Explanation Given: V = 4 3 π r 3 S = 4 π r 2 Calculation: Consider that y = f ( x ) The relative change in y is given by Δ y y And Δ y = f ' ( x ) Δ x The volume and surface area of a tumor of radius r is given by V = 4 3 π r 3 S = 4 π r 2 Then, Δ V = V ' Δ r Δ S = S ' Δ r Differentiate V with respect to R, V = 4 3 π r 3 V ' = 3 × 4 3 π r 2 V ' = 4 π r 2 Then, Δ V = 4 π r 2 Δ r Now find Δ V V , Δ V V = 4 π r 2 Δ r 4 3 π r 3 Δ V V = 3 Δ r r As diameter is double of radius, d = 2r, therefore, Δ r r = Δ d d Δ V V = 3 Δ d d Now, the diameter is 19mm and increase in tumor is 1mm, therefore, Δ d d = 1 19 Hence, Δ V V = 3 Δ d d Δ V V = 3 × 1 19 Δ V V = 0

Bundle: Biocalculus: Calculus, Probability, And Statistics For The Life Sciences + Webassign Printed Access Card For Stewart/day's Biocalculus: Calculus For Life Sciences, 1st Edition, Multi-term

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ISBN: 9781305779037

Author: James Stewart, Troy Day

Publisher: Cengage Learning

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Chapter 3.8, Problem 18E

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The relative changes in the volume and the surface area.

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Chapter 3.8, Problem 17E

Chapter 3.8, Problem 19E

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Chapter 3 Solutions

Bundle: Biocalculus: Calculus, Probability, And Statistics For The Life Sciences + Webassign Printed Access Card For Stewart/day's Biocalculus: Calculus For Life Sciences, 1st Edition, Multi-term

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The relative changes in the volume and the surface area.

Solution Summary: The author explains the relative changes in volume and surface area of a tumor of radius r.

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