Mark M. answered • 03/23/18

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Let V = volume of water in the trough at time t

Let h = depth of water at time t

Let x = length of the surface of the water at the triangular end of the trough at time t

Given: dV/dt = 11 Find: dh/dt when h = 8 in = 2/3 ft

Since the triangular end is isosceles, the altitude bisects the base.

Area of triangular cross section of the filled portion of the trough = (1/2)(x)(h)

By similar triangles (comparing the triangular end of the trough to the triangle formed by the filled portion), we have 2/1 = (x/2)/h. So, x = 4h.

So, Area of triangular cross section = (1/2)(x)(h) = (1/2)(4h)(h) = 2h

^{2}V = (Area of triangular cross section)(length of trough) = (2h

^{2})(10) = 20h^{2}dV/dt = 40h(dh/dt)

Substituting the given information, we have 11 = 40(2/3)(dh/dt)

11 = (80/3)(dh/dt)

dh/dt = 33/80 ft/min