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Example 2 - Chapter 6 Class 12 Application of Derivatives
Last updated at December 16, 2024 by Teachoo
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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 = šš Also it is given that Volume of cube is increasing at rate of 9 cubic cm/sec. Therefore š š½/š š = 9 Putting V = šš (ćš(š„ć^3))/šš” = 9 ćšš„ć^3/šš” . šš„/šš„ = 9 ćšš„ć^3/šš„ . šš„/šš” = 9 3šš . š š/š š = 9 šš„/šš” = 9/ć3š„ć^2 š š/š š = š/š^š Now, We need to find fast is the surface area increasing when the length of an edge is 10 centimeters i.e. š šŗ/š š for x = 10 We know that Surface area of cube = 6 Ć Side2 S = 6š„2 Finding š šŗ/š š šš/šš” = (š(6š„^2))/šš” = (š(6š„2))/šš” . šš„/šš„ = 6. (š(š„2))/šš„ . šš„/šš” = 6 . (2x) . šš„/šš” = 12š„ . š š/š š = 12š„ . š/šš = šš/š For š„= 10 cm šš/šš” = 36/10 šš/šš” = 3.6 Since surface area is in cm2 & time is in seconds, šš/šš” = 3.6 šš2/š š šŗ/š š = 3.6 cm2 /s
