Baby Peter is pulling a toy wagon on a flat ground which is modelled by the $xy$-coordinate plane. Peter is initially at the origin $O$ and the toy wagon is located on the $y$-axis at $W_0(0,6)$. The toy wagon is tethered to him with a rope of length 6 units. As he moves in the direction of the positive $x$-axis (given by point $P$), he pulls the toy wagon (given by the point $W_1$) using the rope $W_1P$ which is kept taut throughout the motion. The path of the toy wagon is given by curve $C$. It is known that the rope is always tangent to $C$.

1. Find the horizontal distance between $W_1$ and $P$ using Pythagoras' theorem.

Let the coordinates of $P$ be $(x_P, 0)$. In the right-angled triangle formed by $W_1(x, y)$, its projection $(x, 0)$ and $P(x_P, 0)$, the hypotenuse is the rope length $6$.
$(x_P - x)^2 + y^2 = 6^2$
Since $P$ is moving in the direction of the positive $x$-axis, $P$ is to the right of $W_1$, so $x_P > x$, giving:
$x_P - x = \sqrt{36 - y^2}$

2. Formulate the gradient of the curve using the slope of the tangent segment $W_1P$.

The rope $W_1P$ is tangent to the curve $C$ at $W_1$. The gradient of the curve at $W_1$ is thus the gradient of the line segment $W_1P$.
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{0 - y}{x_P - x}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-y}{\sqrt{36 - y^2}}$
Comparing this with $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\alpha y}{\sqrt{36-y^2}}$, we get $\alpha = -1$.

1. Establish the specific differential equation from part (a).

The problem states the rope of length $6$ is tangent to the curve $C$.
Let the horizontal position of $P$ be $X$. The coordinates of $W_1$ and $P$ are $(x, y)$ and $(X, 0)$ respectively.
The length of the rope is given by $\sqrt{(X-x)^2 + y^2} = 6 \implies X-x = \sqrt{36-y^2}$ (since $P$ is ahead of $W_1$, so $X > x$).
The gradient of the tangent is $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{0 - y}{X - x} = -\frac{y}{\sqrt{36-y^2}}$.
(This determines that $\alpha = -1$ for the differential equation in part (a)).

2. Differentiate the substitution equation implicitly with respect to $x$.

Given the substitution $w^2 = 36 - y^2$, differentiate both sides implicitly with respect to $x$:
$2w \frac{\mathrm{d}w}{\mathrm{d}x} = -2y \frac{\mathrm{d}y}{\mathrm{d}x}$
$\frac{\mathrm{d}w}{\mathrm{d}x} = -\frac{y}{w} \frac{\mathrm{d}y}{\mathrm{d}x}$

3. Substitute the expression for $\frac{\mathrm{d}y}{\mathrm{d}x}$ to reduce the differential equation entirely in terms of $w$.

Since $w > 0$, we have $w = \sqrt{36-y^2}$.
Substitute $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{y}{\sqrt{36-y^2}} = -\frac{y}{w}$ into the derivative relation:
$\frac{\mathrm{d}w}{\mathrm{d}x} = -\frac{y}{w} \left(-\frac{y}{w}\right) = \frac{y^2}{w^2}$
From the substitution $w^2 = 36 - y^2$, we can rewrite $y^2 = 36 - w^2$. Replacing $y^2$ in the numerator yields:
$\frac{\mathrm{d}w}{\mathrm{d}x} = \frac{36-w^2}{w^2} \quad \text{(shown)}$

1. Set up the integral by separating variables from the differential equation.

Starting from the result in (b):
$\frac{\mathrm{d}w}{\mathrm{d}x} = \frac{36-w^2}{w^2}$
Separate the variables to integrate with respect to $x$ and $w$:
$\int 1 \, \mathrm{d}x = \int \frac{w^2}{36-w^2} \, \mathrm{d}w$

