Problem 63 - Biocalculus Calculus for the Life Sciences

Question

Growing degree days The rate of development of many plant species depends on the temperature of the environment in such a way that maturity is reached only after a certain number of "degree-days." Suppose that temperature \(T\) as a function of time \(t\) is given by $$T(t)=15\left(1+\sin \frac{2 \pi t}{60}\right) \quad 0 \leqslant t \leqslant 60$$ where \(t\) is measured in days. If maturity is reached on day\(t=20,\) what is the number of degree-days required? \(\left[\) Hint: What are the units for \(\int_{0}^{20} T(t) d t ?\)]$

Step-by-Step Solution

Answer

The required number of degree-days is approximately 227.

Step 1: Understanding the Temperature Function

The given temperature function is \( T(t) = 15(1 + \sin \left(\frac{2\pi t}{60}\right)) \). This describes how the temperature changes over time from day 0 to day 60.

Step 2: Setting Up the Integral for Degree-Days

Degree-days is found by integrating the temperature function over time. We need to calculate this from day 0 to day 20, so we set up the integral \( \int_{0}^{20} T(t) \, dt \).

Step 3: Integrating the Temperature Function

We substitute the temperature function into the integral: \( \int_{0}^{20} 15(1 + \sin \left(\frac{2\pi t}{60}\right)) \, dt \). This can be separated into two integrals: \( 15\int_{0}^{20} 1 \, dt + 15\int_{0}^{20} \sin \left(\frac{2\pi t}{60}\right) \, dt \).

Step 4: Calculating the First Integral

The first integral, \( 15\int_{0}^{20} 1 \, dt = 15[t]_{0}^{20} = 15(20 - 0) = 300 \).

Step 5: Calculating the Second Integral

The second integral is \( 15\int_{0}^{20} \sin \left(\frac{2\pi t}{60}\right) \, dt \). We perform a substitution: let \( u = \frac{2\pi t}{60} \), \( \frac{du}{dt} = \frac{2\pi}{60} \Rightarrow dt = \frac{60}{2\pi} du \). Therefore, the integral becomes \( 15 \cdot \frac{60}{2\pi} \int \sin(u) \, du \).

Step 6: Finding the Antiderivative

The antiderivative of \( \sin(u) \) is \( -\cos(u) \). Thus, the integral becomes \( 15 \cdot \frac{60}{2\pi} [-\cos(u)] \bigg|_0^{\frac{2\pi \cdot 20}{60}} \).

Step 7: Evaluating the Definite Integral

At \( t = 20 \), \( u = \frac{2\pi \cdot 20}{60} = \frac{2\pi}{3} \) and at \( t = 0 \), \( u = 0 \). Therefore, the definite integral is \( 15 \cdot \frac{60}{2\pi} [-\cos(\frac{2\pi}{3}) + \cos(0)] = \frac{450}{\pi} \times \left[\frac{1}{2} - 1\right] = -\frac{450}{2\pi} = -\frac{225}{\pi} \).

Step 8: Combining Results

Adding the results from both integrals: \( 300 - \frac{225}{\pi} \approx 227.14 \). Thus, the number of degree-days required is approximately 227.

Key Concepts

Integration

Integration is a fundamental concept in calculus used to find various quantities such as area, volume, and total accumulation over time or space. In this context, integration is employed to calculate the total degree-days, which quantifies the thermal time a plant is exposed to a certain temperature range.
To find the degree-days from a given temperature function, we need to integrate the function over a specified period. In this case, the interval is from day 0 to day 20. The integration of the temperature function effectively accumulates the temperature units (degree-days) over the 20-day period, giving us the total exposure to heat.
Consider the integral \( \int_{0}^{20} T(t) \, dt \), where \( T(t) \) is the temperature function. This integral evaluates the continuous sum of temperatures over time, capturing the total degree-days needed for plant development.

Temperature function

A temperature function provides a mathematical model that describes how temperature changes over time. This function helps predict environmental conditions which are crucial for various biological processes such as plant growth.
In this exercise, the temperature function given is \( T(t) = 15(1 + \sin \left(\frac{2\pi t}{60}\right)) \). Here, \( T(t) \) represents the temperature at any time \( t \), measured in days. The function is periodic, incorporating the sine wave, reflecting the natural fluctuations in temperature over the 60-day cycle.
The constant '15' acts as a coefficient which scales the sine function, ensuring that the temperature oscillates between daily minimums and maximums. Understanding this function allows you to analyze how temperature influences plant development over time.

Degree-days

Degree-days are a measure used in biocalculus to evaluate the effect of temperature on the development of plants over time. They are calculated by integrating the temperature function over the period of interest.
Degree-days provide a cumulative total of thermal exposure necessary for the plant to reach maturity. It considers the daily average temperature's contribution to plant development. In this case, degree-days are calculated by evaluating the integral \( \int_{0}^{20} T(t) \, dt \), incorporating both linear and periodic components of the temperature function.
This concept is essential for understanding how various species mature over different climatic conditions. By obtaining a degree-days measure, we can predict optimal planting times and assess the growth period needed for different crops.

Plant development

Plant development is intricately linked to the environmental conditions, especially temperature, which is often measured in degree-days. This concept describes the sequence of growth phases, from germination to maturity, influenced by accumulated thermal time.
As plants are sensitive to temperature changes, calculating degree-days helps in optimizing planting schedules and maximizing growth rates. By determining the degree-days, researchers and agriculturists can understand and predict when plants will reach various growth milestones.
In this exercise, the plant reaches maturity on day 20, necessitating a specific number of degree-days. The calculated degree-days provide insights into how temperature variations influence the growth cycle, helping people involved in horticulture and agriculture make informed decisions about crop management.