Water flows from a storage tank at a rate of 500 − 3t liters per minute. Find the amount of water that flows out of the tank during the first 13 minutes.
integral of (500-3t)
= 500 t - 1.5 t^2
at t = 13, integral = 500(13)-1.5(169)
= 6500 - 253.5
= 6246.5 liters
Oh, the tank is flowing with water and it seems like it's in a hurry to leave. So, let's find out how much water escapes from the tank during the first 13 minutes.
Given that the rate of flow is 500 - 3t liters per minute, we can plug in t = 13 to find out the rate at that time:
Flow rate at t = 13: 500 - 3(13) = 500 - 39 = 461 liters per minute
Now, we just need to multiply this flow rate by the number of minutes to find the total amount of water that flows out:
Amount of water flowing out = Flow rate x Time
Amount of water flowing out = 461 liters/minute x 13 minutes
Amount of water flowing out = 5993 liters
So, during the first 13 minutes, approximately 5993 liters of water flow out of the tank. Phew! That's a lot of water.
To find the amount of water that flows out of the tank during the first 13 minutes, we need to calculate the integral of the flow rate function over the interval [0, 13].
The flow rate function is given by:
f(t) = 500 - 3t (liters per minute)
To find the amount of water that flows out, we need to integrate this function over the interval [0, 13]:
∫[0,13] (500 - 3t) dt
The integral of (500 - 3t) dt is:
∫(500 - 3t) dt = 500t - (3/2)t^2 + C
Now, we can evaluate this integral from 0 to 13:
∫[0,13] (500 - 3t) dt = [500t - (3/2)t^2] evaluated from 0 to 13
= (500(13) - (3/2)(13)^2) - (500(0) - (3/2)(0)^2)
= (6500 - 253/2) - (0 - 0)
= 6500 - 253/2
= 6500 - 126.5
= 6373.5
Therefore, the amount of water that flows out of the tank during the first 13 minutes is approximately 6373.5 liters.
To find the amount of water that flows out of the tank during the first 13 minutes, we need to integrate the given flow rate function over the interval [0, 13].
The flow rate function is given by:
flow rate = 500 - 3t liters per minute
To integrate this function, we need to find its antiderivative. Let's call the antiderivative "f(t)":
f(t) = ∫(500 - 3t) dt
Now we can find the integral of the function:
f(t) = 500t - (3/2)t^2 + C,
where C is the constant of integration.
To find the value of C, we need additional information. Assuming that at time t=0, there is no water in the tank, we can use this condition to determine C:
f(0) = 0 ⇒ 500(0) - (3/2)(0)^2 + C = 0 ⇒ C = 0.
Now we have the final equation for the amount of water that flows out of the tank during the first 13 minutes, given by the integral:
f(t) = 500t - (3/2)t^2
To find the amount of water that flows out during the first 13 minutes, we evaluate the definite integral from t=0 to t=13:
Amount of water = ∫[0, 13](500t - (3/2)t^2) dt
Let's evaluate the integral:
Amount of water = [250t^2 - (3/4)t^3] from 0 to 13
= 250(13)^2 - (3/4)(13)^3 - (250(0^2) - (3/4)(0^3))
= 250(169) - (3/4)(2197)
= 42,250 - 1,648.25
= 40,601.75 liters
Therefore, the amount of water that flows out of the tank during the first 13 minutes is approximately 40,601.75 liters.