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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?

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

05:19

Alex Lee

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 9

Related Rates

Derivatives

Differentiation

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In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

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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.

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