Check sibling questions

Finding rate of change

Ex 6.1, 2 - Volume of a cube is increasing at 8 cm3/s. How fast

Ex 6.1,2 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,2 - Chapter 6 Class 12 Application of Derivatives - Part 3
Ex 6.1,2 - Chapter 6 Class 12 Application of Derivatives - Part 4
Ex 6.1,2 - Chapter 6 Class 12 Application of Derivatives - Part 5


Transcript

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 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‘ Given that Volume of cube is increasing at rate of 8 cm3/sec. Therefore 𝒅𝑽/𝒅𝒕 = 8 Putting V = π’™πŸ‘ (〖𝑑(π‘₯γ€—^3))/𝑑𝑑 = 8 〖𝑑π‘₯γ€—^3/𝑑𝑑 . 𝑑π‘₯/𝑑π‘₯ = 8 〖𝑑π‘₯γ€—^3/𝑑π‘₯ . 𝑑π‘₯/𝑑𝑑 = 8 3π’™πŸ . 𝑑π‘₯/𝑑𝑑 = 8 𝒅𝒙/𝒅𝒕 = πŸ–/γ€–πŸ‘π’™γ€—^𝟐 Now, We need to find fast is the surface area increasing when the length of an edge is 12 centimeters i.e. 𝒅𝑺/𝒅𝒕 for x = 12 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Ÿ‘π’™πŸ = πŸ‘πŸ/𝒙 For π‘₯= 12 cm 𝑑𝑆/𝑑𝑑 = 32/12 (From (1): 𝒅𝒙/𝒅𝒕 = πŸ–/(πŸ‘π’™^𝟐 )) 𝑑𝑆/𝑑𝑑 = 8/3 Since surface area is in cm2 & time is in seconds, 𝒅𝑺/𝒅𝒕 = πŸ–/πŸ‘ cm2 /s

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.