The mass, M M\,M milligrams, of a substance produced in a chemical reaction is modeled by the equation
M=1200e0.4t5+e0.4tt≥0 M = \frac{1200e^{0.4t}}{5 + e^{0.4t}} \quad t \ge 0 M=5+e0.4t1200e0.4tt≥0where t t\,t is the time in hours after the reaction begins.
Determine the initial mass of the substance produced.
Find the upper limit for the mass of the substance according to this model.
Calculate the time, after the start of the reaction, when the mass reaches 900 mg. Give your answer in hours and minutes to the nearest minute.
Show that
dMdt=Ke0.4t(5+e0.4t)2 \frac{dM}{dt} = \frac{Ke^{0.4t}}{(5 + e^{0.4t})^2} dtdM=(5+e0.4t)2Ke0.4twhere K K\,K is a constant to be determined.
Given that at time t=Tt = Tt=T, the rate of production is dMdt=24\displaystyle \frac{dM}{dt} = 24dtdM=24 mg/h, find the value of T T\,T to one decimal place. (Solutions relying entirely on calculator technology are not acceptable.)
717 exam-style questions on Edexcel A Level Maths Differentiation, covering 9.1 Differentiating sin x and cos x, 9.2 Differentiating exponentials and logarithms, 9.3 The Chain Rule, 9.4 The Product Rule, 9.5 The Quotient Rule, 9.6 Differentiating Trigonometric Functions, 9.7 Parametric Differentiation, 9.8 Implicit Differentiation, 9.9 Using Second Derivatives, and 9.10 Rates of Change. Each one has a worked solution and a mark scheme showing where the marks go.