A conical tank has height 3 m and radius 2 m at the top. Water flows in at a rate of 1.8m^3/min. How fast is the water level rising when it is 2.7 m?
What is the radius of the tank at the bottom?
volumetank=PI*rad^2*h
volume of airspace=PI*rad^2*(3-h)^2/3^2 *(3-h)
volumewater=volumetank-volumeairspace
= PI(rad^2*h-rad^2*(3-h)^2/9 * (3-h)
take derivative wrespect to time, you know dVwater/dh=1.8, and h=2.7, solve for dh/dt
r = 2/3 h
v = 1/3 h π r^2 = 1/3 h π 4/9 h^2 =
... 4/27 π h^3
dv/dt = 4/27 π 3 h^2 dh/dt = 4π/9 h^2 dh/dt
dh/dt = 1.8 / [(4π/9) 2.7^2]
To find how fast the water level is rising, we need to calculate the rate of change of the water level with respect to time, which is denoted as dH/dt. Here's how we can solve the problem step by step:
Step 1: Draw a diagram:
Let's draw a diagram to visualize the conical tank:
/|
/ |
R / | H
/ |
/_____|
In this diagram, R represents the radius of the water level at height H.
Step 2: Set up the given information:
We are given that the height of the tank is 3m and the radius at the top is 2m. Also, we know that the water is flowing in at a rate of 1.8m^3/min.
Step 3: Find the formula to calculate the volume of a cone:
The volume of a cone is given by the formula V = (1/3) * π * r^2 * h, where V is the volume, r is the radius, and h is the height of the cone.
Step 4: Express the volume of the water in terms of the height:
Since the radius of the cone is changing with respect to the height, we need to express the radius in terms of the height. Using similar triangles, we can set up the following proportion:
R / H = 2 / 3
Solving for R, we get:
R = (2/3) * H
Now we can substitute this equation for the radius into the volume formula:
V = (1/3) * π * [(2/3) * H]^2 * H
V = (4/27) * π * H^3
Step 5: Differentiate both sides of the equation with respect to time:
To find how fast the volume is changing over time, we differentiate both sides of the equation with respect to time:
dV/dt = d/dt [(4/27) * π * H^3]
The left side represents the rate at which the volume is changing, and the right side represents the rate at which the height is changing.
Step 6: Calculate dV/dt:
We are given that the water is flowing in at a rate of 1.8m^3/min, so we can substitute dV/dt with 1.8:
1.8 = d/dt [(4/27) * π * H^3]
Step 7: Substitute the given height and solve for dH/dt:
We are given that the height of the water level is 2.7m, so we can substitute H with 2.7:
1.8 = d/dt [(4/27) * π * (2.7)^3]
Simplifying the equation:
1.8 = d/dt [(4/27) * π * 19.683]
1.8 = d/dt [26.056π]
Since the variable is dH/dt, we can solve for it by canceling out the terms on the right side:
1.8 = dH/dt * 26.056π
Solving for dH/dt:
dH/dt = 1.8 / 26.056π
Step 8: Calculate the numerical value of dH/dt:
Now we can substitute the value of π (pi) and calculate the numerical value of dH/dt:
dH/dt = 1.8 / 26.056 * 3.14159
dH/dt ≈ 0.0214 m/min
Therefore, the water level is rising at a rate of approximately 0.0214 meters per minute when it is 2.7 meters.
To find the rate at which the water level is rising, we can use related rates.
Let's call the height of the water in the tank "h" and the radius of the water at any given height "r".
We are given that the height of the tank is 3 m and the radius at the top is 2 m. The volume of a cone can be calculated using the formula V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height.
The rate at which the water level is rising can be found by differentiating the volume equation with respect to time (t). Therefore, we have dV/dt = (1/3)(π(2r)(dh/dt) + (πr^2)(dh/dt)).
Since we are given that the water is flowing in at a rate of 1.8 m^3/min, we have dV/dt = 1.8.
To determine the value of r at a height of 2.7 m, we can use similar triangles. We know that the ratio of the height to the radius is constant, so we can set up the proportion (3-2.7)/2 = h/r and solve for r.
(3-2.7)/2 = 0.3/2 = h/r
0.3/2 = 2/3r
0.15 = 2/3r
r ≈ 0.45 m
Substituting the values into the volume formula, we get V = (1/3)π(0.45^2)h.
Now we can substitute all the known values into the rate equation and solve for dh/dt.
1.8 = (1/3)(π(2(0.45))(dh/dt) + (π(0.45^2))(dh/dt))
Simplifying this equation, we have:
1.8 = (1/3)(2π(0.45)(dh/dt) + π(0.45^2)(dh/dt))
Multiplying by 3 to eliminate the fraction, we get:
5.4 = (2π(0.45)(dh/dt) + π(0.45^2)(dh/dt))
Combining like terms:
5.4 = (2π(0.45) + π(0.45^2))(dh/dt)
Calculating the value inside the parentheses:
5.4 = (2π(0.45) + π(0.45^2))(dh/dt)
5.4 = (0.9π + 0.202π)(dh/dt)
5.4 = (1.102π)(dh/dt)
Finally, solving for dh/dt, we have:
dh/dt = 5.4 / (1.102π)
≈ 1.54 m/min
So, the water level is rising at a rate of approximately 1.54 m/min when the water level is 2.7 m.