A Machinist's Imperfect Disk












2














Exercise




A machinist is required to manufacture a circular metal disk with area $1000$ cm$^2$.




  1. What radius produces such a disk?


  2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?


  3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?







Solution



Note: I've decided to solve part (3) before part (2), as it helps set the stage of the solution of (2).




1. What radius produces such a disk?




$pi r_0^2 = 1000 implies r_0 = sqrt{frac{1000}{pi}} approx 17.8412$




3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?




$a = r_0 approx 17.8412$



$L = pi r_0^2 = 1000$



$x = r$



$f(x) = pi r^2$



$|r - r_0| < delta$



$|pi r^2 - pi r_0^2| < epsilon$




2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?




$|pi r^2 - pi r_0^2| < epsilon$
$implies |pi r^2 - 1000| < 5$
$implies -5 < pi r^2 - 1000 < 5$
$implies 995 < pi r^2 < 1005$
$implies frac{995}{pi} < r^2 < frac{1005}{pi}$
$implies sqrt{frac{995}{pi}} < r < sqrt{frac{1005}{pi}}$
$implies 17.7966 < r < 17.8858$



$|r - r_0| < delta implies |r - 17.8412| < delta$



$|17.7966 - 17.8412| < delta implies 0.0446 < delta$



$|17.8858 - 17.8412| < delta implies 0.0446 < delta$



$delta = min(0.0446, 0.0446) = 0.0446$





Answer




1. What radius produces such a disk?




$$r_0 = 17.8412 text{ cm}$$




2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?




$$delta = 0.0446 text{ cm}$$




3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?




$$|r - r_0| < delta$$



$$|pi r^2 - pi r_0^2| < epsilon$$





Request



Is my answer correct? If not, in what part of my solution did I make a mistake?










share|cite|improve this question





























    2














    Exercise




    A machinist is required to manufacture a circular metal disk with area $1000$ cm$^2$.




    1. What radius produces such a disk?


    2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?


    3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?







    Solution



    Note: I've decided to solve part (3) before part (2), as it helps set the stage of the solution of (2).




    1. What radius produces such a disk?




    $pi r_0^2 = 1000 implies r_0 = sqrt{frac{1000}{pi}} approx 17.8412$




    3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?




    $a = r_0 approx 17.8412$



    $L = pi r_0^2 = 1000$



    $x = r$



    $f(x) = pi r^2$



    $|r - r_0| < delta$



    $|pi r^2 - pi r_0^2| < epsilon$




    2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?




    $|pi r^2 - pi r_0^2| < epsilon$
    $implies |pi r^2 - 1000| < 5$
    $implies -5 < pi r^2 - 1000 < 5$
    $implies 995 < pi r^2 < 1005$
    $implies frac{995}{pi} < r^2 < frac{1005}{pi}$
    $implies sqrt{frac{995}{pi}} < r < sqrt{frac{1005}{pi}}$
    $implies 17.7966 < r < 17.8858$



    $|r - r_0| < delta implies |r - 17.8412| < delta$



    $|17.7966 - 17.8412| < delta implies 0.0446 < delta$



    $|17.8858 - 17.8412| < delta implies 0.0446 < delta$



    $delta = min(0.0446, 0.0446) = 0.0446$





    Answer




    1. What radius produces such a disk?




    $$r_0 = 17.8412 text{ cm}$$




    2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?




    $$delta = 0.0446 text{ cm}$$




    3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?




    $$|r - r_0| < delta$$



    $$|pi r^2 - pi r_0^2| < epsilon$$





    Request



    Is my answer correct? If not, in what part of my solution did I make a mistake?










    share|cite|improve this question



























      2












      2








      2


      0





      Exercise




      A machinist is required to manufacture a circular metal disk with area $1000$ cm$^2$.




      1. What radius produces such a disk?


      2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?


      3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?







      Solution



      Note: I've decided to solve part (3) before part (2), as it helps set the stage of the solution of (2).




      1. What radius produces such a disk?




      $pi r_0^2 = 1000 implies r_0 = sqrt{frac{1000}{pi}} approx 17.8412$




      3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?




      $a = r_0 approx 17.8412$



      $L = pi r_0^2 = 1000$



      $x = r$



      $f(x) = pi r^2$



      $|r - r_0| < delta$



      $|pi r^2 - pi r_0^2| < epsilon$




      2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?




