Ex 6.1,11 - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

Transcript

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)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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