The air in a 57 cubic metre kitchen is initially clean, but when David burns his toast while making breakfast, smoke is mixed with the room's air at a rate of 0.06 mg per second. An air conditioning system exchanges the mixture of air and smoke with clean air at a rate of 8 cubic metres per minute. Assume that the pollutant is mixed uniformly throughout the room and that burnt toast is taken outside after 46 seconds. Let S(t) be the amount of smoke in mg in the room at time t (in seconds) after the toast first began to burn. a. Find a differential equation obeyed by S(t). b. Find S(t) for 0 S t S 46 by solving the differential equation in (a) with an appropriate initial condition. c. What is the level of pollution in mg per cubic meter after 46 seconds? d. How long does it take forthe level of pollution to fall to 0.007 mg per cubic metre after the toast is taken outside? You can confirm that you are on the right track by checking numerical answers to some parts dS(t) a. The differential equation is dt : 0.06-S(t)/427.5 a n [E] (Enter your expression using Maple syntax and use S(t) rather than just 8.) b. As a check that your solution is correct, test one value. S(ll) : 06515811351 9 n [E mg (Enter your answer correct to at least '10 signicant figures. Do not include units.) c. Check the level of pollution in mg per cubic metre after 46 seconds by entering your answer here. correct to at least 10 signicant gures (do not include the units): 00568846637 . n E mg m73 d. The time, in seconds, when the level of pollution falls to 0.007 mg per cubic metre is _. [E seconds (correct to at least 10 significant figures). Note that this check asks for the time since t : 0 but the question part ((1) asks for a time since the toast was taken outside