Directed curvature of a curve
So I have this following exercise
Consider the curve given by the graph of the sine function $t rightarrow (t,sin(t))$. Determine the directed curvature at each point of this curve.
Supposing that directed curvatue and oriented curvature is the same thing there is a definition saying that oriented curvature is given by the identity
$$T'(t_0)=left(kappa_{alpha}(t_0) cdot parallelalpha'(t_0)parallel right)N(t_0)$$
If the turning angle of the curve $theta(t)$ is known, then $kappa_{alpha}(t)= displaystylefrac{theta'(t)}{parallel alpha'(t) parallel}$.
Can someone help me?
differential-geometry curves curvature
asked Dec 12 '18 at 14:47
Alim TeacherAlim Teacher
5012819
add a comment |
So I have this following exercise
Consider the curve given by the graph of the sine function $t rightarrow (t,sin(t))$. Determine the directed curvature at each point of this curve.
Supposing that directed curvatue and oriented curvature is the same thing there is a definition saying that oriented curvature is given by the identity
$$T'(t_0)=left(kappa_{alpha}(t_0) cdot parallelalpha'(t_0)parallel right)N(t_0)$$
If the turning angle of the curve $theta(t)$ is known, then $kappa_{alpha}(t)= displaystylefrac{theta'(t)}{parallel alpha'(t) parallel}$.
Can someone help me?
differential-geometry curves curvature
asked Dec 12 '18 at 14:47
Alim TeacherAlim Teacher
5012819
1
Help you do what? Do the calculations! But, be careful. In your definition, $N$ is not the principal normal vector, but rather a unit vector perpendicular to $T$ so that $T,N$ make a right-handed frame at each point.
– Ted Shifrin
Dec 12 '18 at 22:30
Okay so for a curve parametrized by arc length $parallel alpha'(t) parallel=1$ then $T'(t_0)=kappa_{alpha}(t_0)N(t_0)$ and $T'(t)=alpha''(t)$. Then I take the derivative $alpha'(t)=(1,cos (t))$ and $alpha''(t)=(0,-sin (t))$. Hereafter I am stuck.
– Alim Teacher
Dec 13 '18 at 9:27
I found out how to do it now
– Alim Teacher
Dec 13 '18 at 14:57
add a comment |
So I have this following exercise
Consider the curve given by the graph of the sine function $t rightarrow (t,sin(t))$. Determine the directed curvature at each point of this curve.
Supposing that directed curvatue and oriented curvature is the same thing there is a definition saying that oriented curvature is given by the identity
$$T'(t_0)=left(kappa_{alpha}(t_0) cdot parallelalpha'(t_0)parallel right)N(t_0)$$
If the turning angle of the curve $theta(t)$ is known, then $kappa_{alpha}(t)= displaystylefrac{theta'(t)}{parallel alpha'(t) parallel}$.
Can someone help me?
differential-geometry curves curvature
asked Dec 12 '18 at 14:47
Alim TeacherAlim Teacher
5012819
So I have this following exercise
Consider the curve given by the graph of the sine function $t rightarrow (t,sin(t))$. Determine the directed curvature at each point of this curve.
Supposing that directed curvatue and oriented curvature is the same thing there is a definition saying that oriented curvature is given by the identity
$$T'(t_0)=left(kappa_{alpha}(t_0) cdot parallelalpha'(t_0)parallel right)N(t_0)$$
If the turning angle of the curve $theta(t)$ is known, then $kappa_{alpha}(t)= displaystylefrac{theta'(t)}{parallel alpha'(t) parallel}$.
Can someone help me?
differential-geometry curves curvature
differential-geometry curves curvature
asked Dec 12 '18 at 14:47
Alim TeacherAlim Teacher
5012819
asked Dec 12 '18 at 14:47
Alim TeacherAlim Teacher
5012819
asked Dec 12 '18 at 14:47
Alim TeacherAlim Teacher
5012819
asked Dec 12 '18 at 14:47
Alim TeacherAlim Teacher
5012819
asked Dec 12 '18 at 14:47
Alim TeacherAlim Teacher
5012819
5012819
1
Help you do what? Do the calculations! But, be careful. In your definition, $N$ is not the principal normal vector, but rather a unit vector perpendicular to $T$ so that $T,N$ make a right-handed frame at each point.
