# Example 43 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

Example 43 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 α=tan^(−1)(0.5) So, 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 So, 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 (As 𝑟 = ℎ/2 from (1)) 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(3ℎ^2 )/12 . 𝑑ℎ/𝑑𝑡 𝑑𝑉/𝑑𝑡=(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 = 4𝑚) Putting h = 4m 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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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.