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anonymous

  • one year ago

How do you approach a problem like this ? GIVEN f[x] = a E^(r x) Calculate in terms of r the instantaneous growth rate of f[x] at any point x. Do they mean work out the derivative, and then re-arrange for r= ?

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  1. anonymous
    • one year ago
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    or is it like a limit problem, where I drop the x altogether, with the idea that x is a constant if I am trying to take the derivative in terms of r ?

  2. anonymous
    • one year ago
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    Im guessing a chain rule is applied here somehow

  3. anonymous
    • one year ago
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    yep its asking for the derivative instantaneous growth of (anything) would be like d(anything) / dt

  4. anonymous
    • one year ago
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    I can see how I would get f'[x] but I think it's asking for f'[r] how would things change in that case?

  5. anonymous
    • one year ago
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    its asking f'(x) in terms of r f'(x) represent the instantaneous rate of change of f(x) at x whereas f'(r) represent the instantaneous rate of change of f(x) at r

  6. anonymous
    • one year ago
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    so you get it?

  7. anonymous
    • one year ago
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    so f'(x) in terms of r is different to f'(r) ?

  8. anonymous
    • one year ago
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    yes..

  9. anonymous
    • one year ago
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    well I understand how to calculate f'[x].. I get f'[x] = a r E^(r x) but I'm not sure what they mean by 'in terms of r..

  10. anonymous
    • one year ago
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    in terms of 'r' means as suppose an arbit situation f(x) = r^2-r the f(x) = terms in r

  11. Astrophysics
    • one year ago
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    You have the right idea, a simple example is saying solve x in terms of y, this essentially means x = f(y), so if you have something such as 5x-y=12, you would solve for x, x = (12+y)/5

  12. anonymous
    • one year ago
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    okay.. so something like r = Log[1/a] / x

  13. perl
    • one year ago
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    \[ \Large{ f(x) = ae^{rx} \\~\\ f'(x) = ae^{rx}\cdot r \\~\\ f'(x) = f(x)\cdot r \\~\\ \frac{f'(x)}{f(x)} = r }\]

  14. Astrophysics
    • one year ago
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    Not sure, what exactly you did, but perl did it right :)

  15. anonymous
    • one year ago
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    thank you perl

  16. anonymous
    • one year ago
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    Ahhh ! that's interesting how that works out.

  17. Astrophysics
    • one year ago
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    It can be tricky at first with all the wording, but notice it's just a game of substitution in a way :P

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