Question

A model for tumor growth is given by the Gompertz equation dV dt = a (In b-, In V) V where a >0 and b >0 are constants and V is the volume of the tumor measured in mm3 Find the solution for tumor volume as a function of time that has an initial tumor volume of V(0) 1 mm3.

Answer 1

Answer:

$V(t) = b^{(1-e^{-at})}$

Step-by-step explanation:

We are given the following information in the question:

$\displaystyle\frac{dV}{dt} = a (In b - In V) V$

where a > 0 and b > 0.

$\displaystyle\frac{dV}{dt} = a (In b-In V) V\\\\\displaystyle\frac{dV}{dt} = -aV(ln\frac{V}{b})\\\\\frac{dV}{V(ln\frac{V}{b})} = (-a)dt\\\\\text{Put } ln\frac{V}{b} = z\\\\\text{Integrating both sides}\\\\\int \frac{dV}{V(ln\frac{V}{b})} = \int (-a)dt\\\\\text{We get}\\\\\int \frac{dz}{z} = \int (-a)dt\\\\\\\text{where C is the constant of integration}$

$V(0) = 1~ mm^3\\V(t) = b.e^{e^{-at+C}}\\\text{Putting t =0, V(0) = 1}\\\\V(0) = 1 = b.e^{e^{C}}\\\\V(t) = b^{(1-e^{-at})}$

where v(t) is the required tumor volume as a function of time that has an initial tumor volume of V(0) = 1 cubic mm.