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

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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 the 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 Thus, 𝑑𝑥﷮𝑑𝑡﷯ = 3cm/sec We need to calculate How fast volume of cube increasing w. r. t time when edge is 10 cm i.e. we need to calculate 𝑑𝑉﷮𝑑𝑡﷯ when 𝑥 = 10 cm We know that Volume of cube = (edge)3 V = 𝑥3 Differentiate w.r.t time 𝑑𝑣﷮𝑑𝑡﷯ = 𝑑(𝑥3)﷮𝑑𝑡﷯ 𝑑𝑣﷮𝑑𝑡﷯ = 𝑑(𝑥3)﷮𝑑𝑡﷯ × 𝑑𝑥﷮𝑑𝑥﷯ 𝑑𝑣﷮𝑑𝑡﷯ = 𝑑(𝑥3)﷮𝑠𝑥﷯ × 𝑑𝑥﷮𝑑𝑡﷯ 𝑑𝑣﷮𝑑𝑡﷯ = 3x2 . 𝒅𝒙﷮𝒅𝒕﷯ 𝑑𝑣﷮𝑑𝑡﷯ = 3𝑥2 × 3 𝑑𝑣﷮𝑑𝑡﷯ = 9𝑥2 When 𝑥 = 10 𝑑𝑣﷮𝑑𝑡﷯﷯﷮𝑥 =10﷯ = 9(10)2 𝑑𝑣﷮𝑑𝑡﷯﷯﷮𝑥 =10﷯ = 900 Since value is in cm3 & time is in sec 𝑑𝑣﷮𝑑𝑡﷯ = 900 𝑐 𝑚﷮3﷯﷮𝑠𝑒𝑐﷯ 𝒅𝒗﷮𝒅𝒕﷯ = 900 cm3/sec Hence, volume of a cube is increasing at the rate of 900 cm3/sec when edge is 10cm