Question

How much water flows out of the tank during the first 2 minutes if a valve in a full 6000 gallon water tank is slowly opening and water flows out of the tank through the valve.at a rate in gallons per hour is given by the function \[f\left( t \right) = 300 \cdot {t^2}\] where t is in minutes?

Answer 1

Hint: An average rate of change function is a process that calculates the amount of change in one item divided by the corresponding amount of change in another. To solve the given function, we need to consider the given data, that water flows out of the tank during the first 2 minutes, so we need to integrate the given function between \[t = 0\](the start) and \[t = 2\]min, and then apply the limits to the function to get the quantity of water that flows out of the tank during the first 2 minutes.

Formula used:
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\]

Complete step by step solution:
Given,
water tank = 6000 gallon.
\[f\left( t \right) = 300 \cdot {t^2}\]
Hence, integrate the given function between \[t = 0\](the start) and \[t = 2\]min,
\[f\left( t \right) = 300 \cdot {t^2}\]
We know that, \[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + c\], hence applying it to the given function for\[\int {{t^2}dx = \dfrac{{{t^3}}}{3}} \], we have:
\[ \Rightarrow \int\limits_0^2 {\left( {300 \cdot {t^2}} \right)dt = 300\left[ {\dfrac{{{t^3}}}{3}} \right]} _0^2\]
Now, apply the limits to the terms, as:
\[ = 300\left( {\dfrac{{{2^3}}}{3} - \dfrac{{{0^3}}}{3}} \right)\]
\[ = 300\left( {\dfrac{8}{3} - 0} \right)\]
Evaluating the terms, we get:
\[ = 300\left( {\dfrac{8}{3}} \right)\]
\[ = \dfrac{{2400}}{3}\]
Hence, we get:
\[ = 800\] gallons.
Therefore, 800 gallons of water flows out of the tank during the first 2 minutes and here we have used the given function as rate of gallons per minutes.

Note: The key point to solve the given question, is that we need to consider the time that water flows out of the tank during the first 2 minutes in a 6000 gallons water tank, as to find the average rate of change, we divide the change in the output value by the change in the input value. The average rate of change is constant for a linear function.