AP Calculus BC Quiz
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PARAMETRIC, POLAR, AND VECTOR FUNCTIONS
Question 1
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For
x
=
cos
(
2
t
)
x=\cos(2t)
x
=
cos
(
2
t
)
and
y
=
sin
(
3
t
)
y=\sin(3t)
y
=
sin
(
3
t
)
on
0
≤
t
≤
π
0\le t\le\pi
0
≤
t
≤
π
, which integral sets up arc length?
A
∫
0
π
(
−
2
sin
(
2
t
)
)
2
+
(
3
cos
(
3
t
)
)
2
d
t
\displaystyle \int_{0}^{\pi}\sqrt{(-2\sin(2t))^2+(3\cos(3t))^2}\,dt
∫
0
π
(
−
2
sin
(
2
t
)
)
2
+
(
3
cos
(
3
t
)
)
2
d
t
B
∫
0
π
(
2
cos
(
2
t
)
)
2
+
(
3
sin
(
3
t
)
)
2
d
t
\displaystyle \int_{0}^{\pi}\sqrt{(2\cos(2t))^2+(3\sin(3t))^2}\,dt
∫
0
π
(
2
cos
(
2
t
)
)
2
+
(
3
sin
(
3
t
)
)
2
d
t
C
∫
0
π
(
−
2
cos
(
2
t
)
)
2
+
(
3
sin
(
3
t
)
)
2
d
t
\displaystyle \int_{0}^{\pi}\sqrt{(-2\cos(2t))^2+(3\sin(3t))^2}\,dt
∫
0
π
(
−
2
cos
(
2
t
)
)
2
+
(
3
sin
(
3
t
)
)
2
d
t
D
∫
0
π
(
cos
(
2
t
)
)
2
+
(
sin
(
3
t
)
)
2
d
t
\displaystyle \int_{0}^{\pi}\sqrt{(\cos(2t))^2+(\sin(3t))^2}\,dt
∫
0
π
(
cos
(
2
t
)
)
2
+
(
sin
(
3
t
)
)
2
d
t
E
∫
0
π
(
−
2
sin
(
2
t
)
)
2
+
(
3
sin
(
3
t
)
)
2
d
t
\displaystyle \int_{0}^{\pi}\sqrt{(-2\sin(2t))^2+(3\sin(3t))^2}\,dt
∫
0
π
(
−
2
sin
(
2
t
)
)
2
+
(
3
sin
(
3
t
)
)
2
d
t
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