This is a problem involving parametric equations and finding tangents, derivatives, and implicit derivatives. Let's break down the tasks and solve them step by step.

Part a:

We are given the parametric equations:

$x(t) = \sqrt{t} \ln(t) - 4$ $y(t) = 3 \sqrt{t}, \quad t > 1$

i) Finding the tangent line at the point $(2e - 4, 3e)$

First, we need to calculate the values of $x(t)$ and $y(t)$ at the given $t$-values to find the coordinates where the tangent is.

We'll calculate the derivative $\frac{dy}{dx}$, which is the slope of the tangent at any point. To do this, we use the chain rule:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

• First, differentiate $x(t)$ and $y(t)$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt} (\sqrt{t} \ln(t) - 4)$ $\frac{dy}{dt} = \frac{d}{dt} (3 \sqrt{t})$

Then, calculate the tangent slope at the point $t$ corresponding to $x(t) = 2e - 4$ and $y(t) = 3e$.

ii) Determining $t$ for which the tangent line has a slope of 1

Here, we need to solve for $t$ when the slope $\frac{dy}{dx} = 1$.

I'll start solving for the derivatives and find the values of $t$ for these tasks. Let me compute the necessary derivatives and proceed with solving. sympy import symbols, sqrt, log, diff, Eq, solve

Define variable

t = symbols('t')

Define the parametric equations

x_t = sqrt(t) * log(t) - 4 y_t = 3 * sqrt(t)

Compute the derivatives dx/dt and dy/dt

dx_dt = diff(x_t, t) dy_dt = diff(y_t, t)

Find dy/dx (slope of the tangent)

dy_dx = dy_dt / dx_dt

Calculate the derivative expressions

dx_dt, dy_dt, dy_dx