91Ó°ÊÓ

It takes 3 months for a malignant lung tumor to double in size. At the time a lung tumor was detected in a patient, its mass was 10 grams. a. If untreated, determine the size in grams of the tumor at each of the listed times in the table. Find a formula to express the tumor mass \(M\) (in grams) at any time \(t\) (in months). $$ \begin{array}{cc} \hline t, \text { Time (months) } & M, \text { Mass }(\mathrm{g}) \\ \hline 0 & 10 \\ 3 & \\ 6 & \\ 9 & \\ 12 & \\ \hline \end{array} $$ b. Lung cancer is fatal when a tumor reaches a mass of 2000 grams. If a patient diagnosed with lung cancer went untreated, estimate how long he or she would survive after the diagnosis. c. By what percentage of its original size has the 10 -gram tumor grown when it reaches 2000 grams?

Step by step solution

01

- Understanding the Doubling Time

The tumor doubles in size every 3 months. This means if the tumor is 10 grams at month 0, it will be 20 grams at month 3, 40 grams at month 6, and so on.

02

- Writing the General Formula

Since the tumor doubles every 3 months, we can express the tumor mass as a function of time mathematically. The general formula to represent exponential growth is: \[ M(t) = M_0 \times 2^{(t/3)} \] where \( M_0 \) is the initial mass (10 grams) and \( t \) is the time in months.

03

- Calculating Mass at Given Times

Using the formula \[ M(t) = 10 \times 2^{(t/3)} \]we can calculate the mass at each given time: \( t = 0 \): \( M(0) = 10 \) grams \( t = 3 \): \( M(3) = 10 \times 2^1 = 20 \) grams \( t = 6 \): \( M(6) = 10 \times 2^2 = 40 \) grams \( t = 9 \): \( M(9) = 10 \times 2^3 = 80 \) grams \( t = 12 \): \( M(12) = 10 \times 2^4 = 160 \) grams

04

- Estimating Time to Reach Fatal Mass

To find out when the tumor will reach 2000 grams, use the formula: \[ 2000 = 10 \times 2^{(t/3)} \] Divide both sides by 10: \[ 200 = 2^{(t/3)} \] Taking the logarithm base 2 of both sides: \[ \frac{t}{3} = \text{log}_2(200) \] Since \( \text{log}_2(200) \approx 7.64 \), \[ \frac{t}{3} = 7.64 \] Thus, \( t = 7.64 \times 3 \approx 22.92 \) months. Therefore, the patient would survive approximately 22.92 months if untreated.

05

- Calculating Percentage Growth

The initial mass is 10 grams and the final mass is 2000 grams. The percentage growth can be calculated by: \[ \text{Percentage Growth} = \frac{\text{Final Mass} - \text{Initial Mass}}{\text{Initial Mass}} \times 100 \] Substitute the values: \[ \text{Percentage Growth} = \frac{2000 - 10}{10} \times 100 = 19900\text{%} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

doubling time

Doubling time is the period of time it takes for a quantity to double in size or value. In this exercise, the lung tumor doubles in size every 3 months. This kind of growth is common in many natural processes like population growth or viral infections.
The concept is simple: if a tumor starts at 10 grams, after 3 months, it will be 20 grams. After another 3 months, it will be 40 grams, and so on.
Understanding doubling time helps in predicting future growth and making necessary interventions for exponential growth processes.

logarithms

A logarithm is the inverse operation to exponentiation. While exponentiation involves raising a number to a power, a logarithm answers the question: to what exponent must the base be raised, to produce a given number?
In the exercise, we use logarithms to determine the time it takes for the tumor to reach a certain size. Using base 2 logarithms (\text{log}_2), we can solve for the time \(t\) using the equation: \ 200 = 2^{(t/3)} \ Taking logarithms on both sides, we get: \ \ \frac{t}{3} = \text{log}_2(200). \ Therefore, \(t = 7.64 \times 3 = 22.92\) months.
Logarithms help simplify solving exponentials by converting them into a linear form that is easier to work with.

exponential functions

Exponential functions are mathematical functions of the form \(f(x) = a \times b^{x}\), where the base \(b\) is a constant, and \(x\) is the exponent. They are used to model situations where growth is multiplicative, meaning the rate of growth is proportional to the current value.
In our lung tumor growth scenario, the exponential function is: \(M(t) = M_0 \times 2^{(t/3)}\). Here, \(M_0 = 10\) grams, and the base \(2\) indicates that the tumor doubles every 3 months.
Exponential growth can quickly lead to large numbers and often requires immediate attention or intervention in contexts like medicine, finance, or population studies.

percentage growth

Percentage growth measures the rate of increase from the original value to the new value. It is calculated using the formula: \( \text{Percentage Growth} = \frac{\text{Final Mass} - \text{Initial Mass}}{\text{Initial Mass}} \times 100\).
In the problem, the initial mass of the tumor is 10 grams and the final mass is 2000 grams. So, the percentage growth is: \( \frac{2000 - 10}{10} \times 100 = 19900\text{\%}\).
This means the tumor grows by 19900% from its initial size when it reaches 2000 grams. This massive increase underscores the urgency of addressing exponential growth, particularly in medical conditions like cancer.