A biochemist is studying the production of a specific enzyme in a bioreactor. The mass of the enzyme, MMM mg, in the reactor, ttt hours after the reaction begins, is modelled by the equation
M=1200e0.5t5+e0.5tt≥0 M = \frac{1200e^{0.5t}}{5 + e^{0.5t}} \quad t \ge 0 M=5+e0.5t1200e0.5tt≥0Determine the initial mass of the enzyme in the bioreactor.
According to this model, find the limiting value of the enzyme's mass as ttt becomes very large.
Calculate the time elapsed since the start of the reaction when the mass of the enzyme is exactly 900 mg. Give your answer in hours and minutes to the nearest minute.
Show that
dMdt=Ke0.5t(5+e0.5t)2 \frac{dM}{dt} = \frac{Ke^{0.5t}}{(5 + e^{0.5t})^2} dtdM=(5+e0.5t)2Ke0.5twhere KKK is a constant to be determined.
At time t=Tt = Tt=T, the rate of enzyme production is 40 mg/h. Find the value of TTT, giving your answer to one decimal place. (Solutions relying entirely on calculator technology are not acceptable.)
333 exam-style questions on WJEC A Level Maths 3.6 Differentiation (A-level only), covering 3.6.1 Differentiation (A-level only), 3.6.2 Differentiation (A-level only), 3.6.3 Differentiation (A-level only), 3.6.4 Differentiation (A-level only), 3.6.5 Differentiation (A-level only), 3.6.6 Differentiation (A-level only), 3.6.7 Differentiation (A-level only), and 3.6 Differentiation (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.