💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # 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?

## (a) $S^{\prime}(1)=8 \pi \mathrm{ft}^{2} / \mathrm{ft}$(b) $S^{\prime}(2)=16 \pi \mathrm{ft}^{2} / \mathrm{ft}$(c) $S^{\prime}(3)=24 \pi \mathrm{ft}^{2} / \mathrm{ft}$

Derivatives

Differentiation

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##### Top Calculus 1 / AB Educators   ##### Kristen K.

University of Michigan - Ann Arbor ##### Samuel H.

University of Nottingham

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### Video Transcript

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 Oregon State University

#### Topics

Derivatives

Differentiation

##### Top Calculus 1 / AB Educators   ##### Kristen K.

University of Michigan - Ann Arbor ##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp