Example 4 - Length x of a rectangle is decreasing at rate

Example 4 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 4 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 4 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 4 - Chapter 6 Class 12 Application of Derivatives - Part 5 Example 4 - Chapter 6 Class 12 Application of Derivatives - Part 6 Example 4 - Chapter 6 Class 12 Application of Derivatives - Part 7

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  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 = 10 cm and y = 6 cm, 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 length ๐‘ฅ is decreasing at the rate of 3 cm/minute ๐’…๐’™/๐’…๐’• = โ€“ 3 cm/ min and width y is increasing at the rate of 2 cm/min ๐’…๐’š/๐’…๐’• = 2 cm /min (i) Finding rate of change of Perimeter 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‘ท/๐’…๐’•= 2 [๐’…๐’™/๐’…๐’•+ ๐’…๐’š/๐’…๐’•] 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‘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(ii) Finding rate of change of Area Let A be the Area of rectangle. We need to calculate Rate of change of area when ๐‘ฅ = 10cm & ๐‘ฆ = 6 cm i.e. we need to calculate ๐’…๐‘จ/๐’…๐’• 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 ๐‘ฆ + ๐‘‘๐‘ฆ/๐‘‘๐‘ก ๐‘ฅ 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‘จ/๐’…๐’• = โ€“ ๐Ÿ‘๐’š+๐Ÿ๐’™ Putting ๐’™ = 10 & ๐’š = 6 cm ๐‘‘๐ด/๐‘‘๐‘ก = โ€“ 3(6) + 2(10) = โ€“ 18 + 20 = 2 Since Area is in cm2 & time is in time in minute 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 10 years. He provides courses for Maths and Science at Teachoo.