Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Let \(f\) be the function defined by $$ f(x)=1-x^{2} \quad-1 \leq x \leq 1 $$ Let \(F\) be the extension of \(f\) with period 2 . a. Draw a graph of \(f\). b. Draw a graph of \(F\). c. Evaluate \(F(1), F(2), F(3), F(12), F(31)\) and \(F(1002)\). d. Find the amplitude of \(F\).

Step-by-Step Solution

Answer

The graph of \( F \) is periodic and repeats every 2 units; \( F \) has an amplitude of 0.5.\( F(1) = 0, F(2) = 1, F(3) = 0, F(12) = 1, F(31) = 0, F(1002) = 1. \)

Step 1: Graph Function f(x)

The given function is \( f(x) = 1 - x^2 \) in the interval \( -1 \leq x \leq 1 \). This is a downward-opening parabola with vertex at \( (0,1) \). The graph will be a part of the parabola only for the interval from \(-1\) to \(1\). Plot points such as \( f(-1) = 0 \), \( f(0) = 1 \), \( f(1) = 0 \), and draw the parabola between these points.

Step 2: Extend Function Periodically to Form F(x)

Since \( F \) is defined as the extension of \( f \) with period 2, the function \( F(x) \) will repeat the shape of \( 1 - x^2 \) over intervals of length 2. This means from \( x = -1 \) to \( x = 1 \), \( x = 1 \) to \( x = 3 \), and so on in both directions, forming a periodic wave-like pattern.

Step 3: Evaluate F(x) at Given Points

To evaluate \( F(x) \) at specific points like \( F(1), F(2), F(3), F(12), F(31), F(1002) \), first find the equivalent \( x \) in the interval \([-1,1]\) using the modulo operation. The value of \( x_{0} \) will be equivalent to \( x \mod 2 \):- \( F(1) = f(1) = 0 \) since \( 1 \) is in the original interval.- \( F(2) = f(0) = 1\) because \( 2 \equiv 0 \mod 2 \).- \( F(3) = f(1) = 0 \) since \( 3 \equiv 1 \mod 2 \).- \( F(12) = f(0) = 1 \) because \( 12 \equiv 0 \mod 2 \).- \( F(31) = f(-1) = 0\) since \( 31 \equiv 1 \mod 2 \).- \( F(1002) = f(0) = 1 \) as \( 1002 \equiv 0 \mod 2 \).

Step 4: Determine Amplitude of F(x)

The amplitude of a periodic function is the distance between the maximum and minimum values divided by 2. For \( F(x) \), the maximum value is \( 1 \) and the minimum is \( 0 \). Thus, the amplitude is \( \frac{1 - 0}{2} = 0.5 \).

Key Concepts

Understanding Amplitude in Periodic Functions

Periodic functions are those that repeat their values in regular intervals or periods. The amplitude of a periodic function, such as the extended function \( F(x) \) from the exercise, is an important trait. It measures how much the function varies above and below its central value, often referred to as the equilibrium or rest position.

The amplitude is calculated by taking the difference between the maximum and minimum values of the function, then dividing this by 2. For the function \( F(x) = 1 - x^2 \), which is extended periodically, the highest value reached is 1, and the lowest is 0. Therefore, the amplitude is calculated as:
\[\text{Amplitude} = \frac{1 - 0}{2} = 0.5\]

This value tells us that the periodic wave swings 0.5 units up and 0.5 units down from its central value. Recognizing amplitude helps us understand the scale of oscillation in periodic functions.

Exploring Parabolas and Their Properties

A parabola is a U-shaped curve that can open either upwards or downwards. In the context of the given function \( f(x) = 1 - x^2 \), we encounter a downward-opening parabola. Here are some key properties of parabolas:

• Vertex: The peak (or Lowest point if it opens up) of the parabola. For \( f(x) \), the vertex is at \((0, 1)\).
• Axis of symmetry: A vertical line that runs through the vertex, dividing the parabola into two symmetrical halves. For \( f(x) \), this line is \(x = 0\).
• Direction of opening: Determined by the sign of the coefficient of the \(x^2\) term. A positive sign means it opens upwards; a negative sign means downwards. Since \( f(x) = 1 - x^2 \) has a negative coefficient, it opens downwards.

The essence of solving problems involving parabolas often lies in identifying these characteristics, allowing us to plot them accurately over the specified interval.

The Role of the Modulo Operation in Periodic Functions

The modulo operation plays a crucial part when dealing with periodic functions such as \( F(x) \). It finds the remainder when one number is divided by another, essentially "wrapping around" numbers to fit them within a specified range of values. This property is utilized to evaluate \( F(x) \) at various points, ensuring they fall appropriately within the function's period.

For instance, to determine \( F(1002) \), we calculate \( 1002 \mod 2 \) which gives us 0, aligning it with an equivalent point in the base period of the original function \( f(x) \), in this case, \( f(0) \). Consequently, \( F(1002) = f(0) = 1 \).

Using the modulo operation simplifies identifying which part of the repeating pattern a given \(x\) value corresponds to. It's a valuable tool for managing continuous extensions of functions and maintaining consistent evaluation across each cycle.