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

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Ex 6.1, 11 A particle moves along the curve 6𝑦 = 𝑥3 +2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the 𝑥−coordinate. Given that A particular Moves along the curve 6𝑦 = 𝑥3 + 2 We need to find points on the curve at which 𝑦 coordinate is changing 8 times as fast as the 𝑥 – coordinate i.e. We need to find (𝑥,𝑦) for which 𝑑𝑦/𝑑𝑡= 8 𝑑𝑥/𝑑𝑡 From (1) 6𝑦 = 𝑥3 +2 Diff Both Side w.r.t 𝑡 𝑑(6𝑦)/𝑑𝑡=𝑑(𝑥^3 + 2)/𝑑𝑡 6 𝑑𝑦/𝑑𝑡=𝑑(𝑥^3 )/𝑑𝑡+𝑑(2)/𝑑𝑡 6 𝑑𝑦/𝑑𝑡=𝑑(𝑥^3 )/𝑑𝑡 × 𝑑𝑥/𝑑𝑥+0 6 𝑑𝑦/𝑑𝑡=𝑑(𝑥^3 )/𝑑𝑥 × 𝑑𝑥/𝑑𝑡 6 𝑑𝑦/𝑑𝑡=3𝑥^2 . 𝑑𝑥/𝑑𝑡 𝑑𝑦/𝑑𝑡=(3𝑥^2)/6 𝑑𝑥/𝑑𝑡 𝑑𝑦/𝑑𝑡=𝑥^2/2 𝑑𝑥/𝑑𝑡 We need to find point for which 𝑑𝑦/𝑑𝑡= 8 𝑑𝑥/𝑑𝑡 Putting 𝑑𝑦/𝑑𝑡=𝑥^2/2 . 𝑑𝑥/𝑑𝑡 𝑥^2/2 . 𝑑𝑥/𝑑𝑡=8 𝑑𝑥/𝑑𝑡 𝑥^2/2=8 𝑥^2=8 × 2 …(2) (From (2)) 𝑥^2=16 𝑥=±√16 𝑥=± 4 𝑥=4 , −4 Putting values of 𝑥 in (1) 6𝑦 = 𝑥3 +2 When 𝒙=𝟒 6𝑦=(4)^3+2 6𝑦=64+2 6𝑦=66 𝑦=66/6 = 11 Points is (4 , 11) When 𝒙=− 𝟒 6𝑦=(− 4)^3+2 6𝑦=− 64+2 6𝑦=− 62 𝑦=(− 62)/6=(− 31)/3 Points is (− 4, (− 31)/3) Hence Required points on the curve are (𝟒 , 𝟏𝟏) & (− 𝟒, (− 𝟑𝟏)/𝟑)