微積分章節，相關變率問題，請問圈起來的題目如何解?(過程)?

(a)
正弦函數的週期 = 2π
2π = π t / 6
t = 12

(b)
因為 x(t) = (1/2)*sin( π t / 6 )
所以, x 的最大值 = (1/2)*1 = 1/2
因為 x^2 + y^2 = 1^2 , x > 0 , y > 0
所以, 當 x 達到最大, y 則達到最小.
故 y 的最小值 = √(1^2 - x^2) = √[1 - (1/2)^2] = √(3/4) = √3 / 2

(c)
y^2 = 1^2 - x^2
等式兩邊對 t 微分, 等式恆成立:
2y dy/dt = - 2x dx/dt
y dy/dt = - x dx/dt ..... (1)

y = √(1^2 - (1/4)^2) = √(15/16) = √15 / 4

x = (1/2)*sin( π t / 6 ) = 1/4
sin( π t / 6 ) = 1/2
π t / 6 = π / 6
t = 1

dx/dt
= d [ (1/2)*sin( π t / 6 ) ] / dt
= (1/2)(π/6)cos( π t / 6 )
= (π/12)cos( π*1 / 6 )
= (π/12) * √3 / 2
= √3 π / 24

將以上數值代入(1)式:
y dy/dt = - x dx/dt
( √15 / 4 )(dy/dt) = - (1/4)( √3 π / 24 )

dy/dt
= - √3 π / ( 24 * √15 )
= - π / ( 24 * √5 )
= - √5 π / ( 24 * 5 )
= - √5 π / 120

y軸端速率
= ∣ - √5 π / 120 ∣
= √5 π / 120

(28a) T = ?

x(t) = 0.5*sin(πt/6)

x(0) = 0

x(3) = 0.5 = max

x(6) = 0

x(9) = -0.5 = min

x(12) = 0 = one cycle

=> T = 12"


(28b) y(min) = ?

弓股弦定理:

y(t)^2 = 1 - x(t)^2 = 1 - 0.25*sin(πt/6)^2

2yy' = -0.5*sin(πt/6)cos(πt/6) = -0.25*sin(πt/3)

Set y' = 0

=> sin(πt/3) = 0

=> t = 0, 3, 6, 9, 12, ...

=> min = y(3) = 1 = √(1 - 0.25) = √3/2



(28c) Vy = ?

x = 1/4 = 0.5*sin(πt/6)

=> sin(πt/6) = 1/2

=> πt/6 = π/6 or 5π/6

=> t = 1" or 5"


y = √(1 - x^2)

= √(1 - 1/16)

= (√15)/4



Vy(1) = y'(1)

= -sin(πt/3)/8y

= -sin(π/3)*4/8√15

= -(√3/2)/2√15

= -1/4√5

= -√5/20




Vy(5) = y'(5)

= -sin(π5/3)/8y

= -sin(-π/3)*4/8√15

= √5/20