# The average rate of change of volume of sphere V with respect to when r , when r changes from i. 5 to 8 μm ii. 5 to 6 μm iii. 5 to 5.1 μm

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 3.8, Problem 16E

(a)

To determine

## To find: The average rate of change of volume of sphere V with respect to when r, when r changes from i. 5 to 8 μm ii. 5 to 6 μm iii. 5 to 5.1 μm

Expert Solution

1. i. Therefore, the average rate of change of the volume of a sphere is 172πμm3/μm_.
2. ii. Therefore, the average rate of change of the volume of a sphere is 121.333πμm3/μm_.
3. iii. Therefore, the average rate of change of the volume of a sphere is 102.013πμm3/μm_.

### Explanation of Solution

Calculation:

Volume of a sphere is as below.

V(r)=43πr3

Calculate the average rate of change of the volume of a sphere, when the radius of sphere changes from 5μm to 8μm.

V(r)=V(8)V(5)85

V(r)=43π(8)343π(5)33V(r)=43π(512)43π(125)3V(r)=172πμm3/μm

Therefore, the average rate of change of the volume of a sphere is 172πμm3/μm_.

Calculate the average rate of change of the volume of a sphere, when the radius of sphere changes from 5μm to 6μm.

V(r)=V(6)V(5)65V(r)=43π(6)343π(5)31V(r)=43π(216)43π(125)1V(r)=121.333πμm3/μm

Therefore, the average rate of change of the volume of a sphere is 121.333πμm3/μm_.

Calculate the average rate of change of the volume of a sphere, when the radius of sphere changes from 5μm to 5.1μm.

V(r)=V(5.1)V(5)5.15V(r)=43π(5.1)343π(5)30.1V(r)=43π(132.651)43π(125)0.1V(r)=102.013πμm3/μm

Therefore, the average rate of change of the volume of a sphere is 102.013πμm3/μm_.

(b)

To determine

### To find: The instantaneous rate of change when radius r is 5 μm.

Expert Solution

The instantaneous rate of change is 100πμm3/μm_.

### Explanation of Solution

Calculate the instantaneous rate of change V when r is 5μm

V(r)=43πr3

Differentiate the above volume equation.

V'(r)=43×3×πr2V'(r)=4πr2

Substitute the value 5 for r.

V'(5)=4π(5)2V'(5)=100πμm3/μm

Therefore, when the value of r is 5 the instantaneous rate of change is 100πμm3/μm_.

(c)

To determine

### To find: The rate of change of the volume of a sphere with respect to its radius is equal to the surface area of the volume and why it is geometrically true.

Expert Solution

It is geometrically true. Hence, the resulting change in volume ΔV if the value of change in radius Δr is small is 4πr2.

### Explanation of Solution

Surface area of the sphere is given as below.

S(r)=4πr2 (1)

First derivative of the volume expression is as given below.

V'(r)=4πr2 (2)

Compare the equations (1) and (2).

S(r)=V'(r)=4πr2

Show that if the change in radius Δr is small, then the change in the circle of area is approximately equal to its circumference. Prove the statement through geometry.

Approximate the resulting change in volume ΔV of the sphere of radius r, if change in radius Δr is small.

Write the ΔA expression as below.