



Finding rate of change
Ex 6.1, 1 Deleted for CBSE Board 2022 Exams
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Finding rate of change
Last updated at April 19, 2021 by Teachoo
Ex 6.1, 2 The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?Let π be length of side V be Volume t be time per second We know that Volume of cube = (Side)3 V = ππ Given that Volume of cube is increasing at rate of 8 cm3/sec. Therefore π π½/π π = 8 Putting V = ππ (γπ(π₯γ^3))/ππ‘ = 8 γππ₯γ^3/ππ‘ . ππ₯/ππ₯ = 8 γππ₯γ^3/ππ₯ . ππ₯/ππ‘ = 8 3ππ . ππ₯/ππ‘ = 8 π π/π π = π/γππγ^π Now, We need to find fast is the surface area increasing when the length of an edge is 12 centimeters i.e. π πΊ/π π for x = 12 We know that Surface area of cube = 6 Γ Side2 S = 6π₯2 Finding π πΊ/π π ππ/ππ‘ = (π(6π₯^2))/ππ‘ = (π(6π₯2))/ππ‘ . ππ₯/ππ₯ = 6. (π(π₯2))/ππ₯ . ππ₯/ππ‘ = 6 . (2x) . ππ₯/ππ‘ = 12π₯ . π π/π π = 12π₯ . π/πππ = ππ/π For π₯= 12 cm ππ/ππ‘ = 32/12 (From (1): π π/π π = π/(ππ^π )) ππ/ππ‘ = 8/3 Since surface area is in cm2 & time is in seconds, π πΊ/π π = π/π cm2 /s