Problem 69
Question
Suppose that water is stored in a cylindrical tank of radius 5 \(\mathrm{m}\). If the height of the water in the tank is \(h\), then the volume of the water is \(V=\pi r^{2} h=\left(25 \mathrm{~m}^{2}\right) \pi h=25 \pi h \mathrm{~m}^{2} .\) If we drain the water at a rate of 250 liters per minute, what is the rate at which the water level inside the tank drops? (Note that 1 cubic meter contains 1000 liters.)
Step-by-Step Solution
Verified Answer
The rate at which the water level drops is \( \frac{-1}{100 \pi} \) meters per minute.
1Step 1: Understanding the Problem
We need to determine the rate at which the height (or water level) of the water in the cylindrical tank decreases when water is drained at a rate of 250 liters per minute.
2Step 2: Convert Drainage Rate to Cubic Meters
Since water is drained at a rate of 250 liters per minute and 1 cubic meter contains 1000 liters, we first convert 250 liters to cubic meters. Thus, \( 250 \text{ liters} = \frac{250}{1000} = 0.25 \text{ cubic meters} \).
3Step 3: Relate Volume Change to Height Change
We know from the problem statement that the volume of water in the cylindrical tank is given by \( V = 25 \pi h \). The volume is decreasing at a rate of 0.25 cubic meters per minute, so we know \( \frac{dV}{dt} = -0.25 \).
4Step 4: Express \(\frac{dh}{dt}\) Using Derivatives
We differentiate both sides of the volume equation \( V = 25 \pi h \) with respect to time \( t \). This gives us \( \frac{dV}{dt} = 25 \pi \frac{dh}{dt} \).
5Step 5: Calculate \(\frac{dh}{dt}\)
Since \( \frac{dV}{dt} = -0.25 \), we substitute this into the equation: \( -0.25 = 25 \pi \frac{dh}{dt} \). Solving for \( \frac{dh}{dt} \), we find \( \frac{dh}{dt} = \frac{-0.25}{25 \pi} \). This simplifies to \( \frac{dh}{dt} = \frac{-1}{100 \pi} \) meters per minute.
Key Concepts
Rate of changeVolume of a cylinderRelated rates
Rate of change
When we talk about the rate of change in mathematics, we refer to how one quantity changes in relation to another over time. In the context of calculus, it is typically described using derivatives. Imagine you are riding a bicycle and want to know how fast you are going at any given moment. That immediate speed is the rate of change of your position over time.
In the original exercise about draining a tank, the rate of change is used to determine how quickly the water level (height) drops as water exits the tank. We expressed this as \( \frac{dh}{dt} \), which is the rate of change of height over time. This is essential because it measures the speed of the height decrease as water leaves the tank at a constant rate of 250 liters per minute.
By understanding how to calculate rates of change, we can apply this knowledge to various problems in physics, engineering, economics, and any other field where understanding how quantities change over time is crucial.
In the original exercise about draining a tank, the rate of change is used to determine how quickly the water level (height) drops as water exits the tank. We expressed this as \( \frac{dh}{dt} \), which is the rate of change of height over time. This is essential because it measures the speed of the height decrease as water leaves the tank at a constant rate of 250 liters per minute.
By understanding how to calculate rates of change, we can apply this knowledge to various problems in physics, engineering, economics, and any other field where understanding how quantities change over time is crucial.
Volume of a cylinder
The volume of a cylinder is a measure of how much space is inside the cylinder. It's important in many real-world scenarios, such as determining how much water a tank can hold, or how much food a canister can store. To calculate the cylinder's volume, we use the formula \( V = \pi r^2 h \). Here, \( r \) is the radius of the cylinder's base, \( h \) is the height, and \( \pi \) is approximately 3.14159.
In our specific problem, the volume formula becomes \( V = 25\pi h \), since the radius is given as 5 meters. The part \( 25\pi \) represents the area of the circular base. Multiplying this area by the height \( h \) gives the total volume of the cylinder. When the height changes, the total volume changes accordingly.
Understanding the volume formula is crucial when dealing with problems that involve storage, transportation, and any scenario where capacity plays a role. It's also a foundational concept that forms the basis for more advanced studies in calculus and physics.
In our specific problem, the volume formula becomes \( V = 25\pi h \), since the radius is given as 5 meters. The part \( 25\pi \) represents the area of the circular base. Multiplying this area by the height \( h \) gives the total volume of the cylinder. When the height changes, the total volume changes accordingly.
Understanding the volume formula is crucial when dealing with problems that involve storage, transportation, and any scenario where capacity plays a role. It's also a foundational concept that forms the basis for more advanced studies in calculus and physics.
Related rates
Related rates is a method used in calculus to determine the relationship between the rates of change of different variables. It often involves taking derivatives with respect to time. This method is used when two or more quantities are continuously changing at the same rate, and we want to understand how they affect each other.
In the exercise, related rates help us find the connection between the rate at which the volume of water decreases and the rate at which the water level (height) drops in the tank. The derivative \( \frac{dV}{dt} \) represents the rate of change of volume with time, and \( \frac{dh}{dt} \) is the rate of change of height. By expressing \( \frac{dV}{dt} = 25\pi \frac{dh}{dt} \) after differentiating the volume formula, we establish a link between how fast the volume is decreasing and how fast the height is decreasing.
Using related rates is powerful for solving practical problems where multiple changing aspects need to be considered simultaneously. Whether it’s mixing fluids, inflating balloons, or moving objects, this concept is essential in comprehending how one change influences another in real-time scenarios.
In the exercise, related rates help us find the connection between the rate at which the volume of water decreases and the rate at which the water level (height) drops in the tank. The derivative \( \frac{dV}{dt} \) represents the rate of change of volume with time, and \( \frac{dh}{dt} \) is the rate of change of height. By expressing \( \frac{dV}{dt} = 25\pi \frac{dh}{dt} \) after differentiating the volume formula, we establish a link between how fast the volume is decreasing and how fast the height is decreasing.
Using related rates is powerful for solving practical problems where multiple changing aspects need to be considered simultaneously. Whether it’s mixing fluids, inflating balloons, or moving objects, this concept is essential in comprehending how one change influences another in real-time scenarios.
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