Problem 55 - Calculus Early Transcendentals

Question

Suppose that the growth of a population \(y=y(t)\) is given by the logistic equation $$ y=\frac{60}{5+7 e^{-t}} $$ (a) What is the population at time \(t=0\) ? (b) What is the carrying capacity \(L ?\) (c) What is the constant \(k ?\) (d) When does the population reach half of the carrying capacity? (e) Find an initial-value problem whose solution is \(y(t)\)

Step-by-Step Solution

Answer

(a) Population at \( t=0 \) is 5. (b) Carrying capacity \( L \) is 12. (c) \( k \) is 1. (d) Time to reach half capacity is \( \ln \left( \frac{7}{5} \right) \). (e) Initial-value problem: \( \frac{dy}{dt} = y \left( 1 - \frac{y}{12} \right), y(0) = 5 \).

Step 1: Calculate Population at Time t=0

To find the population at time \( t=0 \), substitute \( t=0 \) into the given logistic equation: \[ y(0) = \frac{60}{5 + 7e^0} \] Simplifying gives: \[ y(0) = \frac{60}{5 + 7} = \frac{60}{12} = 5 \] So, the population at time \( t=0 \) is 5.

Step 2: Determine the Carrying Capacity

The carrying capacity \( L \) is the value that the population approaches as \( t \to \infty \). For the logistic model \( y = \frac{c}{a + b e^{-kt}} \), the carrying capacity is equal to \( \frac{c}{a} \). In this case, \( c = 60 \) and \( a = 5 \), so: \[ L = \frac{60}{5} = 12 \] The carrying capacity is 12.

Step 3: Find the Constant k

To find the constant \( k \), compare the given equation with the logistic equation form \( y = \frac{c}{a + be^{-kt}} \). The exponent \(-kt\) indicates that \( -7 \) is related to \(-k\). Therefore, \( -k \times 1 = -7 \), which gives \( k = 1 \). In combination, \( be^1 = 7 \), meaning \( b = 7 \) and confirms \( k = 1 \).

Step 4: Find the Time for Half Carrying Capacity

Half of the carrying capacity is \( \frac{L}{2} = 6 \). Set \( y(t) = 6 \) and solve for \( t \): \[ \frac{60}{5 + 7e^{-t}} = 6 \] \[ 60 = 6(5 + 7e^{-t}) \] \[ 60 = 30 + 42e^{-t} \] \[ 30 = 42e^{-t} \] \[ e^{-t} = \frac{30}{42} = \frac{5}{7} \] \[ -t = \ln \left( \frac{5}{7} \right) \] \[ t = -\ln \left( \frac{5}{7} \right) \] \[ t = \ln \left( \frac{7}{5} \right) \] So, \( t = \ln \left( \frac{7}{5} \right) \).

Step 5: Formulate the Initial-Value Problem

The form of the logistic differential equation is \( \frac{dy}{dt} = ky(1 - \frac{y}{L}) \). With \( k = 1 \) and \( L = 12 \), the equation is: \[ \frac{dy}{dt} = y \left( 1 - \frac{y}{12} \right) \] Given \( y(0) = 5 \), the initial-value problem is: \[ \frac{dy}{dt} = y \left( 1 - \frac{y}{12} \right), \quad y(0) = 5 \]

Key Concepts

Population Dynamics

Population dynamics is a fascinating field of study that focuses on the changes in population size and composition over time. It involves understanding how populations grow, shrink, or maintain equilibrium. Logistic growth is a common model used in this area of study.
In logistic growth, the population increases rapidly at first but then slows down as it approaches a maximum limit known as the carrying capacity. This model reflects real-world situations where resources such as food, space, and water are limited.

• Initially, when resources are abundant, the population grows exponentially.
• As resources become scarce, growth slows down.
• Eventually, the population stabilizes at a constant level - the carrying capacity.

Understanding these dynamics helps in fields like ecology, resource management, and environmental policy.

Differential Equations

Differential equations are equations that involve functions and their derivatives. They are crucial in modeling the dynamics of changing systems, such as population growth.
The logistic growth model is often illustrated using a first-order differential equation: \[ \frac{dy}{dt} = ky \left( 1 - \frac{y}{L} \right) \]
Here, \[ y \] represents the population size, \[ t \] is time, \[ k \] is the growth rate constant, and \[ L \] is the carrying capacity.

• The term \[ ky \] models exponential growth without limits.
• The factor \[ 1 - \frac{y}{L} \] reduces the growth rate as the population size approaches the carrying capacity.

Solving such equations helps predict future population sizes and can aid in making informed decisions regarding resource allocation.

Carrying Capacity

Carrying capacity is a vital concept in understanding population dynamics. It refers to the maximum population size that an environment can sustain indefinitely.
In our logistic model, carrying capacity \[ L \] is determined by the equation parameters and can be calculated as \[ L = \frac{c}{a} \], where \[ c \] is the maximum rate of population increase and \[ a \] is a constant in the model.

• As the population size reaches carrying capacity, growth ceases.
• At carrying capacity, resources are optimized, neither wasted nor overly scarce.
• Any change in environmental conditions can alter the carrying capacity.

Understanding carrying capacity helps manage natural resources and sustainable development initiatives.

Initial Value Problem

An initial value problem involves finding a function that satisfies a differential equation and meets specific initial conditions.
In the logistic growth scenario, we are given the differential equation: \[ \frac{dy}{dt} = y \left( 1 - \frac{y}{12} \right) \], and must also satisfy \[ y(0) = 5 \].
This means the initial population size is 5 when time \[ t = 0 \].

• The initial condition \[ y(0) = 5 \] ensures the specific solution fits the real-world scenario being studied.
• Such problems are fundamental in predicting future states of dynamic systems.
• By combining the equation and initial values, it is possible to model and visualize population changes over time.

Solving initial value problems is critical in fields like engineering, physics, and any area dealing with dynamic systems.