Problem 3.4 A patient is diagnosed to have Po cancer cells at time t - 0 and the

Question

Problem 3.4 A patient is diagnosed to have Po cancer cells at time t - 0 and these cells grow subsequently according to uP(M-P) where P(t) is the number of the cells at time t while and M are two physiologically significant positive constants, e.g., M is some sort of threshold the number of cancel cells one can tolerate (a generally healthier person may have a higher M value). The patient will die when the number of cancer cells approaches infinity. Compute the condition between the values of Po and M when the person will die because of this cancer. Find the time the patient has left to live starting from the day of the diagnosis, in this case dP dt

Explanation / Answer

given

dP/dt =-aP(M-P)

dP/(P(M-P))=-a dt

let 1/(P(M-P))=A/P +B/(M-P)

1/(P(M-P))=(A(M-P) +BP)/(P(M-P))

1/(P(M-P))=(AM+ P(-A+ B))/(P(M-P))

AM=1 =>A=1/M , -A+B=0 =>B =A=1/M

1/(P(M-P))=(1/M)(1/P) +(1/M)(1/(M-P))

[(1/M)(1/P) +(1/M)(1/(M-P))]dP=-a dt

[(1/P) +(1/(M-P))]dP=-aM dt

integrate on both sides

[(1/P) +(1/(M-P))]dP=-aM dt

lnP- ln(M-P)=-aMt +c

ln(P/(M-P))=-aMt +c

(P/(M-P))=e-aMt +c

(P/(M-P))=Ce-aMt

(M-P)/P=CeaMt

(M/P)-1=CeaMt

(M/P)=1+CeaMt

P=M/(1+CeaMt)

Po at t=0

Po=M/(1+CeaM*0)

Po=M/(1+C)

1+C=M/Po

C=(M/Po)-1

P=M/(1+((M/Po)-1)eaMt)

(1+((M/Po)-1)eaMt) =M/P

((M/Po)-1)eaMt=((M/P)-1)

eaMt=((M/P)-1)/((M/Po)-1)

aMt=ln[((M/P)-1)/((M/Po)-1)]

t=(1/(Ma))ln[((M/P)-1)/((M/Po)-1)]

lime left for patient to live =(1/(Ma))ln[((M/P)-1)/((M/Po)-1)]

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.