# Quick diffrential equation question.

• Jan 1st 2009, 08:34 AM
AshleyT
Quick diffrential equation question.
Quote:

When a mint is spherical mint is sucked, a simple model gives the rate of decrease of its radius as inversely proportional to the square of the radius. Initially, the radius of a mint is 5mm and after 5 mins the radius is 4mm
Okey, my question is, does the differential equation for this start:
$\displaystyle \frac{dr}{dt} = \frac{r^2}{-k}$
I wasn't at school when this was taught(=( ) and there's no examples in my book on inverse proportion.
Thank-you :)
• Jan 1st 2009, 08:55 AM
Jester
Quote:

Originally Posted by AshleyT
Okey, my question is, does the differential equation for this start:
$\displaystyle \frac{dr}{dt} = \frac{r^2}{-k}$
I wasn't at school when this was taught(=( ) and there's no examples in my book on inverse proportion.
Thank-you :)

Try $\displaystyle \frac{dr}{dt} = \frac{k}{r^2}$
• Jan 1st 2009, 09:03 AM
AshleyT
Quote:

Originally Posted by danny arrigo
Try $\displaystyle \frac{dr}{dt} = \frac{k}{r^2}$

But its decreasing, won't it be negative?
• Jan 1st 2009, 09:13 AM
skeeter
k will be a negative value ...

$\displaystyle \frac{dr}{dt} = \frac{k}{r^2}$

$\displaystyle r^2 \, dr = k \, dt$

$\displaystyle \frac{r^3}{3} = kt + C$

when t = 0, r = 5

$\displaystyle \frac{125}{3} = k(0) + C$

$\displaystyle \frac{r^3}{3} = kt + \frac{125}{3}$

when t = 5 , r = 4

$\displaystyle \frac{64}{3} = 5k + \frac{125}{3}$

$\displaystyle -\frac{61}{15} = k$