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標題: [A.Math.] 關於 limit and derivative [打印本頁]

作者: 細薯 時間: 10/6/2005 04:30 PM
標題: [A.Math.] 關於 limit and derivative
我有點東東想問．．

1) 幾時先可以斷定limit does not exist ?

2) derivative 係咩? (我指 derivative ge meaning ) 有冇實質的東西是derivative ?

3) dy / dx , dy & dx 分別代表咩?

作者: 小耀 時間: 10/6/2005 07:44 PM

Originally posted by smallpotato226 at 2005-6-10 04:30 PM:
我有點東東想問．．

1) 幾時先可以斷定limit does not exist ?

2) derivative 係咩? (我指 derivative ge meaning ) 有冇實質的東西是derivative ?

3) dy / dx , dy & dx 分別代表咩?

y=2x^2+3x-4
d咗佢...那dy就係叫你d咗2x^2+3x-4,而dx係因為y=嘅嘢係乜嘢x,所以dx...
所以,u=6t^2+3t+8就會係du/dt

作者: 細薯 時間: 10/6/2005 08:05 PM

Originally posted by 小耀 at 2005-6-10 07:44 PM:

y=2x^2+3x-4
d咗佢...那dy就係叫你d咗2x^2+3x-4,而dx係因為y=嘅嘢係乜嘢x,所以dx...
所以,u=6t^2+3t+8就會係du/dt

係咪：

d y / d x
即係在 y = f(x) 的 equation 內，將 independent variable x 的值， tend to a centain value ?

作者: 抹茶可樂 時間: 10/6/2005 09:00 PM
1) LIMIT tends to infinity then it is said to be "not exist"
e.g.
lim　(x^2+1)/x
x→0
=lim　(x+1/x)
x→0
=lim　x　+　lim　1/x
x→0　　　　x→0
=0 +　lim　1/x
　　　x→0
Not exist as 1/x→∞

2) Derivative is also a function for SLOPE of a function
e.g.
dy/dx=d(x^2)/dx=2x
To find the slope at P(3,9)
put P into 2x
d(x^2)/dx|(3,9)
=2(3)
=6
We say the slope of y=x^2 at P(3,9) is 6

Derivative of a horizontal line is always 0 (dk/dx=0)
In mechanics,
s=ut+(1/2)at^2
we know that the slope of the graph means the velocity of motion
in fact we can use differentiation to find the equation for v
v
=ds/dt
=lim　Δs/Δt
Δt→0
=lim　[u(t+Δt) +(1/2)a(t+Δt)^2 -ut -(1/2)at^2] /Δt
Δt→0
=lim　[ut+uΔt +(1/2)at^2 +atΔt +(1/2)a(Δt)^2 -ut -(1/2)at^2] /Δt
Δt→0
=lim　[uΔt +atΔt +(1/2)a(Δt)^2] /Δt
Δt→0
=lim　[u +at +(1/2)aΔt]
Δt→0
=u+at

You also know that the equation is for UNIFORMED accelerated motion,then,
again,the slope of the graph means the acceleration of motion
v=u+at represents a straight line
Derivative of a straight line is a constant
for y=mx+c
the derivative is m

3)
dy=　lim　Δy
　　Δy→0
dx=　lim　Δx
　　Δx→0
dy/dx can also be written as d/dx(y)

作者: 細薯 時間: 10/6/2005 09:25 PM
THX AR~~
Question 1 仲有少少問題..

1) LIMIT tends to infinity then it is said to be "not exist"
e.g.
lim　(x^2+1)/x
x→0
=lim　(x+1/x)
x→0
=lim　x　+　lim　1/x
x→0　　　　x→0
=0 +　lim　1/x
　　　x→0
Not exist as 1/x→∞

有時條數，我唔知計到哪steps 才可斷定佢does not exist，怎樣才知到哪steps ??

作者: 抹茶可樂 時間: 10/6/2005 09:32 PM

Originally posted by smallpotato226 at 2005-6-10 09:25 PM:
THX AR~~
Question 1 仲有少少問題..

有時條數，我唔知計到哪steps 才可斷定佢does not exist，怎樣才知到哪steps ??

簡化後剩返下面的野無其他野，或者有幾個相加但全部都係下面的野
就 not exist

lim　k/x^n
x→0

lim　kx^n
x→∞

lim　kx^n
x→-∞

(for all rational values of n)

[ Last edited by 落雷 on 2005-6-10 at 09:34 PM ]

作者: 細薯 時間: 10/6/2005 09:43 PM
red >> 我未教diffrentiation....只係教左first principle...不過老師教講過derivative of y = x^2 係 2x..

green >> 即係用 differentiation , s=ut+(1/2)at^2 可以prove 到 v=u+at ?

blue >> 未係太明.....不過都明明地...

