Example 44
Last updated at Dec. 8, 2016 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 ∴ 𝑑𝑙𝑑𝑡 = 5km/h We need to find rate at which the length of his shadow increases. i.e. we need to find 𝑑𝑠𝑑𝑡 From (1) & (2) 61000 𝑙 + 𝑠 = 21000𝑠 6𝑙 + 𝑠 = 2𝑠 6s = 2l + 2s 6s – 2s = 2l 4s = 2l 2s = l l = 2s We need to find 𝑑𝑠𝑑𝑡 Diff w.r.t t 𝑑𝑙𝑑𝑡= 𝑑2𝑠𝑑𝑡 𝒅𝒍𝒅𝒕= 2.𝑑𝑠𝑑𝑡 5 = 2 𝑑𝑠𝑑𝑡 𝑑𝑠𝑑𝑡 = 52 So, 𝒅𝒔𝒅𝒕 = 𝟓𝟐 km/hr.
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Chapter 6 Class 12 Application of Derivatives
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.