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Last updated at Jan. 7, 2020 by Teachoo

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Ex 6.1, 11 A particle moves along the curve 6π¦ = π₯3 +2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the π₯βcoordinate. Given that A particular Moves along the curve 6π¦ = π₯3 + 2 We need to find points on the curve at which π¦ coordinate is changing 8 times as fast as the π₯ β coordinate i.e. We need to find (π₯,π¦) for which ππ¦/ππ‘= 8 ππ₯/ππ‘ From (1) 6π¦ = π₯3 +2 Diff Both Side w.r.t π‘ π(6π¦)/ππ‘=π(π₯^3 + 2)/ππ‘ 6 ππ¦/ππ‘=π(π₯^3 )/ππ‘+π(2)/ππ‘ 6 ππ¦/ππ‘=π(π₯^3 )/ππ‘ Γ ππ₯/ππ₯+0 6 ππ¦/ππ‘=π(π₯^3 )/ππ₯ Γ ππ₯/ππ‘ 6 ππ¦/ππ‘=3π₯^2 . ππ₯/ππ‘ ππ¦/ππ‘=(3π₯^2)/6 ππ₯/ππ‘ ππ¦/ππ‘=π₯^2/2 ππ₯/ππ‘ We need to find point for which ππ¦/ππ‘= 8 ππ₯/ππ‘ Putting ππ¦/ππ‘=π₯^2/2 . ππ₯/ππ‘ π₯^2/2 . ππ₯/ππ‘=8 ππ₯/ππ‘ π₯^2/2=8 π₯^2=8 Γ 2 β¦(2) (From (2)) π₯^2=16 π₯=Β±β16 π₯=Β± 4 π₯=4 , β4 Putting values of π₯ in (1) 6π¦ = π₯3 +2 When π=π 6π¦=(4)^3+2 6π¦=64+2 6π¦=66 π¦=66/6 = 11 Points is (4 , 11) When π=β π 6π¦=(β 4)^3+2 6π¦=β 64+2 6π¦=β 62 π¦=(β 62)/6=(β 31)/3 Points is (β 4, (β 31)/3) Hence Required points on the curve are (π , ππ) & (β π, (β ππ)/π)

Chapter 6 Class 12 Application of Derivatives

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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.