Problem 21 - Calculus for Biology and Medicine

Question

We can approximate the monthly rainfall in Los Angeles using the formula \(p(t)=1.6+1.6 \cos 2 \pi t\), where \(t\) is the fraction of the year elapsed since January 1 and \(p(t)\) is the rainfall measured in inches/month. (a) Explain (without doing any calculations) that there will be a value of \(t\) for which the monthly rainfall is exactly equal to the annual average rainfall. (b) Find this value of \(t\) (there may be more than one). (c) Assuming there are 12 months in a year and each month has the same duration, show that the total rainfall in one year is: $$ P_{\text {total }}=12 \int_{0}^{1} p(t) d t $$ Explain in particular why the factor 12 is needed and what the units of \(P_{\text {total }}\) are. (d) Calculate \(P_{\text {total }}\) -

Step-by-Step Solution

Answer

The total annual rainfall is 19.2 inches.

Step 1: Understand the Definition

The function given for monthly rainfall in Los Angeles is \(p(t)=1.6+1.6 \cos 2 \pi t\). This function models periodic variation in rainfall over the course of a year with period 1 (since it repeats every year). The term \(1.6 \cos 2 \pi t\) oscillates between -1.6 and 1.6, and it represents the variability throughout the year, while the constant 1.6 shifts the function vertically.

Step 2: Conceptualize Value of t for Average Rainfall (a)

Without calculating exactly, the average value of a cosine function over one period (from 0 to 1 in this case) is zero because the positive and negative areas under the curve cancel out. Therefore, the average value of \(p(t)\) over a year will just be constant term 1.6. There exists a point \(t\) at which \(p(t)\) is equal to this average of 1.6, because the cosine function oscillates above and below zero.

Step 3: Solve for Value of t (b)

We want to find the \(t\) where \(p(t) = 1.6\). Substitute this into the equation: \[1.6 + 1.6 \cos 2 \pi t = 1.6\Rightarrow \cos 2 \pi t = 0\Rightarrow 2 \pi t = \frac{\pi}{2} + n \pi, n \in \mathbb{Z} \Rightarrow t = \frac{1}{4}, \frac{3}{4}.\] This shows the rainfall is equal to its annual average at \(t = \frac{1}{4}\) and \(t = \frac{3}{4}\).

Step 4: Formulate Integral for Total Rainfall (c)

The expression \[P_{\text{total}} = 12 \int_0^1 p(t) dt\] computes the total annual rainfall by integrating the monthly rainfall function over one full cycle (year) and then multiplying by 12. The factor 12 is needed because the result of the integral is the average rainfall per month across the year, and multiplying by 12 converts this to total annual rainfall. \(P_{\text{total}}\) is in units of inches per year.

Step 5: Calculate the Integral (d)

Calculate \[\int_0^1 p(t) \, dt = \int_0^1 (1.6 + 1.6 \cos 2 \pi t) \, dt = \int_0^1 1.6 \, dt + 1.6 \int_0^1 \cos(2 \pi t) \, dt.\] For the first term, it's \[1.6 \int_0^1 \, dt = 1.6 [t]_0^1 = 1.6.\] For the second term, since \[\int_0^1 \cos (2 \pi t) \ dt = 0\] (as it's a complete cycle of cosine over one period), the integral evaluates to 0. So, the integral of \(p(t)\) over 0 to 1 is \(1.6\).Thus, \[P_{\text{total}} = 12 \times 1.6 = 19.2 \, \text{inches per year}.\]

Key Concepts

Periodic Functions

In this scenario, a periodic function describes the monthly rainfall in Los Angeles using the formula \(p(t)=1.6+1.6 \cos 2 \pi t\). Periodic functions repeat their values at regular intervals or periods. Here, the interval is one year, representing the cyclical nature of seasons.
This function comprises two main components:

• A constant value of 1.6 inches per month, indicating a base level of rainfall throughout the year.
• An oscillating component, \(1.6 \cos 2 \pi t\), which varies between -1.6 and 1.6 inches, modeling the fluctuation due to seasonal changes.

The period is clear: a full cycle of the cosine function (from 0 to 1) represents one year. This cyclic behavior is central in ecology and biology as it helps model phenomena like growth cycles and ecological systems.

Average Value of a Function

The average value of a function like our rainfall function is crucial in understanding its overall behavior across its entire period. The average value is determined by taking the integral of the function over one period and dividing by the length of that period.

For the cosine part of our function, the average value over \(t\) from 0 to 1 is zero. Since the cosine has equal areas above and below the horizontal axis over one full cycle, the positives and negatives cancel out. Hence, the average value of the function \(p(t)\) is the constant term of 1.6. This implies that despite fluctuations throughout the year, the steady average rainfall is 1.6 inches per month.

Definite Integrals

Definite integrals are used to calculate the total value of a function over an interval. In this context, they help determine the total rainfall over the course of a year.

The integral of \(p(t)\) over the interval from 0 to 1 gives the average monthly rainfall. Incorporating a factor of 12 converts this monthly average into a total annual value due to 12 months per year. Mathematically, it's written as:
\[P_{\text{total}} = 12 \int_0^1 p(t) dt\]
This approach is used in calculus to derive aggregate measures like total mass, energy, or, in our case, rainfall, demonstrating its importance in biological computations.

Unit Conversion in Calculations

Unit conversion is a fundamental aspect of ensuring that calculations reflect real-world measures accurately. In this exercise, it applies to the transition from average monthly to total annual rainfall.

To achieve this, we use the integral value from 0 to 1 (which gives us an average monthly rainfall) and multiply by 12 months. This arithmetic step ensures the units reflect one complete year's worth of rainfall, converting from inches per month to inches per year.
Such conversions are crucial in biology-related calculus problems where calculations might span different timeframes or measurement units, ensuring the results are practical and meaningful.