Ex 6.1, 7 - The length x of a rectangle is decreasing at rate

Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 3

Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 6

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order 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 = 8 cm & 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’™/๐’…๐’• = โ€“ 5 cm/ min And width of rectangle is increasing at the rate of 4 cm/min I.e. ๐’…๐’š/๐’…๐’• = 4 cm/min Let P be the perimeter of rectangle We need to find rate of change of perimeter when ๐‘ฅ = 8 cm & y = 6 cm i.e. Finding ๐‘‘๐‘ƒ/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) 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‘ท/๐’…๐’•= 2 [๐’…๐’™/๐’…๐’•+ ๐’…๐’š/๐’…๐’•] From (1) & (2) ๐‘‘๐‘ฅ/๐‘‘๐‘ก = โ€“5 & ๐‘‘๐‘ฆ/๐‘‘๐‘ก = 4 ๐‘‘๐‘ƒ/๐‘‘๐‘ก= 2(โ€“5 + 4) ๐‘‘๐‘ƒ/๐‘‘๐‘ก= 2 (โ€“1) ๐’…๐‘ท/๐’…๐’•= โ€“2 Since perimeter is in cm & time is in minute ๐‘‘๐‘ƒ/๐‘‘๐‘ก = โ€“ 2 cm/min Therefore, perimeter is decreasing at the rate of 2 cm/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 when ๐‘ฅ = 8 & ๐‘ฆ = 6 cm i.e. ๐’…๐‘จ/๐’…๐’• when ๐‘ฅ = 8 cm & 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‘ฅ. ๐‘ฆ))/๐‘‘๐‘ก ๐‘‘๐ด/๐‘‘๐‘ก = ๐‘‘๐‘ฅ/๐‘‘๐‘ก . ๐‘ฆ + ๐‘‘๐‘ฆ/๐‘‘๐‘ก . ๐‘ฅ. From (1) & (2) ๐‘‘๐‘ฅ/๐‘‘๐‘ก = โ€“5 & ๐‘‘๐‘ฆ/๐‘‘๐‘ก = 4 ๐‘‘๐ด/๐‘‘๐‘ก = (โ€“5)๐‘ฆ + (4)๐‘ฅ ๐‘‘๐ด/๐‘‘๐‘ก = 4๐‘ฅ โ€“ 5๐‘ฆ Putting ๐‘ฅ = 8 cm & ๐‘ฆ = 6 cm ๐‘‘๐ด/๐‘‘๐‘ก = 4 (8) โ€“ 5(6) Using product rule in x . y as (u.v)โ€™ = uโ€™ v + vโ€˜ u ๐‘‘๐ด/๐‘‘๐‘ก = 32 โ€“ 30 ๐’…๐‘จ/๐’…๐’• = 2 Since Area is in cm2 & time is in minute 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 10 years. He provides courses for Maths and Science at Teachoo.