# Example 49 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 49 (Method 1) A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm. Let r be the radius of a circular disc & y be the area of a circular disc. We know that Area of a circle = πr2 𝑦=𝜋𝑟2 Diff w.r.t 𝑥 𝑑𝑦𝑑𝑥 = 𝑑𝜋𝑥2𝑑𝑥 𝑑𝑦𝑑𝑥 = π 2𝑥 Given that Due to expansion its radius increases at the rate of 0.05 cm/s i.e. ∆𝑥 = 0.05 when 𝑥= 3.2 cm We have to find approximate expansion in area when 𝑥 = 3.2 cm ∆y = 𝑑𝑦𝑑𝑥 ∆𝑥 Putting values ∆y = 2π𝑥 . ∆𝑥 ∆y = 2π 3.2 × 0.05 ∆y = 0.320 π Since area in cm2 & time is in seconds Thus the approximate change in volume is 0.320π cm2/s. Example 49 (Method 2) A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at the rate of 0.05 cm/s. Find the rate at which its area is increasing when radius is 3.2 cm. Let A be the area & radius of circular disc A = πr2 Given radius increases at 0.05 cm/s 𝑑𝑟𝑑𝑡 = 0.05cm/s We need to find the rate at which its area is increasing i.e. 𝑑𝐴𝑑𝑡 Now, 𝑑𝐴𝑑𝑡 = 𝑑𝜋𝑟2𝑑𝑡 = π . 𝑑𝑟2𝑑𝑡 = π . 𝑑𝑟2𝑑𝑟 . 𝑑𝑟𝑑𝑡 = π 2𝑟 𝑑𝑟𝑑𝑡 We need to find rate at which area is increasing when radius = 3.2cm 𝑑𝐴𝑑𝑡 = π . 2 × 3.2 ×0.05 = 0.320π Since area is in cm2 & time is in seconds 𝒅𝑨𝒅𝒕 = 0.320π cm2/s

Examples

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

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 You are here

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 has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.