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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

Expert verified
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

01

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.
02

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.
03

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.
04

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.
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

The lapse rate is the rate at which the temperature in Earth's atmosphere decreases with altitude. For example, a lapse rate of \(6.5^{\circ}\) Celsius / km means the temperature decreases at a rate of \(6.5^{\circ} \mathrm{C}\) per kilometer of altitude. The lapse rate varies with location and with other variables such as humidity. However, at a given time and location, the lapse rate is often nearly constant in the first 10 kilometers of the atmosphere. A radiosonde (weather balloon) is released from Earth's surface, and its altitude (measured in kilometers above sea level) at various times (measured in hours) is given in the table below. $$\begin{array}{lllllll} \hline \text { Time (hr) } & 0 & 0.5 & 1 & 1.5 & 2 & 2.5 \\ \text { Altitude (km) } & 0.5 & 1.2 & 1.7 & 2.1 & 2.5 & 2.9 \\ \hline \end{array}$$ a. Assuming a lapse rate of \(6.5^{\circ} \mathrm{C} / \mathrm{km},\) what is the approximate rate of change of the temperature with respect to time as the balloon rises 1.5 hours into the flight? Specify the units of your result and use a forward difference quotient when estimating the required derivative. b. How does an increase in lapse rate change your answer in part (a)? c. Is it necessary to know the actual temperature to carry out the calculation in part (a)? Explain.

Determine whether the following statements are true and give an explanation or counterexample. a. For any equation containing the variables \(x\) and \(y,\) the derivative \(d y / d x\) can be found by first using algebra to rewrite the equation in the form \(y=f(x).\) b. For the equation of a circle of radius \(r, x^{2}+y^{2}=r^{2},\) we have \(\frac{d y}{d x}=-\frac{x}{y},\) for \(y \neq 0\) and any real number \(r>0.\) c. If \(x=1\), then by implicit differentiation, \(1=0.\) d. If \(x y=1,\) then \(y^{\prime}=1 / x.\)

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