Problem 70 - Calculus - AP Edition

Question

Suppose that the time \(T\) (in days) for a cancerous tumor to double in volume (in \(\mathrm{cm}^{3}\) ) is given by $$T=\frac{10 \ln 2}{\ln V_{1}-\ln V_{0}}\left(V_{0}

Step-by-Step Solution

Answer

The error in \(T\) is approximately \(\pm 0.352\) days.

Step 1: Understand the Given Formula

The formula given is \(T=\frac{10 \ln 2}{\ln V_1-\ln V_0}\). This formula calculates the time \(T\) it takes for the tumor to double in volume. \(V_0\) is the initial volume and \(V_1\) is the volume after a certain period (10 days in this case). We need to estimate the error in \(T\) when there is a measurement error in \(V_1\).

Step 2: Calculate \(T\) with Given Values

Substitute \(V_0 = 20 \text{ cm}^3\) and \(V_1 = 33 \text{ cm}^3\) into the equation:\[T = \frac{10 \ln 2}{\ln 33 - \ln 20} = \frac{10 \cdot 0.6931}{\ln 33 - \ln 20}\]First calculate \(\ln 33\) and \(\ln 20\):\(\ln 33 \approx 3.496\), \(\ln 20 \approx 2.996\).Then, \(T = \frac{10 \times 0.6931}{3.496 - 2.996} \approx \frac{6.931}{0.5} = 13.862\).

Step 3: Determine the Maximum Error in \(V_1\)

The volume \(V_1\) is measured with a potential error of \(\pm 0.4 \text{ cm}^3\). This means \(V_1\) could range from \(32.6 \text{ cm}^3\) to \(33.4 \text{ cm}^3\). We need to find how this variation affects \(T\).

Step 4: Estimate Maximum and Minimum Values of \(T\)

Calculate \(T\) for both extremes of \(V_1\):- For \(V_1 = 32.6 \text{ cm}^3\), \(\ln 32.6 \approx 3.486\), then\[T_{min} = \frac{10 \times 0.6931}{3.486 - 2.996} \approx \frac{6.931}{0.49} \approx 14.143\].- For \(V_1 = 33.4 \text{ cm}^3\), \(\ln 33.4 \approx 3.509\), then\[T_{max} = \frac{10 \times 0.6931}{3.509 - 2.996} \approx \frac{6.931}{0.513} \approx 13.510\].

Step 5: Calculate the Error in \(T\)

The error in \(T\) is given by the difference between \(T_{max}\) and \(T_{min}\). Calculate:\[\text{Error} = T_{max} - T = 13.510 - 13.862 \approx -0.352\].\[\text{Error} = T - T_{min} = 13.862 - 14.143 \approx -0.281\].Therefore, the estimated error is approximately \(\pm 0.352\) days.

Key Concepts

Tumor Growth Calculation

Tumor growth calculation models predict how cancerous tumors change in size over time. In our problem, we use a formula to determine the time it takes for a tumor to double in volume. This is important for understanding the tumor's behavior and planning treatment.
Our given formula is:\[T = \frac{10 \ln 2}{\ln V_1 - \ln V_0}\]This equation uses the natural logarithm, where:

• \(T\) represents the time it takes for the tumor to double in size, expressed in days.
• \(V_0\) is the initial volume of the tumor.
• \(V_1\) is the tumor's volume after a specific period, here considered as 10 days later.

By understanding these components, you can determine how quickly a tumor grows, which is vital for timely medical intervention.

Logarithmic Functions

Logarithmic functions are incredibly useful in calculus for simplifying complex expressions, especially when dealing with exponential growth, such as tumor growth. A logarithm, in simple terms, is the inverse of exponentiation.
For instance, in our formula, the terms \(\ln V_1\) and \(\ln V_0\) represent the natural logarithm of the tumor volumes at two different times. Logarithms help break down multiplication operations into simpler addition or subtraction processes.
Natural logarithms, denoted as \(\ln\), have a base of \(e\), where \(e\) is approximately 2.71828. This makes them particularly suitable for continuous growth or decay problems. In the context of our problem, the natural logarithm helps us handle the exponential nature of tumor volume growth over time.
By using logarithms, the calculation becomes more manageable, allowing for precise solutions in exponential scenarios like the one given.

Measurement Error Estimation

Measurement error estimation is critical in most scientific calculations, especially in medical scenarios dealing with tumor growth. Inaccuracies in measuring the tumor’s volume can significantly affect the predicted growth period.
Here, the volume \(V_1\) might have an error margin of \(\pm 0.4 \text{ cm}^3\). This means that the actual volume could range from 32.6 cm³ to 33.4 cm³. To estimate how this measurement error impacts the calculation of \(T\), we perform the calculation at both these boundaries.

• Minimum error: By using the lower end of the volume range, you calculate the minimum possible time \(T\).
• Maximum error: Using the upper volume range gives you the maximum time \(T\).

Finally, the range between these extrema provides an estimate of the potential error in the time calculation. This ensures that any decisions based on this growth period are made with an understanding of its precision limits.