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August 27th, 2013, 12:35 AM   #1
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Integration and antideriving given a rate equation.

Integration and antideriving given a rate equation of growing bacteria.?

Bacteria are dying at a rate of dN/dt=e^0.5t (per day). After 4 days, the death rate becomes 0.25t per hour. Will there be a time when the bacteria will be 0? Initially there was 100 bacteria.
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August 27th, 2013, 05:34 AM   #2
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Re: Integration and antideriving given a rate equation.

You are given that . Since you titled this "Integration" (you don't really need to add "antideriving". Integration and anti-derivation are the same thing) can you integrate to find N as a function of t? (Are you sure there is not supposed to be a "-" in that?) Since that formula is only true for four days, evaluate N(4). Then you are told that "the death rate becomes 0.25t per hour". (Is there really a "t" in that? Since t measures the hours, "t per hour" doesn't really make sense. I interpret that as meaning that we now have where "t" is now in hours. Solve for N(t) (t is "hours since the end of the four days") using the previous "N(4)" as the initial value. Solve N(t)= 0 for t. Be sure to state the answer clearly, not just a value of t since the meaning of "t" has changed.

(There are so many oddities in what you have given, the lack of negatives, etc., that I recommend you go back and reread the problem carefully.)
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