Example 2 - Chapter 6 Class 12 Application of Derivatives

Last updated at August 13, 2026 by

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part 2 - Example 2 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives
part 3 - Example 2 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 4 - Example 2 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives part 5 - Example 2 - Examples - Serial order wise - Chapter 6 Class 12 Application of Derivatives

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Example 2 The volume of a cube is increasing at a rate of 9 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 10 centimeters ?Let š’™ be length of side V be Volume t be time per second We know that Volume of cube = (side)3 V = š’™šŸ‘ Also it is given that Volume of cube is increasing at rate of 9 cubic cm/sec. Therefore š’…š‘½/š’…š’• = 9 Putting V = š’™šŸ‘ (怖š‘‘(š‘„怗^3))/š‘‘š‘” = 9 怖š‘‘š‘„怗^3/š‘‘š‘” . š‘‘š‘„/š‘‘š‘„ = 9 怖š‘‘š‘„怗^3/š‘‘š‘„ . š‘‘š‘„/š‘‘š‘” = 9 3š’™šŸ . š’…š’™/š’…š’• = 9 š‘‘š‘„/š‘‘š‘” = 9/怖3š‘„怗^2 š’…š’™/š’…š’• = šŸ‘/š’™^šŸ Now, We need to find fast is the surface area increasing when the length of an edge is 10 centimeters i.e. š’…š‘ŗ/š’…š’• for x = 10 We know that Surface area of cube = 6 Ɨ Side2 S = 6š‘„2 Finding š’…š‘ŗ/š’…š’• š‘‘š‘†/š‘‘š‘” = (š‘‘(6š‘„^2))/š‘‘š‘” = (š‘‘(6š‘„2))/š‘‘š‘” . š‘‘š‘„/š‘‘š‘„ = 6. (š‘‘(š‘„2))/š‘‘š‘„ . š‘‘š‘„/š‘‘š‘” = 6 . (2x) . š‘‘š‘„/š‘‘š‘” = 12š‘„ . š’…š’™/š’…š’• = 12š‘„ . šŸ‘/š’™šŸ = šŸ‘šŸ”/š’™ For š‘„= 10 cm š‘‘š‘†/š‘‘š‘” = 36/10 š‘‘š‘†/š‘‘š‘” = 3.6 Since surface area is in cm2 & time is in seconds, š‘‘š‘†/š‘‘š‘” = 3.6 š‘š‘š2/š‘  š’…š‘ŗ/š’…š’• = 3.6 cm2 /s

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