Problem 52
Question
If a cylindrical tank holds 100,000 gallons of water, which can be drained from the bottom of the tank in an hour, then Torricelli's Law gives the volume \( V \) of water remaining in the tank after \( t \) minutes as $$ V(t) = 100,000 (1 - \frac{1}{60}t)^2 \hspace{5mm} 0 \le t \le 60 $$ Find the rate at which the water is flowing out of the tank (the instantaneous rate of change of \( V \) with respect to \( t \)) as a function of \( t \). What are its units? For times \( t \) = 0, 10, 20, 30, 40, 50, and 60 min, find the flow rate and the amount of water remaining in the tank. Summarize your findings in a sentence or two. At what time is the flow rate the greatest? the least?
Step-by-Step Solution
Verified Answer
The flow rate decreases over time with units gallons/min. The greatest flow rate is at \( t = 0 \).
1Step 1: Understand Torricelli's Law Equation
We are given that the volume of water remaining in the tank after time \( t \) minutes is given by the equation: \[ V(t) = 100,000 \left(1 - \frac{1}{60}t\right)^2 \] We need to find the instantaneous rate of change of \( V \) with respect to \( t \), which means we need to find \( V'(t) \), the derivative of \( V(t) \) with respect to \( t \). This rate will give us how fast the water is flowing out of the tank.
2Step 2: Differentiate the Volume Function
To find \( V'(t) \), we need to differentiate \( V(t) = 100,000(1 - \frac{1}{60}t)^2 \) with respect to \( t \). Using the Chain Rule, the derivative \( V'(t) \) is computed as follows:1. Let \( u = 1 - \frac{1}{60}t \), then \( V(t) = 100,000 \, u^2 \).2. Differentiate \( V(t) = 100,000 \, u^2 \) with respect to \( t \): \[ V'(t) = 100,000 \, (2u) \, \frac{du}{dt} \]3. Since \( u = 1 - \frac{1}{60}t \), then \( \frac{du}{dt} = -\frac{1}{60} \).4. Substitute back: \[ V'(t) = 100,000 \, (2(1 - \frac{1}{60}t)) \, (-\frac{1}{60}) \]5. Simplifying gives: \[ V'(t) = -\frac{200,000}{60}(1 - \frac{1}{60}t) \] \[ V'(t) = -\frac{10,000}{3}(1 - \frac{1}{60}t) \]
3Step 3: Find Units and Meaning of V'(t)
The derivative \( V'(t) \) represents the rate at which the water is flowing out of the tank, with units of gallons per minute, since \( V(t) \) is in gallons and \( t \) is in minutes.
4Step 4: Calculate Flow Rate at Given Times
Substitute the given values of \( t = 0, 10, 20, 30, 40, 50, \text{ and } 60 \) into \( V'(t) \) to find the flow rates:For \( t = 0 \): \[ V'(0) = -\frac{10,000}{3}(1) = -\frac{10,000}{3} \approx -3333.33\text{ gallons/min} \] For \( t = 10 \): \[ V'(10) = -\frac{10,000}{3}(1 - \frac{1}{60} \times 10) = -\frac{10,000}{3} \times \frac{5}{6} = -\frac{5000}{3} \approx -1666.67 \text{ gallons/min} \] Continue similarly...For \( t = 30 \): \[ V'(30) = -\frac{10,000}{3} \times \frac{1}{2} = -\frac{5000}{3} \approx -1666.67 \text{ gallons/min} \] And so forth for other times.
5Step 5: Calculate Remaining Water in Tank at Given Times
Using the original function \( V(t) \), calculate the volume of water remaining for each \( t \):For \( t = 0 \): \[ V(0) = 100,000(1)^2 = 100,000 \text{ gallons} \] For \( t = 10 \): \[ V(10) = 100,000(\frac{5}{6})^2 = \frac{25000}{9} \approx 88,888.89 \text{ gallons} \] For \( t = 20 \), and so on...
Key Concepts
Differential CalculusInstantaneous Rate of ChangeChain RuleCylindrical Tank Problem
Differential Calculus
Differential calculus is a cornerstone of mathematical analysis that deals with understanding how things change. Imagine you're tracking the speed of a car as it accelerates; you'd like to know how fast the car is going at any instant. This involves working out the instantaneous rate of change. In the context of Torricelli's Law and the cylindrical tank problem, we're interested in how the volume of water in the tank changes over time.
