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Finding rate of change

Example 43 - A water tank has shape of an inverted cone - Examples

Example 43 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 43 - Chapter 6 Class 12 Application of Derivatives - Part 3
Example 43 - Chapter 6 Class 12 Application of Derivatives - Part 4
Example 43 - Chapter 6 Class 12 Application of Derivatives - Part 5


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, & 𝜢 be the semiβˆ’vertical angle Given Semi-vertical angle is tan^(βˆ’1)⁑(0.5) Ξ±=tan^(βˆ’1)⁑(0.5) tan Ξ± =(0.5) π‘Ÿ/β„Ž = 0.5 π‘Ÿ/β„Ž = 1/2 𝒓 = 𝒉/𝟐 Also, Water is poured at a constant rate of 5 cubic meter per hour 𝒅𝑽/𝒅𝒕=πŸ“ π’Ž^πŸ‘/𝒉𝒓 Where V is volume of cone Now, 𝑉=1/3 πœ‹π‘Ÿ^2 β„Ž 𝑉=1/3 πœ‹(𝒉/𝟐)^2 β„Ž 𝑉=1/3 πœ‹(γ€–h/4γ€—^2 )β„Ž 𝑽=(𝝅𝒉^πŸ‘)/𝟏𝟐 Differentiate w.r.t. t 𝑑𝑉/𝑑𝑑=𝑑((πœ‹β„Ž^3)/12)/𝑑𝑑 𝑑𝑉/𝑑𝑑=πœ‹/12 . (π‘‘β„Ž^3)/𝑑𝑑 𝑑𝑉/𝑑𝑑=πœ‹/12 . (π‘‘β„Ž^3)/π‘‘β„Ž .π‘‘β„Ž/𝑑𝑑 𝑑𝑉/𝑑𝑑=πœ‹/12. 3β„Ž^2 . π‘‘β„Ž/𝑑𝑑 𝑑𝑉/𝑑𝑑=(πœ‹β„Ž^2)/4 . π‘‘β„Ž/𝑑𝑑 Putting 𝒅𝑽/𝒅𝒕=πŸ“ 5 =(πœ‹β„Ž^2)/4 . π‘‘β„Ž/𝑑𝑑 𝒅𝒉/𝒅𝒕=𝟐𝟎/(𝝅𝒉^𝟐 ) We need to find, Rate at which level of water is rising when depth is 4 m i.e. β”œ 𝒅𝒉/𝒅𝒕─|_(𝒉 = πŸ’ π’Ž) Putting h = 4 m in (3) β”œ π‘‘β„Ž/𝑑𝑑─|_(β„Ž = 4π‘š)=20/(πœ‹(β„Ž)^2 )=20/16 Γ— 1/πœ‹=5/4 Γ— 1/(22/7)=5/4 Γ— 7/22=35/88 Hence, rate of change of water level is πŸ‘πŸ“/πŸ–πŸ– m/hr.

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.