1. Polar
Coordinates

2. Polar curves Starter: KUS objectives
BAT convert between Cartesian and Polar equations of curves
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
Starter:

3. Cartesian to
Polar

4. a) 𝑟=5 b) 𝑟=6𝑐𝑜𝑠𝑒𝑐𝜃 a) 𝑟=5 𝑟 2 =25 𝑥 2 + 𝑦 2 =25 𝑦=6 b) 𝑟=6𝑐𝑜𝑠𝑒𝑐𝜃
WB4 Find a Cartesian equation of the following curves
a) 𝑟= b) 𝑟=6𝑐𝑜𝑠𝑒𝑐𝜃
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
a) 𝑟=5
b) 𝑟=6𝑐𝑜𝑠𝑒𝑐𝜃
𝑟 2 =25
𝑟= 6 𝑠𝑖𝑛𝜃
Cartesian equation
𝑟𝑠𝑖𝑛𝜃=6
𝑥 2 + 𝑦 2 =25
Cartesian equation
𝑦=6

5. Multiply by r2 (this will allow us to replace the trigonometric part)
WB5
Find a Cartesian equation of each of the following curve
𝑟=2+𝑐𝑜𝑠2𝜃
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
double angle formulae from C3 gives
𝑟=2+( 2𝐶𝑜𝑠 2 𝜃−1)
𝑟=1+ 2𝐶𝑜𝑠 2 𝜃
Trig Identity
𝑐𝑜𝑠2𝜃=2 𝑐𝑜𝑠 2 𝜃−1
Multiply by r2 (this will allow us to replace the trigonometric part)
𝑟 3 = 𝑟 𝑟 2 𝐶𝑜𝑠 2 𝜃
𝑥 2 + 𝑦 =
𝑥 2 + 𝑦 2 +
2 𝑥 2
𝑥 2 + 𝑦 = 3𝑥 2 + 𝑦 2
𝒙 𝟐 + 𝒚 𝟐 𝟑 𝟐 = 𝟑𝒙 𝟐 + 𝒚 𝟐
One of the key advantages of Polar equations is that they can explain very complicated Cartesian equations in a much simpler way!

6. WB5
b) Find a Cartesian equation of the following curve
𝑟 2 =𝑆𝑖𝑛2𝜃
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
double angle formulae from C3 gives
𝑟 2 =2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
Multiply by r2
𝑟 4 =2 𝑟 2 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
𝑟 =2𝑟𝑠𝑖𝑛𝜃𝑟𝑐𝑜𝑠𝜃
𝑥 2 + 𝑦 =2𝑥𝑦
𝒙 𝟐 + 𝒚 𝟐 𝟐 =
𝟐𝒙𝒚

7. WB6 Polar Equation to Cartesian Equation I
𝑟 2 = 𝑎 2 sin 2𝜃 0≤𝜃<𝜋
𝑥 2 + 𝑦 2 = 𝑎 2 2 sin 𝜃 cos 𝜃
𝑟 2 =2𝑎 𝑟 cos 𝜃
𝑥 2 + 𝑦 2 =2 𝑎 2 𝑦 𝑟 𝑥 𝑟
𝑥 2 + 𝑦 2 =2𝑎 𝑥
(𝑥−𝑎) 2 + 𝑦 2 = 𝑎 2
𝑟 2 𝑥 2 + 𝑦 2 =2 𝑎 2 𝑥𝑦
Circle centre (a, 0)
𝑥 2 + 𝑦 =2 𝑎 2 𝑥𝑦
𝑟=2+2 𝑐𝑜s 2𝜃 0≤𝜃<𝜋
𝑟=1+2 𝑐𝑜𝑠 2 𝜃
𝑟 3 = 𝑟 2 +2 𝑟 2 𝑐𝑜𝑠 2 𝜃
𝑥 2 + 𝑦 2 3/2 = 𝑥 2 + 𝑦 2 +2 𝑥 2
𝑥 2 + 𝑦 2 3/2 =3 𝑥 2 + 𝑦 2

8. 𝑦=𝑎 WB6 Cartesian Equation to Polar Equation II
𝑟=𝑎 cos 2𝜃 − 𝜋≤𝜃≤ 𝜋
𝑟=2 sin 𝜃 0≤𝜃<𝜋
HINT: multiply both sides by r
𝑥 2 + 𝑦 = 𝑎 𝑥 2 − 𝑦
𝑥 2 + 𝑦 2 =2𝑦
𝑟= 4 3+ 𝑐𝑜s 𝜃 <𝜃<𝜋
𝑟=𝑎 𝑐𝑜sec 𝜃 0<𝜃<𝜋
𝑦=𝑎
8 𝑥 2 +8𝑥+9 𝑦 2 =16

