Calculus Word Problem?

Question

Consider a biochemical reaction in which a certain substance is both produced
and consumed. The concentration of this substance at time t is defined to be
$c (t)$ $c(t)$ . Assume that the function c obeys the following differential equation:

$\frac{d c}{d t}$ $(dc)/dt$ = $K_{max}$ $K_max$ $\frac{c}{k + c}$ $c/(k+c)$ $- r c$ $−rc$

Where $K_{max}$ $K_max$ , $k$ $k$ , and $r$ $r$ are all positive constants. The first term on the right
hand side of this equation denotes the concentration-dependent production
and the second denotes the consumption.

What is the maximal rate at which the substance is produced?

At what concentration is the production rate 50% of this maximum value?

If the production is turned off, the substance decays. How long will it
take for the concentration to drop by 50%?

At what concentration does production balance consumption?

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Answer 1 · 2016-12-16T18:21:39

See below.

1) The rate of produced substance is maximum when

$K_{max} \frac{c}{k + c}$ $K_(max)c/(k+c)$ is maximum. This function is monotonic increasing so it's maximum is

$lim c \to \infty K_{max} \frac{c}{k + c} = K_{max}$ $lim_(c->oo)K_(max)c/(k+c)=K_max$

2) Choosing

$0.5 K_{max} = K_{max} \frac{c}{k + c}$ $0.5K_(max)=K_(max)c/(k+c)$

and solving for $c$ $c$ we obtain

$c = k$ $c = k$

3) If production is turned off the the concentration evolution is given by

$\frac{d c}{d t} = - r c$ $(dc)/(dt)=-r c$ with solution

$c = C_{0} e^{- r t}$ $c = C_0 e^(-r t)$ so the concentration will drop by $50 %$ $50%$ when

$\frac{1}{2} = e^{- r t}$ $1/2=e^(-rt)$ or after solving for $t$ $t$

$t = \frac{{log}_{e} (2)}{r}$ $t = log_e(2)/r$ seconds.

4) The balance is attained when

$K_{max} \frac{c}{k + c} - r c = 0$ $K_(max)c/(k+c)-rc=0$ or when production equals consumption.

$c = \frac{K_{max} - k r}{r}$ $c = (K_(max) - k r)/r$