A water heater that has the shape of a right cylindrical tank ...

Question

A water heater that has the shape of a right cylindrical tank with a radius of 1 ft and a height of 4 ft is being drained. How fast is water draining out of the tank (in ft³/min) if the water level is dropping at 6 min/in?

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says a cylindrical swimming pool with a radius of 2 m and a depth of 5 m is being emptied. If the water level is dropping at a rate of 3 centimeters per minute, at what rate is the water leaving the pool in cubic meters per minute? We give 4 possible choices as our answers. For choice A, we have the derivative of V with respect to T is equal to minus 3 pi divided by 25, cubic meters per minute. For choice B, we have the derivative of V with respect to T is equal to 3 pi divided by 25, cubic meters per minute. For choice C, we have the derivative of V, with respect to T is equal to 4 divided by 25 cubic meters per minute. And for choice D, we have the derivative of V with respect to T is equal to -4 divided by 25 cubic meters per minute. Now, we need to find the rate at which the water is leaving the pool in cubic meters per minute. And so, if we're looking for the rate at which the water is pool, we're looking at the rate at which our volume of water is changing. So we need to determine the volume of our water in the swimming pool, and since we have a cylindrical swimming pool, that means that our volume here is going to be that of a cylinder. So recall your formula for the volume of a cylinder, which is V is equal to pi R2 multiplied by H, where V here is the volume, R is the radius of the cylinder, and H is going to be the depth of our water. Now, we need to find the rate at which the volume is changing, so we'll take a derivative with respect to T of both sides of our equation here. So we'll get the derivative of V with respect to T. is equal to and on the right hand side pi is a constant so we can move through the derivative. R here our radius is also a constant, so it can move through the derivative. However, our height of our water is going to be changing, so we will have the derivative of H with respect to T. That means on the right hand side we're going to have i R squared multiplied by the derivative of H with respect to T. So now we need to use the information that we're given on the problem to identify values for the quantities on the right hand side of this expression. So, the first quantity we need to look at is our radius R, and this is going to be equal to 2 m, since that is the value that was given in the problem. We are also given the value for our derivative of H with respect to T. And if we look at our problem, it says that the water level is dropping at a rate of 3 centimeters per minute. So dropping there indicates that we're going to have a negative change in our height. So this means that our derivative of H with respect to T is going to be equal to minus 3 centimeters per minute. However, we want everything in units of meters, not centimeters, since our final answer needs to be in cubic meters per minute. So we're going to convert centimeters into meters by dividing this quantity by 100. So this is going to be equal to -3 divided by 100, and that will give us units of meters per minute. So now we can substitute these values back into our expression for the derivative of V with respect to T, and we will have the derivative of V with respect to T. is equal to pi multiplied by the quantity of 2 m in quantity squared multiplied by The quantity of -3 divided by 100. Meters per minute. So, simplifying this expression, this is going to be equal to, we'll have 2 squad, which is 4, and 4 will go into 125 times. So this is going to be equal to minus 3 pi. Divided by 25 and then the units will be meters cubed per minute. So, this is the answer that we are actually looking for, and when we can compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

A water heater that has the shape of a right cylindrical tank with a radius of 1 ft and a height of 4 ft is being drained. How fast is water draining out of the tank (in ft³/min) if the water level is dropping at 6 min/in?

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