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The radius of a sphere is increasing at a rate of $ 4 mm/s. $ How fast is the volume increasing when the diameter is $ 80 mm? $

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04:12

Alex Lee

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 9

Related Rates

Derivatives

Differentiation

Missouri State University

Campbell University

Oregon State University

Idaho State University

Lectures

04:40

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.

44:57

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.

05:00

The radius of a sphere is …

00:30

If the radius of a sphere …

04:33

The volume of a spherical …

The diameter of a certain …

03:44

At what rate is the volume…

01:52

The radius of a spherical …

06:31

The radius $r$ and height …

mhm. In this question we're asked that given the radius of the spheres increase, we want to find how fast the volume increases at a surge in diameter. We first need to remember what the formula for the volume of the sphere is. We know that the volume of the sphere is 4/3 times pi times r cubed. But remember that the volume and the radius are both changing as functions of time. So it's more correct to write it as as functions of time. Now we'll just differentiate both sides with respect to time. Using implicit differentiation. We know that the rate of change of the volume is the prime of T equaling four thirds times pi tons. Now this are of T cubed requires a change rule. We first need to use the power rule to bring the power down. So we're going to have three times our prime times R f T squared multiplied by the derivative of R. Of T. Which is our prime of T. Yeah. And now we can plug in. Our rates were given that the volume Is increasing if the diameter is 80 That means that the radius has to be 40. Yeah. And we're also given that the rate of increase is four mm/s. So we can plug all of this information in. And also this three and three we'll cancel. So we'll have this is equal to four times pi times 40 squared times four. And when we evaluate this, we get that. This is eagle to 25,000 600 pie and this is in millimeters cubed per second. That's how you do this question.

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