– Ted Shifrin
Dec 12 '18 at 22:30
Okay so for a curve parametrized by arc length $parallel alpha'(t) parallel=1$ then $T'(t_0)=kappa_{alpha}(t_0)N(t_0)$ and $T'(t)=alpha''(t)$. Then I take the derivative $alpha'(t)=(1,cos (t))$ and $alpha''(t)=(0,-sin (t))$. Hereafter I am stuck.
– Alim Teacher
Dec 13 '18 at 9:27
I found out how to do it now
– Alim Teacher
Dec 13 '18 at 14:57
add a comment |
1
Help you do what? Do the calculations! But, be careful. In your definition, $N$ is not the principal normal vector, but rather a unit vector perpendicular to $T$ so that $T,N$ make a right-handed frame at each point.
– Ted Shifrin
Dec 12 '18 at 22:30
Okay so for a curve parametrized by arc length $parallel alpha'(t) parallel=1$ then $T'(t_0)=kappa_{alpha}(t_0)N(t_0)$ and $T'(t)=alpha''(t)$. Then I take the derivative $alpha'(t)=(1,cos (t))$ and $alpha''(t)=(0,-sin (t))$. Hereafter I am stuck.
– Alim Teacher
Dec 13 '18 at 9:27
I found out how to do it now
– Alim Teacher
Dec 13 '18 at 14:57
1
1
Help you do what? Do the calculations! But, be careful. In your definition, $N$ is not the principal normal vector, but rather a unit vector perpendicular to $T$ so that $T,N$ make a right-handed frame at each point.
– Ted Shifrin
Dec 12 '18 at 22:30
Help you do what? Do the calculations! But, be careful. In your definition, $N$ is not the principal normal vector, but rather a unit vector perpendicular to $T$ so that $T,N$ make a right-handed frame at each point.
– Ted Shifrin
Dec 12 '18 at 22:30
Okay so for a curve parametrized by arc length $parallel alpha'(t) parallel=1$ then $T'(t_0)=kappa_{alpha}(t_0)N(t_0)$ and $T'(t)=alpha''(t)$. Then I take the derivative $alpha'(t)=(1,cos (t))$ and $alpha''(t)=(0,-sin (t))$. Hereafter I am stuck.
– Alim Teacher
Dec 13 '18 at 9:27
Okay so for a curve parametrized by arc length $parallel alpha'(t) parallel=1$ then $T'(t_0)=kappa_{alpha}(t_0)N(t_0)$ and $T'(t)=alpha''(t)$. Then I take the derivative $alpha'(t)=(1,cos (t))$ and $alpha''(t)=(0,-sin (t))$. Hereafter I am stuck.
– Alim Teacher
Dec 13 '18 at 9:27
I found out how to do it now
– Alim Teacher
Dec 13 '18 at 14:57
I found out how to do it now
– Alim Teacher
Dec 13 '18 at 14:57
add a comment |
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Help you do what? Do the calculations! But, be careful. In your definition, $N$ is not the principal normal vector, but rather a unit vector perpendicular to $T$ so that $T,N$ make a right-handed frame at each point.
– Ted Shifrin
Dec 12 '18 at 22:30
Okay so for a curve parametrized by arc length $parallel alpha'(t) parallel=1$ then $T'(t_0)=kappa_{alpha}(t_0)N(t_0)$ and $T'(t)=alpha''(t)$. Then I take the derivative $alpha'(t)=(1,cos (t))$ and $alpha''(t)=(0,-sin (t))$. Hereafter I am stuck.
– Alim Teacher
Dec 13 '18 at 9:27
I found out how to do it now
– Alim Teacher
Dec 13 '18 at 14:57