Ex 6.1, 7 - The length x of a rectangle is decreasing at rate - Ex 6.1

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.1,7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm 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 & width of rectangle = ๐‘ฆ Given that Length of rectangle is decreasing at the rate of 5 cm/min i.e. ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐‘ก๏ทฏ = โ€“ 5 cm/ min And width of rectangle is increasing at the rate of 4 cm/min I.e. ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ก๏ทฏ = 4 cm/min Let p be the perimeter of rectangle We need to find rate of change of perimeter w. r. t time when ๐‘ฅ = 8 cm & y = 6 cm i.e. find ๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ when ๐‘ฅ = 8 cm & ๐‘ฆ = 6 cm We know that Perimeter of rectangle = 2 (length + width) p = 2 (๐‘ฅ + ๐‘ฆ) Different w. r. t time ๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ = ๐‘‘ 2 ๐‘ฅ+๐‘ฆ๏ทฏ๏ทฏ๏ทฎ๐‘‘๐‘ก๏ทฏ ๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ = 2 ๐‘‘ (๐‘ฅ + ๐‘ฆ)๏ทฎ๐‘‘๐‘ก๏ทฏ ๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ = 2. ๐‘‘๐‘ฅ๏ทฎ๐‘‘๐‘ก๏ทฏ + ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ก๏ทฏ๏ทฏ From (1) & (2) ๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ = 2 ( โ€“ 5 + 4 ) ๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ = 2 ( โ€“ 1) ๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ = โ€“ 2 Since perimeter is in cm & time is in sec. ๐‘‘๐‘๏ทฎ๐‘‘๐‘“๏ทฏ = โˆ’ 2 ๐‘๐‘š๏ทฎ๐‘ ๐‘’๐‘๏ทฏ ๐‘‘๐‘๏ทฎ๐‘‘๐‘ก๏ทฏ = โ€“ 2 cm/sec Hence, perimeter is decreasing at the rate of 2cm/sec when ๐‘ฅ = 8 cm & y = 6 cm (ii) Let A be the Area of rectangle we need to find rate of change of area w.r.t time when ๐‘ฅ = 8 & ๐‘ฆ = 6 cm i.e. ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ when ๐‘ฅ = 8 cm & ๐‘ฆ = 6 cm We know that Area of rectangle = length ร— width A = ๐‘ฅ . ๐‘ฆ Differentiate w.r.t time ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = ๐‘‘(๐‘ฅ . ๐‘ฆ)๏ทฎ๐‘‘๐‘ก๏ทฏ ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = ๐‘‘ ๐‘ฅ๏ทฏ๏ทฎ๐‘‘๐‘ก๏ทฏ . ๐‘ฆ + ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ก๏ทฏ . ๐‘ฅ. ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = โ€“ 5 . ๐‘ฆ + 4 . ๐‘ฅ ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = 4๐‘ฅ โ€“ 5๐‘ฆ when ๐‘ฅ = 8 cm & ๐‘ฆ = 6 cm ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = 4 (8) โ€“ 5(6) ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = 32 โ€“ 30 ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = 2 Since Area is in cm2 & time is in sec So ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = 2๐‘๐‘š2๏ทฎ๐‘ ๐‘’๐‘๏ทฏ ๐‘‘๐ด๏ทฎ๐‘‘๐‘ก๏ทฏ = 2 cm2/ sec Hence, Area is increasing at the rate of 2cm2/sec when ๐‘ฅ = 80cm & ๐‘ฆ = 60 cm

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