Proof of curvature of vector and orthogonality of vector derivative to itself











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In the proof of curvature of a vector r(t) we take the first derivative of r(t) to be orthogonal to the vector itself. But isn't it true only for r(t) with constant magnitude?






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  • Yes. In general $r'(t)$ is not orthogonal to $r(t)$ itself. It happens if and only if the norm of $r(t)$ is constant.
    – Dog_69
    yesterday










  • Then how do we apply it to a curvature where r(t) changes?
    – Srilakshmidaran
    yesterday










  • The general expression for the curvature is $$kappa(t)=frac{|r'(t)times r''(t)|}{|r'(t)|^3} .$$
    – Dog_69
    yesterday















up vote
0
down vote

favorite












In the proof of curvature of a vector r(t) we take the first derivative of r(t) to be orthogonal to the vector itself. But isn't it true only for r(t) with constant magnitude?






calculus







share|cite|improve this question







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Srilakshmidaran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Yes. In general $r'(t)$ is not orthogonal to $r(t)$ itself. It happens if and only if the norm of $r(t)$ is constant.
    – Dog_69
    yesterday










  • Then how do we apply it to a curvature where r(t) changes?
    – Srilakshmidaran
    yesterday










  • The general expression for the curvature is $$kappa(t)=frac{|r'(t)times r''(t)|}{|r'(t)|^3} .$$
    – Dog_69
    yesterday













up vote
0
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up vote
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down vote

favorite











In the proof of curvature of a vector r(t) we take the first derivative of r(t) to be orthogonal to the vector itself. But isn't it true only for r(t) with constant magnitude?






calculus







share|cite|improve this question







New contributor




Srilakshmidaran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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In the proof of curvature of a vector r(t) we take the first derivative of r(t) to be orthogonal to the vector itself. But isn't it true only for r(t) with constant magnitude?






calculus




calculus






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Srilakshmidaran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Yes. In general $r'(t)$ is not orthogonal to $r(t)$ itself. It happens if and only if the norm of $r(t)$ is constant.
    – Dog_69
    yesterday










  • Then how do we apply it to a curvature where r(t) changes?
    – Srilakshmidaran
    yesterday










  • The general expression for the curvature is $$kappa(t)=frac{|r'(t)times r''(t)|}{|r'(t)|^3} .$$
    – Dog_69
    yesterday


















  • Yes. In general $r'(t)$ is not orthogonal to $r(t)$ itself. It happens if and only if the norm of $r(t)$ is constant.
    – Dog_69
    yesterday










  • Then how do we apply it to a curvature where r(t) changes?
    – Srilakshmidaran
    yesterday










  • The general expression for the curvature is $$kappa(t)=frac{|r'(t)times r''(t)|}{|r'(t)|^3} .$$
    – Dog_69
    yesterday
















Yes. In general $r'(t)$ is not orthogonal to $r(t)$ itself. It happens if and only if the norm of $r(t)$ is constant.
– Dog_69
yesterday




Yes. In general $r'(t)$ is not orthogonal to $r(t)$ itself. It happens if and only if the norm of $r(t)$ is constant.
– Dog_69
yesterday












Then how do we apply it to a curvature where r(t) changes?
– Srilakshmidaran
yesterday




Then how do we apply it to a curvature where r(t) changes?
– Srilakshmidaran
yesterday












The general expression for the curvature is $$kappa(t)=frac{|r'(t)times r''(t)|}{|r'(t)|^3} .$$
– Dog_69
yesterday




The general expression for the curvature is $$kappa(t)=frac{|r'(t)times r''(t)|}{|r'(t)|^3} .$$
– Dog_69
yesterday










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(At first, you don’t prove but define curvature instead.) Curvature deals with orthogonality, sort of: it may be defined as the normal component of the acceleration, divided by the square of the velocity.






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    (At first, you don’t prove but define curvature instead.) Curvature deals with orthogonality, sort of: it may be defined as the normal component of the acceleration, divided by the square of the velocity.






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      (At first, you don’t prove but define curvature instead.) Curvature deals with orthogonality, sort of: it may be defined as the normal component of the acceleration, divided by the square of the velocity.






      share|cite|improve this answer























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        up vote
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        (At first, you don’t prove but define curvature instead.) Curvature deals with orthogonality, sort of: it may be defined as the normal component of the acceleration, divided by the square of the velocity.






        share|cite|improve this answer












        (At first, you don’t prove but define curvature instead.) Curvature deals with orthogonality, sort of: it may be defined as the normal component of the acceleration, divided by the square of the velocity.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










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