# How Fast Does the Water Level Rise in a Conical Tank?

• MHB
• tmt1
In summary, the rate at which the water level is rising when the water is 3 m deep is approximately 0.239 m/min, as found by setting up an equation using the given information and solving for dh/dt. This is due to a small mistake in the volume formula used in the original calculation.
tmt1
A water tank has the shape of an inverted circular cone with base radius 2 m
and height 4 m. If water is being pumped into the tank at a rate of 2 m^3/ min, find the rate at which the water level is rising when the water is 3 m deep.

The answer to this question is $$\frac{8}{9\pi}$$But I got a different answer, here is what I did:

We have

$$\d{V}{t} = 2 m^3 /min$$

and we need

$$\d{h}{t} = ?$$

The volume of a cone is

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

Since $$2/4 = r/h$$, $$r= h/2$$ and I can change the volume formula to:

$$V = \frac{1}{12}\pi h^3$$

we derive this:

$$\d{V}{t} = \frac{\pi}{4}h^2 \d{h}{t}$$

$$2 m^3 /min= \frac{\pi}{4}h^2 \d{h}{t}$$

We isolate dh/dt:

$$\frac{8}{\pi 16} = \d{h}{t}$$

h is equal to 4:

$$\frac{8}{\pi h^2} = \d{h}{t}$$

$$\d{h}{t} = \frac{1}{\pi 2}$$

I've done this question multiple times and I keep getting the same answer which is wrong.

EDIT: I solved it, h is equal to 3 not 4.

Last edited:
So the answer is $$\frac{3}{\pi}$$

Hi there! Thank you for sharing your approach to solving this problem. It looks like you made a small mistake in your calculations. The volume formula for a cone is actually V = (1/3)πr^2h, not V = (1/12)πh^3. This is why your answer is different from the correct one.

To solve the problem, we can use the given information to set up an equation:

V = (1/3)π(2)^2h

We know that dV/dt = 2 m^3/min, so we can substitute this in and solve for dh/dt:

2 = (1/3)π(2)^2(dh/dt)

dh/dt = (3/4π) m/min

Substituting in h = 3 (since the water level is 3 m deep), we get:

dh/dt = (3/4π) m/min = 0.239 m/min

Therefore, the rate at which the water level is rising when the water is 3 m deep is approximately 0.239 m/min. I hope this helps clarify the solution. Keep up the good work!

## 1. What is a related rates problem?

A related rates problem is a type of mathematical problem that involves finding the rate of change of one variable with respect to another variable. It usually involves a scenario where two or more variables are changing simultaneously and are related by a mathematical equation.

## 2. How do you solve a related rates problem?

To solve a related rates problem, you need to follow these steps:

• Identify the variables and their rates of change in the problem.
• Write an equation that relates the variables.
• Take the derivative of the equation with respect to time.
• Substitute in the given values and solve for the unknown rate of change.

## 3. What is a cone question?

A cone question is a type of related rates problem that involves a cone-shaped object. It usually asks for the rate of change of either the height or the radius of the cone, while the other variable is changing.

## 4. What is the formula for the volume of a cone?

The formula for the volume of a cone is V = 1/3πr^2h, where r is the radius and h is the height of the cone.

## 5. Can you give an example of a cone question and how to solve it?

Example: A cone-shaped water tank has a height of 10 feet and a radius of 4 feet. The water level is decreasing at a rate of 2 feet per minute. How fast is the volume of water in the tank decreasing when the water level is 6 feet?

Solution: We know that the volume of a cone is V = 1/3πr^2h and the water level is changing at a rate of -2 feet per minute (since it is decreasing). We want to find the rate of change of the volume (dV/dt) when h = 6 feet.

Substituting the given values, we have: V = 1/3π(4)^2(6) = 32π ft^3. Taking the derivative with respect to time, we get: dV/dt = π(4)(2)(-2) = -16π ft^3/min.

Therefore, the volume of water in the tank is decreasing at a rate of 16π ft^3/min when the water level is 6 feet.

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