Problem 58 - Calculus Early Transcendentals

Question

A \(200-\mathrm{L}\) cistern is empty when water begins flowing into it (at \(t=0\) ) at a rate (in liters/minute) given by \(Q^{\prime}(t)=3 \sqrt{t}\). a. How much water flows into the cistern in 1 hour? b. Find and graph the function that gives the amount of water in the tank at any time \(t \geq 0\). c. When will the tank be full?

Step-by-Step Solution

Answer

b. Graph the function that gives the amount of water in the cistern at any time t ≥ 0. c. When will the cistern be full? #Answer# a. After 1 hour, approximately 296.647 L of water has flowed into the cistern. b. The function Q(t)=2t^(3/2) is a power function. It starts at the origin and increases as time progresses. The graph consists of key points like (0, 0), (1, 2), and (60, 296.647). Note that the time domain is limited to positive values (t ≥ 0). c. The cistern will be full after approximately 31.623 minutes.

Step 1: Find the Antiderivative

To find the antiderivative of \(Q'(t)=3\sqrt{t}\), let's rewrite the function as \(Q'(t)=3t^{\frac{1}{2}}\). Now, calculate the antiderivative: $$ Q(t) = \int 3t^{\frac{1}{2}} dt = 3\int t^{\frac{1}{2}} dt $$

Step 2: Calculate the Integral

Using the power rule for integration, \(\int x^n dx = \frac{x^{n+1}}{n+1}+C\), we get: $$ Q(t) = 3 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + C = 2t^{\frac{3}{2}} + C. $$ The initial condition is that at time \(t=0\), the cistern is empty, thus \(Q(0)=0\). $$ 0 = 2(0)^{\frac{3}{2}} + C. $$ Clearly, C=0. So the amount of water in the cistern at any time \(t\) is given by \(Q(t)=2t^{\frac{3}{2}}\).

Step 3: Calculate the Amount of Water after 1 Hour

In order to find the amount of water after 1 hour, convert 1 hour to minutes (because the rate is given in L/min), then plug in t=60 into the function: $$ Q(60) = 2(60)^{\frac{3}{2}} $$ $$ Q(60) = 2(60\sqrt{60}) \approx 296.647 L $$ After 1 hour, approximately 296.647 L of water has flowed into the cistern.

Step 4: Find the Graph of the Function

The function \(Q(t)=2t^{\frac{3}{2}}\) is a power function. It originates at the origin and increases as time progresses. To graph the function, plot key points (like \((0, 0), (1, 2),\) and \((60, 296.647)\). Note that the time domain is limited to positive values (\(t \geq 0\)).

Step 5: Find when the Tank is Full

To find when the tank is full, set Q(t) equal to the tank's capacity (200 L) and solve for t: $$ 200=2t^{\frac{3}{2}} $$ $$ t^{\frac{3}{2}}=100 $$ $$ t=\sqrt[3]{10000}=10\sqrt{10} \approx 31.623 $$ The cistern will be full after approximately 31.623 minutes.

Key Concepts

Antiderivative

In integral calculus, finding the antiderivative is a crucial step when solving problems involving accumulation, such as determining the amount of water in a cistern over time. An antiderivative, also known as an indefinite integral, is a function whose derivative gives the original function we started with. It helps us understand how quantities change and accumulate.
For example, given a rate function like \( Q'(t) = 3\sqrt{t} \) (which describes the rate of water flowing into a cistern), we seek an antiderivative \( Q(t) \) that represents the total amount of water at any time \( t \). By integrating \( Q'(t) \), we obtain \( Q(t) = \int 3t^{\frac{1}{2}} dt \), leading to the function \( Q(t) = 2t^{\frac{3}{2}} + C \). Here, \( C \) is the constant of integration, determined by initial conditions.
The antiderivative thus transforms a rate of change into a cumulative total, making it an essential tool in solving practical problems.

Power Rule

The power rule is a fundamental technique in calculus used to simplify the process of finding antiderivatives. It states that for any real number \( n eq -1 \), the integral of \( x^n \) is given by:

• \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)

This powerful rule allows us to integrate terms of the form \( t^{\frac{1}{2}} \), commonly seen in rate problems.
For the given rate function \( Q'(t) = 3t^{\frac{1}{2}} \), we apply the power rule to find the antiderivative:

• Integrate \( \int 3t^{\frac{1}{2}} dt \)
• Using the power rule, compute: \( 3 \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + C = 2t^{\frac{3}{2}} + C \)

By simplifying the expression, we find that the power rule significantly streamlines the integration process for expressions involving powers of \( t \), providing a methodical approach to handling integrals.

Function Graph

To better understand a mathematical function, graphing it reveals how it behaves over a range of values. A function graph visually depicts the relationship between the independent variable and the dependent variable. In context, the function \( Q(t) = 2t^{\frac{3}{2}} \) represents how much water has flowed into the cistern at any time \( t \), with \( Q(t) \) being dependent on \( t \).
Graphing this function involves plotting points that satisfy the equation, such as \((0, 0), (1, 2),\) and \((60, 296.647)\). The result is a curve that starts from the origin and increases, illustrating the accumulation of water over time.

• Note: The domain is restricted to \( t \geq 0 \) as time cannot be negative.
• Choose various time intervals to see the progression over the complete time range up to filling the cistern.

The graph not only helps interpret the solution but supports visual learners in grasping the concept of accumulation over time.

Initial Condition

Initial conditions in calculus provide specific values you need to solve problems involving differential equations or finding specific antiderivatives. They help uniquify the solution of an indefinite integral by determining the constant \( C \) in the general solution.

• For the function \( Q(t) = 2t^{\frac{3}{2}} + C \), knowing that the cistern was empty at \( t = 0 \) provides the initial condition \( Q(0) = 0 \).

This lets us solve for the constant \( C \):

• \( 0 = 2(0)^{\frac{3}{2}} + C \)
• So \( C = 0 \)

Applying initial conditions is key to tailoring general solutions to match specific scenarios in real-world problems, ensuring that solutions are relevant and applicable.