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Ex 6.1, 11 - A particle moves along the curve 6y = x^3 + 2. Find point

Ex 6.1,11 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,11 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.1,11 - Chapter 6 Class 12 Application of Derivatives - Part 4

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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 6π’š = 𝒙3 + 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 π’…π’š/𝒅𝒕= 8 𝒅𝒙/𝒅𝒕 From (1) 6𝑦 = π‘₯3 +2 Differentiating both sides w.r.t time 𝒅(πŸ”π’š)/𝒅𝒕=𝒅(𝒙^πŸ‘ + 𝟐)/𝒅𝒕 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 π’…π’š/𝒅𝒕=𝒙^𝟐/𝟐 𝒅𝒙/𝒅𝒕 We need to find point for which 𝑑𝑦/𝑑𝑑= 8 𝑑π‘₯/𝑑𝑑 Putting value of 𝑑𝑦/𝑑𝑑 from (2) 𝒙^𝟐/𝟐 . 𝒅𝒙/𝒅𝒕=πŸ– 𝒅𝒙/𝒅𝒕 π‘₯^2/2 =8 π‘₯^2=8 Γ— 2 π‘₯^2=16 π‘₯=±√16 π‘₯=Β± 4 π‘₯=4 , βˆ’4 To find points, we put values of x in our equation of curve 6𝑦 = π‘₯3 +2 When 𝒙=πŸ’ 6𝑦=(4)^3+2 6𝑦=64+2 6𝑦=66 𝑦=66/6 = 11 Point is (πŸ’ , 𝟏𝟏) When 𝒙=βˆ’ πŸ’ 6𝑦=(βˆ’ 4)^3+2 6𝑦=βˆ’ 64+2 6𝑦=βˆ’62 𝑦=(βˆ’62)/6=(βˆ’πŸ‘πŸ)/πŸ‘ Point is (βˆ’πŸ’, (βˆ’πŸ‘πŸ)/πŸ‘)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.