Answer to Question #93070 in Calculus for Manju H B

Question

Find the angle of intersection of r=a(1-cos(theta)) and r=2acos(theta)

Answer 1

1)Let's find $\theta_0$ an angle of intersection between the curves:

$r_1(\theta)-r_2(\theta) = 0 = a(1-cos(\theta))-2a(cos(\theta))=a(1-3cos(\theta)) \implies \theta_0 = arccos(\cfrac{1}{3})$

2)Now let's calculate $m_1$ and $m_2$ :

$m_1=\cfrac{\cfrac{dr_1(\theta_0)}{d\theta}sin(\theta)+r_1(\theta_0)cos(\theta_0)}{\cfrac{dr_1(\theta_0)}{d\theta_0}cos(\theta_0)-r_1(\theta_0)sin(\theta_0)}=\cfrac{sin^2(\theta_0)+r_1(\theta_0)cos(\theta_0)}{sin(\theta_0)cos(\theta_0)-r_1(\theta_0)sin(\theta_0)}=\cfrac{a(\cfrac{2\sqrt2}{3})^2+\cfrac{2}{9}a}{\cfrac{2\sqrt2}{9}a-\cfrac{4\sqrt2}{9}a}=\cfrac{-5\sqrt2}{2}$

$m_2=\cfrac{\cfrac{dr_2(\theta_0)}{d\theta}sin(\theta)+r_2(\theta_0)cos(\theta_0)}{\cfrac{dr_2(\theta_0)}{d\theta_0}cos(\theta_0)-r_2(\theta_0)sin(\theta_0)}=\cfrac{-2sin^2(\theta_0)+r_2(\theta_0)cos(\theta_0)}{-2sin(\theta_0)cos(\theta_0)-r_2(\theta_0)sin(\theta_0)}=\cfrac{-2a(\cfrac{2\sqrt2}{3})^2+\cfrac{2}{9}a}{\cfrac{-4\sqrt2}{9}a-\cfrac{4\sqrt2}{9}a}=\cfrac{-7\sqrt2}{8}$

Then angle $\alpha$ of intersection between the two curves is:

$\alpha=arctan|\cfrac{m_1-m_2}{1+m_1m_2}|=arctan|\cfrac{-\cfrac{13\sqrt2}{8}}{\cfrac{86}{16}}| =arctan|\cfrac{-13\sqrt2}{43}|$

Answer: $\alpha=arctan|\cfrac{-13\sqrt2}{43}|=arctan(\cfrac{13\sqrt2}{43}).$