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 that flows from the tank during the first 45 minutes.

r(t) = 400 − 8t

If v is the volume, r(t) = dv/dt. So,
v(t) = 400t - 4t^2
v(45) = 9900

Well, let's calculate this in a fun way, shall we?

Since we're looking for the amount of water that flows from the tank during the first 45 minutes, we need to find the integral of the function r(t) over the interval [0, 45]. And by integrating, I don't mean you have to invite a bunch of mathematicians over for a party, though that could be fun too!

Now, let's go ahead and integrate r(t) = 400 - 8t with respect to t over the said interval. But I must warn you, numbers might get a little crazy here, so hold on to your seat!

∫[0, 45] (400 - 8t) dt

Now, integrating 400 with respect to t gives us 400t, and integrating -8t with respect to t gives us -4t^2. But don't forget the limits of integration!

So, let's evaluate the integral using the limits 0 and 45:

[400t - 4t^2] from 0 to 45

Now, I'm just going to plug in the limits and see what we get:

[400(45) - 4(45)^2] - [400(0) - 4(0)^2]

400 * 45 is 18,000, and 45 squared is 2,025. So we have:

[18,000 - 4 * 2,025] - [0 - 0]

Now, let's simplify things a bit:

18,000 - 8,100 - 0

Hey, wait a minute, we have 0's! So, all those scary calculations were just a big joke. The answer is 9,900 liters.

There you have it! The amount of water that flows from the tank during the first 45 minutes is 9,900 liters. I hope my little math comedy routine kept you entertained!

To find the amount of water that flows from the tank during the first 45 minutes, we need to calculate the integral of the rate function from 0 to 45.

The rate function is given as r(t) = 400 - 8t liters per minute.

The integral of a rate function represents the accumulation of the quantity over a given interval.

So, the amount of water that flows from the tank during the first 45 minutes can be found by evaluating the integral of r(t) from 0 to 45:

∫(0 to 45) (400 - 8t) dt

To evaluate the integral, we need to find the antiderivative of the function 400 - 8t.

The antiderivative of 400 is 400t, and the antiderivative of -8t is -4t^2.

So, the antiderivative of (400 - 8t) is 400t - 4t^2.

Now, we can evaluate the definite integral from 0 to 45:

∫(0 to 45) (400 - 8t) dt = [400t - 4t^2] evaluated from 0 to 45

Plugging in the upper limit, we get:

[400(45) - 4(45)^2] - [400(0) - 4(0)^2]

Simplifying further:

[18000 - 4(2025)] - [0 - 0]

[18000 - 8100] - [0]

= 9900 liters

Therefore, the amount of water that flows from the tank during the first 45 minutes is 9900 liters.

To find the amount of water that flows from the tank during the first 45 minutes, we need to integrate the flow rate function over the interval [0, 45].

First, let's rewrite the flow rate function as:

r(t) = 400 - 8t

To find the amount of water that flows during the first 45 minutes, we need to integrate this flow rate function over the interval [0, 45]:

∫[0,45] (400 - 8t) dt

Now, let's evaluate this integral.

∫(400 - 8t) dt = 400t - 4t^2 + C

Now, substitute the limits of integration:

[400t - 4t^2] from 0 to 45

Plugging in the upper limit:

= [400 * 45 - 4 * (45)^2] - [400 * 0 - 4 * (0)^2]

= [18000 - 4 * 2025] - [0 - 0]

Simplifying:

= 18000 - 8100

= 9900 liters

Therefore, the amount of water that flows from the tank during the first 45 minutes is 9900 liters.

From a gentle stand point...

There are only 400 L in the tank.
r(t) = 400 − 8t
If time is 45 minutes then
r(t) = 400 - 8(45)
= 400 - 360
= 40
So 360 Litres has left the tank : )