240 Chapter 10 Polar Coordinates, Parametric Equations

EXAMPLE 10.1.6 Graph r = 2 sin θ. Because the sine is periodic, we know that we
will get the entire curve for values of θ in [0, 2π). As θ runs from 0 to π/2, r increases
from 0 to 2. Then as θ continues to π, r decreases again to 0. When θ runs from π to
2π, r is negative, and it is not hard to see that the first part of the curve is simply traced
out again, so in fact we get the whole curve for values of θ in [0, π). Thus, the curve looks
something like figure 10.1.5. Now, this suggests that the curve could possibly be a circle,
and if it is, it would have to be the circle x2 + (y − 1)2 = 1. Having made this guess, we
can easily check it. First we substitute for x and y to get (r cos θ)2 + (r sin θ − 1)2 = 1;
expanding and simplifying does indeed turn this into r = 2 sin θ.

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−1 0 1

Figure 10.1.5 Graph of r = 2 sin θ.

Exercises 10.1.
1. Plot these polar coordinate points on one graph: (2, π/3), (−3, π/2), (−2, −π/4), (1/2, π),
(1, 4π/3), (0, 3π/2).
Find an equation in polar coordinates that has the same graph as the given equation in
rectangular coordinates.
2. y = 3x ⇒ 3. y = −4 ⇒
2
4. xy = 1 ⇒ 5. x2 + y2 = 5 ⇒
6. y = x3 ⇒ 7. y = sin x ⇒
8. y = 5x + 2 ⇒ 9. x = 2 ⇒
2
10. y = x + 1 ⇒ 11. y = 3x2 − 2x ⇒
12. y = x2 + y2 ⇒

Sketch the curve.

13. r = cos θ 14. r = sin(θ + π/4)
15. r = − sec θ 16. r = θ/2, θ ≥ 0
2
17. r = 1 + θ/π 18. r = cot θ csc θ
1
19. r = 20. r 2 = −2 sec θ csc θ
sin θ + cos θ