Last updated at April 19, 2021 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 = π3 + 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 π π/π π= 8 π π/π π From (1) 6π¦ = π₯3 +2 Differentiating both sides w.r.t time π (ππ)/π π=π (π^π + π)/π π 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‘ π π/π π=π^π/π π π/π π We need to find point for which ππ¦/ππ‘= 8 ππ₯/ππ‘ Putting value of ππ¦/ππ‘ from (2) π^π/π . π π/π π=π π π/π π π₯^2/2 =8 π₯^2=8 Γ 2 π₯^2=16 π₯=Β±β16 π₯=Β± 4 π₯=4 , β4 To find points, we put values of x in our equation of curve 6π¦ = π₯3 +2 When π=π 6π¦=(4)^3+2 6π¦=64+2 6π¦=66 π¦=66/6 = 11 Point is (π , ππ) When π=β π 6π¦=(β 4)^3+2 6π¦=β 64+2 6π¦=β62 π¦=(β62)/6=(βππ)/π Point is (βπ, (βππ)/π)

Ex 6.1

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Chapter 6 Class 12 Application of Derivatives

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About the Author

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.