2) Derivative is also a function for SLOPE of a function
e.g.
dy/dx=d(x^2)/dx=2x
To find the slope at P(3,9)
put P into 2x
d(x^2)/dx|(3,9)
=2(3)
=6
We say the slope of y=x^2 at P(3,9) is 6

Derivative of a horizontal line is always 0 (dk/dx=0)
In mechanics,
s=ut+(1/2)at^2
we know that the slope of the graph means the velocity of motion
in fact we can use differentiation to find the equation for v
v
=ds/dt
=lim　Δs/Δt
Δt→0
=lim　[u(t+Δt) +(1/2)a(t+Δt)^2 -ut -(1/2)at^2] /Δt
Δt→0
=lim　[ut+uΔt +(1/2)at^2 +atΔt +(1/2)a(Δt)^2 -ut -(1/2)at^2] /Δt
Δt→0
=lim　[uΔt +atΔt +(1/2)a(Δt)^2] /Δt
Δt→0
=lim　[u +at +(1/2)aΔt]
Δt→0
=u+at

You also know that the equation is for UNIFORMED accelerated motion,then,
again,the slope of the graph means the acceleration of motion
v=u+at represents a straight line
Derivative of a straight line is a constant
for y=mx+c
the derivative is m

作者: 細薯 時間: 10/6/2005 09:54 PM

Originally posted by 落雷 at 2005-6-10 09:32 PM:

簡化後剩返下面的野無其他野，或者有幾個相加但全部都係下面的野
就 not exist

lim　k/x^n
x→0

lim　kx^n
x→∞

lim　kx^n
x→-∞

(for all rational values of n)

如果係下面le D le?

1)
lim square root(x - 3) / (x^2 - 9 )
x→3

2)
lim　 x^3 /( 2x + x^2 )
x→∞

3)
lim　2x^2 + 3x + 1 / (x + 1)
x→∞

作者: 抹茶可樂 時間: 10/6/2005 10:01 PM

Originally posted by smallpotato226 at 2005-6-10 09:43 PM:
red >> 我未教diffrentiation....只係教左first principle...不過老師教講過derivative of y = x^2 係 2x..

green >> 即係用 differentiation , s=ut+(1/2)at^2 可以prove 到 v=u+at ?

blue >> 未係太明.....不過都明明地...

red:
用diffrentiation同埋first principle做出來的結果無可能唔係一樣的
因為diffrentiation是大大簡化了first principle的步驟
分別於
用diffrentiation不會有 lim 符號
用first principle 就是那公式
所以你可以用first principle找出d(x^2)/dx
而first principle一定要記的
since份CE卷會有4-5分問你點樣做first principle

green:上面已經用first principle證明了。
1987年有一條LQ係prove law of refraction (sinθ1/sinθ2=u/v)

blue:用中文講一次
一條直線的slope在任何一點都固定的
y=mx+c,v=u+at都是直線
那麼可以證明s=ut+(1/2)at^2係a是常數先用到

作者: 抹茶可樂 時間: 10/6/2005 10:09 PM

Originally posted by smallpotato226 at 2005-6-10 09:54 PM:

如果係下面le D le?

1)
lim square root(x - 3) / (x^2 - 9 )
x→3

2)
lim　 x^3 /( 2x + x^2 )
x→∞

3)
lim　2x^2 + 3x + 1 / (x + 1)
x→∞

lim √(x - 3) / (x - 3)(x + 3)
x→3

Let y=x-3 then y→0

lim √y / y(y+6)
y→0

=lim √y / 　lim y　/　 lim　(y+6)
y→0　　　　　y→0　　　　y→0

= 0/ 　lim y　/　 6
　　　　y→0　　　

not exist

其他的自己試下啦
if x→∞
x+2005→∞ and x-1→∞

作者: 細薯 時間: 10/6/2005 10:24 PM

Originally posted by 落雷 at 2005-6-10 10:01 PM:
since份CE卷會有4-5分問你點樣做first principle

blue:用中文講一次
一條直線的slope在任何一點都固定的
y=mx+c,v=u+at都是直線
那麼可以證明s=ut+(1/2)at^2係a是常數先用到

final exam (next week) ar Sir 話用differentiation 做 (no mark)...XD
訓練我地用first principle....

即係 acceleration 一定係constant 先用到?

[ Last edited by smallpotato226 on 2005-6-10 at 10:25 PM ]

作者: 抹茶可樂 時間: 10/6/2005 10:29 PM

Originally posted by smallpotato226 at 2005-6-10 10:24 PM:

final exam (next week) ar Sir 話用differentiation 做 (no mark)...XD
訓練我地用first principle....

即係 acceleration 一定係constant 先用到?

[ Last edited by smallpotato226 on 2005-6-10 at ...

differentiation 做 (no mark)...<+因為俾用的話會有人屈機屈時間嘛

即係 acceleration 一定係constant 先用到? <+physics書講了

作者: SusanCrystal 時間: 14/6/2005 10:56 PM
就咁differentiate的話, 教個小學生都識做啦..."(佢知咩叫index, 就好快識做d基本differentiation)
a.maths考試又點會咁益你

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