Author Topic: help needed in maths, Diff eqn.  (Read 677 times)

Offline immortal

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help needed in maths, Diff eqn.
« on: April 20, 2010, 12:04:19 pm »
Hi,
i'd appriate if any1 cud solve this doubt, found in O/N09  P31


   In a model of the expansion of a sphere of radius r cm, it is assumed that, at time t seconds after the
      start, the rate of increase of the surface area of the sphere is proportional to its volume.  When t = 0,
                  dr
      r = 5 and -      = 2.
                  dt

       (i)  Show that r satis?es the differential equation

                                                      dr          2
                                                       - = 0.08r  .                                                   
                                                      dt
                                                                                                                                        2
            [The surface area A and volume V of a sphere of radius r are given by the formulae A = 4(Pi)r  ,
                 4        3
            V = -  (Pi)r  .]
                 3
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Offline astarmathsandphysics

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Re: help needed in maths, Diff eqn.
« Reply #1 on: April 20, 2010, 12:07:06 pm »
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Offline astarmathsandphysics

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Re: help needed in maths, Diff eqn.
« Reply #2 on: April 20, 2010, 12:35:23 pm »
   In a model of the expansion of a sphere of radius r cm, it is assumed that, at time t seconds after the
      start, the rate of increase of the surface area of the sphere is proportional to its volume.  When t = 0,
                  dr
      r = 5 and -      = 2.
                  dt

       (i)  Show that r satis?es the differential equation

                                                      dr          2
                                                       - = 0.08r  .                                                  
                                                      dt
                                                                                                                                        2
            [The surface area A and volume V of a sphere of radius r are given by the formulae A = 4(Pi)r  ,
                 4        3
            V = -  (Pi)r  .]
                 3

If the rate of increase is proportional to the suraface area
\frac {d(4 \pi r^2)}{dt} =K \frac {4 \pi r^3}{3}
8 \pi r \frac {dr}{dt} = \frac {4 \pi r^3}{3}
\frac {dr}{dt} =K \frac {1}{6} r^2
2=K \frac {1}{6} 5^2
K=12/5^2 =0.48
hence \frac {dr}{dt} =0.48r^2 *1/6 =0.08 r^2
« Last Edit: April 20, 2010, 12:40:35 pm by astarmathsandphysics »

Offline immortal

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Re: help needed in maths, Diff eqn.
« Reply #3 on: April 20, 2010, 12:51:56 pm »
Thanks man! :)
Helped a lot..
Life is short...so live it to da fullest :)