# Related Rates - cone draining into cylinder

• aFk-Al
In summary, the volume of water in the conical tank can be expressed as V = (pi/9)h^3. The rate of change of the volume of water in the conical tank when h = 3 is (pi/3) ft^3/min. The rate of change of the depth of water in the cylindrical tank when h = 3 is (pi/12) ft/min.

## Homework Statement

Water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that has a base with area $$400 \pi$$ square feet. The depth, h, in feet, of the water in the conical tank is changing at the rate of (h-12) feet per minute.

A) Write an expression for the volume of the water in the conincal tank as a function of h.

B) At what rate is the volume of the water in the conical tank changing when h=3?

C) Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h = 3?

## Homework Equations

$$V = \frac{1}{3} \pi r^2 h$$

## The Attempt at a Solution

First I took the derivative of the volume function using the product rule:
$$\frac{dv}{dt} = \frac{2}{3} \ast \pi \ast r \ast \frac{dr}{dt} \ast h + \frac{1}{3} \ast \pi \ast r^2 \ast \frac{dh}{dt}$$
I know that dh/dt = (h-12), but I'm not sure how to find dr/dt and I'm also not sure if I use 4 as the radius (b/c the radius of the water in the cone is constantly changing.) I'm guessing there is another relevant equation I'm missing, or something obvious. I know that the ratio of height of the cone to radius is constant, but I'm not sure if that helps or not.

If anyone could help me figure these out, I'm pretty sure I can do B) and C) on my own. Thanks in advance!

Last edited:
aFk-Al said:

## Homework Statement

Water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that has a base with area $$400 \pi$$ square feet. The depth, h, in feet, of the water in the conical tank is changing at the rate of (h-12) feet per minute.

A) Write an expression for the volume of the water in the conincal tank as a function of h.

B) At what rate is the volume of the water in the conical tank changing when h=3?

C) Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h = 3?

## Homework Equations

$$V = \frac{1}{3} \pi r^2 h$$

## The Attempt at a Solution

First I took the derivative of the volume function using the product rule:
$$\frac{dv}{dt} = \frac{2}{3} \ast \pi \ast r \ast \frac{dr}{dt} \ast h + \frac{1}{3} \ast \pi \ast r^2 \ast \frac{dh}{dt}$$
I know that dh/dt = (h-12), but I'm not sure how to find dr/dt and I'm also not sure if I use 4 as the radius (b/c the radius of the water in the cone is constantly changing.) I'm guessing there is another relevant equation I'm missing, or something obvious. I know that the ratio of height of the cone to radius is constant, but I'm not sure if that helps or not.

If anyone could help me figure these out, I'm pretty sure I can do B) and C) on my own. Thanks in advance!

That's exactly why that asked you first, in (a), to write the volume of the water in the cone as a function of h only! Do that first. That has nothing to do with differentiating! If you are not sure how to do that, draw a side view of the conical tank. The cone formed by the entire tank has height twelve and radius 8 while the cone formed by the water has height h and radius r. Of course the triangles formed are similar triangles so the sides are in the same proportion. Once you have the volume of water in the conical tank as a function of h only, it should be easy to differentiate.

Answer for A) V = (pi/9)h^3

Would you mind solving c? and as for a, I actually got v=(1/27)∏h^3

Work process:

r=4 and h=12 so r/h=1/3 ; r= (1/3)h

v=(1/3)∏((1/3h)^2)h
v=(1/3)∏(1/9h^2)h
v=((1/3)*(1/9))∏h^3
v=(1/27)∏h^3

## 1. How does the rate at which a cone drains into a cylinder affect the levels of the cone and cylinder?

The rate at which a cone drains into a cylinder directly affects the levels of the cone and cylinder. As the cone drains into the cylinder, the level of water in the cone decreases while the level of water in the cylinder increases. This is because the volume of water in the cone is being transferred to the cylinder at a constant rate.

## 2. What is the relationship between the rates of change of the cone and cylinder in this scenario?

In this scenario, the rate of change of the cone and cylinder are directly related. This means that if the rate of change of the cone increases, the rate of change of the cylinder will also increase. Similarly, if the rate of change of the cone decreases, the rate of change of the cylinder will also decrease.

## 3. How can I calculate the related rates of change for the cone and cylinder?

The related rates of change for the cone and cylinder can be calculated using the formula: dV/dt = πr²dh/dt, where dV/dt is the rate of change of volume, r is the radius of the cone or cylinder, and dh/dt is the rate of change of height. This formula relates the rates of change for the cone and cylinder through their shared volume.

## 4. Does the shape of the cone or cylinder affect the related rates in this scenario?

Yes, the shape of the cone and cylinder does affect the related rates in this scenario. For example, if the cone has a wider base or the cylinder has a larger diameter, the rates of change will be greater since there is more volume to be transferred between the two shapes.

## 5. How do I know if the related rates in this scenario are increasing or decreasing?

The related rates in this scenario can be increasing or decreasing depending on the specific values for the cone and cylinder. Generally, if the rate of change of the cone is positive (increasing), and the rate of change of the cylinder is negative (decreasing), this would indicate that the water is draining from the cone into the cylinder. However, if both rates of change are positive or both are negative, this could indicate that the water levels in both shapes are either increasing or decreasing simultaneously. This can be determined by analyzing the specific values for the cone and cylinder in the given scenario.