Question 187283
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You don't say, but I have to presume that the sides of the trough are 1 meter and the measure across the top is 1.2 meters.


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Part a.


Since the triangle is isoceles, the height segment is a perpendicular bisector of the base, forming two right triangles with sides 1, .6, and *[tex \Large sqrt{1 - (0.6)^2} = 0.8]


That means the base varies as the height in the proportion 1.2 to 0.8, or:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{b}{h}\ =\ \frac{1.2}{0.8} \ \ \Rightarrow\ \ b\ =\ \frac{3h}{2}]


Hence, the area of the triangle as a function of the height of the water is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A(h)\ =\ \frac{\frac{3h^2}{2}}{2}\ =\ \frac{3h^2}{4}m^2]


And therefore the volume is just the area times the length of the trough:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(h)\ =\ 2\,\cdot\,\frac{3h^2}{4}\ =\ \frac{3h^2}{2}m^3]


Part b.


We need to find volume per unit time, or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  \frac{V(h)}{t} = 6 L/min = .006 m^3/min]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3h^2}{2t}m^3/min\ =\ .006 m^3/min] 


Solve for <i>h</i>:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h^2\ =\ \frac{.006\,\cdot\,2t}{3}\ =\ .004t \ \ \Rightarrow\ \ h(t) = sqrt{.004t}\ m]




John
*[tex \LARGE e^{i\pi} + 1 = 0]
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