A single bacterium divides every 0.5 hour to produce two complete bacteria. If we start with a colony of 6000 bacteria, after t hours there will be A(t)=6000*2^2t=6000*4^t bacteria. Find A'(t) and A'(1)
please explain steps for best answer. thank you!!
Finding derivatives.?
2015-07-29 5:41 am
回答 (2)
2015-07-29 6:03 am
✔ 最佳答案
A(t) = 6000*4^t ln(A) = ln 6000 + t ln 4
A(t) = e^(ln 6000 + t ln 4)
... deriv. of e^u = (e^u) du
A' (t) = [e^(ln 6000 + t ln 4)] [ ln 4 ]
... remember A(t) = e^(ln 6000 + t ln 4) = 6000*4^t so sub
A' (t) = [ln 4] 6000*4^t <<< answer
A' (1) = [ln 4] 6000*4^(1) = [ln 4] 6000*4 ~ 33,271 bacteria / hour
2015-07-29 6:13 am
Way too many
收錄日期: 2021-05-01 22:44:53
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