Optimizing a parameter of an equation involving 2D vectors

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Let $vecu(theta) = left<costheta, sintheta right>$



Given the formula $ttimes(vecu(theta) - vecK_1)=vecK_2$ with $vecK_1, vecK_2 in mathbbR^2$, $theta in [0, 2pi)$ and $t in mathbbR^+$, how can I find a value of $theta$ that minimizes $t$?



Note that $vecK_1$ and $vecK_2$ have values that guarantee infinitely many solutions to the equation above.



PS. I'm new to this topic and learn on my own, so I couldn't really come up with a good title. Feel free to suggest a better one, I'll gladly change it.




linear-algebra trigonometry




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    up vote
    1
    down vote

    favorite












    Let $vecu(theta) = left<costheta, sintheta right>$



    Given the formula $ttimes(vecu(theta) - vecK_1)=vecK_2$ with $vecK_1, vecK_2 in mathbbR^2$, $theta in [0, 2pi)$ and $t in mathbbR^+$, how can I find a value of $theta$ that minimizes $t$?



    Note that $vecK_1$ and $vecK_2$ have values that guarantee infinitely many solutions to the equation above.



    PS. I'm new to this topic and learn on my own, so I couldn't really come up with a good title. Feel free to suggest a better one, I'll gladly change it.




    linear-algebra trigonometry




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $vecu(theta) = left<costheta, sintheta right>$



      Given the formula $ttimes(vecu(theta) - vecK_1)=vecK_2$ with $vecK_1, vecK_2 in mathbbR^2$, $theta in [0, 2pi)$ and $t in mathbbR^+$, how can I find a value of $theta$ that minimizes $t$?



      Note that $vecK_1$ and $vecK_2$ have values that guarantee infinitely many solutions to the equation above.



      PS. I'm new to this topic and learn on my own, so I couldn't really come up with a good title. Feel free to suggest a better one, I'll gladly change it.




      linear-algebra trigonometry




      share|cite|improve this question











      Let $vecu(theta) = left<costheta, sintheta right>$



      Given the formula $ttimes(vecu(theta) - vecK_1)=vecK_2$ with $vecK_1, vecK_2 in mathbbR^2$, $theta in [0, 2pi)$ and $t in mathbbR^+$, how can I find a value of $theta$ that minimizes $t$?



      Note that $vecK_1$ and $vecK_2$ have values that guarantee infinitely many solutions to the equation above.



      PS. I'm new to this topic and learn on my own, so I couldn't really come up with a good title. Feel free to suggest a better one, I'll gladly change it.





      linear-algebra trigonometry





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      asked 14 hours ago









      Kamil Koczurek

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          Take the derivative of the vector equation (using the chain rule), and find $fracrm dtrm dtheta=0$.



          $$ rm d t , ( vecu - vecK_1) + t , left( fracpartial vecupartial theta rm dtheta right) = 0 $$



          $$ rm dt = frac t , fracpartial vecupartial thetavecK_1 - vecu rm dtheta =0 $$



          which has the solution:



          $$ t,left< -sin theta, cos theta right> = 0$$



          or $$t=0$$






          share|cite|improve this answer























          • What is more interesting is finding $fracrm dthetarm dt=0$ which is $theta = rm atanleft( fracKy_1Kx_1 right) $.
            – ja72
            13 hours ago










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          1 Answer
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          up vote
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          Take the derivative of the vector equation (using the chain rule), and find $fracrm dtrm dtheta=0$.



          $$ rm d t , ( vecu - vecK_1) + t , left( fracpartial vecupartial theta rm dtheta right) = 0 $$



          $$ rm dt = frac t , fracpartial vecupartial thetavecK_1 - vecu rm dtheta =0 $$



          which has the solution:



          $$ t,left< -sin theta, cos theta right> = 0$$



          or $$t=0$$






          share|cite|improve this answer























          • What is more interesting is finding $fracrm dthetarm dt=0$ which is $theta = rm atanleft( fracKy_1Kx_1 right) $.
            – ja72
            13 hours ago














          up vote
          1
          down vote













          Take the derivative of the vector equation (using the chain rule), and find $fracrm dtrm dtheta=0$.



          $$ rm d t , ( vecu - vecK_1) + t , left( fracpartial vecupartial theta rm dtheta right) = 0 $$



          $$ rm dt = frac t , fracpartial vecupartial thetavecK_1 - vecu rm dtheta =0 $$



          which has the solution:



          $$ t,left< -sin theta, cos theta right> = 0$$



          or $$t=0$$






          share|cite|improve this answer























          • What is more interesting is finding $fracrm dthetarm dt=0$ which is $theta = rm atanleft( fracKy_1Kx_1 right) $.
            – ja72
            13 hours ago












          up vote
          1
          down vote










          up vote
          1
          down vote









          Take the derivative of the vector equation (using the chain rule), and find $fracrm dtrm dtheta=0$.



          $$ rm d t , ( vecu - vecK_1) + t , left( fracpartial vecupartial theta rm dtheta right) = 0 $$



          $$ rm dt = frac t , fracpartial vecupartial thetavecK_1 - vecu rm dtheta =0 $$



          which has the solution:



          $$ t,left< -sin theta, cos theta right> = 0$$



          or $$t=0$$






          share|cite|improve this answer















          Take the derivative of the vector equation (using the chain rule), and find $fracrm dtrm dtheta=0$.



          $$ rm d t , ( vecu - vecK_1) + t , left( fracpartial vecupartial theta rm dtheta right) = 0 $$



          $$ rm dt = frac t , fracpartial vecupartial thetavecK_1 - vecu rm dtheta =0 $$



          which has the solution:



          $$ t,left< -sin theta, cos theta right> = 0$$



          or $$t=0$$







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited 13 hours ago









          mathreadler

          13.5k71757




          13.5k71757











          answered 13 hours ago









          ja72

          7,16411641




          7,16411641











          • What is more interesting is finding $fracrm dthetarm dt=0$ which is $theta = rm atanleft( fracKy_1Kx_1 right) $.
            – ja72
            13 hours ago
















          • What is more interesting is finding $fracrm dthetarm dt=0$ which is $theta = rm atanleft( fracKy_1Kx_1 right) $.
            – ja72
            13 hours ago















          What is more interesting is finding $fracrm dthetarm dt=0$ which is $theta = rm atanleft( fracKy_1Kx_1 right) $.
          – ja72
          13 hours ago




          What is more interesting is finding $fracrm dthetarm dt=0$ which is $theta = rm atanleft( fracKy_1Kx_1 right) $.
          – ja72
          13 hours ago












           

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