# Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 27, 2018 by Teachoo

Last updated at Dec. 27, 2018 by Teachoo

Transcript

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 = 8cm & 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. ๐๐ฅ/๐๐ก = โ 5 cm/ min And width of rectangle is increasing at the rate of 4 cm/min I.e. ๐๐ฆ/๐๐ก = 4 cm/min Let p be the perimeter of rectangle We need to find rate of change of perimeter w. r. t time when ๐ฅ = 8 cm & y = 6 cm i.e. find ๐๐/๐๐ก when ๐ฅ = 8 cm & ๐ฆ = 6 cm We know that Perimeter of rectangle = 2 (length + width) p = 2 (๐ฅ + ๐ฆ) Different w. r. t time ๐๐/๐๐ก = (๐ (2(๐ฅ+๐ฆ)))/๐๐ก ๐๐/๐๐ก = 2 (๐ (๐ฅ + ๐ฆ))/๐๐ก ๐๐/๐๐ก = 2.(๐๐ฅ/๐๐ก + ๐๐ฆ/๐๐ก) From (1) & (2) ๐๐/๐๐ก = 2 ( โ 5 + 4 ) ๐๐/๐๐ก = 2 ( โ 1) ๐๐/๐๐ก = โ2 Since perimeter is in cm & time is in minute. ๐๐/๐๐ = (โ2 ๐๐)/๐๐๐ ๐๐/๐๐ก = โ2 cm/min Hence, perimeter is decreasing at the rate of 2cm/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 w.r.t time when ๐ฅ = 8 & ๐ฆ = 6 cm i.e. ๐๐ด/๐๐ก when ๐ฅ = 8 cm & ๐ฆ = 6 cm We know that Area of rectangle = length ร width A = ๐ฅ . ๐ฆ Differentiate w.r.t time ๐๐ด/๐๐ก = (๐(๐ฅ . ๐ฆ))/๐๐ก Using product rule in x . y as (u , v)โ = uโ v + vโ u ๐๐ด/๐๐ก = ๐(๐ฅ)/๐๐ก . ๐ฆ + ๐๐ฆ/๐๐ก . ๐ฅ. ๐๐ด/๐๐ก = โ 5 . ๐ฆ + 4 . ๐ฅ ๐๐ด/๐๐ก = 4๐ฅ โ 5๐ฆ when ๐ฅ = 8 cm & ๐ฆ = 6 cm ๐๐ด/๐๐ก = 4 (8) โ 5(6) ๐๐ด/๐๐ก = 32 โ 30 ๐๐ด/๐๐ก = 2 Since Area is in cm2 & time is in minute So ๐๐ด/๐๐ก = 2๐๐2/๐๐๐ ๐๐ด/๐๐ก = 2 cm2/ min Hence, Area is increasing at the rate of 2cm2/min

Chapter 6 Class 12 Application of Derivatives

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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.