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Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: immortal on April 20, 2010, 12:04:19 pm

Title: help needed in maths, Diff eqn.
Post by: immortal 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
Title: Re: help needed in maths, Diff eqn.
Post by: astarmathsandphysics on April 20, 2010, 12:07:06 pm: Home in half hour
Title: Re: help needed in maths, Diff eqn.
Post by: astarmathsandphysics 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 $\frac {dr}{dt} =0.48r^2 *1/6 =0.08 r^2$
Title: Re: help needed in maths, Diff eqn.
Post by: immortal on April 20, 2010, 12:51:56 pm: Thanks man! :)
Helped a lot..