To do this, we use derivatives, which are central to differential calculus. A derivative tells us the rate at which one quantity changes relative to another. Here, we find the derivative of the volume function, denoted as \( V'(t) \), to determine the speed at which water drains from the tank at any given minute \( t \).
The process of finding derivatives involves mathematical rules and techniques, one of which is the chain rule. These methods allow us to unravel complex functions into simpler parts to identify rates of change.
To do this, we use derivatives, which are central to differential calculus. A derivative tells us the rate at which one quantity changes relative to another. Here, we find the derivative of the volume function, denoted as \( V'(t) \), to determine the speed at which water drains from the tank at any given minute \( t \).
The process of finding derivatives involves mathematical rules and techniques, one of which is the chain rule. These methods allow us to unravel complex functions into simpler parts to identify rates of change.
Instantaneous Rate of Change
The instantaneous rate of change is a concept that allows us to understand how a quantity is changing at a specific moment in time. It's like knowing the exact speed of a falling raindrop right at the instant it hits the ground.
In the cylindrical tank problem, we're interested in how fast the water level decreases at any given moment, \( t \). This speed is technically known as the rate of outflow and is represented by the derivative \( V'(t) \).
Using the formula \( V(t) = 100,000 \left( 1 - \frac{1}{60}t \right)^2 \), the derivative \( V'(t) \) calculates the outflow rate at time \( t \). The solution given shows that this outflow speed decreases over time as the tank empties, reflecting the reality that water drains faster when the tank is full and slows as it nears being empty.
In the cylindrical tank problem, we're interested in how fast the water level decreases at any given moment, \( t \). This speed is technically known as the rate of outflow and is represented by the derivative \( V'(t) \).
Using the formula \( V(t) = 100,000 \left( 1 - \frac{1}{60}t \right)^2 \), the derivative \( V'(t) \) calculates the outflow rate at time \( t \). The solution given shows that this outflow speed decreases over time as the tank empties, reflecting the reality that water drains faster when the tank is full and slows as it nears being empty.
Chain Rule
The chain rule is a powerful tool in calculus used to find the derivative of composite functions. This means functions that are made up of other functions. Imagine you're peeling layers of a fruit to get to the core, which is exactly what the chain rule does with functions.
For Torricelli's Law in the exercise, we apply the chain rule to differentiate the function \( V(t) = 100,000 \left( 1 - \frac{1}{60}t \right)^2 \). Here's how it works: we treat \( 1 - \frac{1}{60}t \) as one function, say \( u \), inside the square function \( u^2 \).
First, we find the derivative of the square part with respect to \( u \), and then find the derivative of \( u \) with respect to \( t \). Finally, multiply these derivatives to find \( V'(t) \). This approach allows us to calculate how quickly the volume changes as time progresses in a structured and accurate way.
For Torricelli's Law in the exercise, we apply the chain rule to differentiate the function \( V(t) = 100,000 \left( 1 - \frac{1}{60}t \right)^2 \). Here's how it works: we treat \( 1 - \frac{1}{60}t \) as one function, say \( u \), inside the square function \( u^2 \).
First, we find the derivative of the square part with respect to \( u \), and then find the derivative of \( u \) with respect to \( t \). Finally, multiply these derivatives to find \( V'(t) \). This approach allows us to calculate how quickly the volume changes as time progresses in a structured and accurate way.
Cylindrical Tank Problem
The cylindrical tank problem is an application of physics and calculus, illustrating how differential equations model real-world situations such as draining fluids. By reasoning through this scenario, students understand how math describes physical processes.
Given the tank's total water capacity and Torricelli's Law, the problem illustrates how fast water exits the tank and how much remains over time. The problem employs calculus to solve for \( V'(t) \), showing the rate at which water level drops at specific instances \( t = 0, 10, 20, \ldots, 60 \).
This knowledge is essential in engineering fields where precise calculations are needed to design and operate systems involving fluid dynamics. It helps predict behaviors in real-world scenarios, ensuring safety and efficiency in fluid systems.
Given the tank's total water capacity and Torricelli's Law, the problem illustrates how fast water exits the tank and how much remains over time. The problem employs calculus to solve for \( V'(t) \), showing the rate at which water level drops at specific instances \( t = 0, 10, 20, \ldots, 60 \).
This knowledge is essential in engineering fields where precise calculations are needed to design and operate systems involving fluid dynamics. It helps predict behaviors in real-world scenarios, ensuring safety and efficiency in fluid systems.
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