Let's use the formula for the volume of a sphere, which is given by:
V = (4/3)πr^3

We want to find the rate at which the water level is rising, which is the rate of change of the height of the water in the tank with respect to time. Let's call this rate dh/dt.

We can also relate the height of the water with the radius of the filled portion of the sphere. Let's call this radius r', and we can establish the following relationship:
r' = r - h

Where:
- r is the radius of the entire sphere (which is 5 meters in this case).
- h is the height of the water in the tank.

Now we can differentiate both sides of the equation with respect to time t:
d(r')/dt = d(r - h)/dt

Since the radius of the entire sphere is constant, the rate of change of the radius with respect to time is zero, so the left side of the equation is zero:
0 = -dh/dt

Now let's differentiate the right side of the equation:
d(r-h)/dt = d(r)/dt - d(h)/dt

The rate of change of the radius with respect to time is zero (as mentioned before), so the first term on the right side is zero:
0 = - d(h)/dt

Now we need to find an expression for the rate of change of the volume with respect to time. We already know the rate at which the tank is being filled, which is 3 cubic meters per minute. So we can differentiate the volume equation with respect to time and substitute the known value for the rate of change of the volume:
dV/dt = 3

Differentiating the volume equation with respect to time, we get:
dV/dt = 4πr^2 (dr/dt)

Now let's solve for dr/dt using the values we know:
3 = 4π(5^2) (dr/dt)
3 = 100π (dr/dt)

Divide both sides of the equation by 100π:
3 / 100π = dr/dt

Finally, substitute the value we are given for the rate of change of the radius:
dr/dt = 3 / 100π ≈ 0.00955 m/min

Therefore, the rate at which the water level is rising when the water is 4 meters deep is approximately 0.00955 meters per minute.

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