Water drips out of a hole at the vertex of an upside down cone at a rate of 3 cm^3 per minute. The cone’s height and radius are 2 cm and 1 cm, respectively. At what rate does the height of the water change when the water level is half a centimeter below the top of the cone? The volume of a cone is V = (π/3)*r^2*h, where r is the radius and h is the height of the cone.

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∻Solution by Steps∻ ‖ step 1 ⋮ Given the volume of a cone $ V = \frac{\pi}{3} r^2 h $, differentiate both sides with respect to time $ t $ to find the related rates. ‖ ‖ step 2 ⋮ Applying the chain rule, we get $ \frac{dV}{dt} = \frac{\pi}{3} (2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}) $. ‖ ‖ step 3 ⋮ Given $ \frac{dV}{dt} = -3 $ cm³/min (since water is dripping out), $ h = 1.5 $ cm, and $ r = \frac{h}{2} $, we substitute these values into the differentiated equation. ‖ ‖ step 4 ⋮ Substituting $ r = \frac{h}{2} $ into the equation, we get $ \frac{dV}{dt} = \frac{\pi}{3} (2 \cdot \frac{h}{2} \cdot h \cdot \frac{dr}{dt} + (\frac{h}{2})^2 \cdot \frac{dh}{dt}) $. ‖ ‖ step 5 ⋮ Since $ r $ is a function of $ h $, $ \frac{dr}{dt} $ can be expressed in terms of $ \frac{dh}{dt} $. Differentiating $ r = \frac{h}{2} $ with respect to $ t $, we get $ \frac{dr}{dt} = \frac{1}{2} \frac{dh}{dt} $. ‖ ‖ step 6 ⋮ Substituting $ \frac{dr}{dt} $ into the equation from step 4, we have $ -3 = \frac{\pi}{3} (2 \cdot \frac{h}{2} \cdot h \cdot \frac{1}{2} \frac{dh}{dt} + (\frac{h}{2})^2 \cdot \frac{dh}{dt}) $. ‖ ‖ step 7 ⋮ Simplify the equation to solve for $ \frac{dh}{dt} $ when $ h = 1.5 $ cm. ‖ ‖ step 8 ⋮ After simplification, we get $ -3 = \frac{\pi}{3} (\frac{3}{4} h^2 \frac{dh}{dt}) $. ‖ ‖ step 9 ⋮ Substitute $ h = 1.5 $ cm into the equation to find $ \frac{dh}{dt} $. ‖ ‖ step 10 ⋮ Solving for $ \frac{dh}{dt} $, we get $ \frac{dh}{dt} = \frac{-3}{\frac{\pi}{3} \cdot \frac{3}{4} \cdot 1.5^2} $. ‖ ‖ step 11 ⋮ Calculate the value of $ \frac{dh}{dt} $ to find the rate at which the height of the water changes. ‖ ‖ step 12 ⋮ The final value of $ \frac{dh}{dt} $ is $ \frac{dh}{dt} = \frac{-3}{\frac{\pi}{3} \cdot \frac{3}{4} \cdot 1.5^2} = \frac{-3}{\frac{\pi}{4} \cdot 2.25} = \frac{-3}{\frac{9\pi}{4}} = \frac{-3 \cdot 4}{9\pi} = \frac{-12}{9\pi} = \frac{-4}{3\pi} $ cm/min. ‖ ∻Answer∻ ⚹ $ \frac{dh}{dt} = \frac{-4}{3\pi} $ cm/min ⚹ ∻Key Concept∻ ⚹Related Rates in Conical Volume Change⚹ ∻Explanation∻ ⚹The rate of change of the height of the water in the cone is found by differentiating the volume formula with respect to time and substituting the given values to solve for $ \frac{dh}{dt} $.⚹◊What is the rate at which the height of the water in the cone is changing when the water level is half a centimeter below the top of the cone?⍭ Generate me a similar question◊

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