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