9. Polar to
Cartesian

10. WB7 Find a Polar equivalent for the following Cartesian equation:
a) 𝑦 2 =4𝑥 b) 𝑥 2 − 𝑦 2 =5
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
a) Replace y and x with equivalents
𝑟 2 𝑠𝑖𝑛 2 𝜃=4𝑟𝑐𝑜𝑠𝜃
𝑟 𝑠𝑖𝑛 2 𝜃=4𝑐𝑜𝑠𝜃
𝒚 𝟐 =𝟒𝒙
𝒓=𝟒𝒄𝒐𝒕𝜽𝒄𝒐𝒔𝒆𝒄𝜽
Polar equations are usually written as ‘r = ‘ or ‘r2 = ‘, so use the equations above to try and achieve this
𝑟= 4𝑐𝑜𝑠𝜃 𝑠𝑖𝑛 2 𝜃
𝑟= 4𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃 × 1 𝑠𝑖𝑛𝜃
𝑟=4𝑐𝑜𝑡𝜃𝑐𝑜𝑠𝑒𝑐𝜃
b) Replace y and x with equivalents
𝑟 2 𝑐𝑜𝑠 2 𝜃− 𝑟 2 𝑠𝑖𝑛 2 𝜃=5
𝑟 2 𝑐𝑜𝑠 2 𝜃− 𝑠𝑖𝑛 2 𝜃 =5
𝒓 𝟐 =𝟓𝒔𝒆𝒄𝟐𝜽
𝑟 2 𝑐𝑜𝑠2𝜃=5
𝑟 2 = 5 𝑐𝑜𝑠2𝜃
𝑟 2 =5𝑠𝑒𝑐2𝜃

11. WB8 Find a Polar equivalent for the following Cartesian equation:
𝑦 3 =𝑥+4
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
𝑟 3 𝑠𝑖𝑛𝜃=𝑟𝑐𝑜𝑠𝜃+4
𝑟 3 𝑠𝑖𝑛𝜃−𝑟𝑐𝑜𝑠𝜃=4
Divide both sides by 2 (do this inside the bracket rather than outside – the reason will become apparent in a moment…)
𝑟 3 𝑠𝑖𝑛𝜃−𝑐𝑜𝑠𝜃 =4
𝑟 𝑠𝑖𝑛𝜃− 1 2 𝑐𝑜𝑠𝜃 =2
𝑟𝑠𝑖𝑛 𝜃− 𝜋 6 =2
Replace the bracket with the expression we found
This next step is tricky to spot, but it is possible to write the bracket using sine only
This again relies on the formulae from C3…
𝑆𝑖𝑛 𝐴−𝐵 ≡𝑆𝑖𝑛𝐴𝐶𝑜𝑠𝐵−𝐶𝑜𝑠𝐴𝑆𝑖𝑛𝐵
Let A = θ and B = π/6
𝑆𝑖𝑛 𝜃− 𝜋 6 ≡𝑆𝑖𝑛𝜃𝐶𝑜𝑠 𝜋 6 −𝐶𝑜𝑠𝜃𝑆𝑖𝑛 𝜋 6
Calculate the parts with π/6
𝑆𝑖𝑛 𝜃− 𝜋 6 ≡ 𝑆𝑖𝑛𝜃− 1 2 𝐶𝑜𝑠𝜃
This is equivalent to the part in the brackets, so we can replace it!
𝑟= 2 𝑠𝑖𝑛 𝜃− 𝜋 6
⟹ 𝑟=2𝑐𝑜𝑠𝑒𝑐 𝜃− 𝜋 6

12. WB8 Find a Polar equivalent for the following Cartesian equation:
𝑦 3 =𝑥+4
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
𝑟= 2 𝑠𝑖𝑛 𝜃− 𝜋 6
⇒ 𝑟=2𝑐𝑜𝑠𝑒𝑐 𝜃− 𝜋 6
𝒚 𝟑 =𝒙+𝟒
𝒓=𝟐𝒄𝒐𝒔𝒆𝒄 𝜽− 𝝅 𝟔
⇒ 𝑟 𝑠𝑖𝑛𝜃− 1 2 𝑐𝑜𝑠𝜃 =2
An expression like this can be simplified in the way we just did
Both the numerical parts need to be able to be rewritten as sin and cos of the same angle (in the previous example, π/6 gave us the answers we needed)
You can manipulate the expression to give values that work
If it is possible you need to consider whether it is an expansion of sin or cos, and also whether it is an addition or a subtraction…

13. WB9 Cartesian Equation to Polar Equation I
𝑦= 𝑥 2
𝑥 2 + 𝑦 =4𝑥𝑦
x =rcos θ
y=rsin θ
𝑦 3 =𝑥+4
𝑥 cos ∝ +𝑦 sin ∝ =𝑝 ,
𝑝>0

14. 𝑥 2 + 𝑦 2 =4 WB9 Cartesian Equation to Polar Equation II
𝑦= 1 𝑥 , 𝑥>0
𝑥 2 + 𝑦 2 =4
(𝑥−1) 2 + (𝑦−1) 2 =2
1 𝑥 + 1 𝑦 = 1 𝑎 , 𝑎, 𝑥, 𝑦>0

