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\
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[THOUGHT] We need to compute the area and the slope using Python to verify and then compute the final expression. [/THOUGHT] [PYTHON] from math import pi from sympy import symbols, integrate
theta = symbols('theta') r = 2 + math.cos(theta) a = integrate(0.5 * r**2, (theta, 0, 2*pi)) print(f"Area a = {a}")
theta_val = math.pi/4 r_val = 2 + math.cos(theta_val) dr_dtheta = -math.sin(theta_val)
dx_dtheta = dr_dtheta * math.cos(theta_val) - r_val * math.sin(theta_val) dy_dtheta = dr_dtheta * math.sin(theta_val) + r_val * math.cos(theta_val)
slope = dy_dtheta / dx_dtheta print(f"Slope = {slope}")
Express slope as b + sqrt(c)
We know from calculation it is -2 + sqrt(2)
b = -2 c = 2 print(f"b = {b}, c = {c}")
result = (a * b * c) / pi print(f"Result = {result}") [/PYTHON]
[THOUGHT] We are correcting the Python code to use sympy.cos and sympy.pi to avoid the type error. [/THOUGHT] [PYTHON] from sympy import symbols, integrate, cos, pi theta = symbols('theta') r = 2 + cos(theta) a = integrate(0.5 * r**2, (theta, 0, 2*pi)) a_val = a.evalf() theta_val = pi/4 r_val = 2 + cos(theta_val) dr_dtheta = -sin(theta_val) dx_dtheta = dr_dtheta * cos(theta_val) - r_val * sin(theta_val) dy_dtheta = dr_dtheta * sin(theta_val) + r_val * cos(theta_val) slope = dy_dtheta / dx_dtheta b = -2 c = 2 result = (a_val * b * c) / pi print(f"a = {a_val}, slope = {slope}, result = {result}") [/PYTHON]
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