$k_s=$ the half$-$saturation constant, which is the substrate level at which uptake is half of the

maximum $($moles$/$L$)$

A$)$ If the initial concentration of the substrate is $S_0$ at $t=0.$ Solve the differential

equation using separation of variables to find substrate concentration as a function

of time. $(5$ Points$)$

B$)$ If $S_0=10$mole$\frac{s}{L},v_m=0\mathrm{.}5$ moles $\frac{?}{?}\frac{L}{d},$ and $k_s=2$mole$\frac{s}{L},$ what is the approximate

concentration of the substrate at $t=10d?$ Convert this into a 'root problem' and

determine the root graphically. $(5$ Points$)$

C$)$ If $S_0=10$mole$\frac{s}{L},v_m=0\mathrm{.}5$ moles $\frac{?}{?}\frac{L}{d},$ and $k_s=2$mole$\frac{s}{L},$ what is the concentration

of the substrate at $t=10d?$ Convert this into a 'root problem' and use the bisection

method until the absolute value of the approximate error falls below the stopping

criterion of $0\mathrm{.}5\%.$ Use lower guess $(x_l)=5$ and upper guess $(x_u)=7.(10$ Points$)$