Problem 45 - Calculus Early Transcendentals

Question

The same exponential growth function can be written in the forms \(y(t)=y_{0} e^{k t}, y(t)=y_{0}(1+r)^{t}\) and \(y(t)=y_{0} 2^{t / T_{2}} .\) Write \(k\) as a function of \(r, r\) as a function of \(T_{2}\), and \(T_{2}\) as a function of \(k\).

Step-by-Step Solution

Answer

Question: Identify the relationships between the parameters k, r, and T₂ for the given exponential growth functions. Answer: The relationships between the parameters are: 1. k = ln(1+r) 2. r = e^(ln(2)/T₂) - 1 3. T₂ = ln(2)/k

Step 1: Write \(k\) as a function of \(r\).

Equating the first two forms of the exponential growth function, we have: \(y_{0}e^{kt} = y_{0}(1+r)^t\). Divide both sides by \(y_0\): \(e^{kt} = (1+r)^t\). Now, take the natural logarithm of both sides: \(kt = t\ln(1+r)\). Divide both sides by \(t\): \(k = \ln(1+r)\). This gives us \(k\) as a function of \(r\): \(k = \ln(1+r)\).

Step 2: Write \(r\) as a function of \(T_{2}\).

Equating the second and third forms of the exponential growth function, we have: \(y_{0}(1+r)^t = y_{0}2^{t/T_{2}}\). Divide both sides by \(y_0\): \((1+r)^t = 2^{t/T_{2}}\). Now, take the natural logarithm of both sides: \(t\ln(1+r) = \frac{t}{T_{2}}\ln(2)\). Divide both sides by \(t\): \(\ln(1+r) = \frac{\ln(2)}{T_{2}}\). To solve for \(r\), use the exponential function: \(1+r = e^{\frac{\ln(2)}{T_{2}}}\). Subtract 1 from both sides: \(r = e^{\frac{\ln(2)}{T_{2}}} - 1\). This gives us \(r\) as a function of \(T_{2}\): \(r = e^{\frac{\ln(2)}{T_{2}}} - 1\).

Step 3: Write \(T_{2}\) as a function of \(k\).

Equating the first and third forms of the exponential growth function, we have: \(y_{0}e^{kt} = y_{0}2^{t/T_{2}}\). Divide both sides by \(y_0\): \(e^{kt} = 2^{t/T_{2}}\). Now, take the natural logarithm of both sides: \(kt = \frac{t}{T_{2}}\ln(2)\). Divide both sides by \(t\): \(k = \frac{\ln(2)}{T_{2}}\). To solve for \(T_{2}\), divide both sides by \(\ln(2)\): \(T_{2} = \frac{\ln(2)}{k}\). This gives us \(T_{2}\) as a function of \(k\): \(T_{2} = \frac{\ln(2)}{k}\). In summary, the relationships between the parameters are: 1. \(k = \ln(1+r)\). 2. \(r = e^{\frac{\ln(2)}{T_{2}}} - 1\). 3. \(T_{2} = \frac{\ln(2)}{k}\).

Key Concepts

Natural Logarithm

Understanding the natural logarithm is crucial when dealing with exponential growth functions. It is denoted by \(\ln\) and is a special logarithm where the base is the mathematical constant \(e\), approximately equal to 2.71828. This number \(e\) arises naturally in mathematics and is the base of the natural logarithm because of its unique properties in calculus, particularly concerning growth processes and interest calculations.

The natural logarithm has the defining property that \(\ln(e) = 1\), and conversely, that \(e\) raised to the power of a natural logarithm of a number \(x\) gives the number itself: \(e^{\ln(x)} = x\). This interplay between exponential and logarithmic functions allows us to write exponential equations in logarithmic form, thus simplifying the process of solving them, as seen in our exercise.

• In the exercise, we took the natural logarithm of both sides to isolate the variable.
• We used the property that the logarithm of a number to a power is the power times the logarithm of the number: \(\ln(a^b) = b\cdot\ln(a)\).
• This allowed us to solve for the parameters of interest: \(k\), \(r\), and \(T_{2}\).

Exponential Equations

Exponential equations are equations in which variables occur as exponents. They take the general form \(y = ab^{x}\), where \(a\) is a constant, \(b\) is the base, and \(x\) is the exponent. In our exercise, we've dealt with the classic form of exponential growth functions, which model situations where growth rate is proportional to the size of the function at any point in time.

These equations can often be transformed into a more manageable form by using logarithms, particularly the natural logarithm. Here's how we applied this approach:

• We started with the equation \(y(t) = y_{0}e^{kt}\) and used the natural logarithm to compare it to other forms of the same function.
• By applying \(\ln\) to both sides, we could linearize the equation, thereby isolating the growth rate constant \(k\) or the rate \(r\).
• This manipulation often simplifies the process, making it easier to solve for the unknown variables.

Exponential and Logarithmic Functions

Exponential and logarithmic functions are mathematical inverses of one another, which means that they can undo each other's operations. Exponential functions deal with situations where a quantity grows at a rate proportional to its current value, which leads to a rapid increase of the quantity over time. Logarithmic functions, on the other hand, help us determine the 'time' it takes for the exponential function to reach a certain level.

In the context of our exponential growth problem:

• We recognized the connections between different forms of the exponential function and successfully transitioned from one form to another using the properties of these functions.
• When we encounter \(y_{0}(1+r)^t\), a compound interest formula, we can relate it to the other forms to extract more information and solve for variables.
• The reciprocal relationship between exponential and logarithmic functions became the key tool in deriving expressions for \(k\), \(r\), and \(T_{2}\) with respect to each other.

Understanding these concepts enables students to tackle a wide range of problems involving exponential growth and decay, as well as to comprehend complex financial calculations and population dynamics.