1)Let's find θ 0 \theta_0 θ 0
r 1 ( θ ) − r 2 ( θ ) = 0 = a ( 1 − c o s ( θ ) ) − 2 a ( c o s ( θ ) ) = a ( 1 − 3 c o s ( θ ) ) ⟹ θ 0 = a r c c o s ( 1 3 ) r_1(\theta)-r_2(\theta) = 0 = a(1-cos(\theta))-2a(cos(\theta))=a(1-3cos(\theta))
\implies \theta_0 = arccos(\cfrac{1}{3}) r 1 ( θ ) − r 2 ( θ ) = 0 = a ( 1 − cos ( θ )) − 2 a ( cos ( θ )) = a ( 1 − 3 cos ( θ )) ⟹ θ 0 = a rccos ( 3 1 )
2)Now let's calculate m 1 m_1 m 1 m 2 m_2 m 2
m 1 = d r 1 ( θ 0 ) d θ s i n ( θ ) + r 1 ( θ 0 ) c o s ( θ 0 ) d r 1 ( θ 0 ) d θ 0 c o s ( θ 0 ) − r 1 ( θ 0 ) s i n ( θ 0 ) = s i n 2 ( θ 0 ) + r 1 ( θ 0 ) c o s ( θ 0 ) s i n ( θ 0 ) c o s ( θ 0 ) − r 1 ( θ 0 ) s i n ( θ 0 ) = a ( 2 2 3 ) 2 + 2 9 a 2 2 9 a − 4 2 9 a = − 5 2 2 m_1=\cfrac{\cfrac{dr_1(\theta_0)}{d\theta}sin(\theta)+r_1(\theta_0)cos(\theta_0)}{\cfrac{dr_1(\theta_0)}{d\theta_0}cos(\theta_0)-r_1(\theta_0)sin(\theta_0)}=\cfrac{sin^2(\theta_0)+r_1(\theta_0)cos(\theta_0)}{sin(\theta_0)cos(\theta_0)-r_1(\theta_0)sin(\theta_0)}=\cfrac{a(\cfrac{2\sqrt2}{3})^2+\cfrac{2}{9}a}{\cfrac{2\sqrt2}{9}a-\cfrac{4\sqrt2}{9}a}=\cfrac{-5\sqrt2}{2} m 1 = d θ 0 d r 1 ( θ 0 ) cos ( θ 0 ) − r 1 ( θ 0 ) s in ( θ 0 ) d θ d r 1 ( θ 0 ) s in ( θ ) + r 1 ( θ 0 ) cos ( θ 0 ) = s in ( θ 0 ) cos ( θ 0 ) − r 1 ( θ 0 ) s in ( θ 0 ) s i n 2 ( θ 0 ) + r 1 ( θ 0 ) cos ( θ 0 ) = 9 2 2 a − 9 4 2 a a ( 3 2 2 ) 2 + 9 2 a = 2 − 5 2
m 2 = d r 2 ( θ 0 ) d θ s i n ( θ ) + r 2 ( θ 0 ) c o s ( θ 0 ) d r 2 ( θ 0 ) d θ 0 c o s ( θ 0 ) − r 2 ( θ 0 ) s i n ( θ 0 ) = − 2 s i n 2 ( θ 0 ) + r 2 ( θ 0 ) c o s ( θ 0 ) − 2 s i n ( θ 0 ) c o s ( θ 0 ) − r 2 ( θ 0 ) s i n ( θ 0 ) = − 2 a ( 2 2 3 ) 2 + 2 9 a − 4 2 9 a − 4 2 9 a = − 7 2 8 m_2=\cfrac{\cfrac{dr_2(\theta_0)}{d\theta}sin(\theta)+r_2(\theta_0)cos(\theta_0)}{\cfrac{dr_2(\theta_0)}{d\theta_0}cos(\theta_0)-r_2(\theta_0)sin(\theta_0)}=\cfrac{-2sin^2(\theta_0)+r_2(\theta_0)cos(\theta_0)}{-2sin(\theta_0)cos(\theta_0)-r_2(\theta_0)sin(\theta_0)}=\cfrac{-2a(\cfrac{2\sqrt2}{3})^2+\cfrac{2}{9}a}{\cfrac{-4\sqrt2}{9}a-\cfrac{4\sqrt2}{9}a}=\cfrac{-7\sqrt2}{8} m 2 = d θ 0 d r 2 ( θ 0 ) cos ( θ 0 ) − r 2 ( θ 0 ) s in ( θ 0 ) d θ d r 2 ( θ 0 ) s in ( θ ) + r 2 ( θ 0 ) cos ( θ 0 ) = − 2 s in ( θ 0 ) cos ( θ 0 ) − r 2 ( θ 0 ) s in ( θ 0 ) − 2 s i n 2 ( θ 0 ) + r 2 ( θ 0 ) cos ( θ 0 ) = 9 − 4 2 a − 9 4 2 a − 2 a ( 3 2 2 ) 2 + 9 2 a = 8 − 7 2
Then angle α \alpha α
α = a r c t a n ∣ m 1 − m 2 1 + m 1 m 2 ∣ = a r c t a n ∣ − 13 2 8 86 16 ∣ = a r c t a n ∣ − 13 2 43 ∣ \alpha=arctan|\cfrac{m_1-m_2}{1+m_1m_2}|=arctan|\cfrac{-\cfrac{13\sqrt2}{8}}{\cfrac{86}{16}}|
=arctan|\cfrac{-13\sqrt2}{43}| α = a rc t an ∣ 1 + m 1 m 2 m 1 − m 2 ∣ = a rc t an ∣ 16 86 − 8 13 2 ∣ = a rc t an ∣ 43 − 13 2 ∣
Answer: α = a r c t a n ∣ − 13 2 43 ∣ = a r c t a n ( 13 2 43 ) . \alpha=arctan|\cfrac{-13\sqrt2}{43}|=arctan(\cfrac{13\sqrt2}{43}). α = a rc t an ∣ 43 − 13 2 ∣ = a rc t an ( 43 13 2 ) .
#339914 on Dec 2023
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