# Example 25

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 25 If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its volume. Let r be the radius of the sphere Given r = 9 cm Error in measurement of radius = ∆r ∆r = 0.03 cm Volume of the sphere = V = 43𝜋𝑟3 We need to find error in calculating the volume i.e ∆V ∆V = 𝑑𝑣𝑑𝑟 ∆r = 𝑑43 𝜋𝑟3𝑑𝑟 ∆r ∆V = 43𝜋𝑑𝑟3𝑑𝑟 ∆r ∆V = 43𝜋 3𝑟2 0.03 = 4𝜋𝑟2(0.03) = 4𝜋(9)2 (0.03) = 9.72𝜋 cm3 Hence, approximately error in calculating the volume is 9.72𝝅 cm3

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Example 14

Example 15

Example 16

Example 17

Example 18

Example 19

Example 20

Example 21

Example 22

Example 23

Example 24

Example 25 You are here

Example 26

Example 27

Example 28

Example 29

Example 30

Example 31

Example 32

Example 33

Example 34

Example 35 Important

Example 36

Example 37 Important

Example 38 Important

Example 39

Example 40 Important

Example 41

Example 42

Example 43

Example 44

Example 45

Example 46 Important

Example 47 Important

Example 48

Example 49

Example 50

Example 51

Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.