Tumor growth The growth of cancer tumors may be modeled by

Chapter 7, Problem 58E

(choose chapter or problem)

QUESTION:

Tumor growth The growth of cancer tumors may be modeled by the Gompertz growth equation. Let M(t) be the mass of the tumor for $t \geq 0$. The relevant initial value problem is

$\frac{d M}{d t}=-a M \ln \left(\frac{M}{K}\right), \quad \ M(0)=M_{0}$

where a and K are positive constants and $0<M_{0}<K$.

a. Graph the growth rate function $R(M)=-a M \ \ln \left(\frac{M}{K}\right)$ assuming a = 1 and K = 4. For what values of M is the growth rate positive? For what value of M is the growth rate a maximum?

b. Solve the initial value problem and graph the solution for a = 1, K = 4, and $M_{0}=1$. Describe the growth pattern of the tumor. Is the growth unbounded? If not, what is the limiting size of the tumor?

c. In the general equation, what is the meaning of K?

Questions & Answers

QUESTION:

Tumor growth The growth of cancer tumors may be modeled by the Gompertz growth equation. Let M(t) be the mass of the tumor for $t \geq 0$. The relevant initial value problem is

$\frac{d M}{d t}=-a M \ln \left(\frac{M}{K}\right), \quad \ M(0)=M_{0}$

where a and K are positive constants and $0<M_{0}<K$.

c. In the general equation, what is the meaning of K?

ANSWER:

Problem 58E

Tumor growth
The growth of cancer tumors may be modeled by the Gompertz growth equation. Let M(t)be the mass of the tumor for t ≥ 0. The relevant initial value problem is
$\frac{dM`}{dt}=-aMln\left({\frac{M}{K}}\right),\ M\left({0}\right)={M}_{0}$
where a and K are positive constants and 0<M0<K.
a. Graph the growth rate function $R\left({M}\right)=-aMln\left({\frac{M}{K}}\right)$ assuming a =1 and K =4. For what values of M is the growth rate positive? For what value of M is the growth rate a maximum?
b. Solve the initial value problem and graph the solution for a = 1, K = 4, and M0 = 1. Describe the growth pattern of the tumor. Is the growth unbounded? If not, what is the limiting size of the tumor?
c. In the general equation, what is the meaning of K?
Solution:
Step 1
$R\left({M}\right)=\ -aMln\left({\frac{M}{K}}\right)$ assuming a =1 and K =4,
$R\left({M}\right)=\ -Mln\left({\frac{M}{4}}\right)$
Graph of the growth function:

For growth rate to be positive,
$-Mln\left({\frac{M}{4}}\right)>0$
$Mln\left({\frac{M}{4}}\right)<0$
$ln\left({\frac{M}{4}}\right)<0$
$ln\left({\frac{M}{4}}\right)<ln(1)$
$\frac{M}{4}<1$
$M<4$
So, for the growth rate to be positive M<4.
The growth rate is maximum when, ${R}^{'}(M)=0$
${R}^{'}(M)=$
${\left({-Mln\left({\frac{M}{4}}\right)}\right)}^{'}=0$
$-ln(\frac{M}{4})-M*\frac{4}{M}*\frac{1}{4}=0$
$-ln(\frac{M}{4})-1=0$
$ln(\frac{M}{4})=-1$
$ln(\frac{M}{4})=-ln(e)$
$ln(\frac{M}{4})=ln(\frac{1}{e})$
$\frac{M}{4}=\frac{1}{e}$
$M=\frac{4}{e}=1.471$
Hence for M=1.471, the growth rate is maximum.