- #1

Purgum

- 6

- 0

Dos my problem have relations with differentiel equations ? How can i find exact growth at t time after all ?

VERY URGENT NEED HELP

Thanks

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- Thread starter Purgum
- Start date

- #1

Purgum

- 6

- 0

Dos my problem have relations with differentiel equations ? How can i find exact growth at t time after all ?

VERY URGENT NEED HELP

Thanks

- #2

Berislav

- 239

- 0

Yes.Dos my problem have relations with differentiel equations ?

"Do you have to use calculus?" That depends on the dependency of m and n. Are they functions of time? If so then to find out how fast the tumor grows you will need to integrate over both m and n over a certain time interval and calculate the difference (hint: use linearity of riemann integrals for a nicer looking equation).

P.S.

You really shouldn't triple post.

- #3

PerennialII

Science Advisor

Gold Member

- 901

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https://www.physicsforums.com/showthread.php?t=77495

- #4

Purgum

- 6

- 0

thank u, but my question is about htree differtn thingsBerislav said:Yes.

"Do you have to use calculus?" That depends on the dependency of m and n. Are they functions of time? If so then to find out how fast the tumor grows you will need to integrate over both m and n over a certain time interval and calculate the difference (hint: use linearity of riemann integrals for a nicer looking equation).

P.S.

You really shouldn't triple post.

- #5

Purgum

- 6

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thanks fo sharingPerennialII said:

https://www.physicsforums.com/showthread.php?t=77495

- #6

Purgum

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- #7

LeonhardEuler

Gold Member

- 860

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However, this is not the sort of growth you would expect from tumor cells. You would expect that if all of the cells are dividing indepently of one another, the rate of growth would depend on the current population. For example, if one cell divides once in one hour, and you only have one cell, the rate of growth for one hour would be 1 cell per hour. Now suppose you have a thousand cells. After one hour they each divide once, so you now have two thousand cells. This means the rate of growth for that hour is 1,000 cells per hour. Notice that the growth rate is propotional to the current population. I have been looking at the growth rate over finite intervalt by taking [tex]\frac {\Delta P} {\Delta t}[/tex] but with more and more cells dividing indepently of one another, growth rates over smaller intervals start making sense and we can make the approximation [tex]\frac {\Delta P}{\Delta t} \approx \frac {dP} {dt}[/tex] Now, if this is the sort of growth we're talking about, then "m" probably doesn't measure new cells per hour, but more likely new cells per hour per existing cell. Likewise "n" would be in cells lost per hour per existing cell. The net growth rate would then be (m-n) cells per hour per existing cell. The growth rate is proportional to the number of existing cells. Since the growth rate is [tex]\frac {dP} {dt}[/tex], the equation that expresses this idea is [tex]\frac {dP} {dt}= (m-n) \times P[/tex]

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