Approximating error in area of a circle from its radius












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$begingroup$


Given a circle with area 15 and an error in its area of .2, what would the maximum error in its radius be?



Given that $$A=pi r^2$$ $$A'=2pi r$$ it can be found that the radius when A = 15 must be $$r=sqrt{frac{15}{pi}} $$ so using differentials $$A(r+Delta r)approx A(r)+A'(r)Delta r $$ $$15approxfrac{sqrt{15}}{pi}+2pisqrt{frac{15}{pi}}Delta r$$



so $$Delta rapproxfrac{15-sqrt{frac{15}{pi}}}{2pi sqrt{frac{15}{pi}}}$$



Is this correct?






calculus ordinary-differential-equations







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  • 1




    $begingroup$
    $A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
    $endgroup$
    – Dando18
    Dec 11 '18 at 23:47












  • $begingroup$
    Yes, edited to show that.
    $endgroup$
    – ovil101
    Dec 11 '18 at 23:51












  • $begingroup$
    Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
    $endgroup$
    – LutzL
    Dec 12 '18 at 8:56
















1












$begingroup$


Given a circle with area 15 and an error in its area of .2, what would the maximum error in its radius be?



Given that $$A=pi r^2$$ $$A'=2pi r$$ it can be found that the radius when A = 15 must be $$r=sqrt{frac{15}{pi}} $$ so using differentials $$A(r+Delta r)approx A(r)+A'(r)Delta r $$ $$15approxfrac{sqrt{15}}{pi}+2pisqrt{frac{15}{pi}}Delta r$$



so $$Delta rapproxfrac{15-sqrt{frac{15}{pi}}}{2pi sqrt{frac{15}{pi}}}$$



Is this correct?






calculus ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
    $endgroup$
    – Dando18
    Dec 11 '18 at 23:47












  • $begingroup$
    Yes, edited to show that.
    $endgroup$
    – ovil101
    Dec 11 '18 at 23:51












  • $begingroup$
    Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
    $endgroup$
    – LutzL
    Dec 12 '18 at 8:56














1












1








1


0



$begingroup$


Given a circle with area 15 and an error in its area of .2, what would the maximum error in its radius be?



Given that $$A=pi r^2$$ $$A'=2pi r$$ it can be found that the radius when A = 15 must be $$r=sqrt{frac{15}{pi}} $$ so using differentials $$A(r+Delta r)approx A(r)+A'(r)Delta r $$ $$15approxfrac{sqrt{15}}{pi}+2pisqrt{frac{15}{pi}}Delta r$$



so $$Delta rapproxfrac{15-sqrt{frac{15}{pi}}}{2pi sqrt{frac{15}{pi}}}$$



Is this correct?






calculus ordinary-differential-equations







share|cite|improve this question











$endgroup$




Given a circle with area 15 and an error in its area of .2, what would the maximum error in its radius be?



Given that $$A=pi r^2$$ $$A'=2pi r$$ it can be found that the radius when A = 15 must be $$r=sqrt{frac{15}{pi}} $$ so using differentials $$A(r+Delta r)approx A(r)+A'(r)Delta r $$ $$15approxfrac{sqrt{15}}{pi}+2pisqrt{frac{15}{pi}}Delta r$$



so $$Delta rapproxfrac{15-sqrt{frac{15}{pi}}}{2pi sqrt{frac{15}{pi}}}$$



Is this correct?






calculus ordinary-differential-equations




calculus ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 23:50







ovil101

















asked Dec 11 '18 at 23:44









ovil101ovil101

373




373








  • 1




    $begingroup$
    $A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
    $endgroup$
    – Dando18
    Dec 11 '18 at 23:47












  • $begingroup$
    Yes, edited to show that.
    $endgroup$
    – ovil101
    Dec 11 '18 at 23:51












  • $begingroup$
    Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
    $endgroup$
    – LutzL
    Dec 12 '18 at 8:56














  • 1




    $begingroup$
    $A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
    $endgroup$
    – Dando18
    Dec 11 '18 at 23:47












  • $begingroup$
    Yes, edited to show that.
    $endgroup$
    – ovil101
    Dec 11 '18 at 23:51












  • $begingroup$
    Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
    $endgroup$
    – LutzL
    Dec 12 '18 at 8:56








1




1




$begingroup$
$A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
$endgroup$
– Dando18
Dec 11 '18 at 23:47






$begingroup$
$A=pi r^2 implies r=sqrt{frac{A}{pi}}$. Did you mean the radius is $sqrt{frac{15}{pi}}$?
$endgroup$
– Dando18
Dec 11 '18 at 23:47














$begingroup$
Yes, edited to show that.
$endgroup$
– ovil101
Dec 11 '18 at 23:51






$begingroup$
Yes, edited to show that.
$endgroup$
– ovil101
Dec 11 '18 at 23:51














$begingroup$
Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
$endgroup$
– LutzL
Dec 12 '18 at 8:56




$begingroup$
Did you replace $A(r+Δr)$ with $15$ and $A(r)$ with $frac{sqrt{15}}pi$? What is the justification for both? Where does the $pm0.2$ enter the formula?
$endgroup$
– LutzL
Dec 12 '18 at 8:56










2 Answers
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Your approach is correct but likely an overkill. The final result is also not really pretty.



If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The errors are related by



    $$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$



    Therefore



    $$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$



    Your formula is incorrect because you had $Delta A = A - r$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Your approach is correct but likely an overkill. The final result is also not really pretty.



      If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Your approach is correct but likely an overkill. The final result is also not really pretty.



        If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Your approach is correct but likely an overkill. The final result is also not really pretty.



          If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.






          share|cite|improve this answer









          $endgroup$



          Your approach is correct but likely an overkill. The final result is also not really pretty.



          If you have a known area $A$ with error $delta$, then $r = sqrt{frac{A pm delta}{pi}}$. Then its error is given by $delta_r = r_max - r_min = sqrt{frac{A + delta}{pi}} - sqrt{frac{A - delta}{pi}}$, exactly. If you want to go further on the approximations, you might want to use Bernoulli's inequality so $delta_r = sqrt{frac{A}{pi}}((1+delta/A)^{1/2} - (1-delta/A)^{1/2}) approx sqrt{frac{A}{pi}}(2*frac{delta}{2A}) approx frac{delta}{sqrt{pi A}}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 2:07









          Lucas HenriqueLucas Henrique

          1,059414




          1,059414























              0












              $begingroup$

              The errors are related by



              $$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$



              Therefore



              $$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$



              Your formula is incorrect because you had $Delta A = A - r$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The errors are related by



                $$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$



                Therefore



                $$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$



                Your formula is incorrect because you had $Delta A = A - r$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The errors are related by



                  $$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$



                  Therefore



                  $$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$



                  Your formula is incorrect because you had $Delta A = A - r$






                  share|cite|improve this answer









                  $endgroup$



                  The errors are related by



                  $$ Delta A = frac{dA}{dr} Delta r = 2pi r Delta r $$



                  Therefore



                  $$ Delta r = frac{Delta A}{2pi r} = frac{0.2}{2pi sqrt{frac{15}{pi}}} = frac{0.1}{sqrt{15pi}} $$



                  Your formula is incorrect because you had $Delta A = A - r$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 6:13









                  DylanDylan

                  12.9k31027




                  12.9k31027






























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