✔ 最佳答案
dθ / dt = ( degree for 1 rotation ) / ( period for 1 rotation ) = 2π / 3
By the pic. below, we have
L = 16 * tan θ
dL / dt
= 16 * d tan θ / dt
= 16 * sec^2 θ * ( dθ / dt )
= 16 * sec^2 θ * ( 2π / 3 )
= ( 32π / 3 ) * sec^2 θ
When θ = 15°
dL / dt
= ( 32π / 3 ) * sec^2 15°
= ( 32π / 3 ) * ( 1 + tan^2 15° )
= ( 32π / 3 ) * ( 1 + 7 - 4√3 ) ..... see the note below
= ( 32π / 3 ) * ( 8 - 4√3 ) ..... Ans
Note.
tan^2 15° = 7 - 4√3
pf :
By the formula :
tan ( β / 2 ) = ± √ [ ( 1 - cos β ) / ( 1 + cos β ) ]
So ,
tan^2 ( β / 2 ) = ( 1 - cos β ) / ( 1 + cos β )
Let β = 30° , we have
tan^2 15°
= ( 1 - cos 30° ) / ( 1 + cos 30° )
= [ 1 - ( √3 / 2 ) ] / [ 1 + ( √3 / 2 ) ]
= [ ( 2 - √3 ) / 2 ] / [ ( 2 + √3 ) / 2 ]
= ( 2 - √3 ) / ( 2 + √3 )
= [ ( 2 - √3 )( 2 - √3 ) ] / [ ( 2 + √3 )( 2 - √3 ) ]
= ( 2 - √3 )^2
= 7 - 4√3