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Example 44 - A man of height 2 meters walks at uniform speed

Example 44 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 44 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 44 - Chapter 6 Class 12 Application of Derivatives - Part 4

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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 Ξ”ASB tan ΞΈ = 𝐴𝐡/𝐴𝑆 tan ΞΈ =πŸ”/(𝒙 + 𝒔) In βˆ† MSN tan ΞΈ = 𝑀𝑁/𝑀𝑆 tan ΞΈ =𝟐/𝒔 From (1) & (2) πŸ”/(𝒙 + 𝒔) = 𝟐/𝒔 6s = 2x + 2s 6s – 2s = 2x 4s = 2x 2s = x x = 2s We need to find 𝑑𝑠/𝑑𝑑 Now, x = 2s Diff w.r.t t 𝑑π‘₯/𝑑𝑑= 𝑑(2𝑠)/𝑑𝑑 𝒅𝒙/𝒅𝒕= 2.𝑑𝑠/𝑑𝑑 5 = 2 𝑑𝑠/𝑑𝑑 𝑑𝑠/𝑑𝑑 = 5/2 So, 𝒅𝒔/𝒅𝒕 = πŸ“/𝟐 km/hr. (Given 𝑑π‘₯/𝑑𝑑 = 5km/hr)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.