      $|pi r^2 - pi r_0^2| < epsilon$
      $implies |pi r^2 - 1000| < 5$
      $implies -5 < pi r^2 - 1000 < 5$
      $implies 995 < pi r^2 < 1005$
      $implies frac{995}{pi} < r^2 < frac{1005}{pi}$
      $implies sqrt{frac{995}{pi}} < r < sqrt{frac{1005}{pi}}$
      $implies 17.7966 < r < 17.8858$



      $|r - r_0| < delta implies |r - 17.8412| < delta$



      $|17.7966 - 17.8412| < delta implies 0.0446 < delta$



      $|17.8858 - 17.8412| < delta implies 0.0446 < delta$



      $delta = min(0.0446, 0.0446) = 0.0446$





      Answer




      1. What radius produces such a disk?




      $$r_0 = 17.8412 text{ cm}$$




      2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?




      $$delta = 0.0446 text{ cm}$$




      3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?




      $$|r - r_0| < delta$$



      $$|pi r^2 - pi r_0^2| < epsilon$$





      Request



      Is my answer correct? If not, in what part of my solution did I make a mistake?










      share|cite|improve this question















      Exercise




      A machinist is required to manufacture a circular metal disk with area $1000$ cm$^2$.




      1. What radius produces such a disk?


      2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?


      3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?







      Solution



      Note: I've decided to solve part (3) before part (2), as it helps set the stage of the solution of (2).




      1. What radius produces such a disk?




      $pi r_0^2 = 1000 implies r_0 = sqrt{frac{1000}{pi}} approx 17.8412$




      3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?




      $a = r_0 approx 17.8412$



      $L = pi r_0^2 = 1000$



      $x = r$



      $f(x) = pi r^2$



      $|r - r_0| < delta$



      $|pi r^2 - pi r_0^2| < epsilon$




      2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?




      $|pi r^2 - pi r_0^2| < epsilon$
      $implies |pi r^2 - 1000| < 5$
      $implies -5 < pi r^2 - 1000 < 5$
      $implies 995 < pi r^2 < 1005$
      $implies frac{995}{pi} < r^2 < frac{1005}{pi}$
      $implies sqrt{frac{995}{pi}} < r < sqrt{frac{1005}{pi}}$
      $implies 17.7966 < r < 17.8858$



      $|r - r_0| < delta implies |r - 17.8412| < delta$



      $|17.7966 - 17.8412| < delta implies 0.0446 < delta$



      $|17.8858 - 17.8412| < delta implies 0.0446 < delta$



      $delta = min(0.0446, 0.0446) = 0.0446$





      Answer




      1. What radius produces such a disk?




      $$r_0 = 17.8412 text{ cm}$$




      2. If the machinist is allowed an error tolerance of $pm 5$ cm$^2$ in the area of the disk, how close to the ideal radius in part (1) must the machinist control the radius?




      $$delta = 0.0446 text{ cm}$$




      3. In terms of the $epsilon$, $delta$ definition of $lim limits_{x to a}{f(x)} = L$ what is $x$? What is $f(x)$? What value of $epsilon$ is given? What is the corresponding value of $delta$?




      $$|r - r_0| < delta$$



      $$|pi r^2 - pi r_0^2| < epsilon$$





      Request



      Is my answer correct? If not, in what part of my solution did I make a mistake?







      calculus limits proof-verification word-problem






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      edited Nov 23 '16 at 0:40







      Fine Man

















      asked Nov 22 '16 at 21:18









      Fine ManFine Man

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      902525






















          1 Answer
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          In part 3 you are asked about four items: $x$, $f(x)$, $epsilon$, and $delta$. Your answer omits these. To make it clear:




          • The independent variable $x$ here is the radius $r$ of the disk

          • The function $f$ maps radius to area, i.e., $rmapsto pi r^2$. In other words, $f(x)=pi x^2$.

          • The given $epsilon$ is $5,text{cm}^2$

          • The corresponding (or rather: a suitable) value of $delta$ is the $approx 0.0446,text{cm}$ you found


          On a sidenote, we may observe that picking $delta$ such that $fracepsilondelta=f'(r_0)$ usually gives a good enough approximation. Indeed, for the problem at hand $f'(x)=2pi x$, so we quickly obtain $delta=frac{epsilon}{2pi r_0}approx 0.0446$.
          Things become even easier with relative errors: Raising to $n$th power multiplies (small) relative errors by $n$. Hence for a desired relative output error of $frac 5{1000}=frac1{200}$, the relative input error should be bounded by $frac{1}{400}$, and indeed $frac1{400}cdot 17.8412approx 0.0446$.






