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| Math Central | Quandaries & Queries |
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Question from Michael, a student: Water is being pumped into a trough that is 4.5m long and has a cross section in the shape of an equilateral triangle 1.5m on a side. If the rate of inflow is 2 cubic meters per minute how fast is the water level rising when the water is 0.5m deep? |
Hi Michael.
The volume of water in the trough is equal to the cross-sectional area submerged times the length of the trough.
The cross-sectional area is an equilateral triangle. If you are given the height of an equilateral triangle as h, you should be able to verify that the area of the triangle is just h2/sqrt(3).
Thus, the volume can be expressed as
V = (4.5)h2/sqrt(3)
Take the derivative of both sides with respect to time using the chain rule, then substitute 2 m/min in for dV/dt, 0.5 m in for h,and solve for dh/dt.
That's the solution.
Hope this helps,
Stephen La Rocque.
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