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A cylindrical tank with radius $ 5 m $ is being filled with water at a rate of $ 3 m^3/min. $ How fast is the height of the water increasing?

$\frac{d h}{d t}=\frac{3}{25 \pi} \mathrm{m} / \mathrm{min}$

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Missouri State University

Oregon State University

Harvey Mudd College

Boston College

in this question, we're asked that given a cylindrical tank, it has a specific Gradus and it's being filled at a specific rate with some liquid. And you want to wonder how fast is the height increasing? So let's deal with some of the issues here. We need to know first the volume of a cylinder, the volume of a cylinder. We know that the volume, it's just going to be the usual two pi r square times age. But since obviously the radius is not expanding, we're also were given such a radius. Yeah. And which is 25 Five squared is 25. So our formula becomes just 25 high times H Yeah, technically, since the volume is changing and the height is changing when we fill the tank, it's more correct to write this as The volume is a function of time equals 25 pi times the height is a function of time. Oh, now differentiate both sides using implicit differentiation with respect to time. When we do so we're going to get that the derivative with respect to time Equals 25 pi times the height with respect to time. And since we're given the rate of change of the volume with respect to time, we're going.