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Mathematics 8 Online
OpenStudy (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= ?

OpenStudy (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 ?

OpenStudy (anonymous):

Im guessing a chain rule is applied here somehow

OpenStudy (anonymous):

yep its asking for the derivative instantaneous growth of (anything) would be like d(anything) / dt

OpenStudy (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?

OpenStudy (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

OpenStudy (anonymous):

so you get it?

OpenStudy (anonymous):

so f'(x) in terms of r is different to f'(r) ?

OpenStudy (anonymous):

yes..

OpenStudy (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..

OpenStudy (anonymous):

in terms of 'r' means as suppose an arbit situation f(x) = r^2-r the f(x) = terms in r

OpenStudy (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

OpenStudy (anonymous):

okay.. so something like r = Log[1/a] / x

OpenStudy (perl):

\[ \Large{ f(x) = ae^{rx} \\~\\ f'(x) = ae^{rx}\cdot r \\~\\ f'(x) = f(x)\cdot r \\~\\ \frac{f'(x)}{f(x)} = r }\]

OpenStudy (astrophysics):

Not sure, what exactly you did, but perl did it right :)

OpenStudy (anonymous):

thank you perl

OpenStudy (anonymous):

Ahhh ! that's interesting how that works out.

OpenStudy (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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