Finding a curve that lies on the arbitrary sphere given $alpha(0)$ and $alpha'(0)$.












0












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Find a curve $alpha : (−ε,ε) → Sigma$ on the sphere which has $alpha(0) = (1,0,0)$ and $alpha′(0) = (0, 5, 6)$.




I'm unsure how to approach this. I know the parametarization of a sphere, and obviously the bases of the tangent space, but I don't know if this will help me?






differential-geometry







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    0












    $begingroup$



    Find a curve $alpha : (−ε,ε) → Sigma$ on the sphere which has $alpha(0) = (1,0,0)$ and $alpha′(0) = (0, 5, 6)$.




    I'm unsure how to approach this. I know the parametarization of a sphere, and obviously the bases of the tangent space, but I don't know if this will help me?






    differential-geometry







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Find a curve $alpha : (−ε,ε) → Sigma$ on the sphere which has $alpha(0) = (1,0,0)$ and $alpha′(0) = (0, 5, 6)$.




      I'm unsure how to approach this. I know the parametarization of a sphere, and obviously the bases of the tangent space, but I don't know if this will help me?






      differential-geometry







      share|cite|improve this question











      $endgroup$





      Find a curve $alpha : (−ε,ε) → Sigma$ on the sphere which has $alpha(0) = (1,0,0)$ and $alpha′(0) = (0, 5, 6)$.




      I'm unsure how to approach this. I know the parametarization of a sphere, and obviously the bases of the tangent space, but I don't know if this will help me?






      differential-geometry




      differential-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 25 '18 at 18:24









      Yadati Kiran

      1,751619




      1,751619










      asked Nov 25 '18 at 18:13









      Beryl1934Beryl1934

      1




      1






















          2 Answers
          2






          active

          oldest

          votes


















          0












          $begingroup$

          In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
          $$ gamma: s mapsto pcos s + Zsin s $$
          parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
            $endgroup$
            – Beryl1934
            Nov 25 '18 at 18:58












          • $begingroup$
            Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
            $endgroup$
            – MisterRiemann
            Nov 25 '18 at 19:14





















          0












          $begingroup$

          Let us place here the conditions for future referencing :



          $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



          There is an obvious candidate




          • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


          • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



          $$V(t)=begin{pmatrix}1&0& 0\
          0&cos(a)&-sin(a)\
          0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
          sin(kt)\
          0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
          sin(a)sin(kt)end{pmatrix} tag{2}$$



          We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



          The first condition is clearly fullfilled.



          It remains to comply to 2nd condition (b).



          As :



          $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



          Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



          $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



          Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



          Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



          Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



          Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.






          share|cite|improve this answer











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            2 Answers
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            2 Answers
            2






            active

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            active

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            0












            $begingroup$

            In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
            $$ gamma: s mapsto pcos s + Zsin s $$
            parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
              $endgroup$
              – Beryl1934
              Nov 25 '18 at 18:58












            • $begingroup$
              Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
              $endgroup$
              – MisterRiemann
              Nov 25 '18 at 19:14


















            0












            $begingroup$

            In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
            $$ gamma: s mapsto pcos s + Zsin s $$
            parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
              $endgroup$
              – Beryl1934
              Nov 25 '18 at 18:58












            • $begingroup$
              Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
              $endgroup$
              – MisterRiemann
              Nov 25 '18 at 19:14
















            0












            0








            0





            $begingroup$

            In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
            $$ gamma: s mapsto pcos s + Zsin s $$
            parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?






            share|cite|improve this answer









            $endgroup$



            In general, for two orthogonal unit vectors $p, Z in mathbb{R}^3$, the map
            $$ gamma: s mapsto pcos s + Zsin s $$
            parametrizes a unit speed curve on the unit circle with $gamma(0) = p$ and $dotgamma(0) = Z.$ Can you adapt this to your case?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 '18 at 18:20









            MisterRiemannMisterRiemann

            5,8451624




            5,8451624












            • $begingroup$
              That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
              $endgroup$
              – Beryl1934
              Nov 25 '18 at 18:58












            • $begingroup$
              Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
              $endgroup$
              – MisterRiemann
              Nov 25 '18 at 19:14




















            • $begingroup$
              That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
              $endgroup$
              – Beryl1934
              Nov 25 '18 at 18:58












            • $begingroup$
              Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
              $endgroup$
              – MisterRiemann
              Nov 25 '18 at 19:14


















            $begingroup$
            That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
            $endgroup$
            – Beryl1934
            Nov 25 '18 at 18:58






            $begingroup$
            That makes a lot of sense, thank you. Is there a justification for it? Obviously I can see why it works but there is no reference to it in my notes and textbooks. I don't want it to appear that I picked it out of thin air!
            $endgroup$
            – Beryl1934
            Nov 25 '18 at 18:58














