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

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  1. 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 = ๐‘ฅ cm & width of rectangle = ๐‘ฆ cm Given the length ๐‘ฅ is decreasing at the rate of 3 cm/ minute i.e. ๐‘ฅ is decreasing w.r.t time i.e. ๏ท๐‘‘๐‘ฅ๏ทฎ๐‘‘๐‘ก๏ทฏ = โ€“ 3 cm/ min & The width y is increasing at the rate of 2 cm/min i.e. ๐‘ฆ is increasing w.r.t time ๏ท๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ก๏ทฏ = 2 cm /min (i) 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 ๏ท๐‘‘๐‘ƒ๏ทฎ๐‘‘๐‘ก๏ทฏ 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 ๏ท๐‘‘๐‘ƒ๏ทฎ๐‘‘๐‘ก๏ทฏ= ๏ท๐‘‘ ๏ท2 ๏ท๐‘ฅ + ๐‘ฆ๏ทฏ ๏ทฏ ๏ทฎ๐‘‘๐‘ก๏ทฏ ๏ท๐‘‘๐‘ƒ๏ทฎ๐‘‘๐‘ก๏ทฏ= 2 ๏ท๏ท๐‘‘๏ท๐‘ฅ + ๐‘ฆ๏ทฏ๏ทฎ๐‘‘๐‘ก๏ทฏ๏ทฏ ๏ท๐‘‘๐‘ƒ๏ทฎ๐‘‘๐‘ก๏ทฏ= 2 ๏ท๏ท๐‘‘๐‘ฅ๏ทฎ๐‘‘๐‘ก๏ทฏ+ ๏ท๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ก๏ทฏ๏ทฏ From (1) & (2) ๏ท๐‘‘๐‘ฅ๏ทฎ๐‘‘๐‘ก๏ทฏ = โ€“3 & ๏ท๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ก๏ทฏ = 2 ๏ท๐‘‘๐‘ƒ๏ทฎ๐‘‘๐‘ก๏ทฏ= 2(โ€“ 3 + 2) ๏ท๐‘‘๐‘ƒ๏ทฎ๐‘‘๐‘ก๏ทฏ= 2 (โ€“1) ๏ท๐‘‘๐‘ƒ๏ทฎ๐‘‘๐‘ก๏ทฏ= โ€“2 Since perimeter is in cm & time is in minute ๏ท๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ = ๏ทโˆ’2 ๐‘๐‘š๏ทฎ๐‘š๐‘–๐‘›๏ทฏ ๏ท๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ = โ€“ 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 ๐‘ฅ = 10cm & ๐‘ฆ = 6 cm i.e. we need to calculate ๏ท๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ when ๐‘ฅ=10 & ๐‘ฆ=6 cm We know that Area of rectangle = length ร— width A = ๐‘ฅ ร— ๐‘ฆ Now, ๏ท๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = ๏ท๐‘‘ (๐‘ฅ๐‘ฆ)๏ทฎ๐‘‘๐‘ก๏ทฏ ๏ท๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = ๏ท๐‘‘(๐‘ฅ)๏ทฎ๐‘‘๐‘ก๏ทฏ. y + ๏ท๐‘‘(๐‘ฆ)๏ทฎ๐‘‘๐‘ก๏ทฏ . x From (1) & (2) ๏ท๐‘‘๐‘ฅ๏ทฎ๐‘‘๐‘ก๏ทฏ = โ€“3 & ๏ท๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ก๏ทฏ = 2 ๏ทdA๏ทฎdt๏ทฏ = (โ€“ 3)๐‘ฆ+2 (๐‘ฅ) ๏ท๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = โ€“ 3๐‘ฆ+2๐‘ฅ Putting ๐‘ฅ = 10 & ๐‘ฆ = 6 cm ๏ท๏ท๏ท๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ๏ทฏ๏ทฎ๐‘ฅ =10, ๐‘ฆ = 6๏ทฏ = โ€“ 3(6) + 2(10) ๏ท๏ท๏ท๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ๏ทฏ๏ทฎ๐‘ฅ =10, ๐‘ฆ = 6๏ทฏ = โ€“ 18 + 20 ๏ท๏ท๏ท๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ๏ทฏ๏ทฎ๐‘ฅ =10, ๐‘ฆ = 6๏ทฏ = 2 Since Area is in cm2 & time is in time in minute ๏ท๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = ๏ท2 ๐‘๐‘š2๏ทฎ๐‘š๐‘–๐‘›๏ทฏ ๏ท๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = 2 cm2/min Hence, Area is increasing at the rate of 2 cm2/min.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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