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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and the width y is increasing at the rate of 2 cm/minute. When x = 10cm and y = 6cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle. Let length of rectangle = ๐‘ฅ cm & width of rectangle = ๐‘ฆ cm Given the length ๐‘ฅ is decreasing at the rate of 3 cm/ minute i.e. ๐‘ฅ is decreasing w.r.t time i.e. ๐‘‘๐‘ฅ/๐‘‘๐‘ก = โ€“ 3 cm/ min & The width y is increasing at the rate of 2 cm/min i.e. ๐‘ฆ is increasing w.r.t time ๐‘‘๐‘ฆ/๐‘‘๐‘ก = 2 cm /min (i) Let P be the perimeter of rectangle. We need to calculate rate of change of perimeter when ๐‘ฅ = 10 cm & ๐‘ฆ = 6 cm i.e. we need to calculate ๐‘‘๐‘ƒ/๐‘‘๐‘ก when ๐‘ฅ = 10 cm & ๐‘ฆ = 6 cm We know that Perimeter of rectangle = 2 (length + width) P = 2 (๐‘ฅ + ๐‘ฆ) Now ๐‘‘๐‘ƒ/๐‘‘๐‘ก= (๐‘‘ (2 (๐‘ฅ + ๐‘ฆ) ) )/๐‘‘๐‘ก ๐‘‘๐‘ƒ/๐‘‘๐‘ก= 2 [๐‘‘(๐‘ฅ + ๐‘ฆ)/๐‘‘๐‘ก] ๐‘‘๐‘ƒ/๐‘‘๐‘ก= 2 [๐‘‘๐‘ฅ/๐‘‘๐‘ก+ ๐‘‘๐‘ฆ/๐‘‘๐‘ก] From (1) & (2) ๐‘‘๐‘ฅ/๐‘‘๐‘ก = โ€“3 & ๐‘‘๐‘ฆ/๐‘‘๐‘ก = 2 ๐‘‘๐‘ƒ/๐‘‘๐‘ก= 2(โ€“ 3 + 2) ๐‘‘๐‘ƒ/๐‘‘๐‘ก= 2 (โ€“1) ๐‘‘๐‘ƒ/๐‘‘๐‘ก= โ€“2 Since perimeter is in cm & time is in minute ๐‘‘๐‘/๐‘‘๐‘ก = (โˆ’2 ๐‘๐‘š)/๐‘š๐‘–๐‘› ๐‘‘๐‘/๐‘‘๐‘ก = โ€“ 2 cm/min Therefore perimeter is decreasing at the rate of 2 cm/min (ii) Let A be the Area of rectangle. We need to calculate Rate of change of area w.r.t when ๐‘ฅ = 10cm & ๐‘ฆ = 6 cm i.e. we need to calculate ๐‘‘๐ด/๐‘‘๐‘ก when ๐‘ฅ=10 & ๐‘ฆ=6 cm We know that Area of rectangle = length ร— width A = ๐‘ฅ ร— ๐‘ฆ Now, ๐‘‘๐ด/๐‘‘๐‘ก = (๐‘‘ (๐‘ฅ๐‘ฆ))/๐‘‘๐‘ก ๐‘‘๐ด/๐‘‘๐‘ก = (๐‘‘(๐‘ฅ))/๐‘‘๐‘ก. y + (๐‘‘(๐‘ฆ))/๐‘‘๐‘ก . x From (1) & (2) ๐‘‘๐‘ฅ/๐‘‘๐‘ก = โ€“3 & ๐‘‘๐‘ฆ/๐‘‘๐‘ก = 2 dA/dt = (โ€“ 3)๐‘ฆ+2 (๐‘ฅ) ๐‘‘๐ด/๐‘‘๐‘ก = โ€“ 3๐‘ฆ+2๐‘ฅ Putting ๐‘ฅ = 10 & ๐‘ฆ = 6 cm โ”œ ๐‘‘๐ด/๐‘‘๐‘กโ”ค|_(๐‘ฅ =10, ๐‘ฆ = 6) = โ€“ 3(6) + 2(10) โ”œ ๐‘‘๐ด/๐‘‘๐‘กโ”ค|_(๐‘ฅ =10, ๐‘ฆ = 6) = โ€“ 18 + 20 โ”œ ๐‘‘๐ด/๐‘‘๐‘กโ”ค|_(๐‘ฅ =10, ๐‘ฆ = 6) = 2 Since Area is in cm2 & time is in time in minute ๐‘‘๐ด/๐‘‘๐‘ก = (2 ๐‘๐‘š2)/๐‘š๐‘–๐‘› ๐‘‘๐ด/๐‘‘๐‘ก = 2 cm2/min Hence, Area is increasing at the rate of 2 cm2/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.