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Let f(t) be the amount of oxygen (in suitable units) in a lake t days after sewage is dumped into the lake, and suppose that f(t) is given approximately by the following. At what time is the oxygen content increasing the fastest?
f(t) = 1 - [(15)/(t+15)] + [(225)/(t+15)²]
Last edited by taylor_2009 (2010-02-25 13:57:00)
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How much have you figured out so far? Conceptually, the first derivative of f(t) gives you the rate at which the oxygen level is increasing at time t, and the second derivative gives you the rate of the rate of increase. So you're looking for the maximum of f'(t), which can be determined by finding a critical point in f''(t).
Last edited by TheDude (2010-02-25 17:59:36)
Wrap it in bacon
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I got that f'(t)= -(15)/(t+15)² - (510)/(t+15)³
I got f"t = (30)/(t+15)³ + (1530)/(t+15)^4
I don't know if those are right and I don't know how to solve them for 0.
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Hi;
I got that f'(t)= -(15)/(t+15)² - (510)/(t+15)³
That is incorrect for f'(t).
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Is f'(t) 15/(t+15)² - (510)/(t+15)³?
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Hi taylor_2009;
No that is not correct. Do you need help on it? Or can you see where you went wrong?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Yeah I need help. I've done the problem like 5 times and I keep getting that as an answer.
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Hi taylor_2009;
Let's do it piece by piece using the chain rule:
If we say
Then the chain rule is:
Now for the next term:
Chain rule again:
Now combine both pieces.
You use the same method to get f''(t), so please give it a try.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Would f"(t) be (-30)/(t+15)³ + 1350/(t+15)^4?
You would set f'(t) equal to 0 and determine whats a maximum to get the answer right? I'm not sure how to solve f'(t).
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Hi taylor_2009;
That is correct for f''(t).
If you are trying to find the max or min of f(t) you set f'(t) = 0, now if you are trying to get the max-min of f'(t) you should set what to 0?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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The get the max min of f'(t) you would set f"(t) equal to 0.
I got f'(t)=15 and f"t=-60.
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Hi taylor_2009;
What did you get for the solution to this?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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-60
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Hi;
But when you plug -60 back in to the equation you don't get 0. You get 4 / 6075. So that is not correct.
Times both sides by ( t + 15 )^4
t = 30
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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