Subscribe to our Youtube Channel - https://you.tube/teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

Example 44 A man of height 2 meters walks at a uniform speed of 5 km/h away from a lamp post which is 6 meters high. Find the rate at which the length of his shadow increases.Let AB be the lamp post & Let MN be the man of height 2m. & Let AM = l meter & MS is the shadow of the man Let length of shadow MS = s Given man walks at speed of 5 km/h ∴ 𝑑𝑙/𝑑𝑡 = 5 km/h We need to find rate at which the length of his shadow increases. i.e. we need to find 𝑑𝑠/𝑑𝑡 In ΔASB tan θ = 𝐴𝐵/𝐴𝑆 tan θ =6/(𝑙 + 𝑠) In ∆ MSN tan θ = 𝑀𝑁/𝑀𝑆 tan θ =2/𝑠 …(2) From (1) & (2) 6/(1000 (𝑙 + 𝑠) ) = 2/1000(𝑠) 6/(𝑙 + 𝑠) = 2/𝑠 6s = 2l + 2s 6s – 2s = 2l 4s = 2l 2s = l l = 2s We need to find 𝑑𝑠/𝑑𝑡 Now, l = 2s Diff w.r.t t 𝑑𝑙/𝑑𝑡= 𝑑(2𝑠)/𝑑𝑡 𝒅𝒍/𝒅𝒕= 2.𝑑𝑠/𝑑𝑡 5 = 2 𝑑𝑠/𝑑𝑡 𝑑𝑠/𝑑𝑡 = 5/2 So, 𝒅𝒔/𝒅𝒕 = 𝟓/𝟐 km/hr. (Given 𝑑𝑙/𝑑𝑡 = 5km/hr)

Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8 Important

Example 9 Important

Example 10

Example 11 Important

Example 12

Example 13 Important

Example 14

Example 15

Example 16

Example 17 Important

Example 18

Example 19

Example 20

Example 21

Example 22

Example 23

Example 24

Example 25

Example 26

Example 27

Example 28 Important

Example 29

Example 30 Important

Example 31

Example 32 Important

Example 33 Important

Example 34

Example 35 Important

Example 36

Example 37 Important

Example 38 Important

Example 39

Example 40 Important

Example 41 Important

Example 42 Important

Example 43 Important

Example 44 Important You are here

Example 45 Important

Example 46 Important

Example 47 Important

Example 48 Important

Example 49

Example 50 Important

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 9 years. He provides courses for Maths and Science at Teachoo.