Question

Find the points of intersection of the graphs of the equations. r = 1 + cos θ r = 1 − sin θ r ≥ 0, 0 ≤ θ < 2π

Answer 1

This question is based on the point of intersection.Therefore, the points of intersection of the graphs of the equations at 0 ≤ θ < 2π are:

$(1-\dfrac{\sqrt{2} }{2} ,\dfrac{3\pi }{4} ) and (1+\dfrac{\sqrt{2} }{2} ,\dfrac{7\pi }{4} )$

Given:

Equations: r = 1 + cos θ                                                                      ...(1)

                  r = 1 − sin  θ                                                                      ...(2)                                                                

Where, r ≥ 0, 0 ≤ θ < 2π

We need to determined the point of intersection of the graphs of the equations.

To obtain the points of intersection, Equate the two equations above as follows;

r = 1 + cos θ = 1 - sin θ

=> 1 + cos θ = 1 - sin θ

Solve further for θ. We get,

1 + cos θ = 1 - sinθ

cos θ  = - sinθ

Now dividing both sides by - cos θ and solve it further,

$\dfrac{cos\;\theta}{-cos\;\theta} =\dfrac{-sin\;\theta}{-cos\;\theta}\\\\tan\;\theta=-1\\\\\theta=tan^{-1}(1)\\\\\theta=45^{0}=\dfrac{-\pi }{4}$

To get the 2nd quadrant value of θ, add π ( = 180°) to the value of θ. i.e

$\dfrac{-\pi }{4} +\pi =\dfrac{3\pi }{4}$

Similarly, to get the fourth quadrant value of θ, add 2π ( = 360° ) to the value of θ. i.e

$\dfrac{-\pi }{4} +2\pi =\dfrac{7\pi }{4}$

Therefore, the values of θ are 3π / 4 and 7π / 4.

Now substitute these values into equations (i) and (ii) as follows;

$When \;\theta=\dfrac{3\pi }{4},$

$r = 1 + cos\;\theta = 1 + cos \dfrac{3\pi }{4} =1+\dfrac{-\sqrt{2} }{2} =1-\dfrac{\sqrt{2} }{2}$

$r = 1 + sin\;\theta = 1 - sin \dfrac{3\pi }{4} =1-\dfrac{\sqrt{2} }{2} =1-\dfrac{\sqrt{2} }{2}$

$When\; \theta=\dfrac{7\pi }{4}$

$r = 1 + cos\;\theta = 1 + cos \dfrac{7\pi }{4} =1+\dfrac{\sqrt{2} }{2} =1+\dfrac{\sqrt{2} }{2}$

$r = 1 + sin\;\theta = 1 - sin \dfrac{7\pi }{4} =1-\dfrac{-\sqrt{2} }{2} =1+\dfrac{\sqrt{2} }{2}$

Represent the results  above in polar coordinates of the form (r, θ). i.e

$(1-\dfrac{\sqrt{2} }{2} ,\dfrac{3\pi }{4} ) and (1+\dfrac{\sqrt{2} }{2} ,\dfrac{7\pi }{4} )$

Therefore, at the pole where r = 0, is also one of the points of intersection.  

Therefore, the points of intersection of the graphs of the equations at 0 ≤ θ < 2π are:

$(1-\dfrac{\sqrt{2} }{2} ,\dfrac{3\pi }{4} ) and (1+\dfrac{\sqrt{2} }{2} ,\dfrac{7\pi }{4} )$

For further details, please prefer this link:

https://brainly.com/question/13373561