Problem 8 - Active Calculus

Question

Suppose that a cylindrical water tank with a hole in the bottom is filled with water. The water, of course, will leak out and the height of the water will decrease. Let \(h(t)\) denote the height of the water. A physical principle called Torricelli's Law implies that the height decreases at a rate proportional to the square root of the height. a. Express this fact using \(k\) as the constant of proportionality. b. Suppose you have two tanks, one with \(k=-1\) and another with \(k=-10 .\) What physical differences would you expect to find? c. Suppose you have a tank for which the height decreases at 20 inches per minute when the water is filled to a depth of 100 inches. Find the value of \(k\). d. Solve the initial value problem for the tank in part (c), and graph the solution you determine. e. How long does it take for the water to run out of the tank? f. Is the solution that you found valid for all time \(t ?\) If so, explain how you know this. If not, explain why not.

Step-by-Step Solution

Answer

a. \ \( \frac{dh}{dt} = k \sqrt{h} \) b. The tank with \( k = -10 \) leaks faster than \( k = -1 \). c. \ \( k = 2 \) d. \ \( h(t) = (10 - t)^2 \) e. 10 minutes. f. Valid for \( 0 \leq t \leq 10 \)

Step 1: Understanding Torricelli's Law

Torricelli's Law indicates that the rate of change in the height of water in the tank is proportional to the square root of the height. Mathematically, this is expressed as \ \( \frac{dh}{dt} = k \sqrt{h} \) where \( k \) is the constant of proportionality.

Step 2: Express Rate of Change

From part (a), express the rate of change explicitly: \ \( \frac{dh}{dt} = k \sqrt{h} \) with \( k \) as the constant of proportionality.

Step 3: Comparing Two Tanks

In part (b), compare two tanks where one has \( k = -1 \) and another has \( k = -10 \). The more negative the value of \( k \), the faster the water will leak out. Hence, the tank with \( k = -10 \) will have water draining much faster than the tank with \( k = -1 \).

Step 4: Finding the Value of k

From part (c), we know the height decreases at 20 inches per minute when the water is 100 inches deep: \ \( \frac{dh}{dt} = 20 \text{ when } h = 100 \). Substitute these values into the proportionality equation: \ \( 20 = k \sqrt{100} \implies 20 = k \times 10 \). Hence, \( k = 2 \).

Step 5: Solving Initial Value Problem

For part (d), the differential equation is: \ \( \frac{dh}{dt} = 2 \sqrt{h} \). Separating variables: \ \( \frac{dh}{\sqrt{h}} = 2 dt \). Integrate both sides: \ \( \int \frac{1}{\sqrt{h}} dh = \int 2 dt \). This yields: \ \( 2\sqrt{h} = 2t + C \).\Considering initial conditions, assume initial height \( h(0) = 100 \), so \ \( 2\sqrt{100} = 2 \times 0 + C \implies C = 20 \). Therefore, \ \( 2\sqrt{h} = 2t + 20 \) or, \ \( \sqrt{h} = t + 10 \). Thus, \ \( h(t) = (t + 10)^2 \).

Step 6: Finding Time Until Water Runs Out

For part (e), determine when the water height \( h \) reaches \( 0 \). Solve \( (t + 10)^2 = 0 \), which gives \( t = -10 \). Since negative time isn’t feasible, the solution indicates it will take 10 minutes for the water to run out.

Step 7: Solution Validity

For part (f), verify the validity: \( h(t) = (10 - t)^2 \) is valid for \( 0 \leq t \leq 10 \). Beyond 10 minutes, the solution is not applicable since negative heights are not physically possible.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In real-world applications, they help us describe various phenomena such as the rate of change of a quantity over time.

• In this exercise, the height of water in a tank over time is modeled using a differential equation.
• Torricelli's Law tells us that the rate at which the water height decreases is proportional to the square root of the height.
• The mathematical representation of this law is \(\frac{dh}{dt} = k \sqrt{h}\), where \(k\) is a constant.

Understanding how to set up and solve these equations is crucial. They allow us to predict future values and analyze behaviors dependent on rates of change.

Initial Value Problems

An initial value problem (IVP) is a differential equation accompanied by specific conditions at the start. These initial conditions help determine the particular solution to the differential equation.

• In this exercise, one such condition is given: the height of the water at time \(t=0\) is 100 inches.
• To solve this IVP, we first find the general solution of the differential equation and then use the initial condition to find the specific solution.
• For example, from the differential equation \(\frac{dh}{dt} = k \sqrt{h}\), the initial height \(h(0) = 100\) helps us solve for the constant after integrating the equation.

By integrating and applying the initial condition, we obtain \(h(t) = (t + 10)^2\), a specific solution tailored to the initial condition provided.

Rate of Change

The rate of change is a fundamental concept that describes how a quantity changes over time. It's key for understanding differential equations and their real-world applications.

• In the exercise, the rate of change of water height, \(\frac{dh}{dt}\), is directly examined.
• Torricelli's Law suggests this rate is proportional to the square root of the current height, indicating that as the height decreases, the rate of change slows down.
• For instance, when \(k = 2\) and the height \(h = 100\), the water level changes at 20 inches per minute.

Understanding the rate of change informs us how quickly the water level drops under different conditions, allowing us to predict the time it takes for the tank to empty. For example, it was deduced that the water would completely drain out in 10 minutes when starting from 100 inches height.