Problem 26 - Calculus

Question

Put the functions in the form \(P=P_{0} e^{k t}\). $$P=10(1.7)^{t}$$

Step-by-Step Solution

Answer

The function in the form \(P = P_{0}e^{kt}\) is \(P = 10e^{0.5306t}\).

Step 1: Identify Given Function

The given function is \(P=10(1.7)^{t}\). This is an exponential function where the current population is given in terms of the initial amount and the growth factor per unit time \(t\).

Step 2: Rewrite The Function in Exponential Form

We want to put the function in the form \(P = P_{0} e^{kt}\). The general exponential function \(a^t\) can be rewritten in terms of \(e\) by using the property of exponents: \(a^t = e^{t \ln a}\).

Step 3: Apply Transformation

The base of our function is \(1.7\). We replace \(1.7^t\) with \(e^{t \ln 1.7}\). Therefore, the function becomes \(P = 10 \cdot e^{t \ln 1.7}\).

Step 4: Simplify the Exponent

Let \(k = \ln 1.7\). Calculate \(k\) using a calculator: \(\ln 1.7 \approx 0.5306\). Hence, the function is \(P = 10 e^{0.5306t}\).

Step 5: Verify Form

Verify that the function now matches the form \(P = P_{0} e^{kt}\). Here, \(P_{0} = 10\) and \(k \approx 0.5306\), which satisfies the required form.

Key Concepts

Exponential Function

Exponential functions are a type of mathematical function where the variable is an exponent. They grow by a constant factor over equal intervals. This unique property makes them perfect for modeling phenomena such as population growth, radioactive decay, and interest calculations.
For any exponential function, you can usually write it in the form \(f(t) = a \, b^{t}\), where:

• \(a\) is a constant
• \(b\) is the base (growth factor)
• \(t\) is the variable

In an exponential growth scenario, \(b\) is greater than 1, indicating growth. Conversely, if \(b\) is between 0 and 1, this signifies a decay.
To link back to our problem, the function \(P=10(1.7)^{t}\) represents exponential growth since the base \(1.7\) is greater than 1.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\), where \(e\) is a mathematical constant approximately equal to 2.71828. It plays a crucial role in converting exponential functions into a format involving \(e\).
The expression \(a^t\) can be rewritten using the natural logarithm as \(e^{t \ln a}\). This transformation helps simplify calculations and solve equations involving exponential functions.
For example, in the original problem, we convert \((1.7)^{t}\) to \(e^{t \ln 1.7}\). Here, \(\ln 1.7\) is found using a calculator and is approximately \(0.5306\). This simplification uses the natural logarithm to express the growth factor in terms of \(e\), illustrating its essential bridging function.

Transformation of Functions

Transformation of functions involves altering the base form of a function to express it in a different but equivalent format. This is common with exponential functions to express them using the base \(e\).

• The transformation focuses on changing the base of an exponential function.
• This makes analysis and solving equations easier, as \(e\) is a natural constant used frequently in calculus.

In our problem, \((1.7)^{t}\) is transformed to \(e^{t \ln 1.7}\). This transformation is key to rewriting the function into the standard form \(P=P_{0} e^{kt}\).
Understanding transformations helps in seeing the equivalence of different mathematical expressions and leveraging tools like the natural logarithm for simplification.

Initial Value

In exponential functions, the initial value is the starting amount before any changes have occurred. It is represented by \(P_{0}\) in the standard exponential form \(P = P_{0} e^{kt}\).
The initial value signifies the quantity at the starting point (when \(t = 0\)). It is crucial for establishing the baseline from which the growth or decay initiates.
In our example, the function \(P = 10(1.7)^{t}\) reveals an initial value \(P_{0} = 10\). As time progresses, this value grows by the factor of \(1.7\) for each unit of \(t\). Thus, the initial value is a fundamental part of understanding how the exponential relationship develops over time.