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

  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 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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.