Example 2 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 15, 2021 by Teachoo
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Transcript
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 π be length of side V be Volume t be time per second We know that Volume of cube = (side)3 V = ππ Also it is given that Volume of cube is increasing at rate of 9 cubic cm/sec. Therefore π π½/π π = 9 Putting V = ππ (γπ(π₯γ^3))/ππ‘ = 9 γππ₯γ^3/ππ‘ . ππ₯/ππ₯ = 9 γππ₯γ^3/ππ₯ . ππ₯/ππ‘ = 9 3ππ . π π/π π = 9 ππ₯/ππ‘ = 9/γ3π₯γ^2 π π/π π = π/π^π Now, We need to find fast is the surface area increasing when the length of an edge is 10 centimeters i.e. π πΊ/π π for x = 10 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/ππ = ππ/π (From (1): π π/π π = π/π^π ) For π₯= 10 cm ππ/ππ‘ = 36/10 ππ/ππ‘ = 3.6 Since surface area is in cm2 & time is in seconds, ππ/ππ‘ = 3.6 ππ2/π π πΊ/π π = 3.6 cm2 /s
