An industrial desalination plant monitors the mass of salt extracted from seawater. The process is modeled by two variables:
The engineers model the rate of salt extraction as being directly proportional to 10−tS\displaystyle \frac{10 - t}{S}S10−t.
After 2 hours of operation, the rate of salt extraction is 803\displaystyle \frac{80}{3}380 kg per hour and the total mass of salt extracted is 120 kg.
Show that SdSdt=400(10−t)\displaystyle S \frac{dS}{dt} = 400(10 - t)SdtdS=400(10−t).
Hence, show that S2=400t(20−t)S^2 = 400t(20 - t)S2=400t(20−t).
The plant began its operation at 08:00. (i) The plant's efficiency drops over time. The operation is paused for a second maintenance cycle when the rate of salt extraction falls below 15 kg per hour. Using the results in parts (a) and (b), calculate the time of day when the operation is paused. (ii) Explain why the model used by the engineers is not valid at the start of the process (08:00).
363 exam-style questions on CCEA A Level Maths 3.6 Integration (A-level only), covering 3.6.1 Integration (A-level only), 3.6.2 Integration (A-level only), 3.6.3 Integration (A-level only), 3.6.4 Integration (A-level only), 3.6.5 Integration (A-level only), 3.6.6 Integration (A-level only), and 3.6.7 Integration (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.