F.4 MATHS

h ∝ 1/√k
故此 h = n/√k，其中 n 為常數（變分常數）。

當 k = 64，h = 15 :
15 = n/√64
15 = n/8
n = 120
所以，h = 120/√k

當 k = ko­，h = ho :
ho = 120/√ko ...... [1]

當 h = ho(1 - 18%) :
ho(1 - 18%) = 120/√k
0.82ho = 120/√k ...... [2]

[1]/[2] :
1/0.82 = (√k)/√ko
k/ko = (1/0.82)²
k = (1/0.82)²k­o
k = (2500/1681)ko

k 的百分變化
= {[(2500/1681)ko - ko]/ko} x 100%
= +(81900/1681)%
= +(48又1212/1681)%
(≈ +48.72%)

h=x/k^2 x為非零常數
k^2=x/h
k=√(x/h)
h的新值=(1-18%)h=0.82h
k的百分變化=[√(x/0.82h)-√(x/h)]/√(x/h)*100%
k的百分變化=10.4%