Question

A remote lake that previously contained no northern pike is stocked with these fish. The population, \(P\), of northern pike after \(t\) years can be determined by the equation \(P=10000(1.035)^{t}\) a) How many northern pike were put into the lake when it was stocked? b) What is the annual growth rate, as a percent? c) How long will it take for the number of northern pike in the lake to double?

Short answer

a) 10000 pike.b) 3.5% annual growth rate.c) Approximately 20.14 years.

Step-by-step solution

Identify the initial population

To determine how many northern pike were initially put into the lake, evaluate the population equation when \( t = 0 \). Substitute \( t = 0 \) into the equation \( P = 10000(1.035)^t \) which gives \( P = 10000(1.035)^0 \). Simplifying this, we find \( P = 10000 \).

Determine the population initially stocked

From Step 1, \(P = 10000\). Therefore, 10000 northern pike were put into the lake when it was stocked.

Identify the annual growth factor

The annual growth factor is given within the equation as the base of the exponent \(1.035\).

Convert the growth factor to a percentage

To convert the annual growth factor to a percentage, subtract 1 from the growth factor and then multiply by 100:-\( 1.035 - 1 = 0.035 \)-\( 0.035 \times 100 = 3.5\% \). Thus, the annual growth rate is 3.5%.

Setup the equation for population doubling

To find the time it takes for the northern pike population to double, set the population equation to twice the initial population. This yields the equation: \(2 \times 10000 = 10000(1.035)^t \). Simplify this to find \( 2 = (1.035)^t \).

Solve for t using logarithms

Take the natural logarithm of both sides to solve for \( t \):-\( \ln(2) = \ln((1.035)^t) \)-Use the logarithm power rule: \( \ln(2) = t \cdot \ln(1.035) \)-Solve for \( t \):\[ t = \frac{\ln(2)}{\ln(1.035)} \approx \frac{0.6931}{0.0344} \approx 20.14 \]Thus, it will take approximately 20.14 years for the northern pike population to double.

Key concepts

Exponential Functions

Exponential functions describe situations where quantities grow or decay at a constant relative rate. In our exercise, we see an exponential function used to model the population of northern pike in a lake. The general form of an exponential function is given by: ewline ewline ewline- ewline - ewline ewline\( P(t) = P_0 e^{rt} \) ewline , ewline ewlinewhere: - P(t) is the population at time t, - P_0 is the initial population, - r is the growth rate, and - t is the time. ewline - . ewlineHowever, our given equation is \(P=10000(1.035)^t\), which all follows the same principles. Here, 1.035 is the base that represents the growth factor per year. In such cases, exponential functions are very useful for modeling population growth, nuclear decay, and many other real-world phenomena. ewlineewline Notice that the population grows exponentially, meaning as time progresses, the population increases faster and faster.

Initial Population

To understand the initial population using the exponential equation, we note that it’s the value of the population at time zero (when \(t=0\)). In our example, the initial population of northern pike is given by evaluating the equation \( P = 10000(1.035)^t\) at \( t=0 \): ewline ewline ewline - \(P(0) = 10000(1.035)^0=10000ewline- ewline - ewline ewline ewline =10000\). ewlineSo 10,000 northern pike were initially stocked in the lake. Tracking the initial population is crucial because it forms the basis for all future growth calculations in exponential growth models. ewlineFor instance, if the initial population or number were different, the results for future projections would change.

Growth Rate

The growth rate in exponential functions determines how quickly the quantity (population in this case) increases over time. The growth rate is typically a decimal in the formula. ewlineIn our exercise, the annual growth rate is embedded in the base 1.035. To find the growth rate as a percentage: ewline- Subtract 1 from the base ewline- \(1.035 - 1 = 0.035 ewline- Multiply the result by 100 to get a percentage ewline- \)0.035 \times 100 = 3.5ewline ewlineSo the growth rate is 3.5% per year. ewlineUnderstanding the growth rate is vital for analyzing how a population or any other quantity changes over time. It shows how the quantity expands or contracts from period to period.

Logarithms

Logarithms are the inverse operations of exponentiation and are extremely useful in solving equations involving exponential growth. When we need to find out the time (t) required for our population to double, we use logarithms. ewlineFor our problem, to find how long it takes for the northern pike population to double, we set up the equation ewline ewline10 ewline2 \times 10000 = 10000(1.035)^t, ewlinewhich simplifies to - (1.035)^t=2. Then, we apply the natural logarithm to isolate t: ewline ewline - ewline \(ewline ln(2) = ln((1.035)^t) newline ewline newline newlineApplying logarithm properties, we get newline}t \times ln(1.035) = ln(2)\) ewline-. ewlineFinally, solve for t ewline - = -ewline \(t = ln(2) / ln(1.035) ≈ 20.14\). It takes about 20.14 years for the population to double. ewlineUnderstanding the concepts of logarithms makes manipulating and solving exponential equations simpler and more intuitive.