Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.1, 11 A particle moves along the curve 6𝑦 = π‘₯3 +2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the π‘₯βˆ’coordinate. Given that A particular Moves along the curve 6𝑦 = π‘₯3 + 2 We need to find points on the curve at which 𝑦 coordinate is changing 8 times as fast as the π‘₯ – coordinate i.e. We need to find (π‘₯,𝑦) for which 𝑑𝑦/𝑑𝑑= 8 𝑑π‘₯/𝑑𝑑 From (1) 6𝑦 = π‘₯3 +2 Diff Both Side w.r.t 𝑑 𝑑(6𝑦)/𝑑𝑑=𝑑(π‘₯^3 + 2)/𝑑𝑑 6 𝑑𝑦/𝑑𝑑=𝑑(π‘₯^3 )/𝑑𝑑+𝑑(2)/𝑑𝑑 6 𝑑𝑦/𝑑𝑑=𝑑(π‘₯^3 )/𝑑𝑑 Γ— 𝑑π‘₯/𝑑π‘₯+0 6 𝑑𝑦/𝑑𝑑=𝑑(π‘₯^3 )/𝑑π‘₯ Γ— 𝑑π‘₯/𝑑𝑑 6 𝑑𝑦/𝑑𝑑=3π‘₯^2 . 𝑑π‘₯/𝑑𝑑 𝑑𝑦/𝑑𝑑=(3π‘₯^2)/6 𝑑π‘₯/𝑑𝑑 𝑑𝑦/𝑑𝑑=π‘₯^2/2 𝑑π‘₯/𝑑𝑑 We need to find point for which 𝑑𝑦/𝑑𝑑= 8 𝑑π‘₯/𝑑𝑑 Putting 𝑑𝑦/𝑑𝑑=π‘₯^2/2 . 𝑑π‘₯/𝑑𝑑 π‘₯^2/2 . 𝑑π‘₯/𝑑𝑑=8 𝑑π‘₯/𝑑𝑑 π‘₯^2/2=8 π‘₯^2=8 Γ— 2 …(2) (From (2)) π‘₯^2=16 π‘₯=±√16 π‘₯=Β± 4 π‘₯=4 , βˆ’4 Putting values of π‘₯ in (1) 6𝑦 = π‘₯3 +2 When 𝒙=πŸ’ 6𝑦=(4)^3+2 6𝑦=64+2 6𝑦=66 𝑦=66/6 = 11 Points is (4 , 11) When 𝒙=βˆ’ πŸ’ 6𝑦=(βˆ’ 4)^3+2 6𝑦=βˆ’ 64+2 6𝑦=βˆ’ 62 𝑦=(βˆ’ 62)/6=(βˆ’ 31)/3 Points is (βˆ’ 4, (βˆ’ 31)/3) Hence Required points on the curve are (πŸ’ , 𝟏𝟏) & (βˆ’ πŸ’, (βˆ’ πŸ‘πŸ)/πŸ‘)

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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.