Question

The population of a culture of cells after \(t\) days is approximated by the function \(P(t)=\frac{1600}{1+7 e^{-0.02 t}},\) for \(t \geq 0\) a. Graph the population function. b. What is the average growth rate during the first 10 days? c. Looking at the graph, when does the growth rate appear to be a maximum? d. Differentiate the population function to determine the growth rate function \(P^{\prime}(t)\) e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?

Short answer

Instructions: First, find the average growth rate during the first 10 days by calculating the rate of change between day 0 and day 10. Second, observe the graph of the given function to estimate when the growth rate is maximum. Finally, differentiate the population function with respect to t to determine the growth rate function and graph it. Locate the time of maximum growth rate and find the population at that time.

Step-by-step solution

Graph the population function

To graph the population function \(P(t)=\frac{1600}{1+7 e^{-0.02 t}}\), one can use a graphing calculator or graphical software like Desmos or Geogebra. Input the given function and display the graph to better visualize the cell population over time.

Calculate the average growth rate during the first 10 days

To calculate the average growth rate during the first 10 days, we will find the difference in population between day 10 and day 0, and then divide by the total days elapsed (10 days). The formula for the average growth rate is given by: $$r_{avg} = \frac{P(10)-P(0)}{10-0}$$Replace t with 10 and 0 in the given function and calculate the average growth rate.

Observe graph for maximum growth rate

By looking at the graph of the population function from step 1, you can estimate when the growth rate appears to be at its maximum. This is the point at which the graph is steepest or the tangent to the curve is the steepest.

Differentiate the population function to determine the growth rate function

We will now differentiate the population function to find the growth rate function \(P^{\prime}(t)\). Function: \(P(t)=\frac{1600}{1+7 e^{-0.02 t}}\) To differentiate \(P(t)\) with respect to \(t\), apply the quotient rule which states, \((\frac{u}{v})^\prime = \frac{u^\prime v - uv^\prime}{v^2} \), where \(u=1600\) and \(v=1+7 e^{-0.02 t}\). Differentiate both \(u\) and \(v\). - \(u^\prime = 0\) - \(v^\prime = -0.02(7)e^{-0.02 t}\) Applying quotient rule: \(P^{\prime}(t)= \frac{0 - 1600(-0.02)(7)e^{-0.02 t}}{(1+7 e^{-0.02 t})^2}\) Simplify the expression: \(P^{\prime}(t)= \frac{224 e^{-0.02 t}}{(1+7 e^{-0.02 t})^2} \) This is the growth rate function for the given population function.

Graph the growth rate function

Using the same graphing calculator or software as before, input the growth rate function \(P^{\prime}(t) = \frac{224 e^{-0.02 t}}{(1+7 e^{-0.02 t})^2}\) and display the graph. Observe for the maximum point on the graph of the growth rate function, which corresponds to the time of maximum growth rate. Additionally, once you find the value for t corresponding to the maximum growth rate, input that t value into the population function \(P(t)\) to find the population at the time when the growth rate is maximum.

Key concepts

Average Growth Rate

Understanding the average growth rate is fundamental when studying population dynamics. It represents the change in population over a specific period of time, giving us an indication of how rapidly a population is increasing or decreasing. Consider the formula for average growth rate: \[ r_{avg} = \frac{P(t_2)-P(t_1)}{t_2-t_1} \]
In the context of our cell culture problem, to find the average growth rate over the first 10 days, we would substitute the population values at day 10 and day 0 into the formula, essentially measuring the slope of the line segment connecting these two points on the population graph. This rate provides an overarching view of the population's change over the entire interval, not accounting for any variability within those 10 days.

Graphing Population Functions

Graphing a population function can provide visual insight into the population's behavior over time. When graphing the function \(P(t)=\frac{1600}{1+7 e^{-0.02 t}}\), we expect to see how the population of cells changes with each day. Initially, when dealing with exponential functions, the graph may start off with a slow increase, then ascend rapidly as the effect of the exponential growth kicks in, before finally levelling off as it approaches the carrying capacity—the maximum population the environment can sustain. By using graphing tools like Desmos or GeoGebra, you can accurately plot this function and observe its nuances, making it easier to identify points of interest, such as when the growth rate is maximum.

Differentiation of Exponential Functions

Differentiation is a powerful tool in calculus for understanding how a function changes at any given point. When we differentiate exponential functions, we reveal the growth rate function, which can tell us how fast the population is growing at any moment in time. In our case, the differentiation of \(P(t)=\frac{1600}{1+7 e^{-0.02 t}}\) using the quotient rule provided us with the growth rate function \(P^{\'}(t)= \frac{224 e^{-0.02 t}}{(1+7 e^{-0.02 t})^2} \). This new function allows us to observe not just how the population is growing on average, but how its rate of change varies at each infinitesimal point in time.

Maximum Growth Rate

The maximum growth rate is a point of key interest as it indicates when the population is increasing most rapidly. It occurs at a point where the derivative of the population function reaches its maximum value. After obtaining the growth rate function \(P^{\'}(t)\), we can graph this function to visually inspect where it reaches its peak. Analytically, finding the maximum growth rate would involve setting the derivative of the growth rate function to zero and solving for \(t\). However, by graphing, we can estimate this point more intuitively. Once the time at which the maximum growth rate occurs is known, substituting this back into the original population function will provide us with the size of the population at that specific moment.