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Question from Roger, a student:

a reservoir has the shape of an inverted cone whose cross section is an equilateral triangle. if water is being pumped out of the reservoir at a rate of 2m^3/sec, at what rate is the depth of the water changing when the depth is 40 meters?

Roger,

My diagram shows the water in the reservoir when the height is $h$ meters. $ABC$ is an equilateral triangle and $D$ is the midpoint of $AB$ so that $r$ is the radius of the circular top of the water.

reservoir

What are the measure of the angles $CAD$ and $DCA?$ Write $r$ in terms of $h.$ The volume of water, $V$ is given by $\frac13 \pi r^2 h.$ Write $V$ as a function of $h$ and use the calculus you know to complete the problem.

Write back if you need more assistance.

Penny

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