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  1. Chapter 6 Class 12 Application of Derivatives
  2. Concept wise

Transcript

Ex 6.1,7 The length x of a rectangle is decreasing at the rate of 5 cm/minute & the width y is increasing at the rate of 4 cm/minute. When x = 8cm & y = 6cm, find the rates of change of (a) the perimeter. Let length of rectangle = ๐‘ฅ & width of rectangle = ๐‘ฆ Given that Length of rectangle is decreasing at the rate of 5 cm/min i.e. ๐‘‘๐‘ฅ/๐‘‘๐‘ก = โ€“ 5 cm/ min And width of rectangle is increasing at the rate of 4 cm/min I.e. ๐‘‘๐‘ฆ/๐‘‘๐‘ก = 4 cm/min Let p be the perimeter of rectangle We need to find rate of change of perimeter w. r. t time when ๐‘ฅ = 8 cm & y = 6 cm i.e. find ๐‘‘๐‘/๐‘‘๐‘ก when ๐‘ฅ = 8 cm & ๐‘ฆ = 6 cm We know that Perimeter of rectangle = 2 (length + width) p = 2 (๐‘ฅ + ๐‘ฆ) Different w. r. t time ๐‘‘๐‘/๐‘‘๐‘ก = (๐‘‘ (2(๐‘ฅ+๐‘ฆ)))/๐‘‘๐‘ก ๐‘‘๐‘/๐‘‘๐‘ก = 2 (๐‘‘ (๐‘ฅ + ๐‘ฆ))/๐‘‘๐‘ก ๐‘‘๐‘/๐‘‘๐‘ก = 2.(๐‘‘๐‘ฅ/๐‘‘๐‘ก + ๐‘‘๐‘ฆ/๐‘‘๐‘ก) From (1) & (2) ๐‘‘๐‘/๐‘‘๐‘ก = 2 ( โ€“ 5 + 4 ) ๐‘‘๐‘/๐‘‘๐‘ก = 2 ( โ€“ 1) ๐‘‘๐‘/๐‘‘๐‘ก = โ€“2 Since perimeter is in cm & time is in minute. ๐‘‘๐‘/๐‘‘๐‘“ = (โˆ’2 ๐‘๐‘š)/๐‘š๐‘–๐‘› ๐‘‘๐‘/๐‘‘๐‘ก = โ€“2 cm/min Hence, perimeter is decreasing at the rate of 2cm/min Ex 6.1,7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of change of (b) the area of the rectangle. Let A be the Area of rectangle we need to find rate of change of area w.r.t time when ๐‘ฅ = 8 & ๐‘ฆ = 6 cm i.e. ๐‘‘๐ด/๐‘‘๐‘ก when ๐‘ฅ = 8 cm & ๐‘ฆ = 6 cm We know that Area of rectangle = length ร— width A = ๐‘ฅ . ๐‘ฆ Differentiate w.r.t time ๐‘‘๐ด/๐‘‘๐‘ก = (๐‘‘(๐‘ฅ . ๐‘ฆ))/๐‘‘๐‘ก Using product rule in x . y as (u , v)โ€™ = uโ€™ v + vโ€˜ u ๐‘‘๐ด/๐‘‘๐‘ก = ๐‘‘(๐‘ฅ)/๐‘‘๐‘ก . ๐‘ฆ + ๐‘‘๐‘ฆ/๐‘‘๐‘ก . ๐‘ฅ. ๐‘‘๐ด/๐‘‘๐‘ก = โ€“ 5 . ๐‘ฆ + 4 . ๐‘ฅ ๐‘‘๐ด/๐‘‘๐‘ก = 4๐‘ฅ โ€“ 5๐‘ฆ when ๐‘ฅ = 8 cm & ๐‘ฆ = 6 cm ๐‘‘๐ด/๐‘‘๐‘ก = 4 (8) โ€“ 5(6) ๐‘‘๐ด/๐‘‘๐‘ก = 32 โ€“ 30 ๐‘‘๐ด/๐‘‘๐‘ก = 2 Since Area is in cm2 & time is in minute So ๐‘‘๐ด/๐‘‘๐‘ก = 2๐‘๐‘š2/๐‘š๐‘–๐‘› ๐‘‘๐ด/๐‘‘๐‘ก = 2 cm2/ min Hence, Area is increasing at the rate of 2cm2/min

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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.