



 
Hi Michael. The volume of water in the trough is equal to the crosssectional area submerged times the length of the trough. The crosssectional 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 h^{2}/sqrt(3). Thus, the volume can be expressed as
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. Hope this helps,  


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