Check sibling questions

Finding rate of change

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


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 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 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.