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 = π‘₯ cm & Width of rectangle = 𝑦 cm Given length π‘₯ is decreasing at the rate of 3 cm/minute 𝒅𝒙/𝒅𝒕 = – 3 cm/ min and width y is increasing at the rate of 2 cm/min π’…π’š/𝒅𝒕 = 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 π‘₯ = 10 cm & 𝑦 = 6 cm i.e. we need to calculate 𝑑𝑃/𝑑𝑑 when π‘₯ = 10 cm & 𝑦 = 6 cm We know that Perimeter of rectangle = 2 (Length + Width) P = 2 (π‘₯ + 𝑦) Now 𝑑𝑃/𝑑𝑑= (𝑑 (2 (π‘₯ + 𝑦) ) )/𝑑𝑑 𝑑𝑃/𝑑𝑑= 2 [𝑑(π‘₯ + 𝑦)/𝑑𝑑] 𝒅𝑷/𝒅𝒕= 2 [𝒅𝒙/𝒅𝒕+ π’…π’š/𝒅𝒕] From (1) & (2) 𝑑π‘₯/𝑑𝑑 = –3 & 𝑑𝑦/𝑑𝑑 = 2 𝑑𝑃/𝑑𝑑= 2(– 3 + 2) 𝑑𝑃/𝑑𝑑= 2 (–1) 𝒅𝑷/𝒅𝒕= –2 Since perimeter is in cm & time is in minute 𝑑𝑃/𝑑𝑑 = – 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 π‘₯ = 10cm & 𝑦 = 6 cm i.e. we need to calculate 𝒅𝑨/𝒅𝒕 when π‘₯=10 & 𝑦=6 cm We know that Area of rectangle = Length Γ— Width A = π‘₯ Γ— 𝑦 Now, 𝑑𝐴/𝑑𝑑 = (𝑑 (π‘₯𝑦))/𝑑𝑑 𝑑𝐴/𝑑𝑑 = 𝑑π‘₯/𝑑𝑑 𝑦 + 𝑑𝑦/𝑑𝑑 π‘₯ From (1) & (2) 𝑑π‘₯/𝑑𝑑 = –3 & 𝑑𝑦/𝑑𝑑 = 2 dA/dt = (–3)𝑦+2 (π‘₯) 𝒅𝑨/𝒅𝒕 = – πŸ‘π’š+πŸπ’™ Putting 𝒙 = 10 & π’š = 6 cm 𝑑𝐴/𝑑𝑑 = – 3(6) + 2(10) = – 18 + 20 = 2 Since Area is in cm2 & time is in time in minute 𝑑𝐴/𝑑𝑑 = 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.