Video: Definite Integrals and the Net Change Theorem to Find the Volume of Water in a Tank

Water is poured in a tank at a rate of π‘Ÿ(𝑑) = 0.01𝑑² βˆ’ 0.02𝑑 + 10 liters per hour. The tank contains a hole from which water leaks at a rate of 𝑠(𝑑) = √(3.1𝑑) liters per hour. Given that the tank is initially empty, find the total amount of water in the tank after 3 hours. Give your answer to the nearest liter.

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Video Transcript

Water is poured in a tank at a rate of π‘Ÿ of 𝑑 is equal to 0.01𝑑 squared minus 0.02𝑑 plus 10 liters per hour. The tank contains a hole from which water leaks at a rate of 𝑠 of 𝑑 is equal to the square root of 3.1𝑑 liters per hour. Given that the tank is initially empty, find the total amount of water in the tank after three hours. Give your answer to the nearest liter.

First, let’s discuss what this question is actually asking us to do. The question is asking us to find the total amount of water in the tank after three hours and to give this value to the nearest liter. Since we were asked to find the volume of water in the tank, let’s define a new function 𝑣 of 𝑑, which is the volume of water in the tank in liters after 𝑑 hours. Since we’re told that the tank is initially empty, the volume of water in the tank after zero hours must be equal to zero. Therefore, we can conclude that 𝑣 of zero must be equal to zero.

Let’s try to visualize what’s happening. We’re given a tank of water and we’re told that the rate of water being pumped into this tank is equal to π‘Ÿ of 𝑑 and that there is a leak in this tank and the rate of water leaking out is equal to 𝑠 of 𝑑. The crucial thing to note here is that we’re given the rate of water which is being pumped in and the rate of water which is being leaked out. We are not given the volume which is leaving. If the rate of water being pumped into the tank is π‘Ÿ of 𝑑 and the rate of water being leaked out of the tank is 𝑠 of 𝑑, then π‘Ÿ of 𝑑 minus 𝑠 of 𝑑 must be the rate of change of the volume of water in the tank. And we know that the rate of change of the volume of water in the tank is the same as saying 𝑣 prime of 𝑑.

We’re now ready to use the net change theorem which states that the integral from π‘Ž to 𝑏 of capital 𝑓 prime of π‘₯ with respect to π‘₯ is equal to capital 𝐹 of 𝑏 minus capital 𝐹 of π‘Ž. Applying the net change theorem to 𝑣 prime of 𝑑 on an interval, π‘Ž to 𝑏, gives us that this integral is equal to the volume after 𝑏 hours minus the volume after π‘Ž hours. Since the question wants us to find the volume after three hours and we know that the tank is initially empty, so we already have that 𝑣 of zero is equal to zero. We will use the bounds of zero and three. So this gives us an expression for the volume of water in the tank after three hours.

We can now use our formula for the rate of change of volume of water in the tank to evaluate this integral. We change our integrand of 𝑣 prime of 𝑑 with π‘Ÿ of 𝑑 minus 𝑠 of 𝑑. We then substitute in that π‘Ÿ of 𝑑 is equal to 0.01𝑑 squared minus 0.02𝑑 plus 10 and that 𝑠 of 𝑑 is equal to the square root of 3.1𝑑. We can then integrate each term individually by using the power rule for integrals where we take care to notice that the square root of 3.1𝑑 can be rewritten as the square root of 3.1 multiplied by 𝑑 to the power of a half. We can then simplify all of these expressions.

We then just have to evaluate this by substituting in our limits of zero and three. What we noticed that when we substitute in the limit of zero, we would just get zero. When we finally calculate this, we will get an answer, which is approximately equal to 23.9 liters, which to the nearest liter is 24 giving us a final answer that after three hours there will be approximately 24 liters of water in the tank.

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