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Last updated at Jan. 7, 2020 by Teachoo

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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)

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Chapter 6 Class 12 Application of Derivatives

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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.