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

Last updated at April 16, 2024 by

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 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))/𝑑𝑥 × 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 cm3/sec Hence, volume of a cube is increasing at the rate of 900 cm3/sec

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.