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 $d\left(t\right)=\left(5-0.22t\right)^2$ .
b. At what time is the tank empty?

b. At what time is the tank empty?

Accepted Answer

Welcome back everyone. In this problem, a liquid chemical substance is stored in a cylindrical vessel with a cross sectional area of 30 square feet. At t equals 0 seconds, a discharge hole at the bottom of the vessel with an area of 3 square feet is opened. At the time t is greater than or equal to 0, the liquid level is given by the function l of t equals 4 minus 0.15t all squared. If the liquid level at t equals 0 seconds is 16 feet, determine the time elapsed when the vessel is empty. For our answer choices, a is 80 thirds seconds, b is 80 seconds, c is 40 thirds seconds and d is 40 seconds. Now, let's try and make sense of our problem here. So we're talking about a cylindrical vessel. Okay. Let me draw somewhat of a cylinder over here and it's storing a liquid chemical substance. And it has a discharge hole at the bottom of the vessel. So this is a chemical. So I'm gonna put this in red. Right? This is like some a substance you definitely don't wanna get. And for that discharge hole at the bottom okay. Let's imagine it looks something like this. Then, at the time t is greater than or equal to 0, the liquid level give is given by a function. L of t equals 4 minus 0.15 t all squared. So, what we're trying to figure out is that if we're given the liquid level at a particular time, we want to know the time elapsed when the vessel is empty. Now, if you think about it, when the vessel is full, right, when the vessel is full, our time is equal to 0. So, in other words, our function l would be l of 0 because that's when it's full. But, when our vessel is empty, when it's all the way down to the bottom of the ground, then, the liquid level at the time will be equal to 0. That makes sense. Right? Because no more liquid would be left in the container. So since since the liquid level when the vessel is empty equals 0, then we can find the time elapsed Because at that point where the liquid level is 0, we can substitute 0 into our expression and solve for the time elapsed. So let's go ahead and do that. So at a lot of t equals 0, that means our function now becomes 0 equals 4 minus 0.15 t all squared. We can find the square root of both sides to cancel the square on the right hand side. So the square cancels the square root, and the square root of 0 is 0. So that means 0 equals 4 minus 0.15t. We can add 0.15t to both sides, which means 0.15t equals 4. Therefore, to get t by itself, we can divide both sides by 0.15. So t equals 4 divided by 0.15. And when we think about it, 0.15, if we were because notice or or answers are written as fractions. So we want to find some way to write what we have here as a fraction. 0.15. Alright. Let me write it here. 0.15 can be written as the fraction 3 20ths. So we can rewrite our equation as t equals 4 divided by 3 20ths, which is 4 multiplied by 20 divided by 3. We can go ahead and multiply. 4 multiplied by 20 is 80. 1 multiplied by 3 is 3. Therefore, the time elapsed would be 80 thirds of seconds. When we look back in our answer choices, that would have been answer choice a. Thanks a lot for watching everyone. I hope this video helped.

Watch next