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In Mathematics / College | 2025-07-07

Water is poured into a vessel in the shape of a right circular cone of vertical angle [tex]$60^{\circ}$[/tex] with the axis vertical at a rate of [tex]$8 cm^3 s^{-1}$[/tex]. At what rate is the water surface rising when the depth of the water is 4 cm?

Asked by derri97

Answer (1)

Express the volume V of the water in the cone in terms of its height h , using the relationship between the radius and height derived from the cone's angle: V = 9 1 ​ π h 3 .
Differentiate the volume equation with respect to time t to relate the rates of change of volume and height: d t d V ​ = 3 1 ​ π h 2 d t d h ​ .
Substitute the given values d t d V ​ = 8 and h = 4 into the differentiated equation.
Solve for d t d h ​ to find the rate at which the water surface is rising: 2 π 3 ​ ​ .

Explanation

Problem Setup We are given a right circular cone with a vertical angle of 6 0 ∘ . Water is poured into the cone at a rate of 8 cm 3 / s . We want to find the rate at which the water surface is rising, d t d h ​ , when the depth of the water is 4 cm .

Volume and Radius Relationship Let V be the volume of the water, h be the depth, and r be the radius of the water surface. The volume of a cone is V = 3 1 ​ π r 2 h . Since the semi-vertical angle is 3 0 ∘ , we have tan ( 3 0 ∘ ) = h r ​ , so r = h tan ( 3 0 ∘ ) = 3 ​ h ​ .

Volume in terms of height Substitute r = 3 ​ h ​ into the volume formula: V = 3 1 ​ π ( 3 ​ h ​ ) 2 h = 9 1 ​ π h 3 .

Differentiating with respect to time Differentiate both sides with respect to time t : d t d V ​ = d t d ​ ( 9 1 ​ π h 3 ) = 3 1 ​ π h 2 d t d h ​ .

Substituting given values We are given d t d V ​ = 8 and h = 4 . Substitute these values: 8 = 3 1 ​ π ( 4 2 ) d t d h ​ = 3 16 π ​ d t d h ​ .

Solving for dh/dt Solve for d t d h ​ : d t d h ​ = 16 π 8 ⋅ 3 ​ = 2 π 3 ​ .

Final Answer The rate at which the water surface is rising when the depth of the water is 4 cm is 2 π 3 ​ cm/s.


Examples
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Answered by GinnyAnswer | 2025-07-07