Problem 75 - Contemporary Precalculus

Question

Deal with functions of the form \(f(x)=P e^{k x}\) where \(k\) is the continuous exponential growth rate (see Example 6 ). One hour after an experiment begins, the number of bacteria in a culture is \(100 .\) An hour later, there are 500 . (a) Find the number of bacteria at the beginning of the experiment and the number three hours later. (b) How long does it take the number of bacteria at any given time to double?

Step-by-Step Solution

Answer

(a) At the beginning of the experiment, there were approximately \(\frac{100}{e^{\ln{5}}}\) bacteria. Three hours later, there were 2500 bacteria. (b) The time it takes for the number of bacteria to double at any given time is \(\frac{\ln{2}}{x}\).

Step 1: Understanding the information given

One hour after the experiment begins, there are 100 bacteria, meaning \(f(1) = 100\). Two hours after the experiment begins, there are 500 bacteria, meaning \(f(2) = 500\). We need to find P and k values.

Step 2: Using given information to find P and k

We have two equations from the given information: 1. \(f(1) = Pe^{k(1)} = 100\) 2. \(f(2) = Pe^{k(2)} = 500\) Divide 2nd equation by the 1st equation to eliminate P: \(\frac{Pe^{k(2)}}{Pe^{k(1)}} = \frac{500}{100}\)

Step 3: Solving for k

After simplifying our equation we get: \(e^{k} = 5\) To solve for k, take the natural logarithm of both sides: \(k = \ln{5}\) Now we have the value of k.

Step 4: Solving for P

Using our found k value, we can plug it into our first equation: \(P e^{k(1)} = 100\) \(P e^{\ln{5}} = 100\) Now, simplify by dividing both sides by \(e^{\ln{5}}\): \(P = \frac{100}{e^{\ln{5}}}\)

Step 5: Finding the number of bacteria at the beginning

Now that we have found P, we can plug in value x=0 into our original function: \(f(0) = Pe^{k(0)} = P\) The number of bacteria at the beginning of the experiment is \(\frac{100}{e^{\ln{5}}}\).

Step 6: Finding the number of bacteria three hours later

To find the number of bacteria three hours later, use our function with the values found for P and k when x=3: \(f(3) = \frac{100}{e^{\ln{5}}} e^{\ln{5}(3)}\) Simplify the equation to get: \(f(3) = 100(5^{2})\) Three hours later, there are 2500 bacteria.

Step 7: Solving for the time it takes for the bacteria to double

To find the time it takes for the bacteria to double at any given time, set up an equation like this: \(2f(x)=f(x+k)\) Use our function with the values we found for P and k: \(2Pe^{kx}=Pe^{k(x+k)}\) Simplify the equation: \(2e^{kx}=e^{k(x+k)}\) Now take the natural logarithm of both sides and solve for k: \(k = \frac{\ln{2}}{x}\) The time it takes for the number of bacteria to double is \(\frac{\ln{2}}{x}\).

Key Concepts

Continuous Exponential Growth Rate

In mathematics, exponential growth is a fundamental concept representing how quantities increase over time, typically at a rate proportional to their current value. When discussing **continuous exponential growth**, we're referring to a scenario where growth occurs all the time, without interruption.
In such cases, the growth is expressed by the function \(f(x) = Pe^{kx}\), where:

• \(P\) is the initial quantity or principal amount,
• \(e\) is the base of natural logarithms (approximately 2.718),
• \(k\) is the continuous growth rate, and
• \(x\) is the time variable.

This equation describes how quickly the quantity grows as time progresses. It’s important to highlight that the continuous growth rate \(k\) significantly influences how rapidly or slowly the quantity expands.

To understand how to apply this in real-life scenarios, consider bacteria growth in a lab experiment. If you monitor how they multiply every moment, you'll see that the growth fits nicely into the model of continuous exponential growth. The growth rate \(k\) helps determine how aggressive or fast the bacteria culture expands over time.

Bacteria Growth

Bacteria cultures in science experiments are a classic example of exponential growth. The question often arises how to quantify this biological process, as bacteria multiply at rapid, consistent rates.

In our exercise, the growth follows the function \(f(x) = Pe^{kx}\). Biological processes like this showcase exponential growth beautifully:

• Bacteria double quickly, making it a speedy, easy-to-observe growth.
• The culture size changes dramatically over just a few hours or days, dependent on conditions.

Initially, you start with a known quantity of bacteria. One hour into our experiment, for instance, we have 100 bacteria. By comparing the number one hour later, at 500 bacteria, we can understand the growth rate. This involved dividing initial value equations like this:

• First equation: \(Pe^{k(1)} = 100\)
• Second equation: \(Pe^{k(2)} = 500\)

Using these equations, researchers identify how much and how fast growth happens. This allows predicting bacteria numbers at any given time throughout the experiment.

Understanding bacteria growth through exponential functions expands both mathematical knowledge and practical applications in fields like microbiology and medicine.

Doubling Time

An intriguing aspect of exponential growth, particularly with bacteria, is **doubling time**. This term defines the period it takes for a quantity to double in size under continuous growth.

To compute this, use the formula derived from our growth function:

• Given our function, doubling doesn’t mean simply multiplying by two. Instead, use \(2f(x)=f(x+k)\).
• This simplified function \(f\) lets experts compute exactly when doubling events occur.

Doubling time hinges heavily on \(k\), the continuous growth rate, which you calculate using logarithms: \(k = \ln{2}/x\). This shows how complex calculations provide profound insights—allowing us to predict how long it takes a population to expand.
This concept is very applicable, not only in lab experiments but at the scale of global populations or financial investments too. Understanding doubling time gives critical insights into trends and expected future outcomes.