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Chapter 6 Class 12 Application of Derivatives

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Last updated at May 6, 2021 by Teachoo

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

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 = 10 cm and y = 6 cm, find the rates of change of (a) the perimeter and (b) the area of the rectangle.Let Length of rectangle = š„ cm & Width of rectangle = š¦ cm Given length š„ is decreasing at the rate of 3 cm/minute š š/š š = ā 3 cm/ min and width y is increasing at the rate of 2 cm/min š š/š š = 2 cm /min (i) Finding rate of change of Perimeter Let P be the perimeter of rectangle. We need to calculate rate of change of perimeter when š„ = 10 cm & š¦ = 6 cm i.e. we need to calculate šš/šš” when š„ = 10 cm & š¦ = 6 cm We know that Perimeter of rectangle = 2 (Length + Width) P = 2 (š„ + š¦) Now šš/šš”= (š (2 (š„ + š¦) ) )/šš” šš/šš”= 2 [š(š„ + š¦)/šš”] š š·/š š= 2 [š š/š š+ š š/š š] From (1) & (2) šš„/šš” = ā3 & šš¦/šš” = 2 šš/šš”= 2(ā 3 + 2) šš/šš”= 2 (ā1) š š·/š š= ā2 Since perimeter is in cm & time is in minute šš/šš” = ā 2 cm/min Therefore, perimeter is decreasing at the rate of 2 cm/min(ii) Finding rate of change of Area Let A be the Area of rectangle. We need to calculate Rate of change of area when š„ = 10cm & š¦ = 6 cm i.e. we need to calculate š šØ/š š when š„=10 & š¦=6 cm We know that Area of rectangle = Length Ć Width A = š„ Ć š¦ Now, šš“/šš” = (š (š„š¦))/šš” šš“/šš” = šš„/šš” š¦ + šš¦/šš” š„ From (1) & (2) šš„/šš” = ā3 & šš¦/šš” = 2 dA/dt = (ā3)š¦+2 (š„) š šØ/š š = ā šš+šš Putting š = 10 & š = 6 cm šš“/šš” = ā 3(6) + 2(10) = ā 18 + 20 = 2 Since Area is in cm2 & time is in time in minute šš“/šš” = 2 cm2/min Hence, Area is increasing at the rate of 2 cm2/min.