Mixing
Generalizing the notion of parallel to curved lines in $mathbb{R}^2$ and $mathbb{R}^3$ for an arc-length...
$begingroup$
Let $a(s)$ be an arclength parameterized plane curve. We can create a "parallel" curve $b(s)$ by setting $b(s)=a(s) + eN$ for some small $e>0$.
I think that this defintion makes sense, because given any $s in D$, $b'(s)$ and $a'(s)$ will be parallel lines.
Now, let us consider $a(s)$ as an arc-length parameterized curve in space. I want to show that we can obtain a "parallel" curve $b(s)$ by taking $b(s)=a(s) + e((cos(theta)N + sin(theta)B$) for an appropriate function $theta$.
So, i'm a bit confused here. How does the idea of two lines being parallel generalize to three dimensions? By showing that their tangent lines are parallel somehow? Or could we get a cross product involved and show that their binormal vectors are always parallel or something?
Furthermore, what is this function $theta(s)$ that i'm looking for? My first thought was to use the $theta(s)$ s.t. $T=cos(theta(s))i+sin(theta(s))j$ which exists because $alpha$ is smooth I believe. But i'm just not sure.
Can anyone offer some general or specific insights into the situation?
Also, I'm looking for an easy way to express the curvature of $b(s)$ in terms of $a(s)$ in the first case where both $a(s)$ and $b(s)$ are plane curves.
Thanks a ton!
differential-geometry
edited Jan 27 at 3:22
J. W. Tanner
3,6831320
asked Jan 26 at 22:00
Mathematical MushroomMathematical Mushroom
848
$endgroup$
add a comment |
$begingroup$
Let $a(s)$ be an arclength parameterized plane curve. We can create a "parallel" curve $b(s)$ by setting $b(s)=a(s) + eN$ for some small $e>0$.
I think that this defintion makes sense, because given any $s in D$, $b'(s)$ and $a'(s)$ will be parallel lines.
Now, let us consider $a(s)$ as an arc-length parameterized curve in space. I want to show that we can obtain a "parallel" curve $b(s)$ by taking $b(s)=a(s) + e((cos(theta)N + sin(theta)B$) for an appropriate function $theta$.
So, i'm a bit confused here. How does the idea of two lines being parallel generalize to three dimensions? By showing that their tangent lines are parallel somehow? Or could we get a cross product involved and show that their binormal vectors are always parallel or something?
Furthermore, what is this function $theta(s)$ that i'm looking for? My first thought was to use the $theta(s)$ s.t. $T=cos(theta(s))i+sin(theta(s))j$ which exists because $alpha$ is smooth I believe. But i'm just not sure.
Can anyone offer some general or specific insights into the situation?
Also, I'm looking for an easy way to express the curvature of $b(s)$ in terms of $a(s)$ in the first case where both $a(s)$ and $b(s)$ are plane curves.
Thanks a ton!
differential-geometry
edited Jan 27 at 3:22
J. W. Tanner
3,6831320
asked Jan 26 at 22:00
Mathematical MushroomMathematical Mushroom
848
$endgroup$
add a comment |
$begingroup$
Let $a(s)$ be an arclength parameterized plane curve. We can create a "parallel" curve $b(s)$ by setting $b(s)=a(s) + eN$ for some small $e>0$.
I think that this defintion makes sense, because given any $s in D$, $b'(s)$ and $a'(s)$ will be parallel lines.
Now, let us consider $a(s)$ as an arc-length parameterized curve in space. I want to show that we can obtain a "parallel" curve $b(s)$ by taking $b(s)=a(s) + e((cos(theta)N + sin(theta)B$) for an appropriate function $theta$.
So, i'm a bit confused here. How does the idea of two lines being parallel generalize to three dimensions? By showing that their tangent lines are parallel somehow? Or could we get a cross product involved and show that their binormal vectors are always parallel or something?
Furthermore, what is this function $theta(s)$ that i'm looking for? My first thought was to use the $theta(s)$ s.t. $T=cos(theta(s))i+sin(theta(s))j$ which exists because $alpha$ is smooth I believe. But i'm just not sure.
Can anyone offer some general or specific insights into the situation?
Also, I'm looking for an easy way to express the curvature of $b(s)$ in terms of $a(s)$ in the first case where both $a(s)$ and $b(s)$ are plane curves.
Thanks a ton!
differential-geometry
edited Jan 27 at 3:22
J. W. Tanner
3,6831320
asked Jan 26 at 22:00
Mathematical MushroomMathematical Mushroom
848
$endgroup$
Let $a(s)$ be an arclength parameterized plane curve. We can create a "parallel" curve $b(s)$ by setting $b(s)=a(s) + eN$ for some small $e>0$.
I think that this defintion makes sense, because given any $s in D$, $b'(s)$ and $a'(s)$ will be parallel lines.
Now, let us consider $a(s)$ as an arc-length parameterized curve in space. I want to show that we can obtain a "parallel" curve $b(s)$ by taking $b(s)=a(s) + e((cos(theta)N + sin(theta)B$) for an appropriate function $theta$.
So, i'm a bit confused here. How does the idea of two lines being parallel generalize to three dimensions? By showing that their tangent lines are parallel somehow? Or could we get a cross product involved and show that their binormal vectors are always parallel or something?
Furthermore, what is this function $theta(s)$ that i'm looking for? My first thought was to use the $theta(s)$ s.t. $T=cos(theta(s))i+sin(theta(s))j$ which exists because $alpha$ is smooth I believe. But i'm just not sure.
Can anyone offer some general or specific insights into the situation?
Also, I'm looking for an easy way to express the curvature of $b(s)$ in terms of $a(s)$ in the first case where both $a(s)$ and $b(s)$ are plane curves.
Thanks a ton!
differential-geometry
differential-geometry
edited Jan 27 at 3:22
J. W. Tanner
3,6831320
asked Jan 26 at 22:00
Mathematical MushroomMathematical Mushroom
848
edited Jan 27 at 3:22
J. W. Tanner
3,6831320
asked Jan 26 at 22:00
Mathematical MushroomMathematical Mushroom
848
edited Jan 27 at 3:22
J. W. Tanner
3,6831320
edited Jan 27 at 3:22
J. W. Tanner
3,6831320
edited Jan 27 at 3:22
J. W. Tanner
3,6831320
3,6831320
asked Jan 26 at 22:00
Mathematical MushroomMathematical Mushroom
848
asked Jan 26 at 22:00
Mathematical MushroomMathematical Mushroom
848
asked Jan 26 at 22:00
Mathematical MushroomMathematical Mushroom
848
848
add a comment |
add a comment |
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