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mcastillo356
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- A metal worker must make a circular metal disc with a certain accuracy. Got a solved example, but what if the disc's radius is not that of the example?
The area of a circular disc with radius ##r## is ##A=\pi r^2\;\mbox{cm}^2##. A smith must make a circular metal disc of ##400\pi\;\mbox{cm}^2## with an accuracy of ##\pm{5}\;\mbox{cm}^2##. Which accuracy range must have a ##20\;\mbox{cm}## disc?
Answer The metal worker wants to obtain ##|\pi r^2-400\pi|<5##, this is
$$400\pi-5<\pi r^2<400\pi+5$$
$$19.96017<r<20.03975$$
Therefore, the smith needs ##|r-20|<0.03975##; so he must check the radius of the disc is less than ##0.04\;\mbox{cm}## the value of ##20\;\mbox{cm}##.
The question: what about if the radius is ##30## or, for example, ##5\;\mbox{cm}##?
PS: I'm not native. Forgive my english
Answer The metal worker wants to obtain ##|\pi r^2-400\pi|<5##, this is
$$400\pi-5<\pi r^2<400\pi+5$$
or
$$\sqrt{400-(5/\pi)}<r<\sqrt{400+(5/\pi)}$$$$19.96017<r<20.03975$$
Therefore, the smith needs ##|r-20|<0.03975##; so he must check the radius of the disc is less than ##0.04\;\mbox{cm}## the value of ##20\;\mbox{cm}##.
The question: what about if the radius is ##30## or, for example, ##5\;\mbox{cm}##?
PS: I'm not native. Forgive my english