Question

Discuss the nature of the following time paths: (a) \(y_{t}=3^{t}+1\) (c) \(y_{1}=5\left(-\frac{1}{10}\right)^{t}+3\) (b) \(y_{t}=2\left(\frac{1}{3}\right)^{t}\) (d) \(y_{t}=-3\left(\frac{1}{4}\right)^{t}+2\)

Short answer

(a) Exponential growth; (b) Exponential decay; (c) Exponential decay with oscillation; (d) Exponential decay with reflection and shift.

Step-by-step solution

Identify the type of time path for (a)

The equation given is \( y_{t} = 3^{t} + 1 \). This is an exponential function because the variable \( t \) is the exponent. The base of the exponent is \( 3 \), which is greater than 1, indicating that this function is exponential growth. Additionally, the function includes a constant \(+1\), which shifts the entire exponential graph upwards.

Identify the type of time path for (b)

For \( y_{t} = 2\left(\frac{1}{3}\right)^{t} \), \( t \) is in the exponent, which indicates an exponential function. The base is \( \frac{1}{3} \), which is less than 1, suggesting this is exponential decay. The function multiplies this decaying term by 2, maintaining the decay characteristic with a decay factor.

Identify the type of time path for (c)

Here, the equation is \( y_{1} = 5\left(-\frac{1}{10}\right)^{t}+3 \). Since the exponent \( t \) is applied to \(-\frac{1}{10}\), which is a negative fraction, this introduces oscillation around the horizontal line, while altering the amplitude due to changes in \( t \). This suggests exponential decay with oscillation, which centers around the value \(+3\).

Identify the type of time path for (d)

The equation \( y_{t} = -3\left(\frac{1}{4}\right)^{t} + 2 \) indicates another exponential function with the fraction \( \frac{1}{4} \) being less than 1, pointing to exponential decay. However, the negative coefficient \(-3\) implies the graph is reflected around the horizontal axis, also including a vertical shift upwards by +2.

Key concepts

Exponential Growth

In exponential growth, a quantity increases rapidly over time. This happens when the base of the exponent, in an equation like \[ y_t = a b^t + c \] is greater than 1. * Base Greater Than 1: In this equation, if \( b > 1 \), it signifies that each unit increase in time leads to a multiplication of the quantity by \( b \).* Rapid Increase: The outcome is an exponential curve that rises sharply, representing growth that accelerates over time.
In step 1 of the original solution, the function \( y_t = 3^t + 1 \) exhibits this growth pattern. The base of the exponent is 3, which is greater than 1, indicating the variable \( y_t \) increases exponentially as \( t \) increases. The graph of this function starts slightly above zero due to the constant +1 and shoots upward rapidly as time progresses. This type of growth is often observed in populations, finance, and areas where compounding affects growth rates significantly.

Exponential Decay

Exponential decay describes how a quantity diminishes rapidly. This occurs when the base of the exponent is between 0 and 1, usually represented as:\[ y_t = a b^t + c, \quad 0 < b < 1 \] * Base Between 0 and 1: Here, if \( 0 < b < 1 \), each increase in time results in the expression being multiplied by a fraction, reducing quickly.* Rapid Decrease: This causes a steep downward curve that approaches zero or another horizontal asymptote, showing sharp decay.
For instance, in step 2, the equation \( y_t = 2\left(\frac{1}{3}\right)^t \) is characterized by decay. The base \( \frac{1}{3} \) results in the value being diminished at each interval, representing a classic decay curve.
Exponential decay frequently appears in radioactive decay, cooling processes, and depreciation of assets where things lose value rapidly over time.

Oscillation

In some exponential functions, negative values in the base of exponentiation introduce oscillation, a situation where a function fluctuates above and below a central value.* Oscillating Factors: When the base is a negative fraction as seen in \( y_t = a (-b)^t + c \), where \( 0 < b < 1 \).* Alternating Pattern: The negative sign causes the value to alternately reflect above and below a certain level, depending on whether \( t \) is odd or even.
In step 3, \( y_1 = 5\left(-\frac{1}{10}\right)^t + 3 \), illustrates oscillation. The base \(-\frac{1}{10}\) results in an alternately positive and negative expression, creating a series of jumps. This oscillation, around the value +3, presents through a fluctuating pattern. Such oscillations can emerge in alternating current signals, or in sequences involving repetitive fluctuation.

Graph Shifts

Graph shifts in exponential functions are due to additional constants added to the function. They change the position of the graph in a vertical direction.* Vertical Shift: Represented by \( +c \) or \( -c \) in expressions like \( y_t = ab^t + c \), it affects the starting point of a graph without altering the basic growth or decay pattern.* Adjusting The Curve: If \( +c \) is positive, the entire graph elevates upwards and vice versa for negative.In step 1 and step 4, graphs of \( y_t = 3^t + 1 \) and \( y_t = -3\left(\frac{1}{4}\right)^t + 2 \) experience vertical shifts. The constant \(+1\) shifts the exponential growth curve upward, starting higher on the y-axis. For \( y_t = -3\left(\frac{1}{4}\right)^t + 2 \), the +2 creates a vertical change, lifting the decaying and inverted graph upwards. However, with this shift, the reflection of the graph due to the negative coefficient will still preserve its core dynamics. Vertical shifts are common in modeling data to account for baseline values or offsets in different scenarios.