Example 4 - Chapter 6 Class 12 Application of Derivatives

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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 = 10cm and y = 6cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle. Let length of rectangle = 𝑥 cm & width of rectangle = 𝑦 cm Given the length 𝑥 is decreasing at the rate of 3 cm/ minute i.e. 𝑥 is decreasing w.r.t time i.e. 𝑑𝑥/𝑑𝑡 = – 3 cm/ min & The width y is increasing at the rate of 2 cm/min i.e. 𝑦 is increasing w.r.t time 𝑑𝑦/𝑑𝑡 = 2 cm /min (i) Let P be the perimeter of rectangle. We need to calculate rate of change of perimeter when 𝑥 = 10 cm & 𝑦 = 6 cm i.e. we need to calculate 𝑑𝑃/𝑑𝑡 when 𝑥 = 10 cm & 𝑦 = 6 cm We know that Perimeter of rectangle = 2 (length + width) P = 2 (𝑥 + 𝑦) Now 𝑑𝑃/𝑑𝑡= (𝑑 (2 (𝑥 + 𝑦) ) )/𝑑𝑡 𝑑𝑃/𝑑𝑡= 2 [𝑑(𝑥 + 𝑦)/𝑑𝑡] 𝑑𝑃/𝑑𝑡= 2 [𝑑𝑥/𝑑𝑡+ 𝑑𝑦/𝑑𝑡] From (1) & (2) 𝑑𝑥/𝑑𝑡 = –3 & 𝑑𝑦/𝑑𝑡 = 2 𝑑𝑃/𝑑𝑡= 2(– 3 + 2) 𝑑𝑃/𝑑𝑡= 2 (–1) 𝑑𝑃/𝑑𝑡= –2 Since perimeter is in cm & time is in minute 𝑑𝑝/𝑑𝑡 = (−2 𝑐𝑚)/𝑚𝑖𝑛 𝑑𝑝/𝑑𝑡 = – 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 𝑥 = 10cm & 𝑦 = 6 cm i.e. we need to calculate 𝑑𝐴/𝑑𝑡 when 𝑥=10 & 𝑦=6 cm We know that Area of rectangle = length × width A = 𝑥 × 𝑦 Now, 𝑑𝐴/𝑑𝑡 = (𝑑 (𝑥𝑦))/𝑑𝑡 𝑑𝐴/𝑑𝑡 = (𝑑(𝑥))/𝑑𝑡. y + (𝑑(𝑦))/𝑑𝑡 . x From (1) & (2) 𝑑𝑥/𝑑𝑡 = –3 & 𝑑𝑦/𝑑𝑡 = 2 dA/dt = (– 3)𝑦+2 (𝑥) 𝑑𝐴/𝑑𝑡 = – 3𝑦+2𝑥 Putting 𝑥 = 10 & 𝑦 = 6 cm ├ 𝑑𝐴/𝑑𝑡┤|_(𝑥 =10, 𝑦 = 6) = – 3(6) + 2(10) ├ 𝑑𝐴/𝑑𝑡┤|_(𝑥 =10, 𝑦 = 6) = – 18 + 20 ├ 𝑑𝐴/𝑑𝑡┤|_(𝑥 =10, 𝑦 = 6) = 2 Since Area is in cm2 & time is in time in minute 𝑑𝐴/𝑑𝑡 = (2 𝑐𝑚2)/𝑚𝑖𝑛 𝑑𝐴/𝑑𝑡 = 2 cm2/min Hence, Area is increasing at the rate of 2 cm2/min.