Asked by Anonymous
Oil flows into a tank according to the rate F(t)=(t^2+1)/(1+t), and at the same time empties out at the rate E(t)=(ln(t+7))/(t+2), with both F(t) and E(t) measured in gallons per minute. How much oil, to the nearest gallon, is in the tank at time t = 12 minutes. You must show your setup but can use your calculator for all evaluations.
I have no idea where to start here, or what to do with this equations. Help please!
I have no idea where to start here, or what to do with this equations. Help please!
Answers
Answered by
Steve
The rate of change of volume is
dV/dt = E(t)-F(t)
Assuming the tank was initially empty, the volume is just
∫[0,12] (t^2+1)/(1+t) - ln(t+7)/(t+2) dt
The first one works well if do a long division first, then the remainder, k/(1+t) will integrate to log(1+t)
The second one does not work using elementary functions. Maybe it's ln((t+7)/(t+2)). If so, it's just the difference of logs, which you must integrate by parts.
dV/dt = E(t)-F(t)
Assuming the tank was initially empty, the volume is just
∫[0,12] (t^2+1)/(1+t) - ln(t+7)/(t+2) dt
The first one works well if do a long division first, then the remainder, k/(1+t) will integrate to log(1+t)
The second one does not work using elementary functions. Maybe it's ln((t+7)/(t+2)). If so, it's just the difference of logs, which you must integrate by parts.
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