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




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          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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                    Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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                    up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
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