Related Rate Problem Including a Cone

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SUMMARY

The discussion focuses on solving a related rates problem involving an inverted cone-shaped reservoir with a top radius of 2 meters and a depth of 6 meters. Wine is poured into the reservoir at a rate of 1 m³/sec, and the goal is to determine the rate at which the depth of the wine increases when the depth reaches 4 meters. The volume of the wine is expressed as V=(1/3)π(R^2)h, where R is eliminated using similar triangles, leading to R=h/3. The next step involves differentiating the volume with respect to time to find dh/dt.

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A reservoir in the shape of an inverted cone has a radius of 2 meters at the top and a depth of 6 meters. Wine is poured into the reservoir at a rate of 1 m^3/sec. At what rate is the depth of the wine increasing when the depth is 4 meters?

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Have you written a formula for the volume of the wine in terms of its depth?
 
dV/dt=1 we are asked to find dh/dt when h is 4m.
The volume and the height are related by:
V=(1/3)*Pi*R^2*h
We have to eliminate R.
Sketch the inverted cone: if h is the wine level (and r is the radius of the cone at that h) and the cone height is 9 you can use similar triangles:
r/h=2/6 and so:
r=h/3
The expression is now:
V=(1/3)*Pi*(h/3)^2 *h
you can now differentiate each side.
 

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