Question 1204151
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An elliptical culvert is 3.2 feet tall and 6.3 feet wide. 
It is filled with water to a depth of 0.85 feet. Find the width of the stream.
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The semiaxes are: 3.2/2 = 1.6 ft long vertical and 6.3/2 = 3.15 ft horizontal.


So, the standard form equation for this ellipse is

    {{{x^2/3.15^2}}} + {{{y^2/1.6^2}}} = 1,


written with the center of the ellipse as the beginning of coordinates.


The level of water is at y = -1.6 + 0.85 = -0.75 ft.


Substitute this value of y into the ellipse equation

    {{{x^2/3.15}}} + {{{(-0.75)^2/1.6^2}}} = 1,

and get

    {{{x^2/3.15^2}}} = 1 - {{{(-0.75)^2/1.6^2}}}

or

    {{{x^2/3.15^2}}} = 0.780273438.


It implies

    {{{x^2}}} = {{{0.780273438*3.15^2}}} = 7.742263189,

    x = {{{sqrt(7.742263189)}}} = 2.782492262.


So, the water surface is from x= -2.782492262  to  x= 2.782492262
.


Thus the wide of the stream is  2*2.782492262 = 5.564984524 ft,  or 5.565 ft  (rounded).


<U>ANSWER</U>.  The width of the stream is 5.565 ft,  rounded.
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Solved.