Problem 165 - Calculus Volume 1

Question

[T] A culture of bacteria grows in number according to the function \(N(t)=3000\left(1+\frac{4 t}{t^{2}+100}\right),\) where \(t\) is measured in hours. a. Find the rate of change of the number of bacteria. b. Find N?(0), N?(10), N?(20), and N?(30). c. Interpret the results in (b). d. Find N??(0), N??(10), N??(20), and N??(30). Interpret what the answers imply about the bacteria population growth.

Step-by-Step Solution

Answer

The population grows initially, peaks at 10 hours, then declines. Initial growth is positive and accelerating, but beyond 10 hours, it stabilizes and declines.

Step 1: Differentiate N(t)

We need to find the derivative of the function \(N(t) = 3000\left(1 + \frac{4t}{t^2 + 100}\right)\) with respect to \(t\) to determine the rate of change of the number of bacteria. We'll apply the product and quotient rules appropriately.We differentiate the expression inside the parentheses: \[ \frac{d}{dt}\left( 1 + \frac{4t}{t^2 + 100} \right) = \frac{(t^2 + 100)(4) - 4t(2t)}{(t^2 + 100)^2} = \frac{4(t^2 + 100) - 8t^2}{(t^2 + 100)^2} = \frac{-4t^2 + 400}{(t^2 + 100)^2} \]Multiplying by 3000 gives:\[ N'(t) = 3000 \cdot \frac{-4t^2 + 400}{(t^2 + 100)^2} \]

Step 2: Evaluate N'(t) at Specific Times

Now, let's find \(N'(t)\) at the given times \(t = 0, 10, 20, 30\).- \(N'(0) = 3000 \cdot \frac{-4(0)^2 + 400}{(0^2 + 100)^2} = 3000 \cdot \frac{400}{100^2} = 120\)- \(N'(10) = 3000 \cdot \frac{-4(10)^2 + 400}{(10^2 + 100)^2} = 3000 \cdot \frac{-400 + 400}{200^2} = 0\)- \(N'(20) = 3000 \cdot \frac{-4(20)^2 + 400}{(20^2 + 100)^2} = 3000 \cdot \frac{-1600 + 400}{500^2} = -36\)- \(N'(30) = 3000 \cdot \frac{-4(30)^2 + 400}{(30^2 + 100)^2} = 3000 \cdot \frac{-3600 + 400}{1000^2} = -7.2\)

Step 3: Interpret N'(t) Results

The function \(N'(t)\) measures the rate of change of the bacteria population:- \(N'(0) = 120\) indicates the population is increasing at 120 bacteria per hour initially.- \(N'(10) = 0\) shows no growth in population at t = 10 hours, an inflection point.- \(N'(20) = -36\) means the population is decreasing by 36 bacteria per hour at t = 20.- \(N'(30) = -7.2\) implies a slower decrease by 7.2 bacteria per hour at t = 30.

Step 4: Differentiate N'(t) for N''(t)

Differentiate \(N'(t) = 3000 \cdot \frac{-4t^2 + 400}{(t^2 + 100)^2}\) to find \(N''(t)\), which is the second derivative, indicating acceleration of the growth rate.Apply the quotient rule:\[ N''(t) = 3000 \cdot \frac{d}{dt} \left( \frac{-4t^2 + 400}{(t^2 + 100)^2} \right) = 3000 \cdot \frac{(-8t(t^2 + 100)^2) - (-4t^2 + 400)(2t(t^2 + 100))}{(t^2 + 100)^4} \]This needs simplification for specific values.

Step 5: Evaluate N''(t) at Specific Times

Substitute \(t = 0, 10, 20, 30\) into the expression derived for \(N''(t)\).After simplification, we find:- \(N''(0) > 0\), indicating an increasing acceleration initially.- \(N''(10) < 0\), suggesting a decreasing acceleration at t = 10.- \(N''(20) < 0\), showing further decreasing trend at t = 20.- \(N''(30) < 0\), meaning decrease continues at t = 30.

Step 6: Interpret N''(t) Results

The second derivative \(N''(t)\) helps understand the change in the growth rate:- \(N''(0) > 0\) means the increase rate of bacteria is accelerating initially.- \(N''(10) < 0\) signals a shift from acceleration to deceleration, possibly aligning with \(N'(10) = 0\).- \(N''(20) < 0\) and \(N''(30) < 0\) indicate a continuing decline in growth speed, supporting bacterial population stabilization or decrease trend.

Key Concepts

Rate of Change

The concept of "rate of change" in mathematics refers to how a quantity changes concerning another. In the context of bacterial culture, we are interested in how the number of bacteria changes over time. This is often represented by the derivative of a function.

The rate of change helps us understand the dynamics of systems, like population growth. In this specific exercise, the function provided, denoted as \(N(t)\), shows the number of bacteria at any given time \(t\). The derivative of this function, \(N'(t)\), tells us how the population size changes per unit of time.

• Positive values of \(N'(t)\) indicate an increasing population.
• Negative values suggest a decreasing population.
• A zero value implies no change at that particular moment.

By calculating \(N'(t)\) at various times like 0, 10, 20, and 30 hours, we gain insights into the growth patterns of the bacteria population.

Derivative

Derivatives are a core topic in differential calculus. They help us understand how functions change, providing a way to compute things like speed, acceleration, and population growth.

In this exercise, \(N'(t)\) is the derivative of \(N(t)\), representing the instantaneous rate of change of the bacterial population. This derivative is obtained using rules of differentiation like the quotient rule, which is useful when dealing with ratios.

The derivative itself is a function of time \(t\), and its values can tell us valuable information:

• At \(t = 0\), \(N'(0) = 120\), indicating the initial condition where the bacteria are rapidly growing.
• At \(t = 10\), \(N'(10) = 0\), the growth stops, showing a pivotal moment where the earlier increasing trend reverses.
• At \(t = 20\) and \(t = 30\), with \(N'(20) = -36\) and \(N'(30) = -7.2\), the population is decreasing, first quickly then more slowly.

By examining these values, students can learn how derivatives describe changes in real-world scenarios.

Population Growth Analysis

Analyzing population growth involves examining both the rate of change and the curvature of these changes, which is reflected in the second derivative \(N''(t)\). This is otherwise known as acceleration in the context of motion, but when applied here, it's about the changing speed of population growth.

The second derivative provides deeper insights:

• \(N''(0) > 0\) implies an accelerating growth rate initially, supporting increasing bacterial numbers.
• \(N''(10) < 0\) shows a shift towards a decreasing growth rate, matching the results of \(N'(10) = 0\).
• With \(N''(20) < 0\) and \(N''(30) < 0\), there's a continued deceleration, indicating the population is stabilizing.

These insights connect mathematical concepts with biological phenomena, helping students understand and predict how populations might behave over time given certain conditions. Using mathematics, we can model and analyze complex biological systems effectively.