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Chapter 8: Problem 3

Explain how the growth rate function can be decreasing while the population function is increasing.

Short Answer

Expert verified
Short Answer: The growth rate function describes how quickly a population is changing, while the population function indicates the number of individuals in a population as a function of time. A decreasing growth rate means the population is growing at a slower pace, but it does not mean the population function is decreasing. When the growth rate function is decreasing yet still positive (for example, when \(0 < t < 5\) in the function \(dP/dt = 500 - 100t\)), the growth rate is diminishing, but the population function is still increasing, as the growth rate is greater than zero during this time interval.

Step by step solution

01

Define the key terms

Growth rate function refers to how quickly a population is changing (growing or shrinking) over time. Population function, on the other hand, indicates the number of individuals in a population as a function of time. In essence, the growth rate function is the derivative of the population function.
02

Clarify the concept of decreasing growth rate

A decreasing growth rate means that the growth rate is becoming smaller over time, but it does not mean it is negative. When the growth rate function is decreasing, the population is still growing; however, it's growing at a slower pace as compared to previous periods.
03

Provide an example

Let the population function be \(P(t) = 1000 + 500t - 50t^2\), where \(P(t)\) represents the population at time t. To examine the growth rate function, we need to find the derivative of \(P(t)\) with respect to time t, which represents the change in population over time.
04

Find the derivative

Using basic differentiation rules, we evaluate the derivative of the population function: \(\frac{dP}{dt} = \frac{d(1000 + 500t - 50t^2)}{dt} = 500 - 100t\). Now we have the growth rate function \(\frac{dP}{dt} = 500 - 100t\).
05

Analyse the growth rate function

We can see that the growth rate function is decreasing with respect to time, as the coefficient of t is negative (-100). However, the population growth is still positive as long as \(\frac{dP}{dt} > 0\). To find the interval where the growth rate function is positive, we can solve the inequality: \(500 - 100t > 0\)
06

Solve the inequality

Solving the inequality, we have: \(500 - 100t > 0\) \(-100t > -500\) \(t < 5\) Thus, for \(0 < t < 5\), the growth rate function is decreasing but still positive. This implies that during this time interval, the growth rate (speed of increase) of the population is diminishing, but the population function is still increasing, as its growth rate is greater than zero.

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Most popular questions from this chapter

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y^{\prime}(x)=\sqrt{\frac{x+1}{y+4}}, y(3)=5$$

Widely used models for population growth involve the logistic equation \(P^{\prime}(t)=r P\left(1-\frac{P}{K}\right),\) where \(P(t)\) is the population, for \(t \geq 0,\) and \(r>0\) and \(K>0\) are given constants. a. Verify by substitution that the general solution of the equation is \(P(t)=\frac{K}{1+C e^{-n}},\) where \(C\) is an arbitrary constant. b. Find that value of \(C\) that corresponds to the initial condition \(P(0)=50\). c. Graph the solution for \(P(0)=50, r=0.1,\) and \(K=300\). d. Find \(\lim _{t \rightarrow \infty} P(t)\) and check that the result is consistent with the graph in part (c).

Consider the first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=A,\) for \(t \geq 0,\) where \(a, b,\) and \(A\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to a horizontal line in the direction field. b. Draw a representative direction field in the case that \(a>0\) Show that if \(A>-b / a,\) then the solution increases for \(t \geq 0\) and if \(A<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\) Show that if \(A>-b / a,\) then the solution decreases for \(t \geq 0\) and if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

Use the method outlined in Exercise 43 to solve the following Bernoulli equations. a. \(y^{\prime}(t)+y=2 y^{2}\) b. \(y^{\prime}(t)-2 y=3 y^{-1}\) c. \(y^{\prime}(t)+y=\sqrt{y}\)

Let \(y(t)\) be the concentration of a substance in a chemical reaction (typical units are moles/liter). The change in the concentration, under appropriate conditions, is modeled by the equation \(\frac{d y}{d t}=-k y^{n},\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1)\), the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\) c. Graph the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\)
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