15. Crucial points (r, θ) origin Initial line
Make sure that you can covert between cartesian and polar coordinates.
cartesian (x, y) polar (r, θ) θ is In the range [-, ] or [0, 2]
2. Remember that cos θ = cos (- θ)
If the equation only involves cos it will be symmetrical in the initial line
3. We can use periodicity of the functions to help us
e.g. tan has a period of . Therefore r = tan 5θ will repeat itself after an angle of

16. Some types Of
Graphs

17. WB10 CIRCLES
and
and
is a circle centre (0,0) radius a
Keeping r fixed, whatever the angle generates a circle
is a half-line through 0
at an angle α with the initial line
Keeping θ fixed, whatever the distance r generates a line with gradient tanα through 0

18. WB10 CIRCLES
and
and
is a circle centre radius
is a circle centre radius

19. WB10 LINES
and
and
is a vertical line through
is a horizontal line through

20. WB11
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
Find the Cartesian equations of a polar functions of the form or
Polar
Cartesian
𝑟= sin 𝜃
𝑟= cos 𝜃
𝑟= sin 2𝜃
𝑟= cos 2𝜃
𝑟= sin 3𝜃
𝑟= 𝑐𝑜s 3𝜃

21. Find the Cartesian equations of a polar functions of the form or
WB9
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
Find the Cartesian equations of a polar functions of the form or
𝑟= sin 𝜃
𝑟= cos 𝜃
𝑟 2 = 𝑟 𝑠𝑖𝑛 𝜃
𝑟 2 = 𝑟 𝑐𝑜𝑠 𝜃
𝑥 2 + 𝑦 2 =𝑦
𝑥 2 + 𝑦 2 =𝑥
𝑥 2 + 𝑦− = 1 4
𝑥− 𝑦 2 = 1 4

22. Find the Cartesian equations of a polar functions of the form
WB9
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
Find the Cartesian equations of a polar functions of the form
𝑟= sin 𝑛𝜃 or 𝑟= cos 𝑛𝜃
𝑟= sin 2𝜃
=2 sin 𝜃 cos 𝜃
𝑟= sin 3𝜃
=sin 2𝜃 cos 𝜃 + sin 𝜃 cos 2θ
𝑟 3 =2𝑟 sin 𝜃 𝑟 cos 𝜃
𝑟=2 sin 𝜃 cos 𝜃 cos 𝜃 + sin 𝜃 𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛 2 𝜃
𝑟 2 3/2 =2𝑥𝑦
𝑟=2 sin 𝜃 𝑐𝑜𝑠 2 𝜃 + sin 𝜃 𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛 3 𝜃
𝑥 2 + 𝑦 2 3/2 =2𝑥𝑦
𝑟=3 sin 𝜃 𝑐𝑜𝑠 2 𝜃 − 𝑠𝑖𝑛 3 𝜃
𝑟 4 =3 𝑟 3 sin 𝜃 𝑐𝑜𝑠 2 𝜃− 𝑟 3 𝑠𝑖𝑛 3 𝜃
𝑟 =3 𝑟 2 𝑐𝑜𝑠 2 𝜃𝑟 sin 𝜃 − 𝑟 3 𝑠𝑖𝑛 3 𝜃
𝑥 2 + 𝑦 =3 𝑥 2 𝑦−3 𝑦 3

23. WB9
𝑥=𝑟𝑐𝑜𝑠𝜃
𝑦=𝑟𝑠𝑖𝑛𝜃
𝜃=𝑎𝑟𝑐𝑡𝑎𝑛 𝑦 𝑥
𝑟 2 = 𝑥 2 + 𝑦 2
Find the Cartesian equations of a polar functions of the form or
Polar
Cartesian
𝑥 2 + 𝑦− = 1 4
𝑟= sin 𝜃
𝑥− 𝑦 2 = 1 4
𝑟= cos 𝜃
𝑥 2 + 𝑦 2 3/2 =2𝑥𝑦
𝑟= sin 2𝜃
𝑟= cos 2𝜃
𝑥 2 + 𝑦 2 3/2 = 𝑥 2 − 𝑦 2
𝑟= sin 3𝜃
𝑥 2 + 𝑦 =3 𝑥 2 𝑦−3 𝑦 3
𝑟= 𝑐𝑜s 3𝜃
𝑥 2 + 𝑦 = 𝑥 3 −3𝑥 𝑦 2

24. Practice
Ex 7B

25. One thing to improve is –
KUS objectives BAT convert between Cartesian and Polar equations of curves
self-assess
One thing learned is –
One thing to improve is –

26. END