The surface area SSS (measured in cm2\text{cm}^2cm2) of a particular fungus culture is observed over time ttt (measured in hours). The growth of the culture is modeled by the differential equation
dSdt=8tS1/2e2t,S≥0,t≥0 \frac{\text{d}S}{\text{d}t} = \frac{8t S^{1/2}}{\text{e}^{2t}}, \quad S \ge 0, \quad t \ge 0 dtdS=e2t8tS1/2,S≥0,t≥0Given that the initial surface area of the fungus is 9 cm2 at t=0t = 0t=0, solve this differential equation to find S1/2S^{1/2}S1/2 in terms of ttt, giving your answer in the form S1/2=g(t)S^{1/2} = g(t)S1/2=g(t).
Hence find the equation of the horizontal asymptote to the curve with equation S1/2=g(t)S^{1/2} = g(t)S1/2=g(t).
480 exam-style questions on AQA A Level Maths 1.11 H: Integration, covering 1.11.1 Fundamental Theorem of Calculus, 1.11.2 Integrating standard functions, 1.11.3 Definite integrals and areas, 1.11.4 Integration as the limit of a sum (A-level only), 1.11.5 Integration by substitution and by parts (A-level only), 1.11.6 Integration using partial fractions (A-level only), 1.11.7 Differential equations with separable variables (A-level only), and 1.11.8 Interpreting solutions of differential equations (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.