Finding rate of change

Chapter 6 Class 12 Application of Derivatives (Term 1)
Concept wise

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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 π 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 β΄ ππ/ππ = 3 cm/sec We need to calculate how fast volume of cube increasing when edge is 10 cm i.e. we need to find ππ½/ππ when π₯ = 10 cm We know that Volume of cube = (Edge)3 V = π₯3 Differentiate w.r.t time ππ½/ππ = (π(ππ))/ππ ππ/ππ‘ = (π(π₯3))/ππ‘ Γ ππ₯/ππ₯ ππ/ππ‘ = (π(π₯3))/ππ₯ Γ ππ₯/ππ‘ ππ/ππ‘ = 3π₯2 . ππ/ππ ππ/ππ‘ = 3π₯2 Γ 3 ππ/ππ‘ = 9π₯2 When π₯ = 10 β ππ/ππ‘β€|_(π₯ =10) = 9(10)2 β ππ/ππ‘β€|_(π₯ =10) = 900 Since value is in cm3 & time is in sec ππ½/ππ = 900 cm3/sec Hence, volume of a cube is increasing at the rate of 900 cm3/sec