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            0














            In part 3 you are asked about four items: $x$, $f(x)$, $epsilon$, and $delta$. Your answer omits these. To make it clear:




            • The independent variable $x$ here is the radius $r$ of the disk

            • The function $f$ maps radius to area, i.e., $rmapsto pi r^2$. In other words, $f(x)=pi x^2$.

            • The given $epsilon$ is $5,text{cm}^2$

            • The corresponding (or rather: a suitable) value of $delta$ is the $approx 0.0446,text{cm}$ you found


            On a sidenote, we may observe that picking $delta$ such that $fracepsilondelta=f'(r_0)$ usually gives a good enough approximation. Indeed, for the problem at hand $f'(x)=2pi x$, so we quickly obtain $delta=frac{epsilon}{2pi r_0}approx 0.0446$.
            Things become even easier with relative errors: Raising to $n$th power multiplies (small) relative errors by $n$. Hence for a desired relative output error of $frac 5{1000}=frac1{200}$, the relative input error should be bounded by $frac{1}{400}$, and indeed $frac1{400}cdot 17.8412approx 0.0446$.






            share|cite|improve this answer


























              0














              In part 3 you are asked about four items: $x$, $f(x)$, $epsilon$, and $delta$. Your answer omits these. To make it clear:




              • The independent variable $x$ here is the radius $r$ of the disk

              • The function $f$ maps radius to area, i.e., $rmapsto pi r^2$. In other words, $f(x)=pi x^2$.

              • The given $epsilon$ is $5,text{cm}^2$

              • The corresponding (or rather: a suitable) value of $delta$ is the $approx 0.0446,text{cm}$ you found


              On a sidenote, we may observe that picking $delta$ such that $fracepsilondelta=f'(r_0)$ usually gives a good enough approximation. Indeed, for the problem at hand $f'(x)=2pi x$, so we quickly obtain $delta=frac{epsilon}{2pi r_0}approx 0.0446$.
              Things become even easier with relative errors: Raising to $n$th power multiplies (small) relative errors by $n$. Hence for a desired relative output error of $frac 5{1000}=frac1{200}$, the relative input error should be bounded by $frac{1}{400}$, and indeed $frac1{400}cdot 17.8412approx 0.0446$.






              share|cite|improve this answer
























                0












                0








                0






                In part 3 you are asked about four items: $x$, $f(x)$, $epsilon$, and $delta$. Your answer omits these. To make it clear:




                • The independent variable $x$ here is the radius $r$ of the disk

                • The function $f$ maps radius to area, i.e., $rmapsto pi r^2$. In other words, $f(x)=pi x^2$.

                • The given $epsilon$ is $5,text{cm}^2$

                • The corresponding (or rather: a suitable) value of $delta$ is the $approx 0.0446,text{cm}$ you found


                On a sidenote, we may observe that picking $delta$ such that $fracepsilondelta=f'(r_0)$ usually gives a good enough approximation. Indeed, for the problem at hand $f'(x)=2pi x$, so we quickly obtain $delta=frac{epsilon}{2pi r_0}approx 0.0446$.
                Things become even easier with relative errors: Raising to $n$th power multiplies (small) relative errors by $n$. Hence for a desired relative output error of $frac 5{1000}=frac1{200}$, the relative input error should be bounded by $frac{1}{400}$, and indeed $frac1{400}cdot 17.8412approx 0.0446$.






                share|cite|improve this answer












                In part 3 you are asked about four items: $x$, $f(x)$, $epsilon$, and $delta$. Your answer omits these. To make it clear:




                • The independent variable $x$ here is the radius $r$ of the disk

                • The function $f$ maps radius to area, i.e., $rmapsto pi r^2$. In other words, $f(x)=pi x^2$.

                • The given $epsilon$ is $5,text{cm}^2$

                • The corresponding (or rather: a suitable) value of $delta$ is the $approx 0.0446,text{cm}$ you found


                On a sidenote, we may observe that picking $delta$ such that $fracepsilondelta=f'(r_0)$ usually gives a good enough approximation. Indeed, for the problem at hand $f'(x)=2pi x$, so we quickly obtain $delta=frac{epsilon}{2pi r_0}approx 0.0446$.
                Things become even easier with relative errors: Raising to $n$th power multiplies (small) relative errors by $n$. Hence for a desired relative output error of $frac 5{1000}=frac1{200}$, the relative input error should be bounded by $frac{1}{400}$, and indeed $frac1{400}cdot 17.8412approx 0.0446$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Hagen von EitzenHagen von Eitzen

                276k21269496




                276k21269496






























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