Explanation Given: The large helium balloon is being inflated at the rate of 120 cubic inches per second. Concept Used: Surface area of a sphere is: 4 π r 2 Volume of a sphere is: 4 3 π r 3 Calculation: To determine how fast the surface area of the balloon is increasing at the instant that the balloon has a radius of 12 inches find rate of change of radius with respect to time; that is d A d t , where A is surface area of balloon and t is time. First note that it’s given that the large helium balloon is being inflated at the rate of 120 cubic inches per second; that is d V d t = 120 cubic inches per second Here V represents volume of balloon, and negative sign represents that the balloon is being inflated; thus it’s volume of balloon is being increased. Since, volume of a sphere is 4 3 π r 3 , thus d V d t = 120 ⇒ d d t [ 4 3 π r 3 ] = 120 ⇒ 4 3 π d d t [ r 3 ] = 120 ⇒ 4 3 π ( 3 r 2 ) d d t r = 120 ⇒ 4 π r 2 d r d t = 120 ⇒ d r d t

14th Edition

ISBN: 9781464151088

Author: Associate Professor Laura Taalman, Peter Kohn

Publisher: W.H. Freeman & Company

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Chapter 3.5, Problem 45A

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To find how fast the surface area of the balloon is increasing when the balloon has a radius of 15 inches. When given that the large helium balloon is being inflated at the rate of 120 cubic inches per second.

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Chapter 3.5, Problem 44A

Chapter 3.5, Problem 46A

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To find how fast the surface area of the balloon is increasing when the balloon has a radius of 15 inches. When given that the large helium balloon is being inflated at the rate of 120 cubic inches per second.

Solution Summary: The author calculates how fast the surface area of the balloon is increasing at the instant that the large helium balloon has a radius of 15 inches.

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To find how fast the surface area of the balloon is increasing when the balloon has a radius of 15 inches. When given that the large helium balloon is being inflated at the rate of 120 cubic inches per second.

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