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

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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 (a) the perimeter, and (b) the area of the rectangle. Let length of rectangle = & width of rectangle = Given that Length of rectangle is decreasing at the rate of 5 cm/min i.e. = 5 cm/ min And width of rectangle is increasing at the rate of 4 cm/min I.e. = 4 cm/min Let p be the perimeter of rectangle We need to find rate of change of perimeter w. r. t time when = 8 cm & y = 6 cm i.e. find when = 8 cm & = 6 cm We know that Perimeter of rectangle = 2 (length + width) p = 2 ( + ) Different w. r. t time = 2 + = 2 ( + ) = 2. + From (1) & (2) = 2 ( 5 + 4 ) = 2 ( 1) = 2 Since perimeter is in cm & time is in sec. = 2 = 2 cm/sec Hence, perimeter is decreasing at the rate of 2cm/sec when = 8 cm & y = 6 cm (ii) Let A be the Area of rectangle we need to find rate of change of area w.r.t time when = 8 & = 6 cm i.e. when = 8 cm & = 6 cm We know that Area of rectangle = length width A = . Differentiate w.r.t time = ( . ) = . + . . = 5 . + 4 . = 4 5 when = 8 cm & = 6 cm = 4 (8) 5(6) = 32 30 = 2 Since Area is in cm2 & time is in sec So = 2 2 = 2 cm2/ sec Hence, Area is increasing at the rate of 2cm2/sec when = 80cm & = 60 cm