An oil storage tank ruptures at time t = 0 and oil leaks from the tank at a rate of r(t) = 105e^−0.02t liters per minute. How much oil leaks out during the first hour? (Round your answer to the nearest liter.)
I got 4974.35 and it was incorrect, any ideas?
the amount leaked in t minutes is
a(t) = ∫r(t) dt
So, during the first 60 minutes, that would be
∫[0,60] 105 e^(-0.02t) dt = 3668.73
What did you do to calculate the solution?
I looked at an example on slader and used the method they used. This was right tho, thank you!
To find out how much oil leaks out during the first hour, we need to integrate the given rate function over the interval [0, 60], where t is measured in minutes.
The integral of r(t) = 105e^(-0.02t) with respect to t over the interval [0, 60] will give us the total amount of oil leaked during this time.
∫[0,60] 105e^(-0.02t) dt
To evaluate this integral, we can use the substitution method. Let's substitute u = -0.02t and du = -0.02dt.
∫ -5250e^u du
Now we can integrate:
-5250 ∫ e^u du
= -5250 * e^u + C
Substituting back the original variable:
= -5250 * e^(-0.02t) + C
To evaluate this integral over the interval [0, 60], we can substitute the upper and lower limits of integration:
= -5250 * e^(-0.02 * 60) - (-5250 * e^(-0.02 * 0))
Simplifying:
= -5250 * e^(-1.2) + 5250
Calculating this expression, we get approximately -67.65 liters.
However, since we are looking for a positive quantity (amount of oil leaked), we take the absolute value:
| -67.65 | ≈ 67.65 liters
So the amount of oil that leaks out during the first hour is approximately 67.65 liters.
To find the amount of oil that leaks out during the first hour, we need to integrate the rate function, r(t), over the interval [0, 60] minutes.
The rate function is given as r(t) = 105e^(-0.02t) liters per minute.
To integrate the rate function, we can use the formula for integrating exponential functions:
∫ e^at dt = (1/a) * e^at + C
Here, a = -0.02 and we want to integrate from t = 0 to t = 60. So, the integral becomes:
∫(0 to 60) 105e^(-0.02t) dt
= -5250 (e^(-0.02t)) ∣ (0 to 60)
Now, let's plug in the limits of integration:
= -5250 (e^(-0.02(60)) - e^(-0.02(0)))
= -5250 (e^(-1.2) - e^(0))
≈ -5250 (0.3012 - 1)
≈ -5250 (-0.6988)
≈ 3663.78
Now, since the amount of oil leaked cannot be negative, we take the absolute value of the result:
≈ |3663.78|
≈ 3664 liters
Therefore, the amount of oil that leaks out during the first hour is approximately 3664 liters.