HELP!! 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.
net flow = F(t) - E(t)
Do I just plug in 10 then for both and then subtract? That seems too easy.
yes that would only give you the rate of flow at time t = 10 minutes the rate is not constant It varies with time.
Wow, ok. I'm gonna post a few more questions. Can you hep with those?
No, I believe there's integration involved.
ive havent answered this one but i;ve just been called away...
@ParthKohli that would make sense. Can you explain?
Since the problem talks about gallons per minute, i.e., rate instead of just gallons, something tells me that those functions given to you are rates of change. I could be very wrong though - I have no trust in people who frame word problems these days. Haha
I think you're right lol
So I believe \(F(10) - E(10)\) would only tell you the rate of change of volume at the instant \(t = 10 min\). For finding the absolute volume in the tank, the expression turns out to be\[V(10) = \int_0^{10}(F(t)-E(t)) dt\]
Ok so I have to find the anti-derivative of each then?
Yes indeed... that's the only way to do it.
For F(t) its 6*ln(1+t) + t - 1*ln(1+t) right? I'm stuck with E(t)
Yeah E(t) isn't gonna be elementary damn
If you're in high school, this problem is not for you. Abort mission.
Yea, I'm in high school. This is one of my assignment problems.
Did these guys not think the problem through?! You'll just have to hope that the problem doesn't want you to integrate them then
Because many (in fact most) functions' antiderivatives cannot be expressed in an elementary form, if you know what I'm talking about.
I do. Oh well. Can you help with a few more?
They are less complicated haha
Yeah sure
But you need to report this one problem to your instructor.
I will. Thanks.
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