Menu Close

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-bacte




Question Number 63424 by Rio Michael last updated on 04/Jul/19
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.
$${A}\:{colony}\:{of}\:{bacteria}\:{if}\:{left}\:{undisturbed}\:{will}\:{grow}\:{at}\:{a}\:{rate} \\ $$$${proportional}\:{to}\:{the}\:{number}\:{of}\:{bacteria},\:{P}\:{present}\:{at}\:{time},{t}. \\ $$$${However},{a}\:{toxic}\:{substance}\:{is}\:{being}\:{added}\:{slowly}\:{such}\:{that} \\ $$$${at}\:{time}\:{t},\:{the}\:{bacteria}\:{also}\:{die}\:{at}\:{the}\:{rate}\:\mu{Pt}\:{where}\:\mu\:{is} \\ $$$${a}\:{positive}\:{constant}. \\ $$$$\left({a}\right)\:\:{Show}\:{that}\:{at}\:{time}\:{t}\:{the}\:{rate}\:{of}\:{growth}\:{of}\:{the}\:{bacteria}\:{in} \\ $$$${the}\:{colony}\:{is}\:{governed}\:{by}\:{the}\:{differential}\:{equation} \\ $$$$\:\frac{{dP}}{{dt}}=\:\left({k}−\mu{t}\right){p}\:{where}\:{k}\:{is}\:{apositive}\:{constant}. \\ $$$${when}\:{t}=\mathrm{0},\:\frac{{dP}}{{dt}}=\mathrm{2}{P}\:{and}\:{when}\:{t}=\mathrm{1},\:\frac{{dP}}{{dt}}=\frac{\mathrm{19}}{\mathrm{10}}{P} \\ $$$$\left({b}\right)\:{show}\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dP}}{{dt}}=\:\frac{\mathrm{1}}{\mathrm{10}}\left(\mathrm{20}−{t}\right){P}. \\ $$$$\:{Sir}\:{Forkum}\:{Michael}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *