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Ex 6.1, 7 - The length x of a rectangle is decreasing at rate

Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 3

Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.1,7 - Chapter 6 Class 12 Application of Derivatives - Part 6

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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’…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‘ƒ/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‘ƒ/š‘‘š‘” = (š‘‘ (2(š‘„ + š‘¦)))/š‘‘š‘” š‘‘š‘ƒ/š‘‘š‘”= 2 [š‘‘(š‘„ + š‘¦)/š‘‘š‘”] š’…š‘·/š’…š’•= 2 [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‘ƒ/š‘‘š‘”= 2(ā€“5 + 4) š‘‘š‘ƒ/š‘‘š‘”= 2 (ā€“1) š’…š‘·/š’…š’•= ā€“2 Since perimeter is in cm & time is in minute š‘‘š‘ƒ/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 Ɨ Width A = š‘„ Ɨ 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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.