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Chapter 5: Problem 26

The size of a certain insect population is given by \(P(t)=300 e^{0.01 t},\) where \(t\) is measured in days. (a) How many insects were present initially? (b) Give a differential equation satisfied by \(P(t)\) (c) At what time will the initial population double? (d) At what time will the population equal 1200?

Short Answer

Expert verified
Initially, there were 300 insects. The differential equation is \( \frac{dP}{dt} = 0.01 P(t) \). The population doubles in approximately 69.3 days and reaches 1200 in about 138.6 days.

Step by step solution

01

- Determine Initial Population

To find the initial population, evaluate the function at time zero. Substitute \( t = 0 \) into the equation: \[ P(0) = 300 e^{0.01 \times 0} = 300 e^{0} = 300 \times 1 = 300 \] Therefore, the initial population is 300 insects.
02

- Find a Differential Equation

A differential equation that describes the population can be found by differentiating \( P(t) \). The given function is \( P(t) = 300 e^{0.01 t} \). Differentiating this function with respect to \( t \), we get: \[ \frac{dP}{dt} = 300 \times 0.01 e^{0.01 t} = 3 e^{0.01 t} = 0.01 P(t). \] Thus, the differential equation is: \[ \frac{dP}{dt} = 0.01 P(t). \]
03

- Determine Time for Population to Double

To find the time when the population doubles, set \( P(t) = 2P(0) \). We know \( P(0) = 300 \), so we set up the equation: \[ 2 \times 300 = 300 e^{0.01 t} \] Simplifying, we get: \[ 600 = 300 e^{0.01 t} \] Dividing both sides by 300, we get: \[ 2 = e^{0.01 t} \] Taking the natural logarithm of both sides, we obtain: \[ \ln(2) = 0.01 t \] Solving for \( t \), we find: \[ t = \frac{\ln(2)}{0.01} \approx 69.3 \text{ days}. \]
04

- Determine Time for Population to Reach 1200

To find the time when the population reaches 1200, set \( P(t) = 1200 \). Set up the equation: \[ 1200 = 300 e^{0.01 t} \] Dividing both sides by 300, we get: \[ 4 = e^{0.01 t} \] Taking the natural logarithm of both sides, we get: \[ \ln(4) = 0.01 t \] Solving for \( t \), we find: \[ t = \frac{\ln(4)}{0.01} \approx 138.6 \text{ days}. \]

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Key Concepts

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

Initial Population
The initial population of an insect colony is the number of insects present at the beginning of our observation. In mathematics, the initial population is denoted by evaluating the population function at time zero. For our problem, the population function is given by \(P(t) = 300 e^{0.01 t}\). To find the initial population, we substitute \(t = 0\). This results in \(P(0) = 300 e^{0} = 300 \times 1 = 300\). Therefore, the initial population is 300 insects.
Differential Equations
Differential equations help us understand how a quantity changes over time. The insect population in this problem is described by the function \(P(t) = 300 e^{0.01 t}\). To find the differential equation that this function satisfies, we need to differentiate \(P(t)\) with respect to \(t\). The derivative of \(300 e^{0.01 t}\) is \(300 \times 0.01 e^{0.01 t} = 3 e^{0.01 t}\), which can be rewritten as \(0.01 P(t)\). Therefore, the differential equation is \(\frac{dP}{dt} = 0.01 P(t)\). This means the rate of change of the population over time is directly proportional to the current population.
Population Doubling
Population doubling time is the time required for a population to grow to twice its original size. In our exercise, we need to find when the initial population of 300 insects doubles to 600 insects. This can be set up with the equation \(2 \times 300 = 300 e^{0.01 t}\), leading to \(2 = e^{0.01 t}\). To solve for \(t\), we take the natural logarithm of both sides: \(\ln(2) = 0.01 t\). Solving for \(t\), we get \(t = \frac{\ln(2)}{0.01}\), which approximately equals 69.3 days.
Natural Logarithm
The natural logarithm is a logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to 2.71828. It is often written as \(\ln\). The natural logarithm appears frequently in calculus, especially when solving growth and decay problems. In this problem, we used \(\ln\) to solve for the time when the population doubles and when it reaches a certain size. For instance, solving \(2 = e^{0.01 t}\) involves taking \(\ln\) of both sides to get \(\ln(2) = 0.01 t\), allowing us to find \(t\) by rearranging to \(t = \frac{\ln(2)}{0.01}\).

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