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Water is poured into a bucket according to the rate F(t)=(t+7)/(t+2) , and at the same time empties out through a hole in the b...Asked by Ke$ha
Water is poured into a bucket according to the rate F(t)=(t+7)/(2+t), and at the same time empties out through a hole in the bottom at the rate E(t)=(ln(t+4))/(t+2), with both F(t) and E(t) measured in pints per minute. How much water, to the nearest pint is in the bucket at time t=5 minutes.
(t+7)/(2+t)-(ln(t+4))/(t+2)
t=5
(5+7)/(2+5)-(ln(5+4))/(5+2)
12/7 - ln(9)/7
(12-ln(9))/7=
1.4 pints
(t+7)/(2+t)-(ln(t+4))/(t+2)
t=5
(5+7)/(2+5)-(ln(5+4))/(5+2)
12/7 - ln(9)/7
(12-ln(9))/7=
1.4 pints
Answers
Answered by
Steve
You have found the net rate of change of the water (since E and F are measured in pints/minute) at t=5, not the amount of water.
dA/dt = F(t)-E(t)
= (t+7)/(2+t) - (ln(t+4))/(t+2)
Unfortunately, ln(t+4)/(t+2) does not integrate using elementary functions. Also, without knowing A(t) for some value of t, we cannot get rid of the integration constant C.
Just for reference,
http://www.wolframalpha.com/input/?i=integral+(t%2B7)%2F(2%2Bt)+-+(ln(t%2B4))%2F(t%2B2)+dt
dA/dt = F(t)-E(t)
= (t+7)/(2+t) - (ln(t+4))/(t+2)
Unfortunately, ln(t+4)/(t+2) does not integrate using elementary functions. Also, without knowing A(t) for some value of t, we cannot get rid of the integration constant C.
Just for reference,
http://www.wolframalpha.com/input/?i=integral+(t%2B7)%2F(2%2Bt)+-+(ln(t%2B4))%2F(t%2B2)+dt
Answered by
Ke$ha
Ok, so exactly how would I find the amount of water? I am confused now.
Answered by
Anonymous
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
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
Answered by
Anonymous
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
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
Answered by
Ke$ha
I am confused on how to do this. Will you please help?
Answered by
Anonymous
Steve has shown you how. I then tried to simplify the integration process The problem is that even if you use Wolfram for the integration of ln (x+2)dx/x you get a messy function. I bet your teacher did not expect you to run into that and therefore suspect an error in the problem statement. If you do it as is, you must use the integration in Wolfram since it contains that Li function.
Answered by
Ke$ha
so we would get 9.05584?
Answered by
Lol
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.
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