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Ch. 9 - Differential Equations
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 9 - Differential EquationsProblem 9.5.11
Chapter 9, Problem 9.5.11

9–14. Growth rate functions Make a sketch of the population function P (as a function of time) that results from the following growth rate functions. Assume the population at time t = 0 begins at some positive value.


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Step 1: Understand the graph provided. The graph shows the growth rate function \(P'(t)\) plotted against the population \(P\). The curve starts at \(P=0\) with \(P'=0\), rises to a maximum positive growth rate at some intermediate population, and then decreases back to \(P'=0\) at \(P=K\).
Step 2: Interpret the meaning of \(P'\) in terms of population growth. Since \(P'(t)\) represents the rate of change of the population \(P\) with respect to time \(t\), positive values of \(P'\) mean the population is increasing, and negative values would mean the population is decreasing. Here, \(P'\) is zero at \(P=0\) and \(P=K\), and positive in between.
Step 3: Sketch the population function \(P(t)\) starting from a positive initial population \(P(0)\). Since \(P'(t)\) is positive for \(0 < P < K\), the population will increase over time until it approaches \(K\). At \(P=K\), the growth rate \(P'\) is zero, so the population stabilizes and stops growing.
Step 4: Consider the shape of \(P(t)\). Because \(P'(t)\) increases then decreases as \(P\) approaches \(K\), the population growth rate accelerates initially, then slows down as it nears the carrying capacity \(K\). This suggests an S-shaped or logistic growth curve for \(P(t)\).
Step 5: Conclude that the population \(P(t)\) will start at a positive value, increase over time with a decreasing growth rate as it approaches \(K\), and eventually level off at \(P=K\), representing a stable equilibrium population.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Relationship Between a Function and Its Derivative

The derivative of a function represents its rate of change. In this problem, P' is the growth rate of the population P. Understanding how the sign and magnitude of P' affect the shape of P is crucial: where P' is positive, P increases; where P' is zero, P has critical points; and where P' is negative, P decreases.
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Interpreting the Graph of the Growth Rate Function

The graph shows P' as a function of P, starting at zero at P=0, rising to a maximum, then returning to zero at P=K. This indicates that the growth rate increases with population up to a point, then decreases to zero at a carrying capacity K, suggesting logistic growth behavior.
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Sketching Population Functions from Growth Rates

To sketch P(t) from P'(P), consider how the growth rate changes with population size. Starting from a positive initial population, P will increase when P' > 0, slow down as P' approaches zero, and stabilize at P=K where growth rate is zero. This results in an S-shaped logistic curve for P(t).
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Real World Application
Related Practice
Textbook Question

What is a separable first-order differential equation?

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33–42. Solving initial value problems Solve the following initial value problems.

y'(t) = 1 + eᵗ, y(0) = 4

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23–26. Loan problems The following initial value problems model the payoff of a loan. In each case, solve the initial value problem, for t≥0 graph the solution, and determine the first month in which the loan balance is zero.


B′(t) = 0.005B − 500, B(0) = 50,000

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21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.

u''(x) = 55x⁹ + 36x⁷ - 21x⁵ + 10x⁻³

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Textbook Question

9–14. Growth rate functions Make a sketch of the population function P (as a function of time) that results from the following growth rate functions. Assume the population at time t = 0 begins at some positive value.


Textbook Question

39–42. Special equations A special class of first-order linear equations have the form a(t)y'(t)+a'(t)y(t)=f(t), where a and f are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form

a(t)y'(t) + a'(t)y(t) = d/dt (a(t)y(t)) = f(t). 

Therefore, the equation can be solved by integrating both sides with respect to t. Use this idea to solve the following initial value problems. 


t³y′(t) + 3t²y = (1 + t)/t, y(1) = 6

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