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 August 27th, 2013, 12:35 AM #1 Newbie   Joined: Aug 2013 Posts: 1 Thanks: 0 Integration and antideriving given a rate equation. Integration and antideriving given a rate equation of growing bacteria.? Question: 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.
 August 27th, 2013, 05:34 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Integration and antideriving given a rate equation. You are given that $\frac{dN}{dt}= e^{0.5t}$. 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 $\frac{dN}{dt}= -e^{0.25t}$ 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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