Okay, let's see. I have this problem about a cylindrical tank being filled with water. The tank is right, so it's a standard cylinder with circular bases. The water is being added at a rate of 20π cubic meters per hour, and the water level rises at a rate of 4 meters per hour. I need to find the radius of the tank. Hmm, alright.

First, let me recall the formula for the volume of a cylinder. The volume V of a cylinder is given by V = πr²h, where r is the radius and h is the height (or in this case, the water level). Since the tank is being filled, the volume is changing over time, so maybe I need to relate the rate of change of volume to the rate of change of height.

Yes, that sounds right. They give me the rate at which the volume is increasing, which is dV/dt = 20π m³/h. And the rate at which the height is increasing is dh/dt = 4 m/h. So, if I can relate these two rates using the formula for the volume, I can solve for the radius.

Let me write down the formula again: V = πr²h. Since the radius r is constant (it's a right cylinder), the only variable here is h, the height of the water. So if I differentiate both sides with respect to time t, I can get an expression involving dV/dt and dh/dt.

Differentiating V with respect to t: dV/dt = πr² * dh/dt. Because r is constant, the derivative of r² with respect to t is zero, so only the h term remains. That makes sense.

So now, plugging in the known values: dV/dt is 20π, and dh/dt is 4. So:

20π = πr² * 4

Hmm, let me check. If I write that equation: 20π = πr² * 4. Then I can divide both sides by π to simplify:

20π / π = r² * 4

Which simplifies to 20 = 4r²

Then divide both sides by 4:

20 / 4 = r²

Which is 5 = r²

So then take the square root of both sides:

r = √5

Wait, that seems straightforward. Let me verify.

Given that the volume rate is 20π, and the height is rising at 4 m/h, so the cross-sectional area (πr²) times the rate of height increase (dh/dt) gives the volume rate. So πr² * 4 = 20π. Solving for r² gives 20π / (4π) = 5, so r²=5, so r=√5. Yep, that checks out.

But wait, let me make sure I didn't skip any steps. Let me go through the differentiation again. The volume is V = πr²h. So dV/dt = πr² dh/dt + πh d(r²)/dt. But since r is constant, d(r²)/dt = 0. So indeed, dV/dt = πr² dh/dt. So substituting the given rates:

20π = πr² * 4

Divide both sides by π:

20 = 4r²

Divide by 4:

5 = r²

So r = √5. Since radius can't be negative, we take the positive root. So the radius is √5 meters. That seems correct.

Is there another way to approach this? Maybe by considering the volume at a particular time. For example, if the water level rises 4 meters in one hour, then the volume added in that hour is the area of the base times the height increase. So the volume added is πr² * 4. And they say that this volume is 20π. So πr² *4 =20π. Then solving for r² gives 5, as before. Yep, same result.

So regardless of the method, whether through calculus or basic volume considerations, the radius comes out to √5. Therefore, I think that's the answer.

**Final Answer**
The radius of the tank is \boxed{\sqrt{5}} meters.
Therefore, the radius of the tank is $\boxed{\sqrt{5}}$ meters.
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