bobpursley
answered
PIr^2 h= dV/dt*time
solve for h, given dV/dt= 3ft^3 /min
Explain Bot
answered
V = π * r^2 * h
Where:
V = Volume
π ≈ 3.14159 (pi)
r = Radius of the circular base
h = Height (depth) of the water
We know that water is being pumped at a rate of 3 cubic feet/minute. This means that the volume of water in the tank is increasing by 3 cubic feet every minute. Therefore, we can set up the following equation to represent the rate of change of volume with respect to time:
dV/dt = 3
Where:
dV/dt = Rate of change of volume with respect to time
t = Time
Let's differentiate the volume formula with respect to time to find dV/dt:
dV/dt = π * [(2r)(dr/dt)] * h + π * r^2 * (dh/dt)
Since the radius of the circular base is given as 4 feet, we can substitute r = 4 into the equation:
dV/dt = π * [(2 * 4)(dr/dt)] * h + π * 4^2 * (dh/dt)
Simplifying the equation, we have:
dV/dt = 8Ï€ * (dr/dt) * h + 16Ï€ * (dh/dt)
We know that dV/dt = 3, so we can substitute it into the equation:
3 = 8Ï€ * (dr/dt) * h + 16Ï€ * (dh/dt)
Now, we can solve this equation to find the relationship between h (depth of water) and t (time).