            $begingroup$
            Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
            $endgroup$
            – MisterRiemann
            Nov 25 '18 at 19:14






            $begingroup$
            Well, I don't know what the best justification is, but could think of it this way: we know that in the plane, the unit circle can be parametrized as $s mapsto (cos s, sin s) = e_1cos s + e_2sin s$ where $e_1$ and $e_2$ are the standard basis vectors of $mathbb{R}^2$. Now in $mathbb{R}^3$, we are considering the plane spanned by the vectors $p$ and $Z$, which intersects the sphere in a great circle, the parametrization of which is in some sense obtained by replacing the basis vectors $e_1$ and $e_2$ by the basis vectors $p$ and $Z$ for our plane.
            $endgroup$
            – MisterRiemann
            Nov 25 '18 at 19:14













            0












            $begingroup$

            Let us place here the conditions for future referencing :



            $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



            There is an obvious candidate




            • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


            • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



            $$V(t)=begin{pmatrix}1&0& 0\
            0&cos(a)&-sin(a)\
            0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
            sin(kt)\
            0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
            sin(a)sin(kt)end{pmatrix} tag{2}$$



            We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



            The first condition is clearly fullfilled.



            It remains to comply to 2nd condition (b).



            As :



            $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



            Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



            $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



            Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



            Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



            Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



            Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              Let us place here the conditions for future referencing :



              $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



              There is an obvious candidate




              • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


              • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



              $$V(t)=begin{pmatrix}1&0& 0\
              0&cos(a)&-sin(a)\
              0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
              sin(kt)\
              0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
              sin(a)sin(kt)end{pmatrix} tag{2}$$



              We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



              The first condition is clearly fullfilled.



              It remains to comply to 2nd condition (b).



              As :



              $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



              Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



              $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



              Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



              Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



              Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



              Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                Let us place here the conditions for future referencing :



                $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



                There is an obvious candidate




                • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


                • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



                $$V(t)=begin{pmatrix}1&0& 0\
                0&cos(a)&-sin(a)\
                0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
                sin(kt)\
                0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
                sin(a)sin(kt)end{pmatrix} tag{2}$$



                We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



                The first condition is clearly fullfilled.



                It remains to comply to 2nd condition (b).



                As :



                $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



                Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



                $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



                Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



                Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



                Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



                Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.






                share|cite|improve this answer











                $endgroup$



                Let us place here the conditions for future referencing :



                $$V(0)=(1,0,0)^T (a) text{and} V'(0)=(0,5,6)^T (b) tag{1}$$



                There is an obvious candidate




                • for the trajectory : a great circle of the sphere, obtained here by a rotation of the equator by an angle $a$ around the $x$ axis.


                • in a second step, its cinematics : how must this trajectory be parametrized ? It suffices clearly, thinking to a uniform angular speed) to take $kt$ instead of $t$ ; all this gives :



                $$V(t)=begin{pmatrix}1&0& 0\
                0&cos(a)&-sin(a)\
                0&sin(a)& cos(a)end{pmatrix}begin{pmatrix}cos(kt)\
                sin(kt)\
                0end{pmatrix}=begin{pmatrix}cos(kt)\ cos(a)sin(kt)\
                sin(a)sin(kt)end{pmatrix} tag{2}$$



                We have thus to determine constants $a$ and $k$ in order that conditions (1) are fulfilled.



                The first condition is clearly fullfilled.



                It remains to comply to 2nd condition (b).



                As :



                $$V'(t)=(-k sin(t), k cos(a) cos(kt), k sin(a)cos(kt))^T.$$



                Taking $t=0$ yields, by identification with the values of the 2nd condition $(0,5,6)^T$ :



                $$k cos(a)=5 (a) text{and} k sin(a)=6 (b). tag{3}$$



                Taking the quotient of (b) by (a) gives $tan(a)=6/5$ : Thus $a$ can be taken as $a_0=mathrm{atan}(6/5)$.



                Taking the square of (a) and (b) and adding them, one gets $k_0= sqrt{61}approx 7.8$ (taking $k=-sqrt{61}$ would make going the other way).



                Thus the solution is given by (2) with $k=k_0$ and $a=a_0$.



                Last thing : The interval of variation for parameter $t$ ? You must not pass twice on the same point, thus, take $t_{lim}$ (your $varepsilon$) such that $sqrt{61}t_{lim}<pi iff t_{lim}<pi/sqrt{61}$, for example $t_{lim}=0.4$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 26 '18 at 9:08

























                answered Nov 26 '18 at 2:03









                Jean MarieJean Marie

                28.9k42049




                28.9k42049






























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                    0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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