<Question>
Convert the point $(0,3)$ in rectangular coordinates to polar coordinates.  Enter your answer in the form $(r,\theta),$ where $r > 0$ and $0 \le \theta < 2 \pi.$
</Question>

<think>
<Plan_1>
Okay, so I need to convert the rectangular coordinate (0,3) to polar coordinates. Hmm, let me recall how to do this.
Polar coordinates are in the form (r, θ), where r is the radius (distance from the origin) and θ is the angle made with the positive x-axis. Right.

First, I think the formula for r is straightforward. It should be the distance from the origin to the point, which uses the Pythagorean theorem. So, r = sqrt(x² + y²).

<prompt_1.1> Calculate the radius \( r \) for the polar coordinates using the formula \( r = \sqrt{x^2 + y^2} \). </prompt_1.1>

Now, the angle θ. Since the point is (0,3), it's located on the y-axis. In rectangular coordinates, x=0 and y=3, so that's straight up along the y-axis. 

I remember that θ is measured from the positive x-axis, going counterclockwise. So, if the point is on the positive y-axis, the angle should be 90 degrees, right?
But in radians, 90 degrees is π/2. So, θ = π/2.

Wait, let me make sure. Let me think again.

<prompt_1.2> Verify the angle \( \theta \) using the relationships \( x = r\cos\theta \) and \( y = r\sin\theta \). </prompt_1.2>
-----

Wait, but let me confirm with the conversion formulas.
The general formulas for converting from rectangular (x, y) to polar (r, θ) are:

r = sqrt(x² + y²)

θ = arctan(y/x)

But in this case, x is 0, so arctan(y/x) is arctan(3/0). Wait, division by zero.

<prompt_1.3> Consider alternative methods and verify the solution using the conversion formulas for polar coordinates. </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
Let me check that. For this point (0,3), x is 0 and y is 3. So plugging into the formula, r = sqrt(0² + 3²) = sqrt(0 + 9) = sqrt(9) = 3. That seems right. So the radius is 3.
</execution_1.1>

<execution_1.2>
If the point was (0,1), then θ would be π/2. Since (0,3) is three units up, but the angle should still be the same regardless of the distance. So, even though r is 3 in this case, the angle is still π/2.
</execution_1.2>

<execution_1.3>
Oh, that's undefined. Hmm. So, how do we handle that?

I recall that when converting, if x is 0, then the point is on the y-axis. If x is 0 and y is positive, then θ is π/2. If x is 0 and y is negative, θ is 3π/2. Similarly, if y is 0 and x is positive, θ is 0, and if y is 0 and x is negative, θ is π. 

So, in this case, since x=0 and y=3 (positive), θ should be π/2.
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.1, execution_1.2:
But let me verify using the conversion formulas properly.

<prompt_2.1> Verify the polar coordinates using the conversion formulas for \( r \) and \( \theta \) based on the above plan. </prompt_2.1>
-----

Based on execution_1.3:
Therefore, the polar coordinates should be (3, π/2). Let me just visualize it.
If you plot (0,3) in rectangular coordinates, it's straight up on the y-axis. In polar coordinates, that's 3 units away from the origin at an angle of π/2 radians from the positive x-axis.
-----

Based on execution_1.2:
But let me check another way. Let's use the conversion formulas again.
x = r cosθ and y = r sinθ. If θ is π/2, then cosθ is 0 and sinθ is 1. So, x = 3 * 0 = 0 and y = 3 * 1 = 3.

Based on execution_1.1, execution_1.2:
Therefore, the polar coordinates are (3, π/2).
-----

Based on execution_1.3:
Alternatively, using the arctangent function. Normally, θ = arctan(y/x). But when x is 0, arctan(y/x) is undefined. However, in limits, as x approaches 0 from the positive side, arctan(y/x) approaches π/2.

<prompt_2.2> Re-evaluate the angle \( \theta \) using the arctangent function and previous results. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
For θ, since x is 0, we can't use the regular arctan(y/x) because that would be division by zero. But as I thought before, since the point is on the y-axis, the angle should be π/2. But maybe I should use the arctan function with consideration to the quadrant. 

Wait, there's a formula where θ = arctan(y/x) + adjustment for quadrants. But in this case, x is 0, so it's on the y-axis.
</execution_2.1>

<execution_2.2>
But since x is 0 here, we have to consider the quadrant. Since x is 0 and y is positive, it's on the positive y-axis, so θ is π/2. If it were negative y-axis, it would be 3π/2.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Since x = 0 and y is positive, θ is π/2. So, θ = π/2.
-----

Based on execution_1.1, execution_1.2, execution_2.2:
So, putting it all together, r is 3 and θ is π/2.
So the polar coordinates are (3, π/2).
</Plan_3>

<Execution_3>
No prompt to execute.
</Execution_3>

<Final_answer>
Therefore, the point $(0,3)$ in rectangular coordinates is $\boxed{(3, \frac{\pi}{2})}$ in polar coordinates.
</Final_answer>
</think>