A colony of bacteria if left undisturbed will grow at a rate proportional to the number of bacteria, P present at time,t. However,a toxic substance is being added slowly such that at time t, the bacteria also die at the rate μPt where μ is a positive constant. (a) Show that at time t the rate of growth of the bacteria in the colony is governed by the differential equation (dP/dt)= (k−μt)p where k is apositive constant. when t=0, (dP/dt)=2P and when t=1, (dP/dt)=((19)/(10))P (b) show that (dP/dt)= (1/(10))(20−t)P. Sir Forkum Michael.


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Question Number 63424 by Rio Michael last updated on 04/Jul/19

$A c o l o n y o f b a c t e r i a i f l e f t u n d i s t u r b e d w i l l g r o w a t a r a t e$ $p r o p o r t i o n a l t o t h e n u m b e r o f b a c t e r i a, P p r e s e n t a t t i m e, t .$ $H o w e v e r, a t o x i c s u b s t a n c e i s b e i n g a d d e d s l o w l y s u c h t h a t$ $a t t i m e t, t h e b a c t e r i a a l s o d i e a t t h e r a t e μ P t w h e r e μ i s$ $a p o s i t i v e c o n s t a n t .$ $(a) S h o w t h a t a t t i m e t t h e r a t e o f g r o w t h o f t h e b a c t e r i a i n$ $t h e c o l o n y i s g o v e r n e d b y t h e d i f f e r e n t i a l e q u a t i o n$ $\frac{d P}{d t} = (k - μ t) p w h e r e k i s a p o s i t i v e c o n s t a n t .$ $w h e n t = 0, \frac{d P}{d t} = 2 P a n d w h e n t = 1, \frac{d P}{d t} = \frac{19}{10} P$ $(b) s h o w t h a t$ $\frac{d P}{d t} = \frac{1}{10} (20 - t) P .$ $S i r F o r k u m M i c h a e l .$
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