Draining a tank (Torricelli’s law)

Question

Draining a tank (Torricelli’s law) A cylindrical tank with a cross-sectional area of $10$ m² is filled to a depth of $25$ m with water. At $t=0$ s, a drain in the bottom of the tank with an area of $1$ m² is opened, allowing water to flow out of the tank. The depth of water in the tank (in meters) at time $t$\geq{0}$$ is $d$\left$(t$\right$)=$\left$(5-0.22t$\right$)^2$ .
c. What is an appropriate domain for $d$ ?

c. What is an appropriate domain for $d$ ?

Accepted Answer

Welcome back everyone. In this problem, a cylindrical container holds a liquid with a level of 49 inches. The container has a cross sectional area of 15 square inches. And the function, l of t equals 7 minus 0.20 t all squared, describes the liquid level at the time t is greater than or equal to 0. What is the appropriate domain for l if a drain with an area of 4 square inches at the bottom of the container is open at t equals 0 seconds? For our answer choices, a is open parenthesis 0 15 close parenthesis. B is open bracket 0 20 closed parenthesis. C is open parenthesis 0 25 closed parenthesis and the d is open bracket 0 35 closed bracket. Now, what are we what are we trying to figure out here? Well, we want to know the appropriate domain for l for our Jane. But, let's make sure we understand what it's really talking about. So, here, we're talking about a cylindrical container, Probably, looks something like this. And let me draw that a little better. And for well, I tried. For our cylindrical container, we know that it has liquid at a level of 49 inches. Okay. And our function l describes the liquid level at a given time. We also know that there's a drain in our cylindrical container. Okay? So here's a drain. Let me put it in blue. Okay. And our drain is 4 square inches and it's releasing water at a rate well, that's not specified here but we know that it'll probably look something like this. So, we also know that for the function, since it disguise the liquid level at any time, then at the start or or when our container is full, then at the beginning, it would have been the liquid level when t equals 0 because t is our dependent variable. Likewise, at any time, say let's put it right here. Then, it would describe the liquid level given the time specified. So, at any point, then it's we have it given by the function l of t. So now, we know what we're talking about. When we say we want to find the appropriate domain for l, what it means is that we want to find all of the possible input values. In other words, we want to find the highest and lowest possible values for the time that describes the liquid level. So the liquid level in the container is dependent on the time. So we already know that at the start, the time would have been 0 seconds because we've just started. So when the liquid level is is or when the container is empty, okay, then the liquid level would have been equal to 0. So So in other words, whatever the function is for the liquid level, it would be equal to 0. So let's say that t e equals the time when the container container is empty. Okay? Then at that point, our domain okay? First, our domain starts at 0 seconds and then it goes up to the time when it's empty, t e. So we already have a part of our domain. It starts at 0 because remember, all of the values, the liquid level we are talking about the liquid level at the time t is greater than or equal to 0. Now, we just need to find the time when the container is empty. So how are we going to figure that out? Well, we already have a piece of the puzzle because we already know that when the container is empty, our function equals 0. So since our function l of t equals 7 minus 0.20tallsquared, then atloftequals0 at our the time when the container is empty, we can substitute those values which tells us that 0 equals 7 minus 0.20 t e, the time when the container is empty, all squared. Now that we have that, we've substituted those values and we realize that we can solve for the time when the container is empty. So let's go ahead and do that. Now, first, I want to get rid of the square so I can find the square root of the right hand side and the left hand side. The square root of 0 equals 0 and our square root would cancel our square and that leaves us with 7 minus 0.20te. If I add 0.20te to both sides, then 0.20te equals 7. Now, to find the time when the container is empty, I can divide both sides by 0.20 and 7 divided by 0.20 equals 35. So that means the time when the container is empty equals 35 seconds. So know that we have that time, therefore, this tells us that the domain of our function is 035. Open and close the bracket. And that makes sense because this is simply telling us that at the time it starts, it's 0 seconds and by the time it gets to 35 seconds, our container would be empty. When we go back to our answer choices, that would have been answer choice d. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Torricelli's Law

Differential Equations

Domain of a Function

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