A conical water tank with vertex down has a radius of 12 feet at the top and is 26 feet high. If water flows into the tank at a rate of 30 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 12 feet deep?
To find the rate at which the depth of water is changing, we need to find the derivative of the depth with respect to time.
Let's denote the depth of the water by h and the radius of the water at height h as r. We'll use similar triangles to find the relationship between h and r.
From the given information, we know that the radius at the top of the tank is 12 feet and the height of the tank is 26 feet. We can use this information to find the equation for the radius of the water at height h.
In a similar triangle, the ratio of the corresponding sides is constant. Therefore, we can write:
r/12 = (h - r)/26
Let's solve this equation for r:
r = (12 * (h - r))/26
r = (12h - 12r)/26
26r = 12h - 12r
38r = 12h
r = (12h)/38
r = (6h)/19
Now we have an equation that relates the depth of the water h to the radius of the water r.
We also know that the rate at which water is flowing into the tank is 30 ft^3/min. This is the rate at which the volume V of the water is changing with respect to time t:
dV/dt = 30
The volume of a cone can be calculated using the formula:
V = (1/3) * π * r^2 * h
We can substitute the value of r from the equation we derived earlier:
V = (1/3) * π * ((6h)/19)^2 * h
Now, we can differentiate this equation with respect to time t:
dV/dt = (1/3) * π * (2 * (6h)/19) * ((6h)/19)^2 * dh/dt
dV/dt = (2π * (18h^3)/19^3) * dh/dt
Now, we can substitute the values given in the problem to find the rate at which the depth of water is changing when the water is 12 feet deep:
30 = (2π * (18(12)^3)/19^3) * dh/dt
30 = (2π * (18 * 1728)/19^3) * dh/dt
Now, we can solve this equation for dh/dt:
dh/dt = 30 / ((2π * (18 * 1728)/19^3))
Calculate the right side of the equation:
dh/dt = 30 / (2π * (18 * 1728)/19^3)
dh/dt ≈ 0.019 feet/minute
Therefore, the depth of the water is increasing at a rate of approximately 0.019 feet per minute when the water is 12 feet deep.
To find the rate at which the depth of water is increasing, we need to relate the change in depth to the change in volume of water in the tank.
Let's denote the depth of the water in the tank as h (in feet) and the volume of water in the tank as V (in cubic feet).
Given:
Radius of the top of the tank, r = 12 feet
Height of the tank, H = 26 feet
Rate of change of volume, dV/dt = 30 ft³/min
We can use similar triangles to relate the height h and the radius r at any given time:
h / r = H / R
h / 12 = H / 12 + dh / dr
We are asked to find the rate at which the depth of water is increasing when the depth is 12 feet. So, h = 12 feet.
We can substitute the value of h = 12 feet into the equation above and solve for dh/dt.
12 / 12 = 26 / 12 + dh / dr
1 = 26 / 12 + dh / dr
1 - 26 / 12 = dh / dr
(12 - 26) / 12 = dh / dr
-14 / 12 = dh / dr
-7 / 6 = dh / dr
Now we can relate the rate of change of volume (dV/dt) to the rate of change of height (dh/dt) using the formula for the volume of a cone:
V = (1/3) * π * r² * h
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = (1/3) * π * [2r * dr/dt * h + r² * dh/dt]
Substituting the given values:
dV/dt = (1/3) * π * [2 * 12 * dr/dt * 12 + 12² * (-7/6)]
Now, we have:
dV/dt = (1/3) * π * [24 * dr/dt * 12 - 84]
Since dV/dt is given as 30 ft³/min:
30 = (1/3) * π * [24 * dr/dt * 12 - 84]
Simplifying the equation:
30 = 8π * (2 * dr/dt * 12 - 7)
30 = 16π * dr/dt * 12 - 56π
Now, let's solve for dr/dt:
16π * dr/dt * 12 = 30 + 56π
dr/dt = (30 + 56π) / (16π * 12)
Finally, substitute the value of π, evaluate the expression, and calculate the rate:
dr/dt = (30 + 56 * 3.14159) / (16 * 3.14159 * 12) = 0.268 ft/min
Therefore, the depth of the water is increasing at a rate of approximately 0.268 feet per minute when the water is 12 feet deep.