Problem 33 - Calculus and its applications

Question

Superman comic book. In August \(2014,\) a 1938 comic book featuring the first appearance of Superman sold at auction for a record price of \(\$ 3.2\) million. The comic book originally cost 104 (\$0.10). Use the two data points \((0, \$ 0.10)\) and \((76, \$ 3,200,000),\) and assume that the value \(V\) of the comic book has grown exponentially, as given by \(\frac{d V}{d t}=k V\). (In the summer of \(2010,\) a family faced foreclosure on their mortgage. As they were packing, they came across some old comic books in the basement, and one of them was a copy of this first Superman comic. They sold it and saved their house.) a) Find the function that satisfies this equation. Assume that \(V_{0}=\$ 0.10\) b) Estimate the value of the comic book in 2020 . c) What is the doubling time for the value of the comic book? d) After what time will the value of the comic book be \(\$ 30\) million, assuming there is no change in the growth rate?

Step-by-Step Solution

Answer

In 2020, the comic book is valued around \$4.01 million. The doubling time is about 4.62 years. It will reach \$30 million in approximately 95 years.

Step 1: Understand the relationship - Exponential Growth

The problem mentions that the value \(V\) of the comic book grew exponentially. This is modeled by the differential equation \(\frac{dV}{dt} = kV\), where \(k\) is a constant. The solution to this differential equation is \(V(t) = V_0 e^{kt}\), where \(V_0\) is the initial value at time \(t = 0\).

Step 2: Determine the Growth Constant – Using Initial Conditions

We are given two points: \((0, \\( 0.10)\) and \((76, \\) 3,200,000)\). Use the formula \(V(t) = V_0 e^{kt}\), with \(V_0 = 0.10\) at \(t=0\) and \(V(76) = 3,200,000\). Set up the equation: \(3,200,000 = 0.10 e^{76k}\).

Step 3: Solve for the growth constant \(k\)

From the equation \(3,200,000 = 0.10 e^{76k}\), take the natural logarithm on both sides to solve for \(k\): \[ \ln(3,200,000) - \ln(0.10) = 76k \]Solve for \(k\):\[ k = \frac{\ln(3,200,000) - \ln(0.10)}{76} \].

Step 4: Find the function \(V(t)\)

After finding \(k\), the growth function is \(V(t) = 0.10 e^{kt}\). Substitute \(k\) from Step 3 to get the specific function for \(V(t)\).

Step 5: Estimate the value in 2020

With \(t = 2020 - 1938 = 82\) years, plug this into the growth function: \[ V(82) = 0.10 e^{k \cdot 82} \] Calculate \(V(82)\) using the specific \(k\) value found earlier.

Step 6: Calculate the Doubling Time

By setting \(V(t) = 2V_0\), we find the doubling time \(T_d\) using the formula: \[ 2 = e^{kT_d} \] Taking natural logs gives: \[ T_d = \frac{\ln(2)}{k} \].

Step 7: Find when the value will be \$30 million

To find the time when \( V(t) = 30,000,000 \), use the equation: \(30,000,000 = 0.10 e^{kt}\). Take the natural log to solve for \(t\):\[ t = \frac{\ln(30,000,000) - \ln(0.10)}{k} \].

Key Concepts

Differential Equations

Differential equations are a fundamental tool in mathematics used to describe the relationship between a function and its derivatives. In the context of exponential growth, these equations model how quantities change over time. In our comic book example, the differential equation \(\frac{dV}{dt} = kV\) expresses how the value \(V\) of the comic book changes relative to itself over time. Here, \(k\) is a growth constant that quantifies the rate of this change. This type of modeling is essential in capturing exponential phenomena, seen in various fields such as biology for population growth, physics for radioactive decay, and even in economics for compound interest calculations.

The general solution to this differential equation is \(V(t) = V_0 e^{kt}\), where \(V_0\) is the initial value of the function, and \(e\) is Euler's number, approximately 2.71828. This provides a formula by which we can predict future values of \(V(t)\) based on its current state and the rate of growth. Understanding and solving these equations is crucial for making accurate predictions in various real-world applications.

Doubling Time

Doubling time is a concept often used in exponential growth to determine how long it takes for a quantity to double its value. In our Superman comic book scenario, we can calculate the doubling time of its value by using the formula \(T_d = \frac{\ln(2)}{k}\). Here, \(k\) is the growth constant that we derived earlier from the differential equation. This formula provides a quick way to find the point in time when the initial value would have doubled under continuous growth conditions.

Understanding doubling time is helpful in many fields:

• In finance, it helps investors calculate how long it will take for an investment or savings to grow to twice its size.
• In demographics, it helps researchers predict population growth milestones.
• In ecology, it assists in estimating when a given species' population might require expanded resources or habitat spaces.

By knowing the doubling time, individuals and businesses can plan ahead and make informed decisions based on forecasted growth patterns.

Financial Mathematics

Financial mathematics includes a range of techniques and formulas to manage, plan, and make decisions about money. Exponential growth is a key concept within financial mathematics, essential for understanding investments, savings, and loans. In our case with the Superman comic book, the exponential growth model shows how an initial small investment of \(\$0.10\) soared to millions of dollars, mimicking processes seen in real estate values, stock growth, and bond yields.

In finance, the formula \(V(t) = V_0 e^{kt}\) is akin to the compound interest formula \(A = P(1 + r/n)^{nt}\), where \(A\) is the amount of money accumulated after \(n\) years, including interest, \(P\) is the principal amount, \(r\) is the annual interest rate, and \(n\) is the number of times that interest is compounded per year. While compound interest and exponential growth models are not identical, they both explain how initial investments can grow significantly over time given a steady growth rate.

By mastering these financial concepts, individuals can better navigate loan agreements, savings plans, and investment strategies, ensuring sound financial decisions and growth.