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

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  1. Chapter 6 Class 12 Application of Derivatives
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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﷯ 𝜋﷐𝑟﷮2﷯ℎ 𝑉=﷐1﷮3﷯ 𝜋﷐﷐﷐h﷮2﷯﷯﷮2﷯ℎ 𝑉=﷐1﷮3﷯ 𝜋﷐﷐﷐h﷮4﷯﷮2﷯﷯ℎ 𝑉=﷐𝜋﷐ℎ﷮3﷯﷮12﷯ Differentiate w.r.t. t ﷐𝑑𝑉﷮𝑑𝑡﷯=﷐𝑑﷐﷐𝜋﷐ℎ﷮3﷯﷮12﷯﷯﷮𝑑𝑡﷯ ﷐𝑑𝑉﷮𝑑𝑡﷯=﷐𝜋﷮12﷯ . ﷐𝑑﷐ℎ﷮3﷯﷮𝑑𝑡﷯ ﷐𝑑𝑉﷮𝑑𝑡﷯=﷐𝜋﷮12﷯ . ﷐𝑑﷐ℎ﷮3﷯﷮𝑑ℎ﷯ .﷐𝑑ℎ﷮𝑑𝑡﷯ ﷐𝑑𝑉﷮𝑑𝑡﷯=﷐𝜋﷐3﷐ℎ﷮2﷯﷯﷮12﷯ . ﷐𝑑ℎ﷮𝑑𝑡﷯ ﷐𝑑𝑉﷮𝑑𝑡﷯=﷐𝜋﷐ℎ﷮2﷯﷮4﷯ . ﷐𝑑ℎ﷮𝑑𝑡﷯ 5=﷐𝜋﷐ℎ﷮2﷯﷮4﷯ . ﷐𝑑ℎ﷮𝑑𝑡﷯ ﷐𝑑ℎ﷮𝑑𝑡﷯=﷐20﷮𝜋﷐ℎ﷮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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