The rate of proliferation (that is, reproduction) for the cells in a tumor varies depending on the size of the tumor. The Gompertz growth model is sometimes used to model this growth. According to the Gompertz model the total number of divisions occurring in 1 hour, R, depends on the number of cells, N, through the formula given below. R(N) = N In The positive constants, d and A, both depend on the type of tumor, whether it is being treated or not, and so on. (a) Assume that d-A=1. Use a table or a graph to show that lim-R(N) = 0. Complete the table of values to show that lim-R(N)=0. (Type an integer or decimal rounded to three decimal places as needed.) N 0.1 0.01 0.001 R(N) O
The rate of proliferation (that is, reproduction) for the cells in a tumor varies depending on the size of the tumor. The Gompertz growth model is sometimes used to model this growth. According to the Gompertz model the total number of divisions occurring in 1 hour, R, depends on the number of cells, N, through the formula given below. R(N) = N In The positive constants, d and A, both depend on the type of tumor, whether it is being treated or not, and so on. (a) Assume that d-A=1. Use a table or a graph to show that lim-R(N) = 0. Complete the table of values to show that lim-R(N)=0. (Type an integer or decimal rounded to three decimal places as needed.) N 0.1 0.01 0.001 R(N) O
Chapter6: Exponential And Logarithmic Functions
Section6.8: Fitting Exponential Models To Data
Problem 3TI: Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to...
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![## Tumor Growth and Cell Proliferation
The rate of proliferation (that is, reproduction) for the cells in a tumor varies depending on the size of the tumor. The Gompertz growth model is sometimes used to model this growth. According to the Gompertz model, the total number of divisions occurring in 1 hour, \( R \), depends on the number of cells, \( N \), through the formula given below:
\[ R(N) = dN \ln\left(\frac{A}{N}\right) \]
The positive constants, \( d \) and \( A \), both depend on the type of tumor, whether it is being treated or not, and so on.
---
### Problem:
(a) Assume that \( d = A = 1 \). Use a table or graph to show that \( \lim_{{N} \to 0} R(N) = 0 \).
Complete the table of values to show that \( \lim_{{N} \to 0} R(N) = 0 \).
(Type an integer or decimal rounded to three decimal places as needed.)
#### Table of values:
\[
\begin{array}{|c|c|}
\hline
N & R(N) \\
\hline
0.1 & \\
\hline
0.01 & \\
\hline
0.001 & \\
\hline
\end{array}
\]
### Explanation:
Here we are provided with a function:
\[ R(N) = N \ln\left(\frac{1}{N}\right) \]
Our task is to fill in the values for \( R(N) \) when \( N \) takes values 0.1, 0.01, and 0.001, and observe the result as \( N \) approaches 0.
For N = 0.1:
\[ R(0.1) = 0.1 \ln\left(\frac{1}{0.1}\right) = 0.1 \ln(10) \]
For N = 0.01:
\[ R(0.01) = 0.01 \ln\left(\frac{1}{0.01}\right) = 0.01 \ln(100) \]
For N = 0.001:
\[ R(0.001) = 0.001 \ln](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F708b5043-edcc-41eb-8461-039a390d64a5%2F59e762e8-07ff-4d16-9622-e8bb5f517674%2Fw6dtt2o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Tumor Growth and Cell Proliferation
The rate of proliferation (that is, reproduction) for the cells in a tumor varies depending on the size of the tumor. The Gompertz growth model is sometimes used to model this growth. According to the Gompertz model, the total number of divisions occurring in 1 hour, \( R \), depends on the number of cells, \( N \), through the formula given below:
\[ R(N) = dN \ln\left(\frac{A}{N}\right) \]
The positive constants, \( d \) and \( A \), both depend on the type of tumor, whether it is being treated or not, and so on.
---
### Problem:
(a) Assume that \( d = A = 1 \). Use a table or graph to show that \( \lim_{{N} \to 0} R(N) = 0 \).
Complete the table of values to show that \( \lim_{{N} \to 0} R(N) = 0 \).
(Type an integer or decimal rounded to three decimal places as needed.)
#### Table of values:
\[
\begin{array}{|c|c|}
\hline
N & R(N) \\
\hline
0.1 & \\
\hline
0.01 & \\
\hline
0.001 & \\
\hline
\end{array}
\]
### Explanation:
Here we are provided with a function:
\[ R(N) = N \ln\left(\frac{1}{N}\right) \]
Our task is to fill in the values for \( R(N) \) when \( N \) takes values 0.1, 0.01, and 0.001, and observe the result as \( N \) approaches 0.
For N = 0.1:
\[ R(0.1) = 0.1 \ln\left(\frac{1}{0.1}\right) = 0.1 \ln(10) \]
For N = 0.01:
\[ R(0.01) = 0.01 \ln\left(\frac{1}{0.01}\right) = 0.01 \ln(100) \]
For N = 0.001:
\[ R(0.001) = 0.001 \ln
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