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Math Help - Error propagation

  1. #1
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    Error propagation

    By approximation what percentage does the volume of a ball change if its surface area is changed by -4%? The volume V and surface area S of a ball of radius r are given by the formulas: v=4/3pir^2 and S=4pir^2
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  2. #2
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    Quote Originally Posted by Tiger View Post
    By approximation what percentage does the volume of a ball change if its surface area is changed by -4%? The volume V and surface area S of a ball of radius r are given by the formulas: v=4/3pir^2 and S=4pir^2
    Correction: \displaystyle V={{4}\over{3}}\pi r^3

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  3. #3
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    Hello, Tiger!

    Is some information missing?
    We should have been given one more fact
    . . like the Surface Area at a certain instant.


    \text{By approximation what percentage does the volume of a ball change}
    \text{if its surface area is changed by }\text{-}4\%?

    V\,=\,\frac{4}{3}\pi r^3,\;\;S\,=\,4\pi r^2

    We have: . S \:=\:4\pi r^2 \quad\Rightarrow\quad r^2 \:=\:\dfrac{S}{4\pi} \quad\Rightarrow\quad r \:=\:\dfrac{S^{\frac{1}{2}}}{2\sqrt{\pi}}

    Then: . V \;=\;\dfrac{4\pi}{3}\left(\dfrac{S^{\frac{1}{2}}}{  2\sqrt{\pi}}\right)^3 \;=\;\dfrac{4\pi}{3}\cdot \dfrac{S^{\frac{3}{2}}}{8\pi^{\frac{3}{2}}} \;=\;\dfrac{1}{6\sqrt{\pi}}\,S^{\frac{3}{2}}


    Take differentials:

    . . \displaystyle dV \;=\;\frac{1}{6\sqrt{\pi}}\cdot\frac{3}{2}S^{\frac  {1}{2}}\cdot dS\;=\;\frac{S^{\frac{1}{2}}}{4\sqrt{\pi}}\cdot dS


    We are given: . dS\:=\:-0.04

    . . Hence: . \displaystyle dV\;=\;\frac{S^{\frac{1}{2}}}{4\sqrt{\pi}}(\text{-}0.04) \;=\;-\frac{\sqrt{S}}{100\sqrt{\pi}}


    Therefore: . \displaystyle \Delta V \;\approx\; -\frac{\sqrt{S}}{100\sqrt{\pi}}

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  4. #4
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    Quote Originally Posted by Tiger View Post
    By approximation what percentage does the volume of a ball change if its surface area is changed by -4%? The volume V and surface area S of a ball of radius r are given by the formulas: v=4/3pir^2 and S=4pir^2
    Or....

    Surface area changes by -4\%

    \Rightarrow\ S(0.96)=4{\pi}\left[0.96r^2\right]=4{\pi}\left[\sqrt{0.96}\;r\right]^2

    New radius is \sqrt{0.96}\;r where r is the original radius.

    New volume is \displaystyle\frac{4}{3}{\pi}\left[\sqrt{0.96}\;r\right]^3=\frac{4}{3}{\pi}r^3\left[0.96^{\frac{3}{2}}\right]

    The percentage change is \displaystyle\left[1-0.96^{\frac{3}{2}}\right]100\%
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