The diagram below shows the Gateway Arch, which is a monument in St. Louis, Missouri, United States. The arch stands at 192 metres tall and is 192 metres wide. The arch can be modelled by part of a curve as shown in the diagram below. The highest point of the curve lies on the $y$-axis and the curve is symmetrical about the $y$-axis. The two endpoints both lie on the $x$-axis. It is known that the curve satisfies the differential equation
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = ak \sqrt{1 + \left(\frac{1}{k}\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}$
for some constants $a$ and $k$.

1. Find an expression for the second derivative $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ in terms of $p$.

Given the substitution $p = \frac{1}{k}\frac{\mathrm{d}y}{\mathrm{d}x}$, we differentiate both sides with respect to $x$:
$\frac{\mathrm{d}p}{\mathrm{d}x} = \frac{1}{k}\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$
Multiplying by $k$ gives:
$\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = k\frac{\mathrm{d}p}{\mathrm{d}x}$

2. Substitute the expressions back into the original differential equation to yield the required reduced form.

Substitute $p$ and $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ into the original differential equation $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = ak \sqrt{1 + \left(\frac{1}{k}\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}$:
$k\frac{\mathrm{d}p}{\mathrm{d}x} = ak \sqrt{1 + p^2}$
Dividing both sides by the constant $k$:
$\frac{\mathrm{d}p}{\mathrm{d}x} = a\sqrt{1 + p^2}$

1. Separate the variables of the differential equation and apply the substitution $p = \tan u$.

From part (i), we have the reduced differential equation $\frac{\mathrm{d}p}{\mathrm{d}x} = a\sqrt{1 + p^2}$. Separating the variables:
$\int \frac{1}{\sqrt{1 + p^2}} \, \mathrm{d}p = \int a \, \mathrm{d}x$
Using the given substitution $p = \tan u$, we differentiate to get $\frac{\mathrm{d}p}{\mathrm{d}u} = \sec^2 u \implies \mathrm{d}p = \sec^2 u \, \mathrm{d}u$.
Substitute this into the integral:
$\int \frac{\sec^2 u}{\sqrt{1 + \tan^2 u}} \, \mathrm{d}u = \int a \, \mathrm{d}x$

2. Simplify the integrand using trigonometric identities and integrate both sides.

Using the identity $1 + \tan^2 u = \sec^2 u$, the denominator is $\sqrt{\sec^2 u}$. Since $-\frac{\pi}{2} < u < \frac{\pi}{2}$, $\sec u > 0$, so $\sqrt{\sec^2 u} = \sec u$.
The integral simplifies to:
$\int \frac{\sec^2 u}{\sec u} \, \mathrm{d}u = \int a \, \mathrm{d}x$
$\int \sec u \, \mathrm{d}u = \int a \, \mathrm{d}x$
Using the standard result for integrating $\sec u$:
$\ln (\sec u + \tan u) = ax + C$
(Note: The absolute value is omitted since $\sec u + \tan u > 0$ for the given domain of $u$.)

3. Determine the initial condition from the physical context of the arch.

The problem states that the highest point of the arch lies on the $y$-axis. At this maximum turning point ($x = 0$), the tangent is horizontal, so $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$.
Given the substitution $p = \frac{1}{k}\frac{\mathrm{d}y}{\mathrm{d}x}$, we have:
$\text{At } x = 0, \quad p = \frac{1}{k}(0) = 0$
When $p = 0$, $0 = \tan u \implies u = 0$.

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

Substitute $x = 0$ and $u = 0$ into the integrated equation:
$\ln (\sec 0 + \tan 0) = a(0) + C$
$\ln (1 + 0) = C \implies C = 0$
Thus, the equation simplifies to:
$\ln (\sec u + \tan u) = ax$
$\sec u + \tan u = \mathrm{e}^{ax}$

5. Use trigonometric identities to isolate $\tan u$ and solve for $p$.

Using the Pythagorean identity $\sec^2 u - \tan^2 u = 1$, we can factorise it as:
$(\sec u + \tan u)(\sec u - \tan u) = 1$
Substitute $\sec u + \tan u = \mathrm{e}^{ax}$ into the identity:
$\mathrm{e}^{ax} (\sec u - \tan u) = 1 \implies \sec u - \tan u = \mathrm{e}^{-ax}$
Subtracting this from our earlier equation ($\sec u + \tan u = \mathrm{e}^{ax}$) gives:
$(\sec u + \tan u) - (\sec u - \tan u) = \mathrm{e}^{ax} - \mathrm{e}^{-ax}$
$2\tan u = \mathrm{e}^{ax} - \mathrm{e}^{-ax}$
Since $p = \tan u$:
$2p = \mathrm{e}^{ax} - \mathrm{e}^{-ax}$
$p = \frac{\mathrm{e}^{ax} - \mathrm{e}^{-ax}}{2} \quad \text{(shown)}$

1. Rearrange the expression from (ii) to find $\frac{\mathrm{d}y}{\mathrm{d}x}$ and substitute the given constants $a$ and $k$.

From parts (i) and (ii), we know that $p = \frac{1}{k}\frac{\mathrm{d}y}{\mathrm{d}x}$ and $p = \frac{\mathrm{e}^{ax} - \mathrm{e}^{-ax}}{2}$.
Equating these expressions:
$\frac{1}{k}\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{e}^{ax} - \mathrm{e}^{-ax}}{2}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{k}{2} (\mathrm{e}^{ax} - \mathrm{e}^{-ax})$
Substitute $a = -0.0329$ and $k = 0.701$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{0.701}{2} (\mathrm{e}^{-0.0329x} - \mathrm{e}^{0.0329x}) = 0.3505 (\mathrm{e}^{-0.0329x} - \mathrm{e}^{0.0329x})$

2. Integrate with respect to $x$ to find the general solution for $y$.

$y = \int 0.3505 (\mathrm{e}^{-0.0329x} - \mathrm{e}^{0.0329x}) \, \mathrm{d}x$
$y = 0.3505 \left( \frac{\mathrm{e}^{-0.0329x}}{-0.0329} - \frac{\mathrm{e}^{0.0329x}}{0.0329} \right) + C$
$y = -10.653 (\mathrm{e}^{-0.0329x} + \mathrm{e}^{0.0329x}) + C$

3. Apply the boundary condition at the highest point to find the constant of integration $C$ and write the final equation.

The arch is $192\text{ m}$ tall, and its highest point lies on the $y$-axis. This gives the boundary condition $x = 0$ when $y = 192$.
$192 = -10.653 (\mathrm{e}^{0} + \mathrm{e}^{0}) + C$
$192 = -21.307 + C \implies C = 213.307$
Substituting $C$ back and rounding to 3 significant figures:
$y = -10.7 (\mathrm{e}^{-0.0329x} + \mathrm{e}^{0.0329x}) + 213$