on the torus.
Alternatively, from the implicit definition
$$
f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1=0,
$$
you know that the tangent space is defined by ${v:vcdotnabla f=0}$. Again, you can translate it to
$$
T_{(x,y,z)}={v:[v-(x,y,z)]cdotnabla f=0}
$$
if you are interested in the affine plane.
answered Nov 21 '18 at 17:02
Federico
4,709514
Call $phi:mathbb R^2tomathbb R^3:(alpha,theta)mapsto(x,y,z)$ your parametrization. Then the vectors $left{frac{partialphi}{partialalpha},frac{partialphi}{partialtheta}right}$ span the tangent space $T$. If you want to find an equation, take a normal vector $n=frac{partialphi}{partialalpha}timesfrac{partialphi}{partialtheta}$, and your tangent space becomes
$$
T={vinmathbb R^3:ncdot v=0}.
$$
If you are interested in the affine tangent plane, you just translate $T$ to the corresponding point on the torus.
Alternatively, from the implicit definition
$$
f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1=0,
$$
you know that the tangent space is defined by ${v:vcdotnabla f=0}$. Again, you can translate it to
$$
T_{(x,y,z)}={v:[v-(x,y,z)]cdotnabla f=0}
$$
if you are interested in the affine plane.
answered Nov 21 '18 at 17:02
Federico
4,709514
edited Nov 21 '18 at 17:06
answered Nov 21 '18 at 17:02
Federico
4,709514
answered Nov 21 '18 at 17:02
Federico
4,709514
answered Nov 21 '18 at 17:02
Federico
4,709514
4,709514
... in just the same way you would do it for the sphere.
– Jean Marie
Nov 21 '18 at 17:05
add a comment |
... in just the same way you would do it for the sphere.
– Jean Marie
Nov 21 '18 at 17:05
... in just the same way you would do it for the sphere.
– Jean Marie
Nov 21 '18 at 17:05
... in just the same way you would do it for the sphere.
– Jean Marie
Nov 21 '18 at 17:05
add a comment |
If you're only looking for the equation of the tangent plane of the torus
$$ z^2+left(sqrt{x^2+y^2}-2right)^2 = 1, $$
then you can consider the function
$$ f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1, $$
so that your torus is the preimage $f^{-1}({0}),$ and so that
the gradient $nabla f(x_0,y_0,z_0)$ is a normal vector to the torus at the point $(x_0, y_0, z_0)$, which will immediately give you the equation you are after. Can you fill in the specific details on your own?
answered Nov 21 '18 at 17:02
MisterRiemann
5,7841624
A little more complex than parametric representation, easier to tackle (see solution by @Federico) in particular for locating special planes like bi-tangent Villarceau planes)
– Jean Marie
Nov 21 '18 at 17:09
add a comment |
If you're only looking for the equation of the tangent plane of the torus
$$ z^2+left(sqrt{x^2+y^2}-2right)^2 = 1, $$
then you can consider the function
$$ f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1, $$
so that your torus is the preimage $f^{-1}({0}),$ and so that
the gradient $nabla f(x_0,y_0,z_0)$ is a normal vector to the torus at the point $(x_0, y_0, z_0)$, which will immediately give you the equation you are after. Can you fill in the specific details on your own?
answered Nov 21 '18 at 17:02
MisterRiemann
5,7841624
A little more complex than parametric representation, easier to tackle (see solution by @Federico) in particular for locating special planes like bi-tangent Villarceau planes)
– Jean Marie
Nov 21 '18 at 17:09
add a comment |
If you're only looking for the equation of the tangent plane of the torus
$$ z^2+left(sqrt{x^2+y^2}-2right)^2 = 1, $$
then you can consider the function
$$ f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1, $$
so that your torus is the preimage $f^{-1}({0}),$ and so that
the gradient $nabla f(x_0,y_0,z_0)$ is a normal vector to the torus at the point $(x_0, y_0, z_0)$, which will immediately give you the equation you are after. Can you fill in the specific details on your own?
answered Nov 21 '18 at 17:02
MisterRiemann
5,7841624
If you're only looking for the equation of the tangent plane of the torus
$$ z^2+left(sqrt{x^2+y^2}-2right)^2 = 1, $$
then you can consider the function
$$ f(x,y,z) = z^2+left(sqrt{x^2+y^2}-2right)^2-1, $$
so that your torus is the preimage $f^{-1}({0}),$ and so that
the gradient $nabla f(x_0,y_0,z_0)$ is a normal vector to the torus at the point $(x_0, y_0, z_0)$, which will immediately give you the equation you are after. Can you fill in the specific details on your own?
answered Nov 21 '18 at 17:02
MisterRiemann
5,7841624
answered Nov 21 '18 at 17:02
MisterRiemann
5,7841624
answered Nov 21 '18 at 17:02
MisterRiemann
5,7841624
answered Nov 21 '18 at 17:02
MisterRiemann
5,7841624
5,7841624
A little more complex than parametric representation, easier to tackle (see solution by @Federico) in particular for locating special planes like bi-tangent Villarceau planes)
– Jean Marie
Nov 21 '18 at 17:09
add a comment |
A little more complex than parametric representation, easier to tackle (see solution by @Federico) in particular for locating special planes like bi-tangent Villarceau planes)
– Jean Marie
Nov 21 '18 at 17:09
A little more complex than parametric representation, easier to tackle (see solution by @Federico) in particular for locating special planes like bi-tangent Villarceau planes)
– Jean Marie
Nov 21 '18 at 17:09
A little more complex than parametric representation, easier to tackle (see solution by @Federico) in particular for locating special planes like bi-tangent Villarceau planes)
– Jean Marie
Nov 21 '18 at 17:09
add a comment |
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