geodesic metric in an ellipse












0












$begingroup$


Consider the ellipse given by $x^2/4 + y^2 = 1$. Fix a point $z_0 = (x_0, y_0)$, where $x_0 > 0$ and $y_0 = sqrt{1 - x_0 ^2/4}$. Given $z = (x, sqrt{1 - x ^2/4})$ with $x > 0$, we put
$$ delta(z,z_0) = left | int_x ^ {x_0} sqrt{1 - t ^2/4} ,dt right |. $$



I'm trying to see if exists a constant $C> 0$ such that $delta(z,z_0) leq C |z-z_0|$, for all $z = (x, sqrt{1 - x ^2/4})$ with $x > 0$.



My attempt:



$$delta(z,z_0) = left | frac{1}{4} left ( sqrt{4 - x^2},x - sqrt{4 - x_0 ^2}, x_0 right ) + arcsin (x/2) - arcsin(x_0/2) right |$$
and
$$ |z - z_0| = left ( (x-x_0)^2 + frac{1}{4} left ( sqrt{4 - x^2} - sqrt{4 - x_0 ^2}right )^2 right )^{1/2}. $$



Is there any relation between the difference of $arcsin$ that might help?
Or another way to approach this problem?



Thank you.






real-analysis trigonometry geodesic







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  • $begingroup$
    Hint: $sqrt{1-frac {t^2}4}le 1$.
    $endgroup$
    – random
    Jan 2 at 15:08
















0












$begingroup$


Consider the ellipse given by $x^2/4 + y^2 = 1$. Fix a point $z_0 = (x_0, y_0)$, where $x_0 > 0$ and $y_0 = sqrt{1 - x_0 ^2/4}$. Given $z = (x, sqrt{1 - x ^2/4})$ with $x > 0$, we put
$$ delta(z,z_0) = left | int_x ^ {x_0} sqrt{1 - t ^2/4} ,dt right |. $$



I'm trying to see if exists a constant $C> 0$ such that $delta(z,z_0) leq C |z-z_0|$, for all $z = (x, sqrt{1 - x ^2/4})$ with $x > 0$.



My attempt:



$$delta(z,z_0) = left | frac{1}{4} left ( sqrt{4 - x^2},x - sqrt{4 - x_0 ^2}, x_0 right ) + arcsin (x/2) - arcsin(x_0/2) right |$$
and
$$ |z - z_0| = left ( (x-x_0)^2 + frac{1}{4} left ( sqrt{4 - x^2} - sqrt{4 - x_0 ^2}right )^2 right )^{1/2}. $$



Is there any relation between the difference of $arcsin$ that might help?
Or another way to approach this problem?



Thank you.






real-analysis trigonometry geodesic







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: $sqrt{1-frac {t^2}4}le 1$.
    $endgroup$
    – random
    Jan 2 at 15:08














0












0








0





$begingroup$


Consider the ellipse given by $x^2/4 + y^2 = 1$. Fix a point $z_0 = (x_0, y_0)$, where $x_0 > 0$ and $y_0 = sqrt{1 - x_0 ^2/4}$. Given $z = (x, sqrt{1 - x ^2/4})$ with $x > 0$, we put
$$ delta(z,z_0) = left | int_x ^ {x_0} sqrt{1 - t ^2/4} ,dt right |. $$



I'm trying to see if exists a constant $C> 0$ such that $delta(z,z_0) leq C |z-z_0|$, for all $z = (x, sqrt{1 - x ^2/4})$ with $x > 0$.



My attempt:



$$delta(z,z_0) = left | frac{1}{4} left ( sqrt{4 - x^2},x - sqrt{4 - x_0 ^2}, x_0 right ) + arcsin (x/2) - arcsin(x_0/2) right |$$
and
$$ |z - z_0| = left ( (x-x_0)^2 + frac{1}{4} left ( sqrt{4 - x^2} - sqrt{4 - x_0 ^2}right )^2 right )^{1/2}. $$



Is there any relation between the difference of $arcsin$ that might help?
Or another way to approach this problem?



Thank you.






real-analysis trigonometry geodesic







share|cite|improve this question











$endgroup$




Consider the ellipse given by $x^2/4 + y^2 = 1$. Fix a point $z_0 = (x_0, y_0)$, where $x_0 > 0$ and $y_0 = sqrt{1 - x_0 ^2/4}$. Given $z = (x, sqrt{1 - x ^2/4})$ with $x > 0$, we put
$$ delta(z,z_0) = left | int_x ^ {x_0} sqrt{1 - t ^2/4} ,dt right |. $$



I'm trying to see if exists a constant $C> 0$ such that $delta(z,z_0) leq C |z-z_0|$, for all $z = (x, sqrt{1 - x ^2/4})$ with $x > 0$.



My attempt:



$$delta(z,z_0) = left | frac{1}{4} left ( sqrt{4 - x^2},x - sqrt{4 - x_0 ^2}, x_0 right ) + arcsin (x/2) - arcsin(x_0/2) right |$$
and
$$ |z - z_0| = left ( (x-x_0)^2 + frac{1}{4} left ( sqrt{4 - x^2} - sqrt{4 - x_0 ^2}right )^2 right )^{1/2}. $$



Is there any relation between the difference of $arcsin$ that might help?
Or another way to approach this problem?



