a water tank is created by revolving the graph y=1/x about the y-axis, with the bottom of the tank at y=1. the volume of the tank is given by [v(h) = the integral from 1 to h of (pi/y^2)dy] where h is the height of the water in the tank. Initially, the tank is empty, but water begins to flow into the tank at a rate of 1.5 cubic feet per minute. Determine how fast the level of the water is rising when the water is 2 feet deep.
Since v(h) = ∫[1,h] π/y^2 dy
dv/dt = π/h^2 dh/dt
1.5 = π/2 dh/dt
3/π = dh/dt
To determine how fast the level of the water is rising when the water is 2 feet deep, we need to find the derivative of the volume function with respect to time and then substitute the given values.
Step 1: Find the derivative of the volume function.
The volume function is given by V(h) = ∫(π/y^2)dy from 1 to h.
We need to differentiate this volume function with respect to h. So, we will use the Fundamental Theorem of Calculus.
dV/dh = d/dh(∫(π/y^2)dy)[1 to h]
Using the Second Fundamental Theorem of Calculus, we can write:
dV/dh = (π/h^2)
Step 2: Substitute the given values.
We are given that the water is flowing into the tank at a rate of 1.5 cubic feet per minute. This rate represents dV/dt (the rate of change of volume with respect to time).
Given: dV/dt = 1.5 cubic feet per minute
We can relate dV/dh and dV/dt using the Chain Rule:
dV/dt = (dV/dh) * (dh/dt)
Substituting the values:
1.5 = (π/((2)^2)) * (dh/dt)
Simplifying further:
1.5 = (π/4) * (dh/dt)
Step 3: Solve for dh/dt.
To find the rate at which the water level is rising, we need to determine the value of dh/dt.
Multiplying both sides by (4/π):
1.5 * (4/π) = dh/dt
Approximating π as 3.14:
1.5 * (4/3.14) = dh/dt
dh/dt ≈ 1.91 feet per minute
Therefore, the level of the water is rising at approximately 1.91 feet per minute when the water is 2 feet deep.