A water tank has the shape of an inverted right circular cone with base radius 3 meters and height 6 meters. Water is being pumped into the tank at the rate of 12 meters3/sec. Find the rate, in meters/sec, at which the water level is rising when the water is 2 meters deep. Give 2 decimal places for your answer. Type your answer in the space below. If your answer is a number less than 1, place a leading "0" before the decimal point (ex: 0.35).
To find the rate at which the water level is rising, we can use related rates. Let's first find an equation that relates the rate of change of the depth of the water to the rate of change of the height of the 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 base radius, and h is the height.
Differentiating both sides of this equation with respect to time, we get:
dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r^2 * dh/dt.
The first term on the right-hand side represents the rate at which the radius is changing, and the second term represents the rate at which the height is changing.
We are given that dV/dt = 12 m^3/sec, r = 3 m, and dh/dt (the rate at which the height is changing) is what we need to find. We also know that when the water is 2 meters deep, the height of the cone is 6 - 2 = 4 meters.
Substituting these values into the equation, we have:
12 = (1/3) * π * (2 * 3 * dr/dt) * 4 + (1/3) * π * 3^2 * dh/dt.
Simplifying, we obtain:
12 = 8π(dr/dt) + 9π(dh/dt).
To find the value of dh/dt, we need to isolate it. Rearranging the equation, we get:
dh/dt = (12 - 8π(dr/dt)) / (9π)
Now, let's plug in the given value dr/dt = 12 m^3/sec and calculate dh/dt:
dh/dt = (12 - 8π(12)) / (9π)
= (12 - 96π) / (9π)
≈ -0.36 m/sec
The negative sign indicates that the water level is decreasing.
So, the rate at which the water level is rising when the water is 2 meters deep is approximately -0.36 m/sec.
To find the rate at which the water level is rising, we can use related rates.
Let's first consider the equation for the volume of a cone:
V = (1/3) * π * r^2 * h
Given that the base radius of the cone is 3 meters and the height is 6 meters, we can substitute these values into the equation:
V = (1/3) * π * (3^2) * 6
V = 18π meters^3
We are given that water is being pumped into the tank at a rate of 12 meters^3/sec. This means that the rate of change of volume with respect to time (dV/dt) is 12 meters^3/sec.
We want to find the rate at which the water level is rising, which is the rate of change of height with respect to time (dh/dt). We can express this in terms of dV/dt.
Since the volume V is given by 18π, we can write:
dV/dt = (dV/dh) * (dh/dt)
To find dV/dh, we differentiate the equation for volume with respect to h:
dV/dh = (1/3) * π * r^2
Substituting the given value for the base radius r:
dV/dh = (1/3) * π * (3^2)
dV/dh = 3π meters^3/meter
Now we can rearrange the equation for dV/dt to solve for dh/dt:
dh/dt = (dV/dt) / (dV/dh)
Substituting the given values:
dh/dt = 12 meters^3/sec / (3π meters^3/meter)
= 4/π meters/sec
To find the rate at which the water level is rising when the water is 2 meters deep, we substitute h = 2 into the equation:
dh/dt = 4/π meters/sec ≈ 1.27 meters/sec
Therefore, the rate at which the water level is rising when the water is 2 meters deep is approximately 1.27 meters/sec.
let the radius of the water level be r m
let the height of the water be h m
make a sketch and by similarity:
h/r = 6/3
h/r = 2
h = 2r or r = h/2
V = (1/3)π r^2 h
= (1/3)π (h^2/4)(h) = (1/12)π h^3
dV/dt = (1/4) h^2 dh/dt
so when h= 2 and dV/dt = 12
12 = (1/4)((4)dh/dt
dh/dt = 12 m/sec
check my arithmetic, should have written it out on paper first.