SOLUTION: A trough is 2 m long, and its ends are triangles with sides of length 1 m, 1 m, and 1.2 m.
a) Find the volume V of the water in the trough as a function of the water level h.
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Question 187283: A trough is 2 m long, and its ends are triangles with sides of length 1 m, 1 m, and 1.2 m.
a) Find the volume V of the water in the trough as a function of the water level h.
b) If water is pumped into the empty trough at the rate of 6 L/min, find the water level h as a function of the time t after the pumping begins. (1 m^3 = 1000L)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
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.
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 ^2} = 0.8)
That means the base varies as the height in the proportion 1.2 to 0.8, or:
Hence, the area of the triangle as a function of the height of the water is:
\ =\ \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:
\ =\ 2\,\cdot\,\frac{3h^2}{4}\ =\ \frac{3h^2}{2}m^3)
Part b.
We need to find volume per unit time, or
}{t} = 6 L/min = .006 m^3/min)
So:
Solve for h:
 = sqrt{.004t}\ m)
John
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