Problem 77 - Calculus for Biology and Medicine

Question

Suppose that the amount of water contained in a plant at time \(t\) is denoted by \(V(t) .\) Due to evaporation, \(V(t)\) changes over time. Suppose that the change in volume at time \(t\), measured over a 24-hour period, is proportional to \(t(24-t)\), measured in grams per hour. To offset the water loss, you water the plant at a constant rate of 4 grams of water per hour. (a) Explain why $$ \frac{d V}{d t}=-a t(24-t)+4 $$ \(0 \leq t \leq 24\), for some positive constant \(a\), describes this situation. (b) Determine the constant \(a\) for which the net water loss over a 24 -hour period is equal to 0 .

Step-by-Step Solution

Answer

The constant \(a\) is \( \frac{1}{24} \).

Step 1: Understanding the Problem

We need to identify how the volume of water, \(V(t)\), changes. The change due to evaporation is proportional to \(t(24-t)\). Additionally, water is added at a constant rate of 4 grams per hour.

Step 2: Representing Change in Water Volume

The differential equation for the change in water volume is given by \( \frac{dV}{dt} = -a t(24-t) + 4 \). The term \(-a t(24-t)\) represents the evaporation rate depending on time \(t\), and \(4\) is the constant rate at which water is added.

Step 3: Setting Up the Net Water Loss Equation

To find the constant \(a\), we know the net water loss over 24 hours should be 0. Thus, we integrate the rate of change \( \frac{dV}{dt}\) over a period from \(t=0\) to \(t=24\), and set the integral equal to zero:

Step 4: Integrating the Rate of Change

\[\int_0^{24} \left( -a t(24-t) + 4 \right) \, dt = 0\]This equation should be solved for \(a\) such that the integral evaluates to 0.

Step 5: Evaluating the Integral

First, break down the integral:\[\int_0^{24} -a t(24-t) \, dt + \int_0^{24} 4 \, dt = 0\]Calculate each part separately. For the first integral:\[\int_0^{24} -a t(24-t) \, dt = -a \int_0^{24} (24t - t^2) \, dt = -a \left[ 12t^2 - \frac{1}{3}t^3 \right]_0^{24}\]Evaluate this at the bounds.

Step 6: Calculate Each Part of the Evaluation

Calculate:\( 12 \times 24^2 - \frac{1}{3} \times 24^3 = 12 \times 576 - \frac{1}{3} \times 13824 = 6912 - 4608 = 2304 \).Therefore, the integral evaluates to \(-a (2304)\).

Step 7: Evaluate the Constant Addition Term

The integral of the constant term is:\[\int_0^{24} 4 \, dt = 4t \big|_0^{24} = 4 \times 24 = 96\]

Step 8: Solve the Zero Net Change Equation

Setting the sum of the two parts equal to zero, we have:\[- a (2304) + 96 = 0\]Solving for \(a\), \(a = \frac{96}{2304}\). Calculate this value.

Step 9: Calculate the Constant a

Perform the division:\[a = \frac{96}{2304} = \frac{1}{24}\].Thus, the constant \(a\) is \( \frac{1}{24} \).

Key Concepts

Evaporation

Evaporation refers to the process where liquid water changes into a gas, resulting in water vapor. In the context of this exercise, evaporation is the primary reason for the changing volume of water in the plant over time.
When thinking about how evaporation affects the plant, it's important to recognize a few key points:

• Evaporation is influenced by various factors such as temperature, humidity, and surface area.
• The rate of evaporation in this scenario is depicted as proportional to the function \( t(24-t) \), which suggests some dependency on time.
• This equation implies evaporation is zero at \( t = 0 \) and \( t = 24 \) and maximizes when \( t = 12 \).

Understanding the evaporation term in the differential equation allows for better grasp of how water disappears from the plant system.

Net Water Loss

Net water loss in this context is the total volume difference resulting from evaporation and the addition of water over the 24-hour period. The problem specifically states the condition for net water loss:
Even though the plant experience evaporation, which decreases water volume, constant watering adds 4 grams per hour. For the whole day, this means:

• The goal is to achieve zero net loss over 24 hours.
• This balance leads to the necessity of solving the integral of the rate of change, where all water coming in and out equals to zero.
• It encourages setting up conditions where the evaporative loss is perfectly counteracted by the watering schedule.

Balancing out these contributions lets us determine the constant 'a' that makes sure the plant doesn’t lose water over the day.

Integration

Integration is the process used to find the total change in the plant's water volume over time, and is necessary to solve for the constant 'a.' Let's break this down:

• Integration helps in calculating the accumulation effect over time of both evaporation and water addition.
• The integral of the rate of change \( \int_0^{24} \left( -a t(24-t) + 4 \right) \, dt = 0 \) represents this accumulation.

By performing this integration, the sum of evaporative loss and water addition is translated into a mathematical form, which is critical in solving for an unknown factor, like the constant 'a'. The integration process guarantees that you consider every moment in time across the given period.

Proportionality

Proportionality describes a direct relationship between two variables, such as time and rate of evaporation here. In this exercise:

• The term \( -a t(24-t) \) represents the evaporation rate being proportional to the time-related factor \( t(24-t) \).
• This means as time progresses within the 0 to 24 hour window, evaporation changes in proportion to those time values.
• The proportional relationship is why determining 'a' is crucial, as it scales this time-specific effect directly.

By understanding proportionality, you'll easily see how and why changes in time directly alter the rate of water lost to evaporation, and conversely, how these relationships help us balance overall water loss.