Question

Could someone give some guidance on how to solve c and d?

1. During the last week of classes before finals, Ryan Rabbitt starts to find coffee a daily necessity Every morning, Ryan makes his pour-over coffee with a flat-bottom filter, which can be modeled as a cylinder with radius 5 centimeters and height 12 centimeters. He pours a liquid coffee mix into the filter, and the liquid slowly flows through the filter into a container below When filtering, the flow rate is proportional to the contact area between the liquid and the filter (The "flow rate" is the rate at which the liquid is leaving the filter; the "contact area" is the area of the part of the coffee filter, including the bottom and part of the side, that is touching the liquid.) (a) Let h be the height (in cm) of the surface of liquid above the bottom of the filter. Set up a differential equation of h with respect to the time t (in minutes) after filtration begins (Hint: Let V be the total volume of coffee mixture in the filter, and observe the chain rule dV dV dh dt dh dt What is dV/dt? What is dV/dh?) parameter(s) 2.2 minutes, or more accurately, (b) Use separation of variables to find a formula for h(t). Your answer may involve unknown (c) Suppose the filter is initially full when filtration begins, and the filtration process takes about tn(29/5) Use this information to determine the parameter(s) for your formula for h(t) above (d) As fellow math students join his study group, Ryan realizes he needs a continuing source of coffee for the group members. To do so, Ryan keeps adding liquid to the filter at a constant rate of 6 cubic centimeters per minute. In the long run, what will be the height of the surface level of the coffee? (You should be able to answer this question without explicitly solving any differential equations!)

Answer 1

As you mention only about the 3rd and 4th one, i'll go briefly for the first and 2nd one.

We can write  $\inline \small \frac{dV}{dt}=k(2 \pi r h+\pi r^2 h)$, where $\inline \small k$ is the proportionality constant. And $\inline \small \frac{dV}{dh}=\pi r^2$ .

So we get the differential equation as

$\inline \small \frac{dh}{dt}=\frac{2k}{r}h+k$ .

After solving this we get  $\inline \small h(t)$ as,

$\inline \small h(t)=\frac{r}{2k}(C e^{\frac{2k}{r}t}-k)$, where r is the radius of the cylinder and C is the unknown arbitrary number, yet to find out.

Now we have to unknown in the equation k and C, to find these we need to two equation. This conditions are given the 3rd problem. Initially the filter was full i.e ate t=0, h=12 cm and when t=2.2 min, h=0 as filtration process was over.

Now inputting the 1st condition in the previous equation we get

$\inline \small \frac{C}{k}=\frac{29}{5}$,

and putting the 2nd condition we get,

$\inline \small C (\frac{29}{5})^{\frac{2k}{5}}=k$, rearranging this equation and putting the value of $\inline \small \frac{C}{k}$ we get $\inline \small k=-\frac{5}{2}$ cm/min.

So $\inline \small C=-\frac{29}{2}$ cm/min.

Getting k a -ve number was quite normal because $\inline \small \frac{dV}{dt}$ becomes a -ve number which means as time increase volume will decrease.

For thew fourth part, as rate of volume is constant the filter tube will maintain a height of $\inline \small h=\frac{6}{\pi r^2}=\frac{6}{25 \pi} cm$ .