Problem 8 - Applied Calculus

Question

The size, \(S,\) of a tumor (in cubic millimeters) is given by \(S=2^{t},\) where \(t\) is the number of months since the tumor was discovered. Give units with your answers. (a) What is the total change in the size of the tumor during the first six months? (b) What is the average rate of change in the size of the tumor during the first six months? (c) Estimate the rate at which the tumor is growing at \(t=6 .\) (Use smaller and smaller intervals.)

Step-by-Step Solution

Answer

(a) 63 mm³, (b) 10.5 mm³/month, (c) 43.45 mm³/month at t=6.

Step 1: Calculate Initial and Final Size at First and Sixth Month

To find the total change in size during the first six months, calculate the size at \( t = 0 \) and \( t = 6 \). Since the function is \( S = 2^t \), at \( t = 0 \), \( S = 2^0 = 1 \) mm³. At \( t = 6 \), \( S = 2^6 = 64 \) mm³.

Step 2: Calculate Total Change in Tumor Size Over Six Months

The change in size is the difference between the initial size \( S(t=0) \) and the final size \( S(t=6) \). Thus, the total change is \( 64 - 1 = 63 \) mm³.

Step 3: Calculate Average Rate of Change Over Six Months

The average rate of change is determined by the formula \( \frac{\text{Change in Size}}{\text{Change in Time}} \). Over the first six months, this is \( \frac{64 - 1}{6 - 0} = \frac{63}{6} = 10.5 \) mm³ per month.

Step 4: Estimate Instantaneous Rate of Change at t = 6

To estimate the instantaneous rate of change, calculate the change in size over smaller intervals around \( t = 6 \). First, consider two small intervals: \( [6, 6.01] \) and \( [5.99, 6] \). For \( t = 6.01 \), \( S = 2^{6.01} \approx 64.435 \) mm³. For \( t = 5.99 \), \( S = 2^{5.99} \approx 63.566 \) mm³. Compute rates: \( \frac{64.435 - 64}{0.01} \approx 43.5 \) mm³/month and \( \frac{64 - 63.566}{0.01} \approx 43.4 \) mm³/month. Averaging these gives an estimated rate of \( 43.45 \) mm³/month at \( t = 6 \).

Key Concepts

Rate of Change

Understanding the rate of change helps us describe how quickly a quantity, like tumor size, changes over time. Think of it like the speed of growth. If a tumor grows faster, the rate is higher. In our exercise, we calculate this by comparing different sizes of the tumor at specific times. For example, if the tumor size triples, this tells us how significant the change is. Studying the rate of change is important because it allows us to:

• Predict future sizes.
• Understand the effectiveness of treatment.
• Make decisions on managing growth.

This concept is fundamental in calculus and appears in various contexts, not just in tumor growth but in velocity or economics as well.

Exponential Growth

Exponential growth refers to situations where a quantity increases at a rate proportional to its current value. This means that as something becomes larger, its rate of growth becomes faster. In our case, the tumor size modeled by the equation \( S = 2^{t} \), is growing exponentially over time.This exponential model is different from linear growth because:

• It accelerates more quickly as time goes on.
• Instead of adding a constant amount, it multiplies by a growth factor.
• This can lead to very rapid increases in size.

Exponential growth is common in natural processes, like populations or investment interest, besides biology.

Derivatives

Derivatives are a powerful tool in calculus used to understand how a function changes. They give us the instantaneous rate of change at any given point, like a speedometer measuring a car's speed at a very specific moment. For tumor growth, using derivatives allows us to see how fast the tumor is growing at exactly six months, even if it changes throughout that month.In our exercise, estimating the derivative by using smaller intervals allowed us to approximate the instantaneous growth rate at \( t = 6 \). Knowing the derivative:

• Helps in diagnosing the severity of growth.
• Provides insights into growth patterns.
• Aids in making real-time decisions in medical treatment.

This mathematical concept extends into fields like physics, engineering, and economics where understanding detailed change is crucial.

Average Rate of Change

The average rate of change gives us a simplified view of growth by looking at the change over a set interval, such as across six months. It provides a sort of "bird's-eye view," averaging the entire span rather than focusing on the details within. For our tumor:We calculated the average rate of change using the formula \( \frac{\text{Change in Size}}{\text{Change in Time}} \). This resulted in a value of \( 10.5 \) mm³ per month for the first six months. Computing average rate:

• Allows us to communicate change over time simply.
• Makes it easier to compare different periods of growth.
• Helps in setting treatment goals by looking at overall trends.

Average rates are useful in many areas like finance for determining growth of investments, or speed over a journey's duration.