Related Rates Coffee Filter Problem

In summary, the depth of coffee in the filter is changing at a rate of 8/(169*Pi) cm/s at a certain time when the coffee is 13 cm deep. For part b, it is not clear what information is given about the coffee pot, so it is difficult to provide an answer.
  • #1
manik
7
0
Coffee drains from a conical filter, with diameter of 15cm and a height of 15 cm into a cylindrical coffee pot, with a diameter of 15 cm at 2 cm^2/ sec. At a certain time, the coffee in the filter is 13 cm deep.

a) How fast is the depth of coffee in the filter changing at that time?
b) How fast is the depth of coffee in the pot changing at that time?

Apparently this is an assignment question. I think that 2cm^2 / sec is a typo and should be replaced by 2cm^3 / sec.

http://img242.imageshack.us/img242/3917/attemptau7.jpg

<< that is my attempt for problem a), is this right? Can someone make any suggestions.. thanks in advance
 
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  • #2
I'm getting 8/(169 * Pi) cm/s as my answer. I think one place where you did something wrong was you did not square h/2 in place of r in the original volume equation. It should have looked something like this.

v = 1/3 * PI * r^2 * h --> r = h/2 --> v = 1/3*Pi*(h/2)^2*h which gives

v = 1/12*Pi*h^3 By differentiation we obtain:

dv = 1/4*Pi*h^2*dh solving for dh we get (4*dv)/(Pi*h^2) or 8/(169*Pi) cm/s.
 
  • #3
thanks i got a, but can someone tell me for b i don't get it
 

1. What is the "Related Rates Coffee Filter Problem"?

The "Related Rates Coffee Filter Problem" is a mathematical problem that involves finding the rate of change of one variable with respect to another variable, using the concept of related rates. In this specific problem, the changing variables are the height of a coffee filter and the volume of coffee in the filter, and the goal is to find the rate at which the height is changing when the volume is changing at a given rate.

2. How is this problem related to real-life situations?

This problem is related to real-life situations because it models the process of making coffee using a coffee filter. The height of the filter and the volume of coffee in it are changing variables, and the goal is to find the rate at which the height changes as the volume changes. This can be applied to real-life situations, such as determining the rate at which the coffee is being brewed or the rate at which the coffee is being consumed.

3. What are the important equations to solve this problem?

The important equations to solve this problem are the formula for the volume of a cone (V=πr²h/3), the formula for the rate of change of volume with respect to time (dV/dt), and the formula for the rate of change of height with respect to time (dh/dt).

4. What are the steps to solve the "Related Rates Coffee Filter Problem"?

The steps to solve the "Related Rates Coffee Filter Problem" are as follows:
1. Draw a diagram and label all the given information
2. Write down the relevant equations
3. Take the derivative of both sides of the volume equation with respect to time
4. Substitute the given values and solve for the rate of change of volume (dV/dt)
5. Use the chain rule to find the derivative of the height equation with respect to time
6. Substitute the given values and the value of dV/dt to solve for the rate of change of height (dh/dt)

5. What are the limitations of this problem?

The limitations of this problem include assuming that the coffee filter is a perfect cone with a constant radius, and that the volume of coffee is changing at a constant rate. In real-life situations, these assumptions may not be true and can affect the accuracy of the solution. Additionally, this problem only considers the rate of change of height and volume, and does not take into account other factors that may affect the brewing process, such as temperature and pressure.

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