A conical water tank with vertex down has a radius of 13 feet at the top and is 27 feet high. If water flows into the tank at a rate of 30 ft^3/min, how fast is the depth of the water increasing when the water is 17 feet deep?
v = 1/3 * π * r^2 * h
r = 13/27 h
v = 1/3 * π * (13/27 h)^2 * h = 1/3 * π * (13/27)^2 * h^3
dv/dt = π * (13/27)^2 * h^2 * dh/dt
30 ft^3/min = π * (13/27)^2 * (17 ft)^2 * dh/dt
To solve this problem, we need to use the related rates formula for a cone:
(Volume of cone) = (1/3) * π * (radius of base)^2 * (height)
Differentiating both sides with respect to time will give us:
(dV/dt) = (1/3) * π * (2 * (radius of base) * (dr/dt) * (height) + (radius of base)^2 * (dh/dt))
Given values:
(radius of base) = 13 feet
(height) = 27 feet
(dV/dt) = 30 ft^3/min
We are looking for (dh/dt) when the water is 17 feet deep, so we need to find (dr/dt).
Using similar triangles, we can set up the following proportion:
(height - 17) / (radius of base) = (17) / (x)
Solving for x, we get:
x = (17 * (radius of base)) / (height - 17)
= (17 * 13) / (27 - 17)
= 221 / 10
= 22.1 feet
Now, we can substitute the known values into the related rates formula and solve for (dh/dt):
30 = (1/3) * π * (2 * 13 * (dr/dt) * 27 + 13^2 * (dh/dt))
30 = (1/3) * π * (2 * 13 * (dr/dt) * 27 + 169 * (dh/dt))
30 = (1/3) * π * (2 * 13 * 27 * (dr/dt) + 169 * (dh/dt))
To isolate (dh/dt), we need to solve for (dr/dt). We can rearrange the equation as follows:
2 * 13 * 27 * (dr/dt) + 169 * (dh/dt) = (3/π) * 30
(26 * 27 * (dr/dt)) + (169 * (dh/dt)) = 90/π
(702 * (dr/dt)) + (169 * (dh/dt)) = 90/π
(dr/dt) = (90/π - 169 * (dh/dt)) / 702
Now, substitute the values we know: (dh/dt) = ?, (dr/dt) = (90/π - 169 * (dh/dt)) / 702)
Plug in the values into the related rates formula:
30 = (1/3) * π * (2 * 13 * 27 * (dr/dt) + 169 * (dh/dt))
30 = (1/3) * π * (2 * 13 * 27 * ((90/π - 169 * (dh/dt)) / 702) + 169 * (dh/dt))
Now, solve for (dh/dt).
To find the rate at which the depth of the water is increasing, we need to use related rates.
Let's denote the radius of the water at a given depth as r (in feet), and the height of the water as h (in feet). We are given that the radius of the tank at the top is 13 feet and the height is 27 feet.
First, let's find the relationship between the radius and height of the water by considering similar triangles.
The ratio of the radius to height remains constant throughout the tank because it is a cone. So, we have:
r / h = (13 / 27) (1)
Next, we need to find the relationship between the volume and the radius (or height) of the water. The volume of a cone is given by the formula:
V = (1/3) * π * r^2 * h (2)
Differentiate both sides of equation (2) with respect to time (t) to find the rate of change of volume (dV/dt):
dV/dt = (1/3) * π * [2r * dr/dt * h + r^2 * dh/dt] (3)
Now, we have to find the value of dr/dt when h = 17 ft. Since water is flowing into the tank at a rate of 30 ft^3/min, the rate of change of volume of the water is given by dV/dt = 30 ft^3/min.
Using equation (1), we can express r in terms of h:
r = (13 / 27) * h
Substitute the values of r and h into equation (3):
30 = (1/3) * π * [2 * (13 / 27) * h * dh/dt + (13 / 27)^2 * dh/dt]
Simplify the equation:
90/π = (26 / 27) * h * dh/dt + (169 / 729) * dh/dt
Since r = (13 / 27) * h, we can rewrite the equation as:
90/π = (26 / 27) * h * dh/dt + (169 / 729) * dh/dt
Now, solve for dh/dt by isolating the term:
dh/dt = (90/π) / [(26 / 27) * h + (169 / 729)]
Substitute h = 17 ft into the equation:
dh/dt = (90/π) / [(26 / 27) * 17 + (169 / 729)]
Now, we can calculate the value of dh/dt.