Suppose we pump water into an inverted right-circular cone tank at the rate of 6 cubic feet per minute. The tank has the height 9 ft and radius on the top is 8 ft. What is the rate at which the water level is rising when the water is 3 ft deep? Leave the answer approximate. You may use the formula for the volume of the right-circular cone of radius r and height h: V=1/3pi*r^2*h
At a time of t minutes, let the
radius of the water level be r
let the height of the water be h
by ratio:
r/h = 8/9
r = 8h/9
V = (1/3)π r^2 h
= (1/3)π(64h^2/81)h
= (64π/243) h^3
dV/dt = (64π/81) h^2 dh/dt
6 = (64π/81) (9) dh/dt
I will let you solve for dh/dt
thanks so much!
To find the rate at which the water level is rising, we need to determine the rate at which the volume of water in the tank is changing with respect to time.
First, let's express the volume of water in the tank as a function of the water height. The tank is an inverted right-circular cone, so its volume (V) can be given by the formula:
V = (1/3) * π * r^2 * h
where r is the radius of the circular cross-section of the tank at that height, h is the height of the water in the tank, and π is a mathematical constant approximately equal to 3.14159.
Given that the radius at the top of the tank is 8 ft and the height is 9 ft, we can write the volume function as:
V(h) = (1/3) * π * (8^2) * h
Next, we need to determine the rate at which the volume of water is changing with respect to time. This can be expressed as dV/dt, where dt represents the change in time and dV represents the change in volume.
To find the rate at which the water level is rising when the water is 3 ft deep, we need to find dV/dt when h = 3.
To do this, we will differentiate the volume equation with respect to time (t):
dV/dt = (1/3) * π * (8^2) * dh/dt
The derivative dh/dt represents the rate at which the height of the water is changing.
Since we know that water is being poured into the tank at a rate of 6 cubic feet per minute, we have dh/dt = 6 ft/min.
Now, let's substitute the known values into our derivative equation:
dV/dt = (1/3) * π * (8^2) * 6
Calculating this, we get:
dV/dt = (1/3) * π * 64 * 6
dV/dt = (64/3) * π * 6
dV/dt ≈ 128π ft^3/min
Therefore, the rate at which the water level is rising when the water is 3 ft deep is approximately 128π ft^3/min.