Water is poured into a bucket according to the rate F(t)=(t+7)/(2+t), and at the same time empties out through a hole in the bottom at the rate E(t)=(ln(t+4))/(t+2), with both F(t) and E(t) measured in pints per minute. How much water, to the nearest pint is in the bucket at time t=5 minutes.
(t+7)/(2+t)-(ln(t+4))/(t+2)
t=5
(5+7)/(2+5)-(ln(5+4))/(5+2)
12/7 - ln(9)/7
(12-ln(9))/7=
1.4 pints
so we would get 9.05584?
For future readers, Ke$ha is correct. The answer is 9.05584; however, make sure to round this to the nearest pint as the question asks.
Steve has shown you how. I then tried to simplify the integration process The problem is that even if you use Wolfram for the integration of ln (x+2)dx/x you get a messy function. I bet your teacher did not expect you to run into that and therefore suspect an error in the problem statement. If you do it as is, you must use the integration in Wolfram since it contains that Li function.
To find the amount of water in the bucket at time t = 5 minutes, we need to evaluate the expression:
F(t) - E(t)
Plugging in t = 5 into both F(t) and E(t):
F(5) = (5+7)/(2+5) = 12/7
E(5) = ln(5+4)/(5+2) = ln(9)/7
Now substitute these values into the expression:
12/7 - ln(9)/7
To simplify, we can combine the terms over a common denominator:
(12 - ln(9))/7
Now, compute the value:
(12 - ln(9))/7 ≈ 1.4
Therefore, the amount of water in the bucket at time t = 5 minutes is approximately 1.4 pints.
You have found the net rate of change of the water (since E and F are measured in pints/minute) at t=5, not the amount of water.
dA/dt = F(t)-E(t)
= (t+7)/(2+t) - (ln(t+4))/(t+2)
Unfortunately, ln(t+4)/(t+2) does not integrate using elementary functions. Also, without knowing A(t) for some value of t, we cannot get rid of the integration constant C.
Just for reference,
http://www.wolframalpha.com/input/?i=integral+(t%2B7)%2F(2%2Bt)+-+(ln(t%2B4))%2F(t%2B2)+dt
Ok, so exactly how would I find the amount of water? I am confused now.
I also hit the wall but took the following steps which might help:
let z = t+2 then t = z-2 and dz = dt
let
Q = amount in there and it is 0 at t = 0
then
dQ/dt = dQ/dt
= (z+5)/z -ln(z+2) /z
so integrate
z dz/z + 5 dz/z - ln(z+2)dz/z
= z + 5 ln z - a mess even with Steve's site + constant
adjust constant to get Q = 0 at t = 0 or at z = 2
I bet you have a typo somewhere but the basic idea is that you are given the RATE of filling = F-E
To find the AMOUNT, you must integrate the rate over time. Then of course you have a constant of integration which is determined by the amount in the bucket at t = 0