Related Rates: Coffee flowing from a conical filter to a right cylinder coffeepot

The first problem:

**Making coffee. **Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of $\displaystyle 10\frac{in^3}{min}$.

**a. **How fast is the level in the pot rising when the coffee in the cone is 5 in. deep?

**b. **How fast is the level in the cone falling then?

Some facts given from the figure:

(1) The conical filter has a diameter of 6 inches at the top, and a depth of 6 inches also. It appears to be in the shape of a right circular cone.

(2) The coffeepot has a diameter 6 inches and in the shape of a right cylinder.

**Attempt:**

To address question (a), the level of in the pot rising would refer to $\displaystyle \frac{dh}{dt}$. I began with the equation that relates the variables V, h, and r and treat them as differentiable functions of t with the exception of r; the radius is constant for a right cylinder.

Therefore,

$\displaystyle V={\pi}r^2h$, the volume of a right cylinder. (1)

$\displaystyle \frac{dV}{dt}={\pi}r^2\frac{dh}{dt}$ (2)

$\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{{\pi}r^2}$, (3)

Now, I make an assertion here by recognizing that volume is conserved between the conical filter and the coffeepot; for, any volume loss from the filter should be transferred to the coffeepot and no where else. This is neglecting any effects of absorption of the coffee onto the filter paper, which I believe was a reasonable assumption in this problem. However, the problem poses a condition "when the coffee in the cone is 5" deep."

From here, I think that $\displaystyle \frac{dV}{dt}$ is dependent on the depth of the coffee in the cone which is sensible considering Torricelli's law, but I'm not sure where to go from here. I arrived at the equation,

$\displaystyle \frac{dV}{dt}=\frac{\pi}{3}[r^2\frac{dh}{dt}+2rh\frac{dr}{dt}] $ which I think I must solve for $\displaystyle \frac{dV}{dt}$ and substitute the value into equation (3). I must have gone awry somewhere.

Help?

Re: Related Rates: Coffee flowing from a conical filter to a right cylinder coffeepot

Quote:

Originally Posted by

**Lambin** The first problem:

**Making coffee. **Coffee is draining from a conical filter into a cylindrical coffeepot at the rate of $\displaystyle 10\frac{in^3}{min}$.

**a. **How fast is the level in the pot rising when the coffee in the cone is 5 in. deep?

**b. **How fast is the level in the cone falling then?

Some facts given from the figure:

(1) The conical filter has a diameter of 6 inches at the top, and a depth of 6 inches also. It appears to be in the shape of a right circular cone.

(2) The coffeepot has a diameter 6 inches and in the shape of a right cylinder.

**Attempt:**

To address question (a), the level of in the pot rising would refer to $\displaystyle \frac{dh}{dt}$. I began with the equation that relates the variables V, h, and r and treat them as differentiable functions of t with the exception of r; the radius is constant for a right cylinder.

Therefore,

$\displaystyle V={\pi}r^2h$, the volume of a right cylinder. (1)

$\displaystyle \frac{dV}{dt}={\pi}r^2\frac{dh}{dt}$ (2)

$\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{{\pi}r^2}$, (3)

Now, I make an assertion here by recognizing that volume is conserved between the conical filter and the coffeepot; for, any volume loss from the filter should be transferred to the coffeepot and no where else. This is neglecting any effects of absorption of the coffee onto the filter paper, which I believe was a reasonable assumption in this problem. However, the problem poses a condition "when the coffee in the cone is 5" deep."

From here, I think that $\displaystyle \frac{dV}{dt}$ is dependent on the depth of the coffee in the cone which is sensible considering Torricelli's law, but I'm not sure where to go from here. I arrived at the equation,

$\displaystyle \frac{dV}{dt}=\frac{\pi}{3}[r^2\frac{dh}{dt}+2rh\frac{dr}{dt}] $ which I think I must solve for $\displaystyle \frac{dV}{dt}$ and substitute the value into equation (3). I must have gone awry somewhere.

Help?

You don't need "Torricelli's law", just the fact that, because of "conservation of mass" and the fact that coffee has constant density, we have "conservation of volume". You have correctly stated that the volume of coffee in cylindrical coffee pot is given by [tex]V_p= \pi r^2 h[/itex] and r is constant so $\displaystyle dV_p/dt= \pi r^2 dh/dt$. You also need to take into account that the volume of water in the filter is given by $\displaystyle V_f= \pi r^2h/3$. Here, h and r are always the same so we can write that as $\displaystyle V_f= \pi r^3/3[/itex] and dV_f= \pi r^2 dr/dt$ . I see a "1/3" in one of your formulas so perhaps you are including that. The point is that the coffee flowing from the cone is the coffee flowing to the pot: $\displaystyle dV_p= -dV_f$. The "-" is there because $\displaystyle dV_p$ is positive (coffee is going into it) while $\displaystyle dV_f$ is negative (coffee is going out of it).

Re: Related Rates: Coffee flowing from a conical filter to a right cylinder coffeepot

Quote:

a. How fast is the level in the pot rising when the coffee in the cone is 5 in. deep?

Part a, stated another way, asks for the rate at which the height of the water in the cylinder is increasing when the height of the water in the conical filter is 5 inches.

Given information: The volume of water in the cylinder is increasing at a rate of $\displaystyle 10\tfrac{\text{in}^3}{\text{min}}$

The volume of a (right) cylinder, as you correctly stated, is given by $\displaystyle V=\pi r^2 h$.

The diameter, and thus the radius, of the cylinder are constant. So, $\displaystyle r=3$.

Thus, the volume of water in the cylinder is given by $\displaystyle V=9\pi h$, whereby $\displaystyle h$ is the height of the water in the cylinder.

The rate at which the volume of water increases in the cylinder is then $\displaystyle \tfrac{d}{dt}V = \tfrac{d}{dt}9\pi h \implies \tfrac{dV}{dt}=9\pi \frac{dh}{dt}$

Recall that the volume of water in the cylinder is increasing at a rate of $\displaystyle 10\tfrac{\text{in}^3}{\text{min}}$

Therefore, $\displaystyle 9\pi \frac{dh}{dt} = 10 \implies \tfrac{dh}{dt}=\tfrac{10}{9\pi} \approx 0.354\tfrac{\text{in}}{\text{min}}$