a. Explanation Given: The radius of sphere is a linear function of time. Initial radius of the sphere is 10 cm and it disappears in 2 hours. Formula used: d d x ( x n ) = n x n − 1 d d x ( a f ( x ) ) = a d d x ( f ( x ) ) Calculations: The radius of sphere is a linear function of time. So, the radius be given by, r ( t ) = a t + b Where a andb are constants. Now, it is given that initial radius of sphere is 10 cm. That is, r ( 0 ) = 10 a ( 0 ) + b = 10 0 + b = 10 b = 10 That is, r ( t ) = a t + 10 Also, it is given that the sphere disappears in 2 hours. That is, r ( 2 ) = 0 a ( 2 ) + 10 = 0 2 a + 10 = 0 2 a = − 10 a = − 5 Hence, the radius of the sphere is given by, r ( t ) = − 5 t + 10 Here t is the time in hours. Now, volume of this sphere is given by, V ( r ) = 4 3 π r 3 Using r ( t ) = − 5 t + 10 , V ( t ) = V ( r ( t ) ) = b. To determine To find: the rate of change of surface area of sphere after 1 hour.

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Question

Chapter 3.5, Problem 46PS

To determine

To find:the rate of change of volume of sphere after 1 hour.

To determine

To find: the rate of change of surface area of sphere after 1 hour.

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Chapter 3.5, Problem 45PS

Chapter 3.5, Problem 47PS

Chapter 3 Solutions

Calculus

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the rate of change of volume of sphere after 1 hour.

Solution Summary: The author explains that the radius of sphere is a linear function of time.