A research scientist is monitoring the internal pressure PPP (in kPa) of a sealed bioreactor during a temperature-controlled experiment. The time ttt, in hours, is measured from the start of the experiment. The scientist models the rate of change of pressure as being directly proportional to 15−tP\displaystyle \frac{15 - t}{P}P15−t.
After 5 hours, the pressure is 250 kPa and the rate of increase of pressure is 20 kPa per hour.
Show that PdPdt=500(15−t)\displaystyle P \frac{\text{d}P}{\text{d}t} = 500(15 - t)PdtdP=500(15−t).
Hence, show that P2=500t(30−t)P^2 = 500t(30 - t)P2=500t(30−t).
The experiment began at 06.00. (i) The researcher stops monitoring the bioreactor when the rate of pressure change drops below 10 kPa per hour. Using the results in parts (a) and (b), determine the earliest time that the researcher stops monitoring. (ii) Explain why the model used by the scientist is not valid at 06.00.
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.