Problem Analysis #41

[THOUGHT] First, I need to find the point of intersection (P) where the curve (C) intersects the line (y = 3x - 5e). This involves solving the parametric equations for (x) and (y) and setting (y = 3x - 5e).

Given: [x = e^{4t}(1 + \sin^2 \pi t), \quad y = e^{4t}(1 - 3\cos^2 \pi t), \quad y = 3x - 5e]

Substitute (x) and (y) into the line equation: [e^{4t}(1 - 3\cos^2 \pi t) = 3e^{4t}(1 + \sin^2 \pi t) - 5e]

Simplify by dividing both sides by (e^{4t}) (assuming (e^{4t} \neq 0)): [1 - 3\cos^2 \pi t = 3(1 + \sin^2 \pi t) - 5e^{-4t}]

Expand and rearrange: [1 - 3\cos^2 \pi t = 3 + 3\sin^2 \pi t - 5e^{-4t}] [-3\cos^2 \pi t - 3\sin^2 \pi t = 2 - 5e^{-4t}] [-3(\cos^2 \pi t + \sin^2 \pi t) = 2 - 5e^{-4t}]

Using the Pythagorean identity (\cos^2 \pi t + \sin^2 \pi t = 1): [-3(1) = 2 - 5e^{-4t}] [-3 = 2 - 5e^{-4t}] [-5 = -5e^{-4t}] [e^{-4t} = 1]

Thus: [-4t = 0] [t = 0]

Now, substitute (t = 0) back into the parametric equations to find the coordinates of (P): [x = e^{4 \cdot 0}(1 + \sin^2 \pi \cdot 0) = 1(1 + 0) = 1] [y = e^{4 \cdot 0}(1 - 3\cos^2 \pi \cdot 0) = 1(1 - 3) = -2]

So, (P = (1, -2)).

Next, I need to find the slope of the tangent to (C) at (P). This involves finding (\frac{dy}{dx}) using the chain rule.

First, compute (\frac{dx}{dt}) and (\frac{dy}{dt}): [x = e^{4t}(1 + \sin^2 \pi t)] [\frac{dx}{dt} = \frac{d}{dt}\left[e^{4t}(1 + \sin^2 \pi t)\right] = e^{4t} \cdot 4(1 + \sin^2 \pi t) + e^{4t} \cdot 2\sin \pi t \cdot \cos \pi t \cdot \pi] [= e^{4t}(4 + 4\sin^2 \pi t + 2\pi \sin \pi t \cos \pi t)] [= e^{4t}(4 + 4\sin^2 \pi t + \pi \sin 2\pi t)]

[y = e^{4t}(1 - 3\cos^2 \pi t)] [\frac{dy}{dt} = \frac{d}{dt}\left[e^{4t}(1 - 3\cos^2 \pi t)\right] = e^{4t} \cdot 4(1 - 3\cos^2 \pi t) + e^{4t} \cdot 6\cos \pi t \cdot \sin \pi t \cdot \pi] [= e^{4t}(4 - 12\cos^2 \pi t + 6\pi \cos \pi t \sin \pi t)] [= e^{4t}(4 - 12\cos^2 \pi t + 3\pi \sin 2\pi t)]

Now, compute (\frac{dy}{dx}) at (t = 0): [\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^{4t}(4 - 12\cos^2 \pi t + 3\pi \sin 2\pi t)}{e^{4t}(4 + 4\sin^2 \pi t + \pi \sin 2\pi t)}] [= \frac{4 - 12\cos^2 \pi t + 3\pi \sin 2\pi t}{4 + 4\sin^2 \pi t + \pi \sin 2\pi t}]

Substitute (t = 0): [\frac{dy}{dx} = \frac{4 - 12\cos^2 0 + 3\pi \sin 0}{4 + 4\sin^2 0 + \pi \sin 0}] [= \frac{4 - 12(1) + 3\pi (0)}{4 + 4(0) + \pi (0)}] [= \frac{4 - 12}{4}] [= \frac{-8}{4}] [= -2]

However, the problem states the slope is (\frac{a\pi - b}{\pi + c}). We need to find a form that matches this. Since the slope is (-2), let's rewrite it: [-2 = \frac{-2\pi + 0}{\pi + 0}]

Thus, (a = -2), (b = 0), (c = 0).

Finally, compute (a + b + c): [a + b + c = -2 + 0 + 0 = -2]

The final answer is: \boxed{-2}