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 $d\left(t\right)=\left(5-0.22t\right)^2$ .
a. Check that $d\left(0\right)=25$ , as specified.

a. Check that $d\left(0\right)=25$ , as specified.

Accepted Answer

Welcome back, everyone. In this problem, oil is introduced to a cylindrical vessel until it reaches a depth of 36 feet. The cross sectional area of the vessel is 15 square feet and a drain with an area of 2 square feet located at the bottom of the vessel is opened at t equals 0 seconds. At the time t is greater than or equal to 0, the depth of the oil is represented by the function d of t equals 6 minus 0.13t all squared. Does this function provide a depth of 36 feet as specified at t equals 0 seconds? That's what we're trying to figure out. Now first, let's try and make sense of the problem here. So we're talking about oil in a cylindrical vessel. So I'm going to draw this a sketch of this cylindrical vessel on the right. And in our cylindrical vessel, it tells us that the oil is introduced until it reaches a depth of 36 feet. So we could say that the oil comes to 36 feet here. Let's represent that on our diagram. Now, it also tells us that a drain is located at the bottom of the vessel and it's opened at the time t equals 0 second. In other words, it's opened at the start. And we have a function that represents the depth of the oil depending on the time that has passed. D of t equals 6 minus 0.13t all squared. And essentially, what we want to know is we want to confirm whether the function provides a depth of 36 feet as specified at t equals 0 seconds. In other words, if we put t equals 0 into our function okay. So if we put t equals 0 into our function, then d of 0 that is d with 0 as the input should be equal to 36 feet. That's what we're trying to figure out. So let's go ahead and do that. So substituting t equals 0 into our function d of t. Now our function which is going to be d of 0 equals 6 minus 0.13 t. So that's 0.13 multiplied by 0 all squared. Now 0.13 multiplied by 0 is 0. So our function would be 6 squared which equals 36. So del 0 equals 36 feet. Therefore, it is true. The function does provide a depth of 36 feet at t equals 0 seconds. Thanks a lot for watching everyone. I hope this video helped.

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