Problem Analysis #21

The distance between (P_1(1+\cos t, \sin t)) and (P_2(2\cos\theta(t), 2\sin\theta(t))) is maximized when (\cos t = -\frac{1}{8}). At this point, (\sin t = \frac{3\sqrt{7}}{8}) (positive, as (t) is in the second quadrant).

The expression to evaluate is: [ \sin t \cdot 2\sin\theta(t) - (1+\cos t) \cdot 2\cos\theta(t) ] Given (\theta(t) = 2t) (since (\theta'(t) = 2)): [ = 2 \left[ \sin t \sin(2t) - (1+\cos t) \cos(2t) \right] ]

Substitute the values:

• (\sin t = \frac{3\sqrt{7}}{8})
• (\cos t = -\frac{1}{8})
• (\sin(2t) = 2 \sin t \cos t = 2 \cdot \frac{3\sqrt{7}}{8} \cdot \left(-\frac{1}{8}\right) = -\frac{6\sqrt{7}}{64} = -\frac{3\sqrt{7}}{32})
• (\cos(2t) = \cos^2 t - \sin^2 t = \left(-\frac{1}{8}\right)^2 - \left(\frac{3\sqrt{7}}{8}\right)^2 = \frac{1}{64} - \frac{63}{64} = -\frac{62}{64} = -\frac{31}{32})

Compute the terms:

• (\sin t \sin(2t) = \frac{3\sqrt{7}}{8} \cdot \left(-\frac{3\sqrt{7}}{32}\right) = -\frac{9 \cdot 7}{256} = -\frac{63}{256})
• ((1 + \cos t) \cos(2t) = \left(1 - \frac{1}{8}\right) \cdot \left(-\frac{31}{32}\right) = \frac{7}{8} \cdot \left(-\frac{31}{32}\right) = -\frac{217}{256})

Now: [ \sin t \sin(2t) - (1 + \cos t) \cos(2t) = -\frac{63}{256} - \left(-\frac{217}{256}\right) = -\frac{63}{256} + \frac{217}{256} = \frac{154}{256} ]

Multiply by 2: [ 2 \cdot \frac{154}{256} = \frac{308}{256} = \frac{77}{64} ]

The fraction (\frac{77}{64}) is in simplest terms, as 77 and 64 are coprime. Thus, (p = 77), (q = 64), and (p + q = 141).

\boxed{141}

[THOUGHT] We are using Python to calculate the expression at the critical point. [/THOUGHT] [PYTHON] import math

sin_t = 3 * math.sqrt(7) / 8 cos_t = -1/8 sin_2t = 2 * sin_t * cos_t cos_2t = cos_t2 - sin_t2 term = sin_t * sin_2t - (1 + cos_t) * cos_2t expr = 2 * term print(expr) [/PYTHON]

\boxed{141}