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: $\boxed{-2}$