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

Last updated at April 16, 2024 by Teachoo

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

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.

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