Consider the curve given in polar coordinates by
with 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 $C$ in the $xy$–coordinate plane.
Let be the value obtained in part (1). Express the answer from part (2) in the form
where and are integers, and let those integers be the definitions of and . Using these values of , , and , compute
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[THOUGHT] To solve part (1), we use the polar area formula: (1/2)∫[r² dθ] from 0 to 2π. Expanding (2 + cosθ)² and integrating term by term gives the area as 9π/2. For part (2), we compute dy/dx using the derivatives of x and y in polar coordinates. After simplifying, the slope at the given point is found to be -2 + √2, leading to b = -2 and c = 2. Finally, we compute (abc)/π using the values a = 9π/2, b = -2, and c = 2.
[PYTHON]
Part (1): Compute the area
from sympy import symbols, integrate, pi, cos
theta = symbols('theta') r = 2 + cos(theta) area_integral = integrate((r**2)/2, (theta, 0, 2*pi)) a = area_integral.evalf()
Part (2): Compute the slope of the tangent line
from sympy import sqrt, sin, cos
theta_val = pi/4 r_val = (4 + sqrt(2))/2 dr_dtheta = -sin(theta_val) numerator = dr_dtheta * sin(theta_val) + r_val * cos(theta_val) denominator = dr_dtheta * cos(theta_val) - r_val * sin(theta_val) slope = numerator / denominator slope_simplified = slope.simplify()
Extract b and c from the slope expression
b = -2 c = 2
Compute abc/pi
result = (a * b * c) / pi print(f"a = {a}, slope = {slope_simplified}, b = {b}, c = {c}, result = {result}")
Since the symbolic computation might not simplify directly, we verify numerically
a_num = 9 * pi / 2 b_num = -2 c_num = 2 result_num = (a_num * b_num * c_num) / pi print(f"Numerical result: {result_num}") [/PYTHON]
[THOUGHT] The Python output confirms that the numerical value of is . This is derived from , , and , and the calculation . The symbolic simplification of the slope as also aligns with the given form where and . Thus, the final result is .