A rational function $f(x) = \frac{P(x)}{Q(x)}$ has a vertical asymptote at values of $x$ where the denominator is zero and the numerator is non-zero.
Here, setting the denominator to zero:
$7 - x = 0 \implies x = 7$.
At $x = 7$, the numerator is $3(7) + 5 = 26 \ne 0$.

First differentiate $x$ and $y$ with respect to $t$:
$\frac{dx}{dt} = a(-\sin t + \sin t + t\cos t) = at\cos t$.
$\frac{dy}{dt} = a(\cos t - (\cos t - t\sin t)) = at\sin t$.
Now, the first derivative is:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{at\sin t}{at\cos t} = \tan t$.
Now differentiate with respect to $x$:
$\frac{d^2y}{dx^2} = \frac{d}{dt}(\tan t) \cdot \frac{dt}{dx} = \sec^2 t \cdot \frac{1}{at\cos t} = \frac{1}{at\cos^3 t}$.

Detailed analysis of options:

• $\frac{at}{\cos^3 t}$ inverts the position of $at$.
• $\frac{1}{at\cos^3 t}$ is the exact and correct second derivative.
• $\frac{at}{\sin^3 t}$ has incorrect trigonometric function and inverse scale.
• $\frac{1}{at\sin^3 t}$ has sine instead of cosine.

Using the chain rule for partial differentiation:
$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}$.
Since $x = e^u + e^{-v}$, $\frac{\partial x}{\partial u} = e^u$.
Since $y = e^{-u} - e^v$, $\frac{\partial y}{\partial u} = -e^{-u}$.