Problem 55 - Biocalculus Calculus for the Life Sciences

Question

Water flows from the bottom of a storage tank at a rate of \(r(t)=200-4 t\) liters per minute, where 0\(\leqslant t \leqslant 50\) . Find the amount of water that flows from the tank during the first 10 minutes.

Step-by-Step Solution

Answer

1800 liters flow from the tank during the first 10 minutes.

Step 1: Understanding the Problem

We are given that water flows from a tank at a rate of \(r(t) = 200 - 4t\) liters per minute. We need to find out how much water flows out of the tank during the first 10 minutes.

Step 2: Integrating the Flow Rate

To find the total amount of water that flows out, we need to integrate the rate of flow \(r(t)\) from \(t = 0\) to \(t = 10\). This will give us the total volume of water that has been released in that time period.

Step 3: Setting Up the Integral

The integral to find the total volume \(V\) of water released is: \[V = \int_{0}^{10} (200 - 4t) \, dt.\]

Step 4: Calculating the Antiderivative

The antiderivative of \((200 - 4t)\) is \(200t - 2t^2\). We use these antiderivatives to evaluate the definite integral from \(t = 0\) to \(t = 10\).

Step 5: Evaluating the Definite Integral

Substitute 10 and 0 into the antiderivative and find the difference:\[V = \left[200(10) - 2(10)^2\right] - \left[200(0) - 2(0)^2\right]\]This simplifies to:\[V = [2000 - 200] - [0] \]so \(V = 1800\) liters.

Key Concepts

Integration

In calculus, integration is a powerful mathematical tool. It is the process of finding the integral, which can be thought of as the reverse operation of differentiation. When we talk about integration, we often refer to finding the area under a curve or the total accumulation of a certain quantity.
In our problem, integration helps determine the total volume of water that flows out of the tank over a certain time interval.
By integrating the given flow rate function, which is a mathematical representation of how the rate changes over time, we can find the total amount of water that has left the tank.

Flow Rate Problems

Flow rate problems often involve real-world scenarios where something is happening at a certain rate over time, like water flowing from a tank at a speed defined by a specific function.
In our example, the flow rate, expressed as a function of time, is given by \( r(t) = 200 - 4t \). It provides the rate at which water leaves the tank in liters per minute, depending on the elapsed time \(t\).
Solving flow rate problems generally requires understanding the relationship between a changing rate and the quantity it affects. Integrating this rate function over a specified interval allows us to determine the total amount of substance, like water, that has moved during that period.

Definite Integral

The definite integral is a core concept in calculus used to calculate the accumulation of quantities, like area or total volume, over an interval. It connects the dots between the rate of change and total change.
In the context of our problem, we calculate the definite integral of the flow rate function \( r(t) = 200 - 4t \) from \( t = 0 \) to \( t = 10 \). This calculation gives us the total amount of water that flows out of the tank during the first ten minutes. The definite integral is represented as \[ \int_{0}^{10} (200 - 4t) \, dt \].
By solving this, we realize its significance in determining total change using the limits of integration, providing a precise value for accumulated flow.

Antiderivative

An antiderivative is essentially the opposite of a derivative. It's a function whose derivative is the given function.
In solving integration problems, finding the antiderivative is a necessary step because it helps us evaluate definite integrals. It's like reverse engineering the problem. For our task, the antiderivative of the function \( 200 - 4t \) is \( 200t - 2t^2 \).
This function allows us to calculate the definite integral by plugging in the upper and lower bounds of our interval, \( t = 0 \) and \( t = 10 \), then finding the difference to get the total amount of water released during this period. Antiderivatives simplify calculating changes over an interval.