Explanation Given: The large helium balloon is being inflated at the rate of 120 cubic inches per second. Concept Used: Volume of a sphere is: 4 3 π r 3 Calculation: To determine how fast the radius of the balloon is increasing when the balloon contains 300 cubic inches of air find rate of change of radius with respect to time; that is d r d t , where r is radius 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 = 120 4 π r 2 ⇒ d r d t = 9

14th Edition

ISBN: 9781464140051

Author: TAALMAN, Laura, Kohn, Peter

Publisher: W. H. Freeman

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Chapter 3.5, Problem 44A

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To find how fast the radius of the balloon is increasing when the balloon contains 300 cubic inches of air. 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 43A

Chapter 3.5, Problem 45A

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To find how fast the radius of the balloon is increasing when the balloon contains 300 cubic inches of air. 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 radius of the large helium balloon is increasing when the balloon contains 300 cubic inches of air.

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To find how fast the radius of the balloon is increasing when the balloon contains 300 cubic inches of air. 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 Solutions