Questions: The growth of the number of students taking at least one online course can be approximated by a logistic function with k=0.0463, where t is the number of years since 2002. In 2002 (when t=0), the number of students enrolled was 1.659 million. Assume that the number will level out at around 7.3 million students. (a) Find the growth function G(t) for the number of students (in millions) enrolled in at least one online course. Find the number of students enrolled in at least one online course and the rate of growth in the number for the following years. (b) 2003 (c) 2010 (o) What happens to the rate of growth over time? (d) 2017

Solution Steps

To solve this problem, we need to use the logistic growth function, which is given by:

\[ G(t) = \frac{L}{1 + e^{-k(t - t_0)}} \]

where:

• \( L \) is the carrying capacity (7.3 million students),
• \( k \) is the growth rate (0.0463),
• \( t \) is the number of years since 2002,
• \( t_0 \) is the initial time (2002, so \( t_0 = 0 \)),
• \( G(t_0) \) is the initial number of students (1.659 million).

We will use this function to find the number of students enrolled in at least one online course for the years 2003, 2010, and 2017. Additionally, we will calculate the rate of growth, which is the derivative of the logistic function.

1. Define the logistic growth function \( G(t) \).
2. Calculate the number of students for the given years by substituting \( t \) values into \( G(t) \).
3. Calculate the rate of growth by finding the derivative of \( G(t) \) and evaluating it at the given years.

The logistic growth function is given by:

\[ G(t) = \frac{L}{1 + \left(\frac{L - G_0}{G_0}\right) e^{-k(t - t_0)}} \]

where:

• \( L = 7.3 \) (carrying capacity in millions),
• \( k = 0.0463 \) (growth rate),
• \( G_0 = 1.659 \) (initial number of students in millions),
• \( t_0 = 0 \) (initial time, 2002).

We need to find \( G(t) \) for the years 2003, 2010, and 2017. The corresponding \( t \) values are:

• \( t = 2003 - 2002 = 1 \)
• \( t = 2010 - 2002 = 8 \)
• \( t = 2017 - 2002 = 15 \)

Using the logistic growth function, we get:

• For 2003: \( G(1) \approx 1.719 \) million
• For 2010: \( G(8) \approx 2.181 \) million
• For 2017: \( G(15) \approx 2.706 \) million

The rate of growth is the derivative of the logistic function:

\[ \frac{dG}{dt} = k G(t) \left(1 - \frac{G(t)}{L}\right) \]

Evaluating this at the given years:

• For 2003: \( \frac{dG}{dt}(1) \approx 0.06085 \) million/year
• For 2010: \( \frac{dG}{dt}(8) \approx 0.07080 \) million/year
• For 2017: \( \frac{dG}{dt}(15) \approx 0.07884 \) million/year

Final Answer