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A spherical balloon is being inflated. Find the rate of increase of the surface area $ (S = 4\pi r^2) $ with respect to the radius $ r $ when $ r $ is (a) 1 ft, (b) 2 ft, and (c) 3 ft. What conclusion can you make?

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00:53

Amrita Bhasin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 7

Rates of Change in the Natural and Social Sciences

Derivatives

Differentiation

Campbell University

Oregon State University

Boston College

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.

01:32

A spherical balloon is bei…

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04:07

Inflating a balloon The vo…

02:25

The volume $V=(4 / 3) \pi …

01:22

05:05

04:53

Suppose that you are blowi…

02:59

in this problem, There's a spherical balloon. And so we're interested in looking at the surface area. And we know the surface area of a sphere is for pyre squared. We're looking at the rate of increase, so we need the derivative s prime of our. So that would be eight pi. Times are. And what we want to do is evaluate that derivative for various radius values. Articles one, articles two and r equals three feet so we can substitute those numbers in and we get s prime of one is a pie times one, which is a pie. And the units for service area would be square feet and the units for radius would be feet so of square feet per foot and then s prime of to we would have a pie times two So 16 pie square feet per foot, and for three s prime of three would be a pi times three so 24 pi feet per foot, square feet per foot. So a very small change in radius gives you a very large change in surface area. That would be my conclusion

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