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Example 2 - Volume of a cube is increasing at a rate of 9 cubic

Example 2 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 2 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 2 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 2 - Chapter 6 Class 12 Application of Derivatives - Part 5

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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 𝒙 be length of side V be Volume t be time per second We know that Volume of cube = (side)3 V = π’™πŸ‘ Also it is given that Volume of cube is increasing at rate of 9 cubic cm/sec. Therefore 𝒅𝑽/𝒅𝒕 = 9 Putting V = π’™πŸ‘ (〖𝑑(π‘₯γ€—^3))/𝑑𝑑 = 9 〖𝑑π‘₯γ€—^3/𝑑𝑑 . 𝑑π‘₯/𝑑π‘₯ = 9 〖𝑑π‘₯γ€—^3/𝑑π‘₯ . 𝑑π‘₯/𝑑𝑑 = 9 3π’™πŸ . 𝒅𝒙/𝒅𝒕 = 9 𝑑π‘₯/𝑑𝑑 = 9/γ€–3π‘₯γ€—^2 𝒅𝒙/𝒅𝒕 = πŸ‘/𝒙^𝟐 Now, We need to find fast is the surface area increasing when the length of an edge is 10 centimeters i.e. 𝒅𝑺/𝒅𝒕 for x = 10 We know that Surface area of cube = 6 Γ— Side2 S = 6π‘₯2 Finding 𝒅𝑺/𝒅𝒕 𝑑𝑆/𝑑𝑑 = (𝑑(6π‘₯^2))/𝑑𝑑 = (𝑑(6π‘₯2))/𝑑𝑑 . 𝑑π‘₯/𝑑π‘₯ = 6. (𝑑(π‘₯2))/𝑑π‘₯ . 𝑑π‘₯/𝑑𝑑 = 6 . (2x) . 𝑑π‘₯/𝑑𝑑 = 12π‘₯ . 𝒅𝒙/𝒅𝒕 = 12π‘₯ . πŸ‘/π’™πŸ = πŸ‘πŸ”/𝒙 (From (1): 𝒅𝒙/𝒅𝒕 = πŸ‘/𝒙^𝟐 ) For π‘₯= 10 cm 𝑑𝑆/𝑑𝑑 = 36/10 𝑑𝑆/𝑑𝑑 = 3.6 Since surface area is in cm2 & time is in seconds, 𝑑𝑆/𝑑𝑑 = 3.6 π‘π‘š2/𝑠 𝒅𝑺/𝒅𝒕 = 3.6 cm2 /s

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Davneet Singh

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.