微積分章節，相關變率問題，請問圈起來的題目如何解?

(a)
令梯腳與房子距離為 B , 梯子與房子接觸點的高度為 H , 則:
B^2 + H^2 = 25^2
H^2 = 625 - B^2
等式兩邊分別對 t 微分, 等式恆成立:
2H dH/dt = - 2B dB/dt
dH/dt = - (B/H) * dB/dt = - (B/H) * 2 = - 2B/H
所求下移速率 v = ∣ dH/dt ∣ = 2B/H
When B = 7 , H = √(25^2 - 7^2) = 24 , v = 2*7/24 = 7/12
When B = 15 , H = 20 , v = 2*15/20 = 3/2
When B = 24 , H = 7 , v = 2*24/7 = 48/7

(b)
令題意之三角形面積為 A
dA/dt
= d [ (1/2)BH ] / dt
= (1/2)( B' H + B H' )
= (1/2)[ 2H + B(-2B)/H ]
= H - B^2/H
= 24 - 7^2/24
= 527/24

(c)
令題意之角度為 θ
tan θ = B/H
等式兩邊分別對 t 微分, 等式恆成立:
sec^2 θ * dθ/dt = ( B' H - B H' ) / H^2
( 25^2 / 24^2 ) * dθ/dt = [ 2*24 - 7*(-7/12) ] / 24^2
dθ/dt = [ 2*24 - 7*(-7/12) ] / 25^2 = ( 48 + 49/12 )/625 = 1/12

Ans:
(a)
當 base = 7 ft , 下移速率為 7/12 ft/sec
當 base = 15 ft , 下移速率為 3/2 ft/sec
當 base = 24 ft , 下移速率為 48/7 ft/sec
(b) 527/24 ft^2/sec
(c) 1/12 rad/sec