The following exercises deal with Fresnel integrals. 252. [T] An equivalent formula for the period of a pendulum with amplitude θ max is T ( θ max ) = 2 2 L g ∫ 0 θ max d θ cos θ − cos ( θ max ) where L is the pendulum length and g is the gravitational acceleration constant. When θ max = π 3 we get 1 cos t − 1 / 2 ≈ 2 ( 1 + t 2 2 + t 4 3 + 181 t 6 720 ) . Integrate this approximation to estimate in terms of L and g . Assuming g = 9.806 meters per second squared, find an approximate length L such that T ( π 3 ) = 2 seconds.
The following exercises deal with Fresnel integrals. 252. [T] An equivalent formula for the period of a pendulum with amplitude θ max is T ( θ max ) = 2 2 L g ∫ 0 θ max d θ cos θ − cos ( θ max ) where L is the pendulum length and g is the gravitational acceleration constant. When θ max = π 3 we get 1 cos t − 1 / 2 ≈ 2 ( 1 + t 2 2 + t 4 3 + 181 t 6 720 ) . Integrate this approximation to estimate in terms of L and g . Assuming g = 9.806 meters per second squared, find an approximate length L such that T ( π 3 ) = 2 seconds.
The following exercises deal with Fresnel integrals.
252. [T] An equivalent formula for the period of a pendulum with amplitude
θ
max
is T(
θ
max
) =
2
2
L
g
∫
0
θ
max
d
θ
cos
θ
−
cos
(
θ
max
)
where L is the pendulum length and g is the gravitational acceleration constant. When
θ
max
=
π
3
we get
1
cos
t
−
1
/
2
≈
2
(
1
+
t
2
2
+
t
4
3
+
181
t
6
720
)
. Integrate this approximation to estimate in terms of L and g. Assuming g = 9.806 meters per second squared, find an approximate length L such that
T
(
π
3
)
= 2 seconds.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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Another term/word that can be used synonymously to
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gradient.
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Common difference →
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What is the inequality sign that represents "at most"?
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