Last updated at April 19, 2021 by Teachoo

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Ex 6.1, 7 The length x of a rectangle is decreasing at the rate of 5 cm/minute & the width y is increasing at the rate of 4 cm/minute. When x = 8 cm & y = 6cm, find the rates of change of (a) the perimeter.Let Length of rectangle = ๐ฅ & Width of rectangle = ๐ฆ Given that Length of rectangle is decreasing at the rate of 5 cm/min i.e. ๐ ๐/๐ ๐ = โ 5 cm/ min And width of rectangle is increasing at the rate of 4 cm/min I.e. ๐ ๐/๐ ๐ = 4 cm/min Let P be the perimeter of rectangle We need to find rate of change of perimeter when ๐ฅ = 8 cm & y = 6 cm i.e. Finding ๐๐/๐๐ก when ๐ฅ = 8 cm & ๐ฆ = 6 cm We know that Perimeter of rectangle = 2 (Length + Width) P = 2 (๐ฅ + ๐ฆ) Now, ๐๐/๐๐ก = (๐ (2(๐ฅ + ๐ฆ)))/๐๐ก ๐๐/๐๐ก= 2 [๐(๐ฅ + ๐ฆ)/๐๐ก] ๐ ๐ท/๐ ๐= 2 [๐ ๐/๐ ๐+ ๐ ๐/๐ ๐] From (1) & (2) ๐๐ฅ/๐๐ก = โ5 & ๐๐ฆ/๐๐ก = 4 ๐๐/๐๐ก= 2(โ5 + 4) ๐๐/๐๐ก= 2 (โ1) ๐ ๐ท/๐ ๐= โ2 Since perimeter is in cm & time is in minute ๐๐/๐๐ก = โ 2 cm/min Therefore, perimeter is decreasing at the rate of 2 cm/min 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 (b) the area of the rectangle.Let A be the Area of rectangle We need to find Rate of change of area when ๐ฅ = 8 & ๐ฆ = 6 cm i.e. ๐ ๐จ/๐ ๐ when ๐ฅ = 8 cm & ๐ฆ = 6 cm We know that Area of rectangle = Length ร Width A = ๐ฅ ร ๐ฆ Now, ๐๐ด/๐๐ก = (๐(๐ฅ. ๐ฆ))/๐๐ก ๐๐ด/๐๐ก = ๐๐ฅ/๐๐ก . ๐ฆ + ๐๐ฆ/๐๐ก . ๐ฅ. From (1) & (2) ๐๐ฅ/๐๐ก = โ5 & ๐๐ฆ/๐๐ก = 4 ๐๐ด/๐๐ก = (โ5)๐ฆ + (4)๐ฅ ๐๐ด/๐๐ก = 4๐ฅ โ 5๐ฆ Putting ๐ฅ = 8 cm & ๐ฆ = 6 cm ๐๐ด/๐๐ก = 4 (8) โ 5(6) Using product rule in x . y as (u.v)โ = uโ v + vโ u ๐๐ด/๐๐ก = 32 โ 30 ๐ ๐จ/๐ ๐ = 2 Since Area is in cm2 & time is in minute ๐๐ด/๐๐ก = 2 cm2/ min Hence, Area is increasing at the rate of 2cm2/min

Ex 6.1

Ex 6.1, 1
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About the Author

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.