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!
To find the amount of oil in the tank at time t = 12 minutes, we need to calculate the net change in the amount of oil in the tank over the interval [0, 12] minutes.
The rate of oil flowing into the tank is given by F(t) = (t^2 + 1)/(1 + t), measured in gallons per minute.
The rate of oil emptying out of the tank is given by E(t) = (ln(t + 7))/(t + 2), measured in gallons per minute.
To calculate the net change, we need to find the integral of F(t) - E(t) over the interval [0, 12].
Let's start by calculating the integral of F(t) - E(t) over the interval [0, 12]:
∫(F(t) - E(t)) dt = ∫((t^2 + 1)/(1 + t) - (ln(t + 7))/(t + 2)) dt.
To solve the integral, we can break it down into two separate integrals:
∫(t^2 + 1)/(1 + t) dt - ∫(ln(t + 7))/(t + 2) dt.
Let's solve each integral separately:
First integral: ∫(t^2 + 1)/(1 + t) dt.
To solve this integral, we can use the method of partial fractions:
(t^2 + 1)/(1 + t) = A + B/(1 + t),
where A and B are constants to be determined.
Multiplying through by (1 + t), we get:
t^2 + 1 = A(1 + t) + B.
Expanding the right side gives:
t^2 + 1 = A + At + B.
Comparing like terms, we have:
A = 1,
A + B = 0.
Solving for B, we find:
B = -1.
Therefore, the first integral becomes:
∫(t^2 + 1)/(1 + t) dt = ∫(1 - 1/(1 + t)) dt
= ∫1 dt - ∫1/(1 + t) dt
= t - ln|1 + t| + C1,
where C1 is the constant of integration.
Second integral: ∫(ln(t + 7))/(t + 2) dt.
To solve this integral, we can use u-substitution:
Let u = t + 7,
Then du = dt.
Substituting into the integral, we get:
∫(ln(u))/(u - 5) du.
Next, we can use integration by parts, with:
dv = 1 du,
v = u,
u = ln(u),
du = du.
Using the formula for integration by parts, we have:
∫(ln(u))/(u - 5) du = u*ln(u) - ∫(u/ (u - 5)) du.
The remaining integral is a simple polynomial integral:
∫(u/ (u - 5)) du = ∫((u - 5 + 5)/ (u - 5)) du
= ∫(1 + (5/ (u - 5))) du
= u + 5ln|u - 5| + C2,
where C2 is the constant of integration.
Substituting back u = t + 7, we get:
∫(ln(u))/(u - 5) du = (t + 7)ln|t + 2| + 5ln|t + 2 - 5| + C2.
Now, we can substitute the two integrals back into the original net change integral:
∫(F(t) - E(t)) dt = ∫(t^2 + 1)/(1 + t) dt - ∫(ln(t + 7))/(t + 2) dt
= t - ln|1 + t| - [(t + 7)ln|t + 2| + 5ln|t + 2 - 5|] + C.
Evaluating the integral over the interval [0, 12], we get:
Net change = [12 - ln|1 + 12| - [(12 + 7)ln|12 + 2| + 5ln|12 + 2 - 5|]
- [(0 - ln|1 + 0| - [(0 + 7)ln|0 + 2| + 5ln|0 + 2 - 5|]].
Calculating the values above using a calculator, we can find the net change in the amount of oil in the tank over the interval [0, 12].
Finally, to determine the amount of oil in the tank at time t = 12 minutes, we need to take the initial amount (at t = 0) and add the net change:
Amount at t = 12 = Initial amount + Net change.
Note: The initial amount is not provided in the problem statement, so we cannot determine the exact amount of oil in the tank at t = 12.
To find the amount of oil in the tank at a given time, we need to determine the net rate at which oil is flowing in or out of the tank.
The net rate, N(t), is given by subtracting the rate at which oil is flowing out (E(t)) from the rate at which it is flowing in (F(t)):
N(t) = F(t) - E(t)
Now, let's substitute the given functions for F(t) and E(t):
N(t) = (t^2 + 1)/(1 + t) - (ln(t + 7))/(t + 2)
To find the amount of oil in the tank at t = 12 minutes, we need to integrate the net rate function from 0 to 12:
Oil in tank at t = 12 minutes = ∫[0,12] N(t) dt
Now, we can use a calculator to evaluate this integral.
However, it's important to note that the integral of N(t) might not have a nice, closed-form solution. In such cases, numerical methods like Simpson's rule or the trapezoidal rule can be used to approximate the integral.
If you have access to a graphing calculator or a numerical integration tool, you can simply plug in the function N(t) and evaluate the integral over the given bounds.
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.