Finding rate of change

Chapter 6 Class 12 Application of Derivatives (Term 1)
Concept wise

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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 & 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 β΄ ππ/ππ = 5 km/h We need to find rate at which length of his shadow increases. i.e. we need to find ππ/ππ In ΞASB tan ΞΈ = π΄π΅/π΄π tan ΞΈ =π/(π + π) In β MSN tan ΞΈ = ππ/ππ tan ΞΈ =π/π From (1) & (2) π/(π + π) = π/π 6s = 2x + 2s 6s β 2s = 2x 4s = 2x 2s = x x = 2s We need to find ππ /ππ‘ Now, x = 2s Diff w.r.t t ππ₯/ππ‘= π(2π )/ππ‘ ππ/ππ= 2.ππ /ππ‘ 5 = 2 ππ /ππ‘ ππ /ππ‘ = 5/2 So, ππ/ππ = π/π km/hr. (Given ππ₯/ππ‘ = 5km/hr)