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 how fast the depth of the water is increasing when the water is 12 feet deep, we need to use related rates.
Let's start by establishing the given values:
- The radius at the top of the conical water tank is 12 feet.
- The height of the conical water tank is 26 feet.
- The rate at which water flows into the tank is 30 ft³/min.
Now, let's establish the variables we need:
- Let "r" represent the radius of the water at a given time.
- Let "h" represent the height of the water at a given time.
- Let "V" represent the volume of the water in the tank at a given time.
Next, we need to establish the formula for the volume of the water in a conical tank. The formula for the volume of a cone is given by:
V = (1/3)πr²h.
Now, we need to take the derivative of the volume with respect to time, since we want to find how fast the depth of the water is increasing. Using the chain rule, we have:
dV/dt = (1/3)π * 2r * dr/dt * h + (1/3)πr² * dh/dt.
Since we are given that water flows into the tank at a rate of 30 ft³/min, we can substitute dV/dt with 30. Also, since the water flows into the tank at a constant rate, the rate of change of the radius is 0. Thus, we have:
30 = (1/3)π * 2r * 0 + (1/3)πr² * dh/dt,
30 = 0 + (1/3)πr² * dh/dt.
Now, we need to find the value of "r" when the water is 12 feet deep. In a cone, similar to the proportions of similar triangles, we have:
r/h = R/H,
where R is the radius at the top of the conical tank (12 feet), and H is the height of the conical tank (26 feet).
Substituting the given values, we have:
r/12 = 12/26,
r = (12 × 12)/26,
r ≈ 5.54 feet.
Substituting this value of "r" into the equation, we have:
30 = (1/3)π(5.54)² * dh/dt.
Now, we can solve for dh/dt, which represents the rate at which the depth of the water is increasing when the water is 12 feet deep:
To find the rate at which the depth of the water is increasing, we need to use related rates. Let's denote the depth of the water as "h" (in feet), the radius of the water surface as "r" (in feet), and the volume of the water as "V" (in cubic feet).
We are given that the water is flowing into the tank at a rate of 30 ft^3/min, which means dV/dt = 30 ft^3/min.
We want to find dh/dt, the rate at which the depth of the water is increasing when the water is 12 feet deep. We need to find an equation that relates h, r, and V.
Since the tank is conical, we know that the volume of a cone is given by V = (1/3)πr^2h.
Differentiating both sides of this equation with respect to time (t), we get:
dV/dt = (1/3)π(2rh dr/dt + r^2 dh/dt)
Now let's substitute the given values: r = 12 ft and h = 12 ft, and solve for dr/dt.
dV/dt = 30 ft^3/min
r = 12 ft
h = 12 ft
30 ft^3/min = (1/3)π(2 * 12 ft * dr/dt + (12 ft)^2 dh/dt)
30 ft^3/min = (1/3)π(24 ft * dr/dt + 144 ft^2 * dh/dt)
Since we are interested in finding dh/dt when h = 12 ft, we can substitute these values into the equation:
30 ft^3/min = (1/3)π(24 ft * dr/dt + 144 ft^2 * dh/dt) [Equation 1]
Now, we need to find dr/dt when h = 12 ft. To find dr/dt, we can use similar triangles.
In the original cone with height 26 ft, the ratio of the radius to the height is constant. Therefore, we have:
12 ft / h = 12 ft / 26 ft
Simplifying this equation, we get:
12/26 = r/12
r = 12 * (12/26) = 144/26 = 72/13 ft
Now we can differentiate this equation with respect to time:
d(12/26)/dt = (72/13) dr/dt
Simplifying, we get:
-(12/26^2) dh/dt = (72/13) dr/dt
Substituting the known values dh/dt = 0 (since the height is not changing) and r = 72/13 ft, we can solve for dr/dt:
-(12/26^2) * 0 = (72/13) dr/dt
0 = (72/13) dr/dt
This means that dr/dt = 0 ft/min when the height is 12 ft.
Now we can substitute this value into Equation 1 and solve for dh/dt:
30 ft^3/min = (1/3)π(24 ft * 0 ft/min + 144 ft^2 * dh/dt)
30 ft^3/min = (1/3)(144 ft^2 * dh/dt)
90 ft^3/min = 144 ft^2 * dh/dt
Dividing both sides by 144 ft^2, we get:
90 ft^3/min / 144 ft^2 = dh/dt
0.625 ft/min = dh/dt
Therefore, the depth of the water is increasing at a rate of 0.625 ft/min when the height of the water is 12 ft.
Make your sketch,
let the radius of the water level be r ft
let the height of the water be h ft
by ratios:
r/h = 12/26
26r = 12h
r = 6h/13
Vol = (1/3)π r^2 h
= (1/3)π (36h^2/169) g
= (12/169)π h^3
d(Vol)/dt = (36/169)π h^2 dh/dt
30 = (36/169)π (144) dh/dt
dh/dt = 30(169/(36π(144))
= 845/(864π) ft/min = appr .3113 ft/min
check my arithmetic, I should have written it down instead of doing on the screen only