Ex 6.1, 2 The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?Let
𝒙 be length of side
V be Volume
t be time per second
We know that
Volume of cube = (Side)3
V = 𝒙𝟑
Given that
Volume of cube is increasing at rate of 8 cm3/sec.
Therefore
𝒅𝑽/𝒅𝒕 = 8
Putting V = 𝒙𝟑
(〖𝑑(𝑥〗^3))/𝑑𝑡 = 8
〖𝑑𝑥〗^3/𝑑𝑡 . 𝑑𝑥/𝑑𝑥 = 8
〖𝑑𝑥〗^3/𝑑𝑥 . 𝑑𝑥/𝑑𝑡 = 8
3𝒙𝟐 . 𝑑𝑥/𝑑𝑡 = 8
𝒅𝒙/𝒅𝒕 = 𝟖/〖𝟑𝒙〗^𝟐
Now,
We need to find fast is the surface area increasing when the length of an edge is 12 centimeters
i.e. 𝒅𝑺/𝒅𝒕 for x = 12
We know that
Surface area of cube = 6 × Side2
S = 6𝑥2
Finding 𝒅𝑺/𝒅𝒕
𝑑𝑆/𝑑𝑡 = (𝑑(6𝑥^2))/𝑑𝑡
= (𝑑(6𝑥2))/𝑑𝑡 . 𝑑𝑥/𝑑𝑥
= 6. (𝑑(𝑥2))/𝑑𝑥 . 𝑑𝑥/𝑑𝑡
= 6 . (2x) . 𝑑𝑥/𝑑𝑡
= 12𝑥 . 𝒅𝒙/𝒅𝒕
= 12𝑥 . 𝟖/𝟑𝒙𝟐
= 𝟑𝟐/𝒙
For 𝑥= 12 cm
𝑑𝑆/𝑑𝑡 = 32/12
(From (1): 𝒅𝒙/𝒅𝒕 = 𝟖/(𝟑𝒙^𝟐 ))
𝑑𝑆/𝑑𝑡 = 8/3
Since surface area is in cm2 & time is in seconds,
𝒅𝑺/𝒅𝒕 = 𝟖/𝟑 cm2 /s

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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