(a) Explanation Given information: The volume of a growing spherical cell is V = 4 3 π r 3 , where the radius r is measured in micro-meters ( 1 μ m = 1 − 6 m ) . Calculation: The average rate of change of the volume of the spherical cell with respect to its radius r as r changes from r 1 to r 2 is given by Δ V Δ r = V ( r 2 ) − V ( r 1 ) r 2 − r 1 = 4 3 π r 2 3 − 4 3 π r 1 3 r 2 − r 1 Where V = 4 3 π r 3 is the volume of a sphere with radius of r . So that, The rate of change of V as r changes from 5 to 8 ( μ m ) V ( 8 ) − V ( 5 ) 8 − 5 = 4 3 π ( 8 3 − 5 3 ) 3 = 172 π = 540 (b) To determine To find: the instantaneous rate of change of V with respect to r when r = 5 μ m . (c) To determine To show: Show that the rate of change of the volume of a cell with respect to its radius is equal to its surface area ( S = 4 π r 2 ) . Explain geometrically why this result is true. Argue by analogy with Exercise 55 (c).

15th Edition

ISBN: 9781305177291

Author: James Stewart, Troy Day

Publisher: Cengage Learning

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Question

Chapter 3.3, Problem 56E

(a)

To determine

To find:average rate of change of V with respect tor as r changes from

5 to 8 μm (ii) 5 to 6 μm (iii) 5 to 5.1 μm

(b)

To determine

To find:the instantaneous rate of change of V with respect to r when r = 5 μm .

(c)

To determine

To show:Show that the rate of change of the volume of a cell with respect to its radius is equal to its surface area (S=4πr2) . Explain geometrically why this result is true. Argue by analogy with Exercise 55 (c).

Expert Solution & Answer

Chapter 3 Solutions

Biocalculus: Calculus For Life Sciences

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average rate of change of V with respect tor as r changes from 5 to 8 μ m (ii) 5 to 6 μ m (iii) 5 to 5.1 μ m

Solution Summary: The author calculates the average rate of change of the volume of a growing spherical cell with respect to its radius r.

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To find:average rate of change of V with respect tor as r changes from

5 to 8 μm (ii) 5 to 6 μm (iii) 5 to 5.1 μm

To find:the instantaneous rate of change of V with respect to r when r = 5 μm .

To show:Show that the rate of change of the volume of a cell with respect to its radius is equal to its surface area (S=4πr2) . Explain geometrically why this result is true. Argue by analogy with Exercise 55 (c).

Chapter 3 Solutions

average rate of change of V with respect tor as r changes from 5 to 8 μ m (ii) 5 to 6 μ m (iii) 5 to 5.1 μ m

Solution Summary: The author calculates the average rate of change of the volume of a growing spherical cell with respect to its radius r.

Concept explainers

Videos

To find:average rate of change of V with respect tor as r changes from 5 to 8 μm (ii) 5 to 6 μm (iii) 5 to 5.1 μm

To find:the instantaneous rate of change of V with respect to r when r = 5 μm .

To show:Show that the rate of change of the volume of a cell with respect to its radius is equal to its surface area (S=4πr2) . Explain geometrically why this result is true. Argue by analogy with Exercise 55 (c).

Chapter 3 Solutions

To find:average rate of change of V with respect tor as r changes from

5 to 8 μm (ii) 5 to 6 μm (iii) 5 to 5.1 μm