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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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.