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anonymous
 one year ago
this one should be some what fun
Water flows into a tank according to the rate F(t) = (6 + t)/(1 + t), and at the same time empties out at the rate E(t) = (ln(t+2))/(t+1) , with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest gallon, is in the tank at time t = 10 minutes. You must show your setup but can use your calculator for all evaluations.
anonymous
 one year ago
this one should be some what fun Water flows into a tank according to the rate F(t) = (6 + t)/(1 + t), and at the same time empties out at the rate E(t) = (ln(t+2))/(t+1) , with both F(t) and E(t) measured in gallons per minute. How much water, to the nearest gallon, is in the tank at time t = 10 minutes. You must show your setup but can use your calculator for all evaluations.

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3presumably the tank is empty to start with..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do i just evaluate both functions at t =10 and then subtract f(10)  E(10)? and Id assume the tank was empty i wasn't given anymore information

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.5\(\frac{dV}{dt} = F(t)  E(t)\)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.5build and solve integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits \frac{ (t+6)(\ln(t+2)) }{ (t+1) }\]?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Looks good, but that wont evaluate into elementary functions so you may simply use wolfram http://www.wolframalpha.com/input/?i=%5Cint_0%5E%2810%29+%5Cfrac%7B+%28t%2B6%29%28%5Cln%28t%2B2%29%29+%7D%7B+%28t%2B1%29+%7D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:( i was hoping it would be more complicated

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3evaluating it is complicated indeed but i think you should not waste time messing with special functions at this point..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah your probably right , anyway thanks once again

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3maybe just double check the problem has no typoes because usually you should expect to see easily workable integrals in these type of problems.. the focus should be on setting up the integral, not evaluating it..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0all seems clear to me , do you see anything out of place ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3integral becomes easy if E(t) is like below : \(E(t) = (\ln(t+\color{red}{1}))/(t+1)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0nope triple checked its t + 2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3then we're good, nearest to gallon, the water in the tank is 18 gallon at t = 10

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 can i ask you something?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why are you so helpfull? im serious you clearly are on a higher level of math, why do you keep helping people with silly linear calculus I problems, you couldn't possibly be learning anything and i'm sure your own work keeps you busy you have infinity medals what keeps you here ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Haha thanks for the good words! But you have no idea, I suck at math, thats the only reason I'm still here ;p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0god , i must be truly awful then

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I think you're doing great!
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