Show dV/dt = -kV^2/3 given.......

[donal1353]

Hi, im having trouble with a "show" style question that forms part 1 of a multi part question, I have no problems with the rest if I take the "show" stage to be absolutly true. Some general guidance regarding the process would be appreciated. The question is as follows:

A spherical pill with volume V and surface area S is swallowed and slowly dissolves in the stomach. Assume that the rate of decrease in the volume V is directly proportional to the pill's surface area.

a)i)
Show dV/dt = -kV^2/3 , where K is the constant of proportionality and k > 0.

 

[khansaheb]

Hi, im having trouble with a "show" style question that forms part 1 of a multi part question, I have no problems with the rest if I take the "show" stage to be absolutly true. Some general guidance regarding the process would be appreciated. The question is as follows:

A spherical pill with volume V and surface area S is swallowed and slowly dissolves in the stomach. Assume that the rate of decrease in the volume V is directly proportional to the pill's surface area.

a)i)
Show dV/dt = -kV^2/3 , where K is the constant of proportionality and k > 0.

Volume of a sphere (of radius = r) = 4/3 * (pi) * r^3 and .......... Surface area of a sphere (of radius = r) = ??

As the pill gets dissolved - does it maintain "spherical" geometry?

 

[mario99]

The volume of a sphere is:
$\displaystyle V = \frac{4}{3}\pi R^3$

The surface area of a sphere is:
$\displaystyle S = 4\pi R^2$

The first attempt to write the differential equation is:
$\displaystyle \frac{dV}{dt} = -k_0S = -k_04\pi R^2$

Use the first formula to find $\displaystyle R$, then plug the result in the differential equation, and combine all constants as $\displaystyle k$ to get the required form.