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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 rearrange for r= ?
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 rearrange for r= ?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or 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
 one year ago
Best ResponseYou've already chosen the best response.0Im guessing a chain rule is applied here somehow

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep its asking for the derivative instantaneous growth of (anything) would be like d(anything) / dt

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0its 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
 one year ago
Best ResponseYou've already chosen the best response.0so f'(x) in terms of r is different to f'(r) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well 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
 one year ago
Best ResponseYou've already chosen the best response.0in terms of 'r' means as suppose an arbit situation f(x) = r^2r the f(x) = terms in r

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0You 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 5xy=12, you would solve for x, x = (12+y)/5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay.. so something like r = Log[1/a] / x

perl
 one year ago
Best ResponseYou've already chosen the best response.2\[ \Large{ f(x) = ae^{rx} \\~\\ f'(x) = ae^{rx}\cdot r \\~\\ f'(x) = f(x)\cdot r \\~\\ \frac{f'(x)}{f(x)} = r }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Not sure, what exactly you did, but perl did it right :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh ! that's interesting how that works out.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0It 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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