Applications of Calculus

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May 28th 2008, 02:53 AM
Flay

Applications of Calculus

I'm not quite sure how to approach this question, could someone help me out here? I'm sure I'll have further questions after this, so check back in this thread for updates. Anyways, the question asks;

A certaing brand of medicine tablet is in the shape of a sphere with diameter 5mm. Ther rate at which the pill dissolves is dr/dt = -k, where r is the radius of the sphere at time t hours, and k is a positive constant.

a) Show that r = 5/2 - kt

b) If the pill dissolves completely in 12 hours, find k.
May 28th 2008, 03:39 AM
Gusbob

i)
$\frac{dr}{dt} = -k$
$r = \int -k \,dt$
$r = -kt + C$
When t = 0, r = 5/2 (initially the tablet has diameter 5, which means radius is 5/2)
$\frac{5}{2} = -k(0) +C$
$C = \frac{5}{2}$
Therefore $r = \frac{5}{2} -kt$

ii)
The pill is dissolved when r = 0, t = 12. Sub these into the equation we derived:

$0 = \frac{5}{2} -12k$
$12k = \frac{5}{2}$
$k = \frac{5}{24}$
May 28th 2008, 03:51 AM
Flay

Thanks, for some reason my mind just didn't make the leap from diameter to radius. (Headbang)
May 28th 2008, 03:54 AM
Flay

Got another one;

Twenty-five wallabies are released on Wombat Island and the population is observed over the next six years. It is found that the rate of increase in the wallaby population is given by $\frac{dP}{dt} = 12t - 3t^2$ , and the population itself is given by $P = 25 + 6t^2 - t^3$ . When does the population increase most rapidly?
May 28th 2008, 04:13 AM
Gusbob

I'm pretty sure the population increase most rapidly means when
$\frac{d^2P}{dt^2}=0$ , an inflexion where concavity changes from positive to negative

Differentiating your rate of increase formula
$\frac{d^2P}{dt^2} = 12-6t$
$12-6t = 0$ for possible points of inflexion
$t = 2$

Checking the concavity of both sides of t =2, we find that concavity changes from positive to negative. Therefore there is an inflexion at t = 2

The population increased most rapidly in the second year (assuming the year they released it is year 0)