Question

Water flows from the bottom of a storage tank at a rate of
r(t) = 400 − 8t
liters per minute, where 0 ≤ t ≤ 50. Find the amount of water (in liters) that flows from the tank during the first 35 minutes.
liters

Answer 1

To find the amount of water that flows from the tank during the first 35 minutes, we need to calculate the integral of the flow rate function \( r(t) = 400 - 8t \) from \( t = 0 \) to \( t = 35 \).

The amount of water \( W \) that flows out of the tank can be found using the definite integral:

\[
W = \int_0^{35} r(t) \, dt = \int_0^{35} (400 - 8t) \, dt
\]

Now, we can compute the integral:

1. **Integrate the function**:

\[
\int (400 - 8t) \, dt = 400t - 4t^2 + C
\]

2. **Evaluate the definite integral from 0 to 35**:

\[
W = \left[ 400t - 4t^2 \right]_0^{35}
\]

Calculating at the upper limit \( t = 35 \):

\[
W(35) = 400(35) - 4(35^2) = 14000 - 4(1225) = 14000 - 4900 = 9100
\]

At the lower limit \( t = 0 \):

\[
W(0) = 400(0) - 4(0^2) = 0
\]

So, the water that flows out during the first 35 minutes is

\[
W = 9100 - 0 = 9100 \text{ liters}
\]

Thus, the amount of water that flows from the tank during the first 35 minutes is **9100 liters**.