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.)
425 exam-style questions on OCR A Level Maths 1.7 Differentiation, covering 1.7.1 Derivative as gradient of the tangent, 1.7.2 Gradient of the tangent at a point, 1.7.3 Sketching the gradient function, 1.7.4 Second derivatives, 1.7.5 Second derivative as rate of change of gradient, 1.7.6 Convex, concave and points of inflection (A-level only), 1.7.7 Differentiation from first principles for powers of x, 1.7.8 Differentiation from first principles for sin x and cos x (A-level only), 1.7.9 Differentiating x^n, 1.7.10 Differentiating e^(kx) and a^(kx) (A-level only), 1.7.11 Differentiating trigonometric functions (A-level only), 1.7.12 Derivative of ln x (A-level only), 1.7.13 Tangents and normals, 1.7.14 Stationary points, 1.7.15 Increasing and decreasing functions, 1.7.16 Points of inflection (A-level only), 1.7.17 Product and quotient rules (A-level only), 1.7.18 Chain rule (A-level only), 1.7.19 Parametric and implicit differentiation (A-level only), 1.7.20 Constructing differential equations (A-level only), and 1.7 Differentiation. Each one has a worked solution and a mark scheme showing where the marks go.