Prove angles of $(vec r), (vec v),(vec a)$ are constant in this trajectory

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Given the trajectory



$$r(t)=[Acos(omega t),Bsin(omega t+alpha)]$$



Prove that the angles between $vecr, vecv, veca$ are constant.



So the first idea is looking at the dot product and then (hopefully) $cos(theta), theta$ being the angle of vectors, wouldn't have a time dependence. However doing all the extensive work it doesn't quite work. How could I do this?




multivariable-calculus physics




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  • two times differentiate concludes $a=r''=-w^2r$ then $aparallel r$.
    – Nosrati
    Aug 18 at 3:42










  • Certainly true for $veca, vecr$ although I don't see it for $vecv$ and the other two
    – user371816
    Aug 18 at 3:44










  • $aparallel r$ so you just need to prove for $r$ and $v$.
    – Nosrati
    Aug 18 at 3:52










  • Exactly, the issue is proving it for $vecr$ and $vecv$ :/
    – user371816
    Aug 18 at 4:12






  • 1




    This appears to be false on the face of it. The path is an ellipse, for which the angle between radial and tangent vectors is not constant.
    – amd
    Aug 18 at 6:24














up vote
0
down vote

favorite












Given the trajectory



$$r(t)=[Acos(omega t),Bsin(omega t+alpha)]$$



Prove that the angles between $vecr, vecv, veca$ are constant.



So the first idea is looking at the dot product and then (hopefully) $cos(theta), theta$ being the angle of vectors, wouldn't have a time dependence. However doing all the extensive work it doesn't quite work. How could I do this?




multivariable-calculus physics




share|cite|improve this question






















  • two times differentiate concludes $a=r''=-w^2r$ then $aparallel r$.
    – Nosrati
    Aug 18 at 3:42










  • Certainly true for $veca, vecr$ although I don't see it for $vecv$ and the other two
    – user371816
    Aug 18 at 3:44










  • $aparallel r$ so you just need to prove for $r$ and $v$.
    – Nosrati
    Aug 18 at 3:52










  • Exactly, the issue is proving it for $vecr$ and $vecv$ :/
    – user371816
    Aug 18 at 4:12






  • 1




    This appears to be false on the face of it. The path is an ellipse, for which the angle between radial and tangent vectors is not constant.
    – amd
    Aug 18 at 6:24












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Given the trajectory



$$r(t)=[Acos(omega t),Bsin(omega t+alpha)]$$



Prove that the angles between $vecr, vecv, veca$ are constant.



So the first idea is looking at the dot product and then (hopefully) $cos(theta), theta$ being the angle of vectors, wouldn't have a time dependence. However doing all the extensive work it doesn't quite work. How could I do this?




multivariable-calculus physics




share|cite|improve this question














Given the trajectory



$$r(t)=[Acos(omega t),Bsin(omega t+alpha)]$$



Prove that the angles between $vecr, vecv, veca$ are constant.



So the first idea is looking at the dot product and then (hopefully) $cos(theta), theta$ being the angle of vectors, wouldn't have a time dependence. However doing all the extensive work it doesn't quite work. How could I do this?





multivariable-calculus physics





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 18 at 3:48









Michael Hardy

205k23187463




205k23187463










asked Aug 18 at 3:27









user371816

304




304











  • two times differentiate concludes $a=r''=-w^2r$ then $aparallel r$.
    – Nosrati
    Aug 18 at 3:42










  • Certainly true for $veca, vecr$ although I don't see it for $vecv$ and the other two
    – user371816
    Aug 18 at 3:44










  • $aparallel r$ so you just need to prove for $r$ and $v$.
    – Nosrati
    Aug 18 at 3:52










  • Exactly, the issue is proving it for $vecr$ and $vecv$ :/
    – user371816
    Aug 18 at 4:12






  • 1




    This appears to be false on the face of it. The path is an ellipse, for which the angle between radial and tangent vectors is not constant.
    – amd
    Aug 18 at 6:24
















  • two times differentiate concludes $a=r''=-w^2r$ then $aparallel r$.
    – Nosrati
    Aug 18 at 3:42










  • Certainly true for $veca, vecr$ although I don't see it for $vecv$ and the other two
    – user371816
    Aug 18 at 3:44










  • $aparallel r$ so you just need to prove for $r$ and $v$.
    – Nosrati
    Aug 18 at 3:52










  • Exactly, the issue is proving it for $vecr$ and $vecv$ :/
    – user371816
    Aug 18 at 4:12






  • 1




    This appears to be false on the face of it. The path is an ellipse, for which the angle between radial and tangent vectors is not constant.
    – amd
    Aug 18 at 6:24















two times differentiate concludes $a=r''=-w^2r$ then $aparallel r$.
– Nosrati
Aug 18 at 3:42




two times differentiate concludes $a=r''=-w^2r$ then $aparallel r$.
– Nosrati
Aug 18 at 3:42












Certainly true for $veca, vecr$ although I don't see it for $vecv$ and the other two
– user371816
Aug 18 at 3:44




Certainly true for $veca, vecr$ although I don't see it for $vecv$ and the other two
– user371816
Aug 18 at 3:44












$aparallel r$ so you just need to prove for $r$ and $v$.
– Nosrati
Aug 18 at 3:52




$aparallel r$ so you just need to prove for $r$ and $v$.
– Nosrati
Aug 18 at 3:52












Exactly, the issue is proving it for $vecr$ and $vecv$ :/
– user371816
Aug 18 at 4:12




Exactly, the issue is proving it for $vecr$ and $vecv$ :/
– user371816
Aug 18 at 4:12




1




1




This appears to be false on the face of it. The path is an ellipse, for which the angle between radial and tangent vectors is not constant.
– amd
Aug 18 at 6:24




This appears to be false on the face of it. The path is an ellipse, for which the angle between radial and tangent vectors is not constant.
– amd
Aug 18 at 6:24















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