中五附加數微分應用一問

s/6 = (s-1)/h
6(s-1) =sh
6s -6 =sh
s=6/ (6-h)

s= 6/ (6-10t+5t^2)

ds/dt =0 當 s是最大或最細
ds/dt = d (6)(6-10t+5t^2)^(-1) / dt =0
(-6)(-10+10t)=0
60-60t=0
t=1

要測試 s 是最大, 要再 d 一次,
d^2 s/dt^2 = d (60-60t) /dt = -60 <0
因為d完 < 0, 所以 s 是最大, 當 t=1

2007-11-19 14:57:33 補充：
ds/dt =d (6)(6-10t┼5t^2)^(-1) /dt =0=d(6)(6-10t ┼5t^2)^(-1) / d(6-10t┼5t^2) * d(6-10t┼5t^2)/ dt =(-6)(6-10t┼5t^2)(-2) *(-10┼10t)=60(1-t) / (6-10t┼5t^2)^(2)60(1-t) =0 是最大或最細t=1

2007-11-19 14:58:03 補充：
再d一次d^2 s/dt^2 = d (60) (1-t) (6-10t┼5t^2)^(-2) / dt=(60)(1-t) [d (6-10t┼5t^2)^(-2) /dt] (60)(6-10t┼5t^2)^(-2) [d (1-t) /dt]=(60)(1-t)(-2)(6-10t┼5t^2)^(-3)(-10 10t) (60)(-1)(6-10t┼5t^2)^(-2)=0 (-60) (6-10┼5)^(-2) = -60 因為　-60

2007-11-19 15:01:45 補充：
因為　-60 細過 0, t=1 是最大s = 6/( 6-10┼ 5)s =6

2007-11-20 12:36:42 補充：
*****修 改**********再d一次d^2 s/dt^2 = d (60) (1-t) (6-10t┼5t^2)^(-2) / dt=(60)(1-t) [d (6-10t┼5t^2)^(-2) /dt]┼ (60)(6-10t┼5t^2)^(-2) [d (1-t) /dt]=(60)(1-t)(-2)(6-10t┼5t^2)^(-3)(-10┼ 10t) (60)(-1)(6-10t┼5t^2)^(-2)=0 (-60) (6-10┼5)^(-2) = -60 因為　-60 細過 0, t=1 是最大

h/6 = (s-1)/s (similar triangle)
h = 6(s-1)/s
So, 6(s-1)/s = 10t – 5t²
6s-6=10st-5st²
5st²-10st+6s=6
s = 6/(5t²-10t+6)
When 5t²-10t+6 is minimum, s is maximum.
5t²-10t+6 = 5(t-1)²+1
So, 5t²-10t+6 has a minimum value 1. (when t=1)
maximum of s = 6/1 = 6.

2007-11-24 15:19:31 補充：
少你又話簡單 多你又話複雜
最諷刺就係最佳回答竟然比我們的還要長十倍..更複離
我懷疑這究竟是不是你的分身...

(s-1)/s = h/6
s = 6/(6-h)
= 6/(6-10t+5t^2)
t = 1 時, 6-10t+5t^2 最細, 即 s 最大
t = 1 時, 6-10t+5t^2 = 1
s 的最大值 = 6 m