Let be the curve given parametrically byLet be the point where intersects the line . Determine the slope of the tangent to at . [3 points]If the slope is , compute .\
[THOUGHT] The intersection occurs where e^{4t} = e, so t = 1/4. Verify this and compute the slope dy/dx = (dy/dt)/(dx/dt) at t=1/4 using sympy for exact computation. [/THOUGHT] [PYTHON] import sympy as sp
t = sp.symbols('t') x = sp.exp(4t) * (1 + sp.sin(sp.pi * t)**2) y = sp.exp(4t) * (1 - 3 * sp.cos(sp.pi * t)**2)
dxdt = sp.diff(x, t) dydt = sp.diff(y, t)
slope = dydt / dxdt slope_at_t = slope.subs(t, sp.Rational(1,4))
print(slope_at_t.simplify()) print(sp.N(slope_at_t)) # numerical check [/PYTHON]
[THOUGHT] The slope is \frac{3\pi - 2}{\pi + 6}, so a=3, b=2, c=6, and a+b+c=11. [/THOUGHT] \boxed{11}
Explanation
The reasoning begins by identifying the intersection point P where the curve C intersects the line y = 3x - 5e. Substituting the parametric equations into the line equation yields e^{4t}(1 + sin²(πt)) for x and e^{4t}(1 - 3 cos²(πt)) for y, leading to e^{4t}(1 - 3 cos²(πt)) = 3 e^{4t}(1 + sin²(πt)) - 5e. Simplifying shows e^{4t} = e at t = 1/4, as the exponential factor aligns with the constant term.
Next, the slope of the tangent is computed as dy/dx = (dy/dt) / (dx/dt) evaluated at t = 1/4. The parametric derivatives are calculated using SymPy: dx/dt from differentiating x = e^{4t}(1 + sin²(πt)) and dy/dt from y = e^{4t}(1 - 3 cos²(πt)). Substituting t = 1/4 into the ratio dydt / dxdt and simplifying produces the exact expression (-2 + 3π)/(π + 6), equivalent to (3π - 2)/(π + 6).
This matches the form \frac{aπ - b}{π + c}, where a = 3, b = 2, and c = 6. Therefore, a + b + c = 3 + 2 + 6 = 11, confirming the result from the Python output.