Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

The mass of a tumor grows at a rate proportional to its size. The first measurement of its mass was \(4.0\) grams. Four months later its mass was \(6.76\) grams. How large was the tumor six months before the first measurement? If the instrument can detect tumors of mass 1 gram or greater, would the tumor have been detected at that time?

Step-by-Step Solution

Answer

The tumor was less than 1 gram six months earlier and not detectable.

Step 1: Understand the model for growth

We are given that the tumor grows at a rate proportional to its size. This indicates continuous exponential growth, which can be modeled by the equation \( m(t) = m_0 e^{kt} \), where \( m(t) \) is the mass at time \( t \), \( m_0 \) is the initial mass, and \( k \) is the growth rate constant.

Step 2: Set up the initial conditions

We have two initial conditions: at the first measurement (\( t = 0 \)), the mass is \( m_0 = 4.0 \) grams, and after four months (\( t = 4 \)), the mass is \( m(4) = 6.76 \) grams. These conditions will help us find the growth rate constant \( k \).

Step 3: Calculate the growth rate constant

Using the equation for exponential growth, we can write \( 6.76 = 4.0 e^{4k} \). Solving for \( k \) involves rearranging and taking the natural logarithm:\[ e^{4k} = \frac{6.76}{4.0} \]\[ 4k = \ln\left(\frac{6.76}{4.0}\right) \]\[ k = \frac{1}{4} \ln\left(\frac{6.76}{4.0}\right) \].

Step 4: Solve for the mass six months earlier

We want to find the mass six months before the first measurement, i.e., at \( t = -6 \). Using our model with \( m_0 = 4.0 \) and the calculated \( k \):\[ m(-6) = 4.0 e^{-6k} \]Substitute \( k \) from the previous step to find \( m(-6) \).

Step 5: Calculate and interpret the result

Substituting the value of \( k \) calculated earlier:\[ m(-6) = 4.0 e^{-6 \left( \frac{1}{4} \ln\left(\frac{6.76}{4.0}\right) \right)} = 4.0 e^{-\frac{3}{2} \ln\left(\frac{6.76}{4.0}\right)} \]This can be simplified to:\[ m(-6) = 4.0 \left(\frac{4.0}{6.76}\right)^{\frac{3}{2}} \]Evaluating this expression gives a mass less than 1 gram, meaning the tumor would not have been detected by the instrument six months before the first measurement.

Key Concepts

Tumor Growth Model

A tumor growth model provides a mathematical way to represent how a tumor increases in size over time. The key idea behind this model is that the growth rate (how fast the tumor grows) is proportional to the size of the tumor itself. This means that larger tumors grow faster than smaller ones.

- In this context, 'proportional' means that if you know the size of a tumor at one point in time, you can predict its size at another time using mathematical tools. - This model assumes continuous and exponential growth, making it particularly useful for understanding how rapidly changing quantities like tumor masses increase.

Using this model, you can start with an initial tumor mass and predict its mass at any given time, either in the future or the past, as long as the growth conditions remain constant. This approach involves using exponential growth equations, allowing for precise calculation based on initial conditions and the growth rate constant.

Growth Rate Constant

The growth rate constant, often represented by the letter \( k \), is a vital part of understanding exponential growth in any biological system, including tumor growth. This constant provides a measure of how quickly the process (in this case, the tumor growing) occurs. Essentially, it determines the steepness of the growth curve.

- A larger growth rate constant means the tumor grows faster.- If \( k \) is small, the tumor will grow more slowly.

To calculate \( k \), you need two different measurements of the tumor's mass at two different points in time. For example, in the equation \( m(t) = m_0 e^{kt} \), using known mass values such as the initial mass \( m_0 \) at time zero and the mass \( m(t) \) at a later time point can help solve for \( k \).

This calculation often involves using the natural logarithm because of the exponential nature of the equation, making it an essential step in solving exponential growth problems.

Exponential Functions

Exponential functions are mathematical tools used to describe situations where a quantity grows or decays at a rate proportional to its current size. They are typically expressed in the form \( m(t) = m_0 e^{kt} \), where:

• \( m(t) \) is the mass at time \( t \).
• \( m_0 \) is the initial mass.
• \( e \) is a constant approximately equal to 2.718.
• \( k \) is the growth rate constant.

Exponential functions are powerful because they can model natural processes such as population growth, radioactive decay, and, as in our example, tumor growth.

The defining feature of exponential functions is how they grow increasingly fast as time passes. The rate of growth accelerates, meaning the time it takes to grow from one size to another can get shorter as the size increases. This makes calculation with exponential functions crucial in predicting future growth or retrodicting past conditions.

Understanding these functions helps in predicting the mass of a tumor at various points in time, simply by solving the equation for the desired \( t \). This is practical for determining whether it was detectable at an earlier stage given a detection threshold, like knowing if it had reached 1 gram six months before.