Example 32 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 32 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 & MN be the man of height 2m. & AM = x meter & MS is the shadow of the man Let length of shadow MS = s meter Given man walks at speed of 5 km/h ∴ 𝒅𝒙/𝒅𝒕 = 5 km/h We need to find rate at which length of his shadow increases. i.e. we need to find 𝒅𝒔/𝒅𝒕 In ΔASB tan θ = 𝐴𝐵/𝐴𝑆 tan θ =𝟔/(𝒙 + 𝒔) In ∆ MSN tan θ = 𝑀𝑁/𝑀𝑆 tan θ =𝟐/𝒔 From (1) & (2) 𝟔/(𝒙 + 𝒔) = 𝟐/𝒔 6s = 2x + 2s 6s – 2s = 2x 4s = 2x 2s = x x = 2s We need to find 𝑑𝑠/𝑑𝑡 Now, x = 2s Diff w.r.t t 𝑑𝑥/𝑑𝑡= 𝑑(2𝑠)/𝑑𝑡 𝒅𝒙/𝒅𝒕= 2.𝑑𝑠/𝑑𝑡 5 = 2 𝑑𝑠/𝑑𝑡 𝑑𝑠/𝑑𝑡 = 5/2 So, 𝒅𝒔/𝒅𝒕 = 𝟓/𝟐 km/hr.
Examples
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 Important
Example 17
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22
Example 23 Important
Example 24
Example 25 Important
Example 26 Important
Example 27
Example 28 Important
Example 29 Important
Example 30 Important
Example 31 Important
Example 32 Important You are here
Example 33 Important
Example 34 Important
Example 35
Example 36 Important
Example 37
Question 1
Question 2
Question 3
Question 4 Important
Question 5
Question 6
Question 7
Question 8
Question 9
Question 10
Question 11
Question 12
Question 13 Important
Question 14 Important
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo