document.write( "Question 667241: A trough of length 6 m has a uniform cross section which is an equilateral triangle with sides 1 m. Water leaks from the bottom of the trough at a constant rate of 0.1 m^3 per minute. Find the rate at which the water level is falling at the instant when it is 20 cm deep?
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Algebra.Com's Answer #414929 by Edwin McCravy(20054) $\"\"$ $\"About$

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A trough of length 6 m has a uniform cross section which is an equilateral triangle with sides 1 m. Water leaks from the bottom of the trough at a constant rate of 0.1 m^3 per minute. Find the rate at which the water level is falling at the instant when it is 20 cm deep?
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document.write( "This is a cross section of the trough.  The blue line is the water\r\n" );
document.write( "level.  The red line h is the height of the water level.  We are\r\n" );
document.write( "looking for the rate at which h is shrinking when h=20cm.   \r\n" );
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document.write( "The volume of the water is the area of the equilateral triangle\r\n" );
document.write( "whose base is the blue line times the trough length of 6m, The\r\n" );
document.write( "area of the triangle is (2x)(h) or xh and multiplying\r\n" );
document.write( "this by the trough length of 6 m, we have\r\n" );
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document.write( "                      V = 6xh\r\n" );
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document.write( "We also know that  = tan(60°) = Ö3.\r\n" );
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document.write( "So , therefore x =  = , so V = 6xh becomes \r\n" );
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document.write( "                     V = 6h\r\n" );
document.write( "or\r\n" );
document.write( "                     V = 2·Ö3·h²\r\n" );
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document.write( "Differentiating with respect to time t\r\n" );
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document.write( "                     = 4·Ö3·h· \r\n" );
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document.write( "We are given that  = -0.1 m³ taken negative because the\r\n" );
document.write( "volume of water is decreasing.  And we want the particular value of\r\n" );
document.write( " when h = 20 cm = 0.2 m.\r\n" );
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document.write( "                    -0.1 = 4·Ö3·0.2·\r\n" );
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document.write( "Solve that for  and we get\r\n" );
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document.write( "                    = \r\n" );
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document.write( "Multiply top and bottom by 10\r\n" );
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document.write( "                    =                    \r\n" );
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document.write( "                    = \r\n" );
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document.write( "Rationalize the denominator:\r\n" );
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document.write( "                    = \r\n" );
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document.write( "                    = \r\n" );
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document.write( "That's in meters/minute, so to change it to centimeters/minute,\r\n" );
document.write( "multiply by 100\r\n" );
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document.write( "                    = \r\n" );
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document.write( "                    = \r\n" );
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document.write( "which is about -7.2 centimeters/minute, which means that the water \r\n" );
document.write( "level is falling at  centimeters/minute or \r\n" );
document.write( "about 7.2 centimeters/minute.\r\n" );
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document.write( "Edwin

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