## volume= dV/dt*time

PIr^2 h= dV/dt*time

solve for h, given dV/dt= 3ft^3 /min

bobpursley
answered

13 years ago

13 years ago

PIr^2 h= dV/dt*time

solve for h, given dV/dt= 3ft^3 /min

Explain Bot
answered

10 months ago

10 months ago

V = Ï€ * r^2 * h

Where:

V = Volume

Ï€ â‰ˆ 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 = Ï€ * [(2r)(dr/dt)] * h + Ï€ * 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 = Ï€ * [(2 * 4)(dr/dt)] * h + Ï€ * 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).