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Is differentiel equation needed ?

  1. Jul 12, 2005 #1
    If a tumor cell grows with rate m, and dies with rate n (m>n), their population number is P, after tme t, how can set up a mathematical fomuler for growth ? If i also have data from 100 patients, is it useful ?
    Dos my problem have relations with differentiel equations ? How can i find exact growth at t time after all ?
  2. jcsd
  3. Jul 12, 2005 #2
    "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).

    You really shouldn't triple post.
  4. Jul 12, 2005 #3


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  5. Jul 12, 2005 #4
    thank u, but my question is about htree differtn things :confused:
  6. Jul 12, 2005 #5
    thanks fo sharing :smile: :smile: :smile: :smile: :smile: :smile: :smile: :smile:
  7. Jul 12, 2005 #6
    but can u explain to me basic things bout that? i don understand why how people can fomute that fomuler.. can u help ???
  8. Jul 14, 2005 #7


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    The formula you need will depend on exactly what you mean by rate. If you mean that the tumor cells grow at a constant rate of "m" cells per second and die at a constant rate of "n" cells per second, then its easy to see that the number of cells "t" seconds later is P(t)= P(initial) (m-n)t, because after every second there will be (m-n) more cells than before.

    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 curent 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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