Question

Water is pumped into an empty tank at a rate of r(t)=20e^(.02t) gallons per minute. Approximately how many gallons of water have been pumped into the tank in the first five minutes?
Answer: 105 gal
Please explain how to come to this answer and with this problem you may use a calculator.

Answer 1

\[\huge\rm r(\color{orangered}{t})=20e^{.02\color{orangered}{t}}\]At time t=5 minutes we have,\[\huge\rm r(\color{orangered}{5})=20e^{(.02\cdot\color{orangered}{5})}\]Throw this into your calculator, this will complete the first half of the problem,
figuring out the `rate` per minute.

Answer 2

Oh the rate is variable.. hmm
So are we supposed to do integration for this?

Answer 3

I think you have to do the integral, unless you are willing to rely on the fact that the rate is pretty close to constant over that range...

Answer 4

r(0) = 20, r(5) = 22.1034 You could use the midpoint value and be well within 1% error compared to the integral

Answer 5

To do it with the integral, you're just integrating the rate equation times \(dt\)

\[V=\int _0^5r(t) dt=\int_0^520e^{0.02t}\,dt =\]

Answer 6

Hmm...I just tried to do that but it didn't come out right. I got 58.9653. Do I need to do something with this value?

Answer 7

show me your work...what do you get for the indefinite integral?

Answer 8

\[\int\limits_{0}^{5}20e^.02t\]
\[20\int\limits_{0}^{5}e^.02dt\]
u=.02t dt=.02dt
\[20 \times.4 \int\limits_{0}^{5}e^udu\]
\[.4(e^u) from 0 \to 5\]
.4(e^5-e^0) = 58.9653

Answer 9

for that 3rd integral that should be .02 instead of .4.

Answer 10

Uh, no.

\[u = 0.02t\]\[du = 0.02\,dt\]\[dt = \frac{du}{0.02}\]
\[20\int e^{0.02t}\,dt = 20\int e^u du = \frac{20}{0.02}\int e^u du = 1000e^u = 1000e^{0.02t}\]

Answer 11

This is a pretty common mistake, I'm afraid...but that's good, because if you remember not to make it, you eliminate a lot of mistakes :-)

Answer 12

hmm I was taught a different way but i see what you did. But if i plug in my 5 it doesn't come out right so i'm kinda confused.

Answer 13

Well, there's a right way, and a wrong way. This is a right way :-)

Answer 14

Remember, you need to change the limits of integration if you change the variable you are integrating

Answer 15

Oh right. Let me see if that works. :)

Answer 16

But evaluating the problem with my indefinite integral:

\[1000e^{0.02*5}-1000e^{0.02*0} = 1000e^{0.1}-1000e^0 \approx 1105.171 - 1000(1) = 105.171\]

Answer 17

Oh my goodness I got it! :) Thank you so much

Answer 18

Glad I could help. By the way, if you look at what @zepdrix proposed doing initially, it actually comes out quite close to the right answer!

\[r(5) = 20e^{0.02*5} = 22.1034\]
Using that rate for 5 minutes, we get \[V = 5*22.1034 = 110.517\]
Not bad for an estimate...

If we do as I suggested, and use the rate halfway through:
\[r(2.5) = 20e^{0.02*2.5} = 21.0254\]
\[V = 5*21.0254 = 105.127\]
\[\frac{105.127-105.171}{105.171}*100\% = -0.042\%\]
Close enough for government work :-)