Water is flowing freely from the bottom of a conical tank which is 12 feet deep and 6 feet in radius at the
top. If the water is flowing at a rate of 2 cubic feet per hour, at what rate is the depth of the water in the
tank going down when the depth is 3 feet.
When the water is x feet deep, and the surface has radius r, then by similar triangles,
x/r = 12/6
x = 2r
r = x/2
So, since the volume of water
v = 1/3 pi r^2 x
= 1/12 pi x^3
dv/dt = pi/4 x^2 dx/dt
-2 = pi/4 (9) dx/dt
dx/dt = -8/(9pi) = -.28 ft^3/hr
To find the rate at which the depth of the water in the tank is decreasing when the depth is 3 feet, we can use related rates.
Related rates involve finding the rate at which one quantity changes with respect to the rate at which another related quantity changes. In this case, we want to find the rate of change of the depth of the water (h) with respect to time (t).
Let's use the formula for the volume of a cone to establish a relationship between the variables involved. The volume (V) of a cone is given by V = (1/3)πr^2h, where r is the radius of the top of the cone and h is the height (or depth).
Since the rate of change of the volume is given as 2 cubic feet per hour, we can differentiate the volume formula with respect to time (t) to find the rate at which the volume is changing:
dV/dt = (1/3)π(2rh)(dh/dt) + (1/3)πr^2(dh/dt)
Now, we need to find the values of r and h when the depth is 3 feet. The radius (r) is given as 6 feet, but we need to find the radius at the height of 3 feet. Since the cone is symmetrical, we can use similar triangles to find the radius:
r/h = 6/12
r = 6h/12
r = h/2
So when the depth (h) is 3 feet, the radius (r) is 3/2 feet.
Now we can substitute these values into the rate of change equation:
2 = (1/3)π((3/2) * 3)(dh/dt) + (1/3)π((3/2)^2)(dh/dt)
Simplifying:
2 = (9/2)π(dh/dt) + (9/4)π(dh/dt)
Combining like terms:
2 = (27/4)π(dh/dt)
Now, we can solve for (dh/dt), which gives us the rate at which the depth of the water is decreasing when the depth is 3 feet:
(dh/dt) = 2 * (4/27π)
(dh/dt) = 8/27π
Therefore, the depth of the water in the tank is decreasing at a rate of 8/27π feet per hour when the depth is 3 feet.