The graph of the polar equation which in the form of
r = a sin nθ
(Or)
r = a cos nθ
will produce rose curves. They are called rose curves because the loops that are formed by resemble petals.
Maximum r value is |a|.
Example 1 :
Draw the graph of r = 4 sin 3θ.
Solution :
a = 4, n = 3 (odd). So the rose curve will have n petals.
That is,
3 petals
Where will be each petal ?
360 / 3 = 120
Example 2 :
Analyzing the polar relation r = 5 sin 3θ
(a) Type of curve
(b) Algebraically find the values of θ where the tips of the petals occur for 0 ≤ θ ≤ 2π
(c) Graph r = 5 sin 3θ over 0 ≤ θ ≤ 2π, using petals tips as guide. Identify the starting point use arrows to indicate the directions of motion and find the number of petals accordingly.
(d) Find the zeroes of r for 0 ≤ θ ≤ 2π.
Solution :
(a) Since the value of n is more than 1, it will give rose curves.
(b)
Like this by applying some more values, we will get another rose petal and that will lie on y axis.
When θ = ∏/12 R = 5 sin(3∏/12) = 5 sin(∏/4) = 5/√2 è 3.53 |
When θ = 2∏/12 R = 5 sin(3∏/6) = 5 sin(∏/2) = 5(1) è 5 |
When θ = 3∏/12 R = 5 sin (3∏/4) = 5 cos (∏/4) = 5/√2 è 3.53 |
When θ = 4∏/12 R = 5 sin(3∏/3) = 5 sin∏ = 5(0) è 0 |
When θ = 5∏/12 R = 5 sin(5∏/4) = 5 (-sin∏/4) = -5/√2 è -3.53 |
When θ = 6∏/12 R = 5 sin(3∏/2) = 5 (-1) è -5 |
When θ = 7∏/12 R = 5 sin(7∏/4) = 5 (-cos(∏/4)) = - 5/√2 è -3.53 |
When θ = 8∏/12 R = 5 sin2∏ = 5 (0) è -3.53 |
(c)
Total number of petals is 3.
(d) r = 5 sin 3θ
Zeroes of r :
5 sin 3θ = 0
sin 3θ = 0
3θ = sin-1(0)
3θ = 0, ∏, 2∏, 3∏,.......
θ = n∏/3 and n ∈ Z (integer)
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