Answer:
i. pole
ii. (1 - , 3π / 4) and;
iii. (1 + , 7π / 4)
Step-by-step explanation:
Given;
r = 1 + cosθ --------------(i)
r = 1 - sinθ --------------(ii)
To find the points of intersection,
(i) Equate the two equations above as follows;
r = 1 + cos θ = 1 - sin θ
=> 1 + cos θ = 1 - sin θ
(ii) Solve for θ in the equation from (i) above;
1 + cos θ = 1 - sinθ
cos θ = - sinθ [<em>divide both sides by -cosθ</em>]
cosθ / -cosθ = -sinθ / -cosθ
-1 = tanθ [<em>remember that sinθ/cosθ = tanθ</em>]
tanθ = -1 [<em>since tanθ is negative, θ is in the 2nd and 4th quadrant</em>]
θ = tan⁻¹(-1)
θ = -45⁰ = - π / 4
To get the 2nd quadrant value of θ, add π ( = 180°) to the value of θ. i.e
- (π / 4) + π = 3π / 4
Similarly, to get the fourth quadrant value of θ, add 2π ( = 360° ) to the value of θ. i.e
- (π / 4) + 2π = 7π / 4
Therefore, the values of θ are 3π / 4 and 7π / 4
(iii) Substitute these values into equations (i) and (ii) as follows;
When θ = 3π / 4
r = 1 + cos θ = 1 + cos (3π / 4) = 1 + ( ) = 1 -
When θ = 3π / 4
r = 1 - sin θ = 1 - sin (3π / 4) = 1 - ( ) = 1 -
When θ = 7π / 4
r = 1 + cos θ = 1 + cos (7π / 4) = 1 + ( ) = 1 +
When θ = 7π / 4
r = 1 - sin θ = 1 - sin (7π / 4) = 1 - ( ) = 1 +
(iv) Represent the results in (iii) above in polar coordinates of the form (r, θ). i.e
(1 - , 3π / 4) and (1 + , 7π / 4)
(v) Also, at the pole where r = 0, is also one of the points of intersection. This can be shown from a sketch of the graphs of the equations.
Therefore, the points of intersection of the graphs of the equations at 0 ≤ θ < 2π are:
i. pole
ii. (1 - , 3π / 4) and;
iii. (1 + , 7π / 4)