Problem 59 - Calculus an Applied Approach

Question

Population Growth A population of bacteria is introduced into a culture. The number of bacteria \(P\) can be modeled by $$P=500\left(1+\frac{4 t}{50+t^{2}}\right)$$ where \(t\) is the time (in hours). Find the rate of change of the population when \(t=2\).

Step-by-Step Solution

Answer

After doing the calculation in Step 5, you find that the rate of change of the bacterial population when \(t=2\) is 1705 bacteria per hour.

Step 1: Identify the function

In this problem, the given function representing bacteria population growth is \(P=500\left(1+\frac{4 t}{50+t^{2}}\right)\). We'll have to calculate the derivative of this function as to find its rate of change.

Step 2: Simplify initial function

The first step in calculating the derivative would be to simplify the function into a form that is easier to differentiate. Multiply 500 through the parentheses to get \( P=500+ \frac{2000t}{50+t^{2}} \)

Step 3: Apply the Quotient Rule

The Quotient Rule states that the derivative of \(\frac{f}{g} = \frac{g*f' - f*g'}{g^2}\). Here \(f = 2000t\) and \(g = 50 + t^2\). Calculate \(f'\) and \(g'\) (the derivatives of \(f\) and \(g\) respectively). We get \(f' = 2000\) and \(g' = 2t\). Applying the quotient rule we get \( \frac{dP}{dt} = \frac{(50+t^{2})*2000 -2000t*2t}{(50+t^{2})^2} \)

Step 4: Simplify the derivative function

Simplify derivative equation to get it into a manageable form. After simplification, we get \(\frac{dP}{dt} = \frac{100000-3000t^2}{(50+t^2)^2}\)

Step 5: Substitute the given time

Finally, substitute \(t = 2\) into the derivative of the function to get the rate of change at \(t = 2\). You get, \(\frac{dP}{dt} = \frac{100000-3000*(2^2)}{(50+(2)^2)^2}\).

Key Concepts

Bacteria Population Modeling

Understanding the dynamics of bacteria growth is a fundamental aspect of biology and environmental science. The simplified model used here is a typical representation of how populations can grow and interact with their environments. It's formulated as
\[\begin{equation} P=500\bigg(1+\frac{4t}{50+t^2}\bigg) \end{equation}\] where the variable \(P\) stands for the population size and \(t\) represents time in hours. Here, the growth of the bacteria is affected both by a linear factor of time \(4t\) and a quadratic term \(t^2\), indicating a more complex growth behavior than simple exponential or logistic models. In reality, bacteria growth can be influenced by numerous factors like availability of nutrients, temperature, and competition. While our model is simplified, it's essential to recognize such models are starting points for understanding how populations can change over time. The integration of calculus, through the derivative, provides us a powerful tool to analyze the rates at which such changes occur.

Calculus Derivative Application

Calculus is widely used in biological sciences to determine rates of change, such as the growth rate of a bacterial population. In this exercise, we're applying calculus by taking the derivative of the population model. To find the derivative, \[\begin{equation} \frac{dP}{dt} \end{equation}\]we look at how a tiny increment in time, \(dt\), affects the change in the population, \(dP\). It gives us the 'instantaneous' rate of growth at any point in time, which is crucial for predicting future population sizes or understanding when a population reaches a certain level. When the derivative is positive, the population is increasing. Conversely, if the derivative is negative, the population is decreasing. If the derivative equals zero, it indicates a stationary point which can signify a peak, trough or inflection point in population growth, thus serving as a valuable indicator of population dynamics over time.

Quotient Rule in Differentiation

The quotient rule is a technique in calculus used when differentiating ratios or fractions of functions. It is expressed as \[\begin{equation} \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{[g(x)]^2} \end{equation}\]where \(f(x)\) represents the numerator and \(g(x)\) the denominator of our function. To use the quotient rule, we differentiate both the numerator and the denominator separately, termed as \(f'(x)\) and \(g'(x)\) respectively, and plug them into the formula. This rule allows us to handle complex rational expressions like those found in our population model.In our exercise, applying the quotient rule was essential to find the rate of change of the bacteria population. While it initially seems daunting, this methodical approach breaks down the problem into simpler steps, making complex differentiation manageable. The calculated derivative provides valuable insights into the behavior of the bacterial population at any given time.