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Very nice question indeed. Lemme write out the solving stepwise then u can ask questions:
PART 1: Alright, let's let the amount of water in the tank at any time be V. Given that the amount of water flowing into the tank doubles every minute, we can write dV/dt = 2t where t is in minutes. Solving for V by integration gives us. V(t) = t^2 + c where c is a constant. To solve for c, we observe that initially the tank is empty (of course) thus V(0) = c = 0 And therefore the equation giving us the amount of water in the tank at any time t is V(t) = t^2 ---------------------------------------------------------------------- PART 2: The tank is full in an hour hence V(60) = 60^2 meaning that the capacity of the tank is 60^2 So we want the time when the tank has (60^2)/2 in it. V(t) = t^2 =(60^2)/2 Solving for t we get t = sqrt [ (60^2)/2 ] = 42.43 minutes
There you go @Donkey . Hope it makes sense.
Well laid out explanation @allank however the users will get more out of your lesson if you were to stop at the final equation and allow them to solve it. Remember that we are here to help spread knowledge and encourage learning in others. The code of conduct, http://openstudy.com/code-of-conduct, asks that we not give answers to help promote education and curtail cheating, which has been witnessed through some of the associated academy schools.
Will do @eSpeX and thanks @mrbarry @Nancy_Lam
Since it has been a year, I am not sure if this is still an open problem, but the answer provided here was wrong on several fronts. The answer provided did not answer the question being asked. The questions states that the water flowing into the tank is doubling and not, as the answer provided solves for, the volume of fluid in the tank. The problem also assumes that the tank starts out empty, and we don't have enough information to determine if that is the case. So for the tank starting out empty, If the problem stated the amount of fluid in the tank at any time between t=0 and t=60 we would have enough information to solve for the integration constant, but that is not present here. So for example, since we are solving for the rate of water flowing into the tanks a perfectly valid solution to the problem as stated could have the tank start out more than half full. In that case we don't have enough information to find out when the tank was half full. To help visualize that take any solution to the rate problem filling a tank of size A, and it is also a valid solution for a tank of size 2A that starts half full. From that we can see that is is also a solution for tanks of size 3A that start out with 2A fluid in them, or for tanks of size 5/4A that start out with 1/4A of fluid in them. There are an infinite set of solutions. Just picking the integration constant to being zero is convenient, but its an arbitrary choice not supported by the problem. If we are talking about the volume of fluid in the tank doubling. Then we can solve for the initial volume of fluid in the tank. Aaron