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Chapter 6 Class 12 Application of Derivatives (Term 1)
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Example 44 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 19, 2021 by

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

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 & 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.