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Differential Equations

Question
CBSEENMA12033083

The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Solution
Let v be volume of spherical balloon of radius r.
 therefore space space space space space space space straight v space equals space 4 over 3 πr cubed                                  ...(1)
From given condition,
               dv over dt equals space straight k space space or space space straight d over dt open parentheses 4 over 3 πr cubed close parentheses space equals space straight k                open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets
therefore space space fraction numerator 4 straight pi over denominator 3 end fraction. space 3 space straight r squared space dr over dt space equals space straight k space space space or space space 4 πr squared dr over dt space equals space straight k
Separating the variables and integrating, we get.
4 straight pi integral straight r squared space dr space equals space straight k space integral space dt space space space or space space space 4 space straight pi space straight r cubed over 3 space equals space straight k space straight t space plus straight c              ...(2)
Now t = 0 when r = 3
therefore space space space space 4 straight pi fraction numerator left parenthesis 3 right parenthesis cubed over denominator 3 end fraction space equals space straight k cross times 0 space plus space straight c space space space rightwards double arrow space straight c space equals space 36 space straight pi                ...(3)
Again   t = 3  when r = 6
therefore space space fraction numerator 4 straight pi over denominator 3 end fraction left parenthesis 6 right parenthesis cubed space equals space 3 straight k space plus space 36 straight pi                                    open square brackets because space space of space left parenthesis 3 right parenthesis close square brackets
therefore space space space space 288 space straight pi space equals space 3 space straight k space plus space 36 space straight pi space space or space space 3 space straight k space equals space 252 space straight pi
therefore space space space straight k space equals space 84 space straight pi
Putting straight k space equals 84 space straight pi comma space space straight c space equals space 36 space straight pi space in space left parenthesis 2 right parenthesis comma space we space get
                  fraction numerator 4 straight pi over denominator 3 end fraction straight r cubed equals space space 84 space straight pi space straight t space plus space 36 space straight pi space space space or space space space straight r cubed over 3 space equals space 21 space straight t space plus space 9
therefore space space space straight r cubed space equals space 63 space straight t space plus space 27 space space space space rightwards double arrow space space space space straight r space equals space left square bracket 9 space left parenthesis 7 space straight t space plus space 3 right parenthesis right square bracket to the power of 1 third end exponent

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