Consider the curve given in polar coordinates bywith the origin as the pole and the positive –axis as the initial line.
- [(1)] Compute the area of the region enclosed by the curve .
- [(2)] At the point find the slope of the tangent line to the curve in the –coordinate plane. Let be the value obtained in part (1). Express the answer from part (2) in the formwhere and are integers, and let those integers be the definitions of and . Using these values of , , and , compute\
[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 . The slope is , so , . Then . [/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).