2. Use algebraic manipulation to rewrite the integrand into an integrable form.

Rewrite the improper algebraic fraction by adding and subtracting 36 in the numerator:
$\frac{w^2}{36-w^2} = \frac{-(36-w^2) + 36}{36-w^2} = -1 + \frac{36}{36-w^2}$
So the integral becomes:
$\int 1 \, \mathrm{d}x = \int \left( -1 + \frac{36}{36-w^2} \right) \, \mathrm{d}w$

3. Integrate using the standard formula for $\frac{1}{a^2-x^2}$.

Using the standard integral result $\int \frac{1}{a^2-w^2} \, \mathrm{d}w = \frac{1}{2a} \ln \left| \frac{a+w}{a-w} \right|$ with $a=6$:
$x = -w + 36 \left( \frac{1}{2(6)} \ln \left| \frac{6+w}{6-w} \right| \right) + C$
$x = -w + 3 \ln \left| \frac{6+w}{6-w} \right| + C$
Since $w = \sqrt{36-y^2} \ge 0$ and from the context $y > 0$, we know $w < 6$. Thus, $\frac{6+w}{6-w} > 0$ and we can drop the absolute value signs:
$x = -w + 3 \ln \left( \frac{6+w}{6-w} \right) + C$

4. Apply the initial condition to find the constant of integration $C$.

Initially, the toy wagon is at $W_0(0,6)$, so $x = 0$ when $y = 6$.
When $y = 6$, $w = \sqrt{36-6^2} = 0$.
Substituting $x = 0$ and $w = 0$ into our general solution:
$0 = -0 + 3 \ln \left( \frac{6+0}{6-0} \right) + C$
$0 = 3 \ln(1) + C \implies C = 0$
Thus, the particular solution is:
$x = -w + 3 \ln \left( \frac{6+w}{6-w} \right)$

5. Substitute $w = \sqrt{36-y^2}$ back to express $x$ purely in terms of $y$.

Replace $w$ with $\sqrt{36-y^2}$:
$x = -\sqrt{36-y^2} + 3 \ln \left( \frac{6+\sqrt{36-y^2}}{6-\sqrt{36-y^2}} \right)$
Optionally, rationalise the argument of the logarithm to present the final answer in a neat, simplified form:
$x = -\sqrt{36-y^2} + 3 \ln \left( \frac{(6+\sqrt{36-y^2})^2}{36 - (36-y^2)} \right)$
$x = -\sqrt{36-y^2} + 3 \ln \left( \frac{(6+\sqrt{36-y^2})^2}{y^2} \right)$
$x = -\sqrt{36-y^2} + 6 \ln \left( \frac{6+\sqrt{36-y^2}}{y} \right)$

1. Determine the expression for $x$ in terms of $y$ (solving part c).

From part (b), we have $\frac{\mathrm{d}x}{\mathrm{d}w} = \frac{w^2}{36-w^2} = -1 + \frac{36}{36-w^2}$.
Integrating with respect to $w$:
$x = \int \left( -1 + \frac{36}{36-w^2} \right) \mathrm{d}w = -w + 3\ln\left(\frac{6+w}{6-w}\right) + C$
Given the initial position $W_0(0,6)$, when $x=0$, $y=6$ and thus $w=\sqrt{36-6^2}=0$.
$0 = 0 + 3\ln(1) + C \implies C=0$
Substituting $w = \sqrt{36-y^2}$ gives:
$x = -\sqrt{36-y^2} + 3\ln\left(\frac{6+\sqrt{36-y^2}}{6-\sqrt{36-y^2}}\right)$

2. Evaluate the limit of $x$ as $y$ approaches 0.

As $y \to 0$, $y^2 \to 0$ and thus $\sqrt{36-y^2} \to 6$.
Consider the term inside the natural logarithm:
$\frac{6+\sqrt{36-y^2}}{6-\sqrt{36-y^2}}$
The numerator approaches $6 + 6 = 12$, and the denominator approaches $6 - 6 = 0$ (from the positive side, since $y>0$).
Hence, $\frac{6+\sqrt{36-y^2}}{6-\sqrt{36-y^2}} \to \infty$.
Since $\ln(v) \to \infty$ as $v \to \infty$, the entire expression for $x$ approaches infinity.
Therefore, $x \to \infty$.