anonymous
  • anonymous
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= ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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 ?
anonymous
  • anonymous
Im guessing a chain rule is applied here somehow
anonymous
  • anonymous
yep its asking for the derivative instantaneous growth of (anything) would be like d(anything) / dt

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anonymous
  • anonymous
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
so you get it?
anonymous
  • anonymous
so f'(x) in terms of r is different to f'(r) ?
anonymous
  • anonymous
yes..
anonymous
  • anonymous
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..
anonymous
  • anonymous
in terms of 'r' means as suppose an arbit situation f(x) = r^2-r the f(x) = terms in r
Astrophysics
  • Astrophysics
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
anonymous
  • anonymous
okay.. so something like r = Log[1/a] / x
perl
  • perl
\[ \Large{ f(x) = ae^{rx} \\~\\ f'(x) = ae^{rx}\cdot r \\~\\ f'(x) = f(x)\cdot r \\~\\ \frac{f'(x)}{f(x)} = r }\]
Astrophysics
  • Astrophysics
Not sure, what exactly you did, but perl did it right :)
anonymous
  • anonymous
thank you perl
anonymous
  • anonymous
Ahhh ! that's interesting how that works out.
Astrophysics
  • Astrophysics
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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