Water flows into a conical tank at a rate of 2 ft³/min. If the...

Question

Water flows into a conical tank at a rate of 2 ft³/min. If the radius of the top of the tank is 4 ft and the height is 6 ft, determine how quickly the water level is rising when the water is 2 ft deep in the tank.

Accepted Answer

Hello, in this video, we are going to be solving the following related rates problem. We are told that a spherical balloon is being inflated at a rate of 4 ft cubed per minute. We want to determine the rate of change at which the radius of the balloon is increasing when the radius is 2 ft. So, let's just go ahead and break down what the problem is telling us. The problem is telling us that we are working with a balloon that is in the shape of a sphere. Now, we are also told that air is being pumped into the sphere at a rate of 4 ft cued per minute. When air is being pumped into the sphere, that is going to expand the volume of the sphere. That means that the rate of change of the volume is 4 ft cubed per minute. And we can represent the rate of change in the volume as a time derivative DVDT. What we want to do If we want to solve for the rate of change of the radius. We can so we can write the rate of change of the radius as the time derivative DRDT, and we want to solve for the rate of change of the radius when the radius is equal to 2. So, our goal is to solve for DRDT of the rate of change of the volume of a sphere. Since we are talking about the volume of the sphere, let's go ahead and write down the equation to the volume of a sphere. That equation is V is equal to 4/3 pi multiplied by radius cubed. What we want to do now is we want to solve for the rate of change of the radius. In order to do this, we can take the derivative of the entire equation with respect to time. Through implicit differentiation that is going to give us DVDT equal to 4/3 pi multiplied by the derivative of r cubed, which is 3R squared through the power rule, then multiplied by DRDT through implicit differentiation. We are going to simplify this equation. We will multiply 4/3, pine, and 3 are squared together. That is going to give us DVDT. Equal to 12 i. Are squared divided by 3, and this will be multiplied by DRDT. This can further simplify to be DVDT equal to 4 pi R squared DRDT. Now, our goal here is to solve for the rate of change of the radius. So our goal is to solve for DRDT. We can replace DVDT with 4, because we are told the rate of change of the volume is 4, and we can replace the radius with 2, since we are interested in the rate of change of the radius, when the radius is equal to 2. That will allow us to further simplify the equation to be 4 equal to 4 multiplied by pi, multiplied by 2 squared, multiplied by DRDT. Now let's simplify the equation further. 2 squared will simplified to be 4, and 4 multiplied by 4 will be 16. So we have 4 equal to 16 pi multiplied by DRDT. And if we divide both sides of the equation by 16 pi, we have the rate of change of the volt of the radius, DRDT equal to 4 divided by 16 pi. This will further simplify to be 1 divided by 4 pi. Which is going to be the rate of change of the radius. This will be 1 divided by 4 ft per minute. And this will be the solution to the problem. And with that being said, the overall solution is going to be B. So, I hope this video helps you in understanding how to approach a related rates problem, and I will go ahead and see you all in the next video.

Water flows into a conical tank at a rate of 2 ft³/min. If the radius of the top of the tank is 4 ft and the height is 6 ft, determine how quickly the water level is rising when the water is 2 ft deep in the tank.

Key Concepts

Related Rates

Volume of a Cone

Implicit Differentiation

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