Problem 30 - Precalculus with Calculus Previews

Question

Find the amplitude and period of the given function. Sketch at least one cycle of the graph. $$ y=5 \cos \left(-\frac{\pi}{2} x\right) $$

Step-by-Step Solution

Answer

Amplitude is 5; Period is 4.

Step 1: Understanding the Function Setup

The given function is a cosine function of the form \( y = a \cos(bx) \) where \( a = 5 \) and \( b = -\frac{\pi}{2} \). The amplitude is determined by \( |a| \) and the period is determined by \( \frac{2\pi}{|b|} \). Before we begin, we need to identify these constants and variables in the function.

Step 2: Calculating the Amplitude

The amplitude of a cosine function \( y = a \cos(bx) \) is the absolute value of \( a \). In this case, \( a = 5 \), so the amplitude is \( |5| = 5 \).

Step 3: Calculating the Period

The period of a cosine function is calculated by the formula \( \frac{2\pi}{|b|} \). Substitute \( b = -\frac{\pi}{2} \), so the period is \( \frac{2\pi}{|-\frac{\pi}{2}|} = \frac{2\pi}{\frac{\pi}{2}} = 4 \).

Step 4: Sketching One Cycle of the Graph

With an amplitude of 5 and a period of 4, the graph of the function will oscillate from 5 to -5 vertically and repeat itself every 4 units horizontally. The cosine function starts at the maximum value, so at \( x = 0 \), \( y = 5 \). At \( x = 1 \), the function reaches 0, at \( x = 2 \), it reaches \( -5 \), at \( x = 3 \), back to 0, and at \( x = 4 \), back to 5, completing one cycle.

Key Concepts

Amplitude

When dealing with trigonometric functions like the cosine function, one of the fundamental properties you'll often need to determine is the amplitude. Amplitude refers to the distance from the horizontal midline of the graph to its peak (or trough). This describes how "tall" or "short" the wave appears. From the mathematical perspective, in the function format \( y = a \cos(bx) \), the amplitude is given by the absolute value of the coefficient \( a \). This means you simply take the modulus of \( a \) without considering any negative sign.

In our original problem, the function is \( y = 5 \cos\left(-\frac{\pi}{2} x\right) \). Here, \( a = 5 \), so the amplitude is \( |5| = 5 \). The amplitude affects all the points on the cosine curve proportionally. This means the peak of the wave will reach 5 units above the midline, and the trough will go 5 units below the midline.

Period

Another essential feature of a cosine function is its period. The period is the horizontal length required for the function to complete one full cycle of its repeating wave. Specifically, it tells you how wide or narrow each cycle stretches along the x-axis.

The period of a cosine function \( y = a \cos(bx) \) is determined using the formula \( \frac{2\pi}{|b|} \). The term \( b \) denotes how many oscillations will happen over a specific range, influencing how squeezed or stretched the waves appear.

• In our exercise, \( b = -\frac{\pi}{2} \).
• To find the period, calculate \( \frac{2\pi}{\left| -\frac{\pi}{2} \right|} \).
• This simplifies to \( \frac{2\pi}{\frac{\pi}{2}} = 4 \).

This means the graph repeats every 4 units along the x-axis, allowing you to sketch its repeating pattern accurately.

Cosine Function

The cosine function is a fundamental trigonometric function, often used to model periodic phenomena like sound waves or tides. It has a characteristic waveform—starting at a maximum, descending through zero to a minimum, and back to zero, completing a periodic cycle.

In the general form \( y = a \cos(bx) \), \( a \) and \( b \) are constants influencing the graph's shape:

• The coefficient \( a \) determines the amplitude, affecting the vertical stretch of the wave.
• The coefficient \( b \) (excluding changes in sign which only affects direction) determines the period which sets the horizontal length of one complete cycle.

The phase shift and vertical shift, handled by other constants in the more extended form, \( y = a \cos(bx - c) + d \), move the entire wave left or right and up or down, respectively.

In our example \( y=5 \cos\left(-\frac{\pi}{2} x\right) \), the graph has no phase or vertical shifts, meaning it will simply oscillate with the derived amplitude and period. Here, it starts at 5 at \( x = 0 \), dips to -5, then returns to 5 by \( x = 4 \), showing a beautifully symmetric wave around the x-axis.