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Last updated at Jan. 7, 2020 by Teachoo

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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 = ๐ฅ cm & width of rectangle = ๐ฆ cm Given the length ๐ฅ is decreasing at the rate of 3 cm/ minute i.e. ๐ฅ is decreasing w.r.t time i.e. ๐๐ฅ/๐๐ก = โ 3 cm/ min & The width y is increasing at the rate of 2 cm/min i.e. ๐ฆ is increasing w.r.t time ๐๐ฆ/๐๐ก = 2 cm /min (i) Let P be the perimeter of rectangle. We need to calculate rate of change of perimeter when ๐ฅ = 10 cm & ๐ฆ = 6 cm i.e. we need to calculate ๐๐/๐๐ก when ๐ฅ = 10 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] ๐๐/๐๐ก= 2 [๐๐ฅ/๐๐ก+ ๐๐ฆ/๐๐ก] From (1) & (2) ๐๐ฅ/๐๐ก = โ3 & ๐๐ฆ/๐๐ก = 2 ๐๐/๐๐ก= 2(โ 3 + 2) ๐๐/๐๐ก= 2 (โ1) ๐๐/๐๐ก= โ2 Since perimeter is in cm & time is in minute ๐๐/๐๐ก = (โ2 ๐๐)/๐๐๐ ๐๐/๐๐ก = 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 ๐ฅ = 10cm & ๐ฆ = 6 cm i.e. we need to calculate ๐๐ด/๐๐ก when ๐ฅ=10 & ๐ฆ=6 cm We know that Area of rectangle = length ร width A = ๐ฅ ร ๐ฆ Now, ๐๐ด/๐๐ก = (๐ (๐ฅ๐ฆ))/๐๐ก ๐๐ด/๐๐ก = (๐(๐ฅ))/๐๐ก. y + (๐(๐ฆ))/๐๐ก . x From (1) & (2) ๐๐ฅ/๐๐ก = โ3 & ๐๐ฆ/๐๐ก = 2 dA/dt = (โ 3)๐ฆ+2 (๐ฅ) ๐๐ด/๐๐ก = โ 3๐ฆ+2๐ฅ Putting ๐ฅ = 10 & ๐ฆ = 6 cm โ ๐๐ด/๐๐กโค|_(๐ฅ =10, ๐ฆ = 6) = โ 3(6) + 2(10) โ ๐๐ด/๐๐กโค|_(๐ฅ =10, ๐ฆ = 6) = โ 18 + 20 โ ๐๐ด/๐๐กโค|_(๐ฅ =10, ๐ฆ = 6) = 2 Since Area is in cm2 & time is in time in minute ๐๐ด/๐๐ก = (2 ๐๐2)/๐๐๐ ๐๐ด/๐๐ก = 2 cm2/min Hence, Area is increasing at the rate of 2 cm2/min.

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Chapter 6 Class 12 Application of Derivatives

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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 9 years. He provides courses for Maths and Science at Teachoo.