Problem Analysis #0

[THOUGHT] First, compute the area a = (1/2) ∫ from 0 to 2π of r² dθ, where r = 2 + cos(θ). Use sympy for exact value. [/THOUGHT] [PYTHON] from sympy import *

theta = symbols('theta') r = 2 + cos(theta) area_expr = (1/2) * integrate(r**2, (theta, 0, 2*pi)) print(area_expr) [/PYTHON]

[THOUGHT] The area $a = \frac{9\pi}{2}$. The slope is $-2 + \sqrt{2}$, so $b=-2$, $c=2$. Then $\frac{abc}{\pi} = \frac{\frac{9\pi}{2} \cdot (-2) \cdot 2}{\pi} = -18$. [/THOUGHT] $\boxed{-18}$

Explanation

The reasoning follows the problem's requirements step-by-step. First, the area $a$ is computed using the polar area formula $A = \frac{1}{2} \int_0^{2\pi} r^2 \, d\theta$, where $r = 2 + \cos \theta$. Substituting gives $r^2 = (2 + \cos \theta)^2 = 4 + 4 \cos \theta + \cos^2 \theta$. Integrating term-by-term: $\int_0^{2\pi} 4 \, d\theta = 8\pi$, $\int_0^{2\pi} 4 \cos \theta \, d\theta = 0$, and $\int_0^{2\pi} \cos^2 \theta \, d\theta = \int_0^{2\pi} \frac{1 + \cos 2\theta}{2} \, d\theta = \pi$. Thus, $A = \frac{1}{2} (8\pi + \pi) = \frac{9\pi}{2}$, matching the Python output of $4.5\pi$.

For part (2), verify the point $(r, \theta) = \left( \frac{4 + \sqrt{2}}{2}, \frac{\pi}{4} \right)$ lies on the curve: $r = 2 + \cos(\pi/4) = 2 + \frac{\sqrt{2}}{2} = \frac{4 + \sqrt{2}}{2}$, which holds. The slope $\frac{dy}{dx}$ in polar coordinates is $\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$, where $x = r \cos \theta$, $y = r \sin \theta$, $\frac{dr}{d\theta} = -\sin \theta$.

Thus, $\frac{dx}{d\theta} = \frac{dr}{d\theta} \cos \theta - r \sin \theta = (-\sin \theta) \cos \theta - r \sin \theta$, and $\frac{dy}{d\theta} = (-\sin \theta) \sin \theta + r \cos \theta$. At $\theta = \pi/4$, $\sin \theta = \cos \theta = \frac{\sqrt{2}}{2}$, so $\frac{dr}{d\theta} = -\frac{\sqrt{2}}{2}$.

Substitute: $\frac{dy}{d\theta} = \left(-\frac{\sqrt{2}}{2}\right) \frac{\sqrt{2}}{2} + r \frac{\sqrt{2}}{2} = -\frac{1}{2} + r \frac{\sqrt{2}}{2}$, $\frac{dx}{d\theta} = \left(-\frac{\sqrt{2}}{2}\right) \frac{\sqrt{2}}{2} - r \frac{\sqrt{2}}{2} = -\frac{1}{2} - r \frac{\sqrt{2}}{2}$.

The slope is $\frac{-\frac{1}{2} + r \frac{\sqrt{2}}{2}}{-\frac{1}{2} - r \frac{\sqrt{2}}{2}} = \frac{-1 + r \sqrt{2}}{-1 - r \sqrt{2}} = \frac{1 - r \sqrt{2}}{1 + r \sqrt{2}}$. Now, $r \sqrt{2} = \frac{4 + \sqrt{2}}{2} \cdot \sqrt{2} = \frac{4\sqrt{2} + 2}{2} = 2\sqrt{2} + 1$.

So, numerator: $1 - (2\sqrt{2} + 1) = -2\sqrt{2}$; denominator: $1 + (2\sqrt{2} + 1) = 2 + 2\sqrt{2}$. Thus, slope = $\frac{-2\sqrt{2}}{2(1 + \sqrt{2})} = \frac{-\sqrt{2}}{1 + \sqrt{2}}$.

Rationalize: $\frac{-\sqrt{2}}{1 + \sqrt{2}} \cdot \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{-\sqrt{2}(1 - \sqrt{2})}{1 - 2} = \frac{-\sqrt{2} + 2}{-1} = -(2 - \sqrt{2}) = \sqrt{2} - 2 = -2 + \sqrt{2}$. Hence, $b = -2$, $c = 2$.

Finally, compute $\frac{abc}{\pi} = \frac{\left(\frac{9\pi}{2}\right) \cdot (-2) \cdot 2}{\pi} = \frac{\frac{9\pi}{2} \cdot (-4)}{\pi} = \frac{-18\pi}{\pi} = -18$.