A gas barrel has the shape of a right cylinder with the radius of 5 ft. and the height of 15 ft. If gas is being pumped into the barrel at the rate of 3 ft.3/min, find the rate at which the gas is rising when the gas is 8 ft. deep.
any help would be great!!
Since the tank is a right circular cylinder, the cross-section does not change.
The volume of a 1-ft-high section is 25pi ft^3, so the liquid rises at a rate of 3/25pi ft/min.
Now, if you want to use calculus, you can say that
v = pi r^2 h
v' = pi(2rr'h + r^2h')
But, r is constant, so r' = 0
v' = pi r^2 h'
3 = pi * 25 * h'
h' = 3/25pi
okay, thank you!!!
To find the rate at which the gas is rising when the gas is 8 ft. deep, we need to use related rates.
Given:
Radius of the gas barrel (r) = 5 ft
Height of the gas barrel (h) = 15 ft
Let V be the volume of the gas in the barrel at any time. We know that the volume of a cylinder is given by the formula V = πr²h.
We are given that the rate at which the gas is being pumped into the barrel is 3 ft³/min. Let's call this rate dV/dt, which represents the rate of change of volume with respect to time.
To find the rate at which the gas is rising when the gas is 8 ft. deep, we need to find dh/dt (the rate at which the height is changing) when h = 8 ft.
We can differentiate the formula for volume with respect to time to get:
dV/dt = π(2rh)(dh/dt)
Now, we can plug in the known values:
3 ft³/min = π(2(5 ft)(8 ft))(dh/dt)
Simplifying further:
3 ft³/min = π(10 ft²)(dh/dt)
Now, solve for dh/dt:
dh/dt = (3 ft³/min) / (π(10 ft²))
Calculate the value:
dh/dt ≈ 0.095 ft/min
Therefore, the gas is rising at a rate of approximately 0.095 ft/min when the gas is 8 ft deep.
To solve this problem, we can apply the concept of related rates, which involve finding the rate of change of one variable with respect to another variable.
Let's define the variables:
- V: volume of the gas in the barrel (in cubic feet)
- r: radius of the gas column (in feet)
- h: height of the gas column (in feet)
We are given:
- The radius, r = 5 ft.
- The rate of change of volume, dV/dt = 3 ft^3/min.
- We need to find the rate at which the gas is rising (dh/dt) when the gas is 8 ft. deep.
First, we need to derive the formula for the volume of a right cylinder:
V = πr^2h
Now, we can differentiate both sides of this equation with respect to time (t) using the chain rule:
dV/dt = d(πr^2h)/dt = 2πrh(dr/dt) + πr^2(dh/dt)
We know that r = 5 ft., and we need to find dh/dt when h = 8 ft., so we substitute these values into the equation:
3 = 2π(5)(8)(dr/dt) + π(5^2)(dh/dt)
Simplifying, we have the equation:
3 = 80π(dr/dt) + 25π(dh/dt)
Now, we need to solve for dh/dt:
dh/dt = (3 - 80π(dr/dt)) / (25π)
Since we are given the rate of change of volume, dr/dt is 0 because the radius remains constant. So we can substitute dr/dt = 0 into the equation:
dh/dt = (3 - 80π(0)) / (25π)
dh/dt = 3 / (25π)
Now we can calculate the value of dh/dt:
dh/dt ≈ 3 / (25π)
dh/dt ≈ 0.038 ft/min
Therefore, the rate at which the gas is rising when the gas is 8 ft. deep is approximately 0.038 ft/min.