Water is being pumped at a 3 cubic feet/minute into a cylindrical storage tank with a circular base of a radius 4 feet. Find the depth of the water (d) as a ntion of time (t)

volume= dV/dt*time

PIr^2 h= dV/dt*time

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

To find the depth of the water (d) as a function of time (t), we can use the formula for the volume of a cylindrical tank. The volume (V) of a cylinder is given by the formula:

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 = 8π * (dr/dt) * h + 16π * (dh/dt)

We know that dV/dt = 3, so we can substitute it into the equation:

3 = 8π * (dr/dt) * h + 16π * (dh/dt)

Now, we can solve this equation to find the relationship between h (depth of water) and t (time).