# Example 4 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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 = cm & width of rectangle = cm Given the length is decreasing at the rate of 3 cm/ minute i.e. is decreasing w.r.t time i.e. = 3 cm/ min & The width y is increasing at the rate of 2 cm/min i.e. is increasing w.r.t time = 2 cm /min (i) Let P be the perimeter of rectangle. We need to calculate rate of change of perimeter when = 10 cm & = 6 cm i.e. we need to calculate when = 10 cm & = 6 cm We know that Perimeter of rectangle = 2 (length + width) P = 2 ( + ) Now = 2 + = 2 + = 2 + From (1) & (2) = 3 & = 2 = 2( 3 + 2) = 2 ( 1) = 2 Since perimeter is in cm & time is in minute = 2 = 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 = 10cm & = 6 cm i.e. we need to calculate when =10 & =6 cm We know that Area of rectangle = length width A = Now, = ( ) = ( ) . y + ( ) . x From (1) & (2) = 3 & = 2 dA dt = ( 3) +2 ( ) = 3 +2 Putting = 10 & = 6 cm =10, = 6 = 3(6) + 2(10) =10, = 6 = 18 + 20 =10, = 6 = 2 Since Area is in cm2 & time is in time in minute = 2 2 = 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.