Problem 8 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Sketch the graphs and label the axes for $$\text { (a) } \quad y=0.2 \cos \left(\frac{2 \pi}{0.8} t\right) \quad \text { and } \quad \text { (b) } \quad y=5 \cos \left(\frac{1}{8} t+\pi / 6\right)$$

Step-by-Step Solution

Answer

Function (a) has amplitude 0.2, period 0.8; function (b) has amplitude 5, period 16, phase shift \(-\frac{4\pi}{3}\). Graph accordingly.

Step 1: Identify the amplitude, period, and phase shift for (a)

For the function \(y=0.2 \cos \left(\frac{2\pi}{0.8} t\right)\), the amplitude is \(0.2\), since it is the coefficient in front of the cosine. The period is calculated using \(\frac{2\pi}{B}\), where \(B=\frac{2\pi}{0.8}\), resulting in a period of \(0.8\). There is no phase shift since there is no horizontal shift term.

Step 2: Sketch graph of (a)

To sketch the graph of \(y=0.2 \cos \left(\frac{2\pi}{0.8} t\right)\), start with the amplitude marking 0.2 and -0.2 on the y-axis. The period is 0.8, so label the x-axis for one complete period as 0 to 0.8. Divide graph into four equal segments for key points: maximum, zero crossing, minimum, zero crossing. Plot these points and draw a smooth cosine curve through them.

Step 3: Identify the amplitude, period, and phase shift for (b)

For the function \(y=5 \cos \left(\frac{1}{8} t + \frac{\pi}{6}\right)\), the amplitude is \(5\), the period is \(\frac{2\pi}{\frac{1}{8}} = 16\), and the phase shift is calculated by setting \(\frac{1}{8} t + \frac{\pi}{6} = 0\); this gives \(t = -\frac{\pi}{6}\times 8 = -\frac{4\pi}{3}\).

Step 4: Sketch graph of (b)

Begin sketching \(y=5 \cos \left(\frac{1}{8} t + \frac{\pi}{6}\right)\) with the amplitude markers at 5 and -5 on the y-axis. Set the period on the x-axis from 0 to 16 and apply the phase shift of \(-\frac{4\pi}{3}\), approximately -4.19. Start the graph \(\frac{4\pi}{3}\) units to the right of where it normally begins and plot key cosine points within the period. Connect the points with a smooth curve.

Step 5: Label axes on the graphs

For both sketches, ensure that the x-axes are labeled with time (t) while the y-axes are labeled with displacement (y). Indicate the key periods for each function on the x-axis: 0 to 0.8 for (a) and 0 to 16 for (b). Both amplitudes should be annotated on the y-axis.

Key Concepts

Amplitude

The concept of amplitude in trigonometric functions is crucial because it tells us how tall the waves are. Amplitude refers to the maximum distance the wave peaks reach from the middle or baseline of the wave. For a cosine or sine wave, which oscillates above and below a central line, the amplitude is simply the coefficient of the sine or cosine function.

In simple terms:

• Amplitude is the 'height' of the wave.
• It affects the vertical stretch of the graph.
• To find the amplitude, look for the number before \cos\ or \sin\ in the equation.

For example, in the function \( y = 0.2 \cos\left(\frac{2\pi}{0.8} t\right) \), the amplitude is \( 0.2 \). Similarly, in \( y = 5 \cos\left( \frac{1}{8}t + \frac{\pi}{6} \right) \), the amplitude is \( 5 \). These values show the maximum and minimum points the wave can reach on the graph, providing a simple way to visualize how far the wave stretches along the y-axis.

Period

The period of a trigonometric function is the distance over which the function repeats its shape. To understand the period, consider how long it takes for a pattern in the graph to start over. The formula to calculate the period for sine and cosine functions is \( \frac{2\pi}{B} \), where \( B \) is the coefficient of \( t \) inside the function.

Key points about period:

• The period dictates how "wide" a cycle of the wave is on the x-axis.
• It's the length for one complete cycle of the wave.
• For a function \( f(t) = \cos(Bt) \), compute the period using: \( \text{Period} = \frac{2\pi}{B} \).

In our exercises, for the first function \( y = 0.2 \cos\left( \frac{2 \pi}{0.8} t \right) \), the period is \( 0.8 \). That means every 0.8 units, the wave begins its cycle again. For the second function \( y = 5 \cos\left( \frac{1}{8}t + \frac{\pi}{6} \right) \), the calculation gives a period of \( 16 \). This means on the graph, the wave pattern stretches over 16 units before repeating. Understanding the period helps in accurately plotting and predicting future points of the wave's cycle.

Phase Shift

Phase shift in trigonometric functions refers to the horizontal movement of the wave along the x-axis. This shift can slide the graph left or right depending on the direction indicated by the function's equation. Simply put, the phase shift indicates where the wave starts.

Important facts about phase shift:

• Phase shift shows how far to move the graph horizontally from its standard position.
• Calculated by resolving \( Bt + C = 0 \) for \( t \) in \( f(t) = \cos(Bt + C) \).
• A positive phase shift moves the graph to the left, while a negative shift moves it to the right.

For instance, in our function \( y = 0.2 \cos\left(\frac{2\pi}{0.8} t \right) \), there is no additional term added to \( t \), so no phase shift occurs. The graph starts from its standard position. However, in \( y = 5 \cos\left( \frac{1}{8}t + \frac{\pi}{6} \right) \), the phase shift is calculated by setting the inside term to zero: \( \frac{1}{8}t + \frac{\pi}{6} = 0 \). Solving this results in a phase shift of \(-\frac{4\pi}{3}\); when converted roughly, this is a rightward shift of about 4.19 units. Phase shifts are essential for marking the wave's correct starting point on a graph.