Ex 6.1

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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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 = 8 cm & y = 6cm, find the rates of change of (a) the perimeter.Let Length of rectangle = š„ & Width of rectangle = š¦ Given that Length of rectangle is decreasing at the rate of 5 cm/min i.e. šš/šš = ā 5 cm/ min And width of rectangle is increasing at the rate of 4 cm/min I.e. šš/šš = 4 cm/min Let P be the perimeter of rectangle We need to find rate of change of perimeter when š„ = 8 cm & y = 6 cm i.e. Finding šš/šš” when š„ = 8 cm & š¦ = 6 cm We know that Perimeter of rectangle = 2 (Length + Width) P = 2 (š„ + š¦) Now, šš/šš” = (š (2(š„ + š¦)))/šš” šš/šš”= 2 [š(š„ + š¦)/šš”] šš·/šš= 2 [šš/šš+ šš/šš] From (1) & (2) šš„/šš” = ā5 & šš¦/šš” = 4 šš/šš”= 2(ā5 + 4) šš/šš”= 2 (ā1) šš·/šš= ā2 Since perimeter is in cm & time is in minute šš/šš” = ā 2 cm/min Therefore, perimeter is decreasing at the rate of 2 cm/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 when š„ = 8 & š¦ = 6 cm i.e. ššØ/šš when š„ = 8 cm & š¦ = 6 cm We know that Area of rectangle = Length Ć Width A = š„ Ć š¦ Now, šš“/šš” = (š(š„. š¦))/šš” šš“/šš” = šš„/šš” . š¦ + šš¦/šš” . š„. From (1) & (2) šš„/šš” = ā5 & šš¦/šš” = 4 šš“/šš” = (ā5)š¦ + (4)š„ šš“/šš” = 4š„ ā 5š¦ Putting š„ = 8 cm & š¦ = 6 cm šš“/šš” = 4 (8) ā 5(6) Using product rule in x . y as (u.v)ā = uā v + vā u šš“/šš” = 32 ā 30 ššØ/šš = 2 Since Area is in cm2 & time is in minute šš“/šš” = 2 cm2/ min Hence, Area is increasing at the rate of 2cm2/min