A conical water tank with vertex down has a radius of 12 feet at the top and is 23 feet high. If water flows into the tank at a rate of 20 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 17 feet deep?
in 2012 i was 5. now i am 21
To solve this problem, we need to use related rates, specifically the formula relating the volume of a cone to its height. The formula is:
V = (1/3) * π * r^2 * h
where V is the volume, r is the radius, h is the height of the cone, and π is Pi (approximately 3.14).
Given that the radius, r, is 12 feet, and the tank is 23 feet high, we can find the volume, V, in terms of the height, h.
V = (1/3) * π * (12)^2 * h
Now, let's differentiate both sides of the equation implicitly with respect to time (t), where t is the time in minutes.
dV/dt = (1/3) * π * (12)^2 * dh/dt
The left side of the equation represents the rate at which the volume is changing with respect to time, which is given as 20 ft^3/min.
20 = (1/3) * π * 12^2 * dh/dt
We are interested in finding how fast the depth of the water is increasing when the water is 17 feet deep, so we substitute h = 17 into the equation.
20 = (1/3) * π * 12^2 * dh/dt
Now, we solve for dh/dt.
dh/dt = (20 * 3) / (π * 12^2)
dh/dt = 60 / (π * 144)
dh/dt ≈ 0.045 ft/min
Therefore, the depth of the water is increasing at a rate of approximately 0.045 feet per minute when the water is 17 feet deep.
To find the rate at which the depth of the water is increasing, we need to use related rates and the formula for the volume of a cone.
First, let's establish some variables:
- Let "h" represent the depth of the water in the cone.
- Let "V" represent the volume of the water in the cone.
- We are given that the rate of change of volume with respect to time is dV/dt = 20 ft^3/min.
Next, let's find the formula for the volume of a cone:
The volume (V) of a cone is given by the formula:
V = (1/3) * π * r^2 * h,
where "r" is the radius of the cone and "h" is the height (or depth) of the water.
In this case, the radius (r) of the cone is given to be 12 feet at the top, which means that as the water level increases, the radius decreases linearly. We can express the radius (r) in terms of the depth (h) using similar triangles.
Let's find the equation relating r and h:
Since the tank is conical with the vertex at the bottom, we can write:
(h + 23) / (r + 12) = h / r
Cross-multiplying, we have:
r * (h + 23) = (r + 12) * h
Expanding this equation, we get:
r * h + 23r = r * h + 12h
Simplifying, we find:
23r = 12h
r = (12h) / 23
Now, we have an expression for "r" in terms of "h".
To find the rate at which the depth of the water is increasing when the water is 17 feet deep (dh/dt), we can use implicit differentiation:
Differentiating both sides of the equation "23r = 12h" with respect to time "t", we get:
23(dr/dt) = 12(dh/dt)
Now, we have an equation relating the rates of change of the radius (dr/dt) and the depth (dh/dt).
We know that dr/dt = 0 because the radius does not change with time.
Substituting dr/dt = 0 and dV/dt = 20 into the equation 23(dr/dt) = 12(dh/dt), we get:
23(0) = 12(dh/dt)
0 = 12(dh/dt)
Therefore, the depth of the water is not changing with time when the water is 17 feet deep. The rate of change of the depth (dh/dt) is 0 ft/min.
at a depth of h ft, the radius of the water surface is r=(12/23)*h ft
v = 1/3 pi r^2 h
= pi/3 * (12/23)^2 * h^3
dv/dt = pi(12/23)^2 h^2 dh/dt
20 = pi(12/23)^2 (17)^2 dh/dt
dh/dt = 2645 / 10404pi = 0.081 ft/min = 0.97 in/min