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:
- Integrate the function:
\[ \int (400 - 8t) , dt = 400t - 4t^2 + C \]
- 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.
2 answers