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Finding rate of change

Ex 6.1, 4 - An edge of a variable cube is increasing at 3 cm/s

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


Transcript

Ex 6.1, 4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of cube increasing when the edge is 10 cm long?Let 𝒙 be the edge of cube. & V be the volume of cube. Given that Edge of cube is increasing at the rate of 3 cm/ sec ∴ 𝒅𝒙/𝒅𝒕 = 3 cm/sec We need to calculate how fast volume of cube increasing when edge is 10 cm i.e. we need to find 𝒅𝑽/𝒅𝒕 when π‘₯ = 10 cm We know that Volume of cube = (Edge)3 V = π‘₯3 Differentiate w.r.t time 𝒅𝑽/𝒅𝒕 = (𝒅(π’™πŸ‘))/𝒅𝒕 𝑑𝑉/𝑑𝑑 = (𝑑(π‘₯3))/𝑑𝑑 Γ— 𝑑π‘₯/𝑑π‘₯ 𝑑𝑉/𝑑𝑑 = (𝑑(π‘₯3))/𝑑π‘₯ Γ— 𝑑π‘₯/𝑑𝑑 𝑑𝑉/𝑑𝑑 = 3π‘₯2 . 𝒅𝒙/𝒅𝒕 𝑑𝑉/𝑑𝑑 = 3π‘₯2 Γ— 3 𝑑𝑉/𝑑𝑑 = 9π‘₯2 When π‘₯ = 10 β”œ 𝑑𝑉/𝑑𝑑─|_(π‘₯ =10) = 9(10)2 β”œ 𝑑𝑉/𝑑𝑑─|_(π‘₯ =10) = 900 Since value is in cm3 & time is in sec 𝒅𝑽/𝒅𝒕 = 900 cm3/sec Hence, volume of a cube is increasing at the rate of 900 cm3/sec

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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.