# Creating an ODE for a growing cylinder

The exact question is: "Expanding peat discs are used to grow seedlings (google for an image!). These discs absorb water through their surface, causing an increase in height, h, while keeping a constant radius, r. Assume the disc is a cylinder and that the density of the disk, p = mass/volume, remains constant.

The rate of increase in the mass of the disc, m, is proportional to its total surface area. Derive a differential equation for the rate of change in m. You may include arbitrary constants, but the RHS should depend only on constants and m."

So far, I've thought of the fact that the mass only depends on the height, since the radius and density are constant. But then it mentions that the mass is proportional to the total surface area, meaning its proportional to height, since radius is constant and surface area depends on r and h.

The question says that the RHS should only depend on m, so does that mean that I have to define dm/dh in terms of m and constants? Or can I include h since it is the independant variable.

Those are my thoughts. I cannot seem to think of a DE for this aside from dm/dt = constant, unless, since dm/dt is directly proportional to surface area, it becomes dm/dt = constant + other_constant*h, as that is the surface area of a cylinder. But my thoughts on that answer is that the RHS does not contain m.

Note that since we do not know how $h$ changes as a function of time, the rate of change of the mass should be a derivative w.r.t. $h$.

You then have $$\rho=\frac{m}{V(h)}=\frac{m}{\pi r^2 h}$$ and $$\frac{dm}{dh}\propto 2\pi r^2+2\pi r h,$$

which is two equations with two variables $m$ and $h$. Can you see what to do now?

Solution:

Isolate $h$ in the first equation and substitute that into the second one to obtain $$\frac{dm}{dh}\propto 2\pi r^2+\frac{2m}{\rho r}.$$

• Wow, it's so much more clear when the equations are right in front of me. Guess I need to start writing out my thoughts more instead of just thinking my thoughts. Thank you very much! – Noah Murad Sep 30 '16 at 4:03
• @NoahMurad Np :) Yes, I find that it almost always helps to write the information known to you down - that way, you free up the working memory you spent just holding that information in your mind, and are able use it to actually think of a solution. – Bobson Dugnutt Sep 30 '16 at 10:22

Since the solution has already been leaked :), let me just explain rigorously how it's done.

Given the density $\rho(x,y,x)$ of any object $C$, it's mass $m$ is calculated as $$m = \int\int\int_{C} \, \rho(x,y,z) \, dx \, dy \, dz$$ Since in your case $C$ has constant density, you end up with $$m = \int\int\int_{C} \, \rho \, \, dx \, dy \, dz = \rho \, \int\int\int_{C} \, \, dx \, dy \, dz = \rho \, \text{Volume}(C) = \rho \, V$$ where by $V$ I have denoted the volume of the cylinder $C$. We assume, that $V=V(t)$ changes with time, while $C$ is absorbing water. Since $\rho$ stays fixed, then the mass of $C$ also changes with time, i.e. $m = m(t) = \rho V(t)$. Fortunately, $C$ is a cylinder of fixed radius, so $V = \pi\,r^2 h$, where $h$ is the height of the cylinder. Then the height $h = h(t) = \frac{1}{\pi r^2} \, V(t)$ aslo changes with time. In terms of the mass $$h(t) = \frac{1}{\pi r^2} \, V(t) = \frac{1}{\pi \rho \, r^2} \, m(t)$$ The area of the cylinder is easy to compute $$S(t) = \pi r^2 + 2 \pi r \, h(t) = \pi r^2 + 2 \pi r \, \frac{1}{\pi \rho \, r^2} \, m(t) = \pi r^2 + \, \frac{2 }{ \rho \, r} \, m(t)$$ and written in terms of the mass. By assumption, $\frac{dm}{dt}(t)$ is proportional to $S(t)$ at any moment of time $t$. This means that there is a constant $k$ such that for all $t$ $$\frac{dm}{dt}(t) = k\, S(t) = k \pi r^2 + \, \frac{2 k}{ \rho \, r} \, m(t)$$ Finally we have derived the very simple linear differential equation $$\frac{d m}{ dt} = k \pi r^2 + \, \frac{2 k}{ \rho \, r} \, m$$ in terms of the mass $m = m(t)$.