Example 31 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 31 A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tan–1 (0.5). Water is poured into it at a constant rate of 5 cubic meter per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4 m.Water tank is in shape of cone Let r be the radius of cone, h be the height of cone, & 𝜶 be the semi−vertical angle Given Semi-vertical angle is tan^(−1)(0.5) α=tan^(−1)(0.5) tan α =(0.5) 𝑟/ℎ = 0.5 𝑟/ℎ = 1/2 𝒓 = 𝒉/𝟐 Also, Water is poured at a constant rate of 5 cubic meter per hour 𝒅𝑽/𝒅𝒕=𝟓 𝒎^𝟑/𝒉𝒓 Where V is volume of cone Now, 𝑉=1/3 𝜋𝑟^2 ℎ 𝑉=1/3 𝜋(𝒉/𝟐)^2 ℎ 𝑉=1/3 𝜋(〖h/4〗^2 )ℎ 𝑽=(𝝅𝒉^𝟑)/𝟏𝟐 Differentiate w.r.t. t 𝑑𝑉/𝑑𝑡=𝑑((𝜋ℎ^3)/12)/𝑑𝑡 𝑑𝑉/𝑑𝑡=𝜋/12 . (𝑑ℎ^3)/𝑑𝑡 𝑑𝑉/𝑑𝑡=𝜋/12 . (𝑑ℎ^3)/𝑑ℎ .𝑑ℎ/𝑑𝑡 𝑑𝑉/𝑑𝑡=𝜋/12. 3ℎ^2 . 𝑑ℎ/𝑑𝑡 𝑑𝑉/𝑑𝑡=(𝜋ℎ^2)/4 . 𝑑ℎ/𝑑𝑡 Putting 𝒅𝑽/𝒅𝒕=𝟓 5 =(𝜋ℎ^2)/4 . 𝑑ℎ/𝑑𝑡 𝒅𝒉/𝒅𝒕=𝟐𝟎/(𝝅𝒉^𝟐 ) We need to find, Rate at which level of water is rising when depth is 4 m i.e. ├ 𝒅𝒉/𝒅𝒕┤|_(𝒉 = 𝟒 𝒎) Putting h = 4 m in (3) ├ 𝑑ℎ/𝑑𝑡┤|_(ℎ = 4𝑚)=20/(𝜋(ℎ)^2 )=20/16 × 1/𝜋=5/4 × 1/(22/7)=5/4 × 7/22=35/88 Hence, rate of change of water level is 𝟑𝟓/𝟖𝟖 m/hr.
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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 and Computer Science at Teachoo