Problem 23 - Thomas Calculus in SI Units

Question

The analysis of tooth shrinkage by C. Loring Brace and colleagues at the University of Michigan's Museum of Anthropology indicates that human tooth size is continuing to decrease and that the evolutionary process did not come to a halt some 30,000 years ago, as many scientists contend. In northern Europeans, for example, tooth size reduction now has a rate of \(1 \%\) per 1000 years. a. If \(t\) represents time in years and \(y\) represents tooth size, use the condition that \(y=0.99 y_{0}\) when \(t=1000\) to find the value of \(k\) in the equation \(y=y_{0} e^{k t} .\) Then use this value of \(k\) to answer the following questions. b. In about how many years will human teeth be \(90 \%\) of their present size? c. What will be our descendants' tooth size 20,000 years from now (as a percentage of our present tooth size)?

Step-by-Step Solution

Answer

k \approx -0.00001005; 90% size in ~10583 years; 81.7% size in 20,000 years.

Step 1: Formulate the equation for tooth size

We know from the problem that the tooth size decreases at a rate of 1% every 1000 years. Given the equation for exponential decay is \( y = y_0 e^{kt} \), and using the condition \( y = 0.99 y_0 \) when \( t = 1000 \), we can plug in these values to find \( k \).

Step 2: Substitute given condition to find value of k

Substitute the known values (\( y = 0.99 y_0 \) and \( t = 1000 \)) into the equation: \( 0.99 y_0 = y_0 e^{1000k} \). Simplifying, this gives \( 0.99 = e^{1000k} \).

Step 3: Solve for k

Take the natural logarithm (\( \ln \)) of both sides: \( \ln(0.99) = 1000k \). Solving for \( k \), we find \( k = \frac{\ln(0.99)}{1000} \approx -0.00001005 \).

Step 4: Calculate years for 90% tooth size

Using \( k \) from Step 3, substitute in the equation \( 0.9 = e^{kt} \) to find \( t \) when \( y = 0.90 y_0 \). We get \( \ln(0.9) = kt \), solving for \( t \) gives \( t \approx \frac{\ln(0.9)}{-0.00001005} \approx 10583 \text{ years} \).

Step 5: Calculate tooth size in 20,000 years

Using the same decay equation, substitute \( t = 20000 \) to find \( y = y_0 e^{20000k} \). Substitute \( k \), we have \( y = y_0 e^{20000 \times -0.00001005} \equiv y = y_0 \cdot 0.817 \). Hence, in 20,000 years, the tooth size will be approximately \(81.7\%\) of the current size.

Key Concepts

Exponential Functions

Exponential functions are a type of mathematical function commonly used to model situations where a constant rate of growth or decay takes place. An exponential function can be represented in the form:

• \( y = y_0 e^{kt} \)

In this equation, \(y\) represents the final amount, \(y_0\) is the initial amount, \(e\) is the base of the natural logarithm, and \(kt\) is the exponent, where \(k\) is the rate of growth or decay, and \(t\) is time. This form of the function indicates either growth or decay, depending on the sign of \(k\).
For instance, in the context of the shrinking tooth size problem, the function models a decreasing trend, with \(k\) being negative due to the decay process. By substituting \(t\) with the given time value and solving for \(k\), one can determine the rate of decay over a specified time period.

Natural Logarithm

The natural logarithm, denoted \(\ln\), is the logarithm to the base \(e\), which is an irrational number approximately equal to 2.71828. It plays a crucial role when dealing with exponential functions, especially when the exponent needs to be isolated or simplified.
When solving exponential decay problems, like determining the rate at which something diminishes, the natural logarithm can help simplify equations. In a scenario where you have \( e^{kt} = 0.99 \), taking the natural logarithm of both sides assists in isolating \(kt\) so that you can solve for \(k\).
For example, if you take the natural logarithm of both sides of the equation \(0.99 = e^{1000k}\), you end up with \(\ln(0.99) = 1000k\). This allows us to rearrange and solve for the decay constant \(k\) efficiently.

Decay Rate

The decay rate is a measure indicating how rapidly a quantity decreases over time. In a scenario where an exponential function models decay, the decay rate \(k\) is critical. A negative \(k\) represents a decrease or shrinkage in the quantity being measured.
In our tooth size example, the decay rate \(k\) was determined by understanding that tooth size decreases by 1% over 1000 years. By using this information, we substituted into the exponential function the given conditions and solved for \(k\) as shown in the solution steps.
This decay rate tells us how quickly the tooth size decreases over any given timeframe. For further use, knowing \(k\) allows us to predict future tooth sizes or find out past sizes by adjusting the time variable \(t\) in the formula. Therefore, understanding and calculating \(k\) is crucial for modeling and predicting decay in various fields, such as biology and physics.