Water is drained out of an inverted cone that has the same dim...

Question

Water is drained out of an inverted cone that has the same dimensions as the cone depicted in Exercise 36. If the water level drops at 1 ft/min, at what rate is water (in ft³/min) draining from the tank when the water depth is 6 ft?

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says a cylindrical tank is being emptied of water at a constant rate. The tank has a height of 10 ft and a radius of 5 ft. If the water level is decreasing at a rate of 0.5 ft per minute, what is the rate at which the volume of water is changing in cubic feet per minute when the water depth is 4 ft? We have 4 possible choices as our answers. For choice A, we have the derivative of V with respect to T is equal to -2.5 multiplied by pibcu per minute. For choice B, we have the derivative of V with respect to T is equal to 2.5 multiplied by i BQ per minute. For choice C, we had the derivative of V with respect to T is equal to minus 12.5 multiplied by pi feet cubed per minute. And for choice D, we have the derivative of V with respect to T is equal to 12.5 multiplied by pi feet cubed per minute. Now, the question is wanting to know what is the rate at which the volume of water is changing in cubic feet per minute, when the water depth is 4 ft. So, the first thing we're going to do is start off by recalling our formula for the volume of a cylinder. So, recall that that is V is equal to pi R2 multiplied by H, where V here is going to be our volume, R is the radius of the cylinder, and H is the height of the cylinder. Now we're wanting to know what is the rate at which the volume of water is changing. So we'll need to take a derivative with respect to time of both sides of this equation. So we'll take a derivative with respect to T of both sides. When I do so, I'll get the derivative of V with respect to T. is equal to on the right hand side, i is a constant, so we can move through the derivative. R here is the radius. It's also a constant, so it can move through the derivative. However, the height of our water is changing. So on the right hand side we'll have pi R2 multiplied by the derivative of H with respect to T. So now we just need to identify the quantities in our equation here. So, R here is the radius, and we're told that The radius in our problem is going to be 5 ft. So, R is equal to 5 ft. We're also given the rate at which the water level is changing, so that is our derivative of H with respect to T. And we're told that the water level is decreasing at a rate of 0.5 ft per minute. So, since it's decreasing, it's going to be a negative value. So our derivative of H with respect to T is going to be equal to minus 0.5 ft per minute. So now we'll substitute these values into our expression for our change in volume with respect to time. So that's gonna be equal to pi. Multiplied by The quantity of 5 ft. In quantity squared multiplied by -0.5 ft per minute. So simplifying this expression here, this is going to be equal to. Minus 12.5, multiplied by pi, and that has units of feet cubed per minute. Now, here, our change in volume with respect to time is a constant. So, this is going to be true at any water depth, so this is going to be true when the water depth is 4 ft. So this here is actually our answer to the problem. And so, if we compare our answer here to our answer choices, we see that the correct answer choice corresponds to answer choice C. So I hope that this explanation has been useful, and I'll see you in the next video.

Water is drained out of an inverted cone that has the same dimensions as the cone depicted in Exercise 36. If the water level drops at 1 ft/min, at what rate is water (in ft³/min) draining from the tank when the water depth is 6 ft?

Key Concepts

Related Rates

Volume of a Cone

Differentiation

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