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Proof that area of odd-petal rose curve is the same as a circle with the petal length as its diameter.
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^^^ Proof that this is a circle with radius 0.5, with its origin at (0, 0.5):
^^^ Proof that this is a circle with radius 0.5, with its origin at (0, 0.5):
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4
Expression 12: "r" equals sine left parenthesis, "n" theta , right parenthesis
r
=
s
i
n
n
θ
domain \theta Minimum: 0
0
less than or equal to theta less than or equal to
≤
θ
≤
domain \theta Maximum: pi
π
12
Expression 13: "n" equals 3
n
=
3
1
1
11
1
1
13
Expression 14: "t" equals 1
t
=
1
0
0
pi
π
14
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23
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Any area highlighted in orange is the derivative of the highlighted shape's area with respect to
"t"
t
.
Notice(using the
"S"
S
slider) how the orange area can be distributed in parts to show how the blue or red area changes with respect to
"t"
t
.
The total area of the blue area, using that
pi "r" squared
π
r
2
formula, is
StartFraction, pi Over 4 , EndFraction
π
4
, so that is the area of an odd-petal rose curve, as shown by this model.
(It is double,
StartFraction, pi Over 2 , EndFraction
π
2
, for even-petal rose curves.)
O
"x"
x
"y"
y
"a" squared
a
2
"a" Superscript, "b" , Baseline
a
b
7
7
8
8
9
9
over
÷
functions
(
(
)
)
less than
<
greater than
>
4
4
5
5
6
6
times
×
| "a" |
|
a
|
,
,
less than or equal to
≤
greater than or equal to
≥
1
1
2
2
3
3
negative
−
A B C
StartRoot, , EndRoot
pi
π
0
0
.
.
equals
=
positive
+
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