Thank you.






real-analysis trigonometry geodesic




real-analysis trigonometry geodesic






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share|cite|improve this question













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edited Jan 2 at 14:00









A. P

1085




1085










asked Jan 2 at 13:26









user 242964user 242964

720412




720412












  • $begingroup$
    Hint: $sqrt{1-frac {t^2}4}le 1$.
    $endgroup$
    – random
    Jan 2 at 15:08


















  • $begingroup$
    Hint: $sqrt{1-frac {t^2}4}le 1$.
    $endgroup$
    – random
    Jan 2 at 15:08
















$begingroup$
Hint: $sqrt{1-frac {t^2}4}le 1$.
$endgroup$
– random
Jan 2 at 15:08




$begingroup$
Hint: $sqrt{1-frac {t^2}4}le 1$.
$endgroup$
– random
Jan 2 at 15:08










1 Answer
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$begingroup$

This direction is pretty straight forward im guessing you probably wanted the other direction and misstyped your question



$$ delta(z,z_0) = left | int_x ^ {x_0} sqrt{1 - t ^2/4} ,dt right | leq left | int_x ^ {x_0} 1 ,dt right |
\ leq |x-x_0|leq |z-z_0|$$



$Rightarrow C=1$ because $ delta(z,z_0)leq 1*|z-z_0|$



For the other direction you only run into problems when either $x=2$ or $x_0=2$. Set $x_0=2$ and then look at $$lim_{xto 2} frac{delta((x,sqrt{1-x^2/4}),(2,0))}{|(x,sqrt{1-x^2/4})-(2,0)|}$$
Using H'lopital you can calculate that limit and if it is 0 your in tough luck. Then there is no $C>0$ so that $ delta(z,z_0)geq C* |z-z_0|forall z,z_0$






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    $begingroup$

    This direction is pretty straight forward im guessing you probably wanted the other direction and misstyped your question



    $$ delta(z,z_0) = left | int_x ^ {x_0} sqrt{1 - t ^2/4} ,dt right | leq left | int_x ^ {x_0} 1 ,dt right |
    \ leq |x-x_0|leq |z-z_0|$$



    $Rightarrow C=1$ because $ delta(z,z_0)leq 1*|z-z_0|$



    For the other direction you only run into problems when either $x=2$ or $x_0=2$. Set $x_0=2$ and then look at $$lim_{xto 2} frac{delta((x,sqrt{1-x^2/4}),(2,0))}{|(x,sqrt{1-x^2/4})-(2,0)|}$$
    Using H'lopital you can calculate that limit and if it is 0 your in tough luck. Then there is no $C>0$ so that $ delta(z,z_0)geq C* |z-z_0|forall z,z_0$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      This direction is pretty straight forward im guessing you probably wanted the other direction and misstyped your question



      $$ delta(z,z_0) = left | int_x ^ {x_0} sqrt{1 - t ^2/4} ,dt right | leq left | int_x ^ {x_0} 1 ,dt right |
      \ leq |x-x_0|leq |z-z_0|$$



      $Rightarrow C=1$ because $ delta(z,z_0)leq 1*|z-z_0|$



      For the other direction you only run into problems when either $x=2$ or $x_0=2$. Set $x_0=2$ and then look at $$lim_{xto 2} frac{delta((x,sqrt{1-x^2/4}),(2,0))}{|(x,sqrt{1-x^2/4})-(2,0)|}$$
      Using H'lopital you can calculate that limit and if it is 0 your in tough luck. Then there is no $C>0$ so that $ delta(z,z_0)geq C* |z-z_0|forall z,z_0$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        This direction is pretty straight forward im guessing you probably wanted the other direction and misstyped your question



        $$ delta(z,z_0) = left | int_x ^ {x_0} sqrt{1 - t ^2/4} ,dt right | leq left | int_x ^ {x_0} 1 ,dt right |
        \ leq |x-x_0|leq |z-z_0|$$



        $Rightarrow C=1$ because $ delta(z,z_0)leq 1*|z-z_0|$



        For the other direction you only run into problems when either $x=2$ or $x_0=2$. Set $x_0=2$ and then look at $$lim_{xto 2} frac{delta((x,sqrt{1-x^2/4}),(2,0))}{|(x,sqrt{1-x^2/4})-(2,0)|}$$
        Using H'lopital you can calculate that limit and if it is 0 your in tough luck. Then there is no $C>0$ so that $ delta(z,z_0)geq C* |z-z_0|forall z,z_0$






        share|cite|improve this answer









        $endgroup$



        This direction is pretty straight forward im guessing you probably wanted the other direction and misstyped your question



        $$ delta(z,z_0) = left | int_x ^ {x_0} sqrt{1 - t ^2/4} ,dt right | leq left | int_x ^ {x_0} 1 ,dt right |
        \ leq |x-x_0|leq |z-z_0|$$



        $Rightarrow C=1$ because $ delta(z,z_0)leq 1*|z-z_0|$



        For the other direction you only run into problems when either $x=2$ or $x_0=2$. Set $x_0=2$ and then look at $$lim_{xto 2} frac{delta((x,sqrt{1-x^2/4}),(2,0))}{|(x,sqrt{1-x^2/4})-(2,0)|}$$
        Using H'lopital you can calculate that limit and if it is 0 your in tough luck. Then there is no $C>0$ so that $ delta(z,z_0)geq C* |z-z_0|forall z,z_0$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 15:27









        A. PA. P

        1085




        1085






























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