Linear ODE's

**Aryth** · October 22nd 2009, 03:04 PM

A substance evaporates at a rate proportional to the exposed surface. If a spherical mothball of radius $\frac{1}{2}$ cm has radius 0.4 cm after 6 months, how long will it take:

(a) For the radius to be $\frac{1}{4}$ cm?

(b) For the volume of the mothball to be half of what it was originally?

**mr fantastic** · October 23rd 2009, 03:45 AM

Originally Posted by Aryth

A substance evaporates at a rate proportional to the exposed surface. If a spherical mothball of radius $\frac{1}{2}$ cm has radius 0.4 cm after 6 months, how long will it take:

(a) For the radius to be $\frac{1}{4}$ cm?

(b) For the volume of the mothball to be half of what it was originally?

Here's a start: $\frac{dV}{dt} = -k \pi r^2 = -k \pi \left(\frac{3V}{4 \pi}\right)^{2/3}$ .

**chisigma** · October 23rd 2009, 04:07 AM

Originally Posted by mr fantastic

Here's a start: $\frac{dV}{dt} = k \pi r^2 = k \pi \left(\frac{3V}{4 \pi}\right)^{2/3}$ .

The title of the thread is : Linear ODE...

... and that could mean that a linear ODE is required. This goal is achieved writing...

$\frac{dV}{dt} = 4\cdot \pi \cdot r^{2} \frac{dr}{dt} = -4\cdot \pi \cdot r^{2} \cdot \alpha \rightarrow \frac{dr}{dt} = - \alpha$ (1)

Very easy!...

Kind regards

$\chi$ $\sigma$

**mr fantastic** · October 23rd 2009, 04:17 AM

Originally Posted by chisigma

The title of the thread is : Linear ODE...

... and that could mean that a linear ODE is required. This goal is achieved writing...

$\frac{dV}{dt} = 4\cdot \pi \cdot r^{2} \frac{dr}{dt} = -4\cdot \pi \cdot r^{2} \cdot \alpha \rightarrow \frac{dr}{dt} = - \alpha$ (1)

Very easy!...

Kind regards

$\chi$ $\sigma$

My DE has the form $\frac{dt}{dV} = f(V)$ . Last time I checked, such DE's were classified as linear. However, I do agree that your DE $\frac{dr}{dt} = - \alpha$ provides for a much simpler solution.

**Aryth** · October 23rd 2009, 09:38 AM

Thanks a lot. The word problems really mess me up.

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