STPM-学校功课討論区

dunwan2tellu

quadratic 那题就 (4 - sqrt[7])/3 = 0.351 ...

let cosh^-1 y = t ，那么 cosh t = y

所以 sinh^-1 x = 2t => x = sinh 2t = 2sinh t cosh t = 2sqrt[y^2-1] * y

square both side 就得到答案

注意到： cosh^2 t - sinh^2 t = 1

img3nius

有什么网址可以找得到完整的note吗?
因为目前我只有部分的note
而且是马来文的 有些是跟人拿来的手抄笔记....
(我是persendirian的)

zipp_882000

O 8)A particle P passes a point with position vector (3i-2j) and its velocity is (i+j).At the same time ,a particle Q which is moving at a constant velocity of (4i-2j) passes the point with position velocity vector (i+4j) >Find the displacement of P relative to Q ar time t second and determine the time before both particle are at their closest distance to each other .Calculate the closest distance.(velocity is measured in ms^-1 and distance in m)
thx


12) two particle A and B ,start to move at the same time from the point with position vectors 3i+4j and 8i+4j respectively .The particles move with constant velocity and the velocity of B is given by i+4j.

a)        If the velocity of A is 5i+2j, show that the closest distance between A and B is sqrt5 m.
b)        If the speed of A is 5 ms^-1 and the particle collide in the movement that follows ,find the direction of the movement of A.

13)The engine of a ship ,A has broken down and the ship drift eastward at a speed of 3 knots At 11 am a rescue boat leaves a port which is situated 40 nautical miles south west of ship A.If the rescue boat sails at speed of 10 knots what is the bearing in which the ship should sail so that it can overtake ship A and when will the overtaking occur ?

14)A ship is sailing on the baering of 210 degree at a speed 12 knots,A patrol boat is sailing on a bearing 315 degree .when patrol boat is 1 nautical mile south of the ship,the patrol boat and the ship are both sailing in the path of collision.what is the speed of the patrol ?If both the ship and the patrol boat do not change course how long does it take before the collision occurs.

15) At 1200 hours ,a submatine is traveling northwards at a speed of 10 knots .At that moment the submarine is 15 nautical miles north –east of a destroyer ,that can travel with a maximum speed of 30 knots .Find the direction in which the destroyer should be steered so that it will collide with the submarine in the shortest time ,calculate
a) the time whne the destroyer collide with the submarine
b)the distance between the destroyer and the submarine at 1215 hours

dunwan2tellu

8)A particle P passes a point with position vector (3i-2j) and its velocity is (i+j).At the same time ,a particle Q which is moving at a constant velocity of (4i-2j) passes the point with position velocity vector (i+4j) >Find the displacement of P relative to Q ar time t second and determine the time before both particle are at their closest distance to each other .Calculate the closest distance.(velocity is measured in ms^-1 and distance in m)

OP =3i - 2j , OQ = i + 4j ==>
At time t , P is at (3i - 2j) + (i + j)t = (3+t)i + (t -2)j
At time t , Q is at (i + 4j) + t(4i - 2j) = (1+4t)i + (4-2t)j
Displacement of P relative to Q :  [(3+t)i + (t -2)j ] - [(1+4t)i + (4-2t)j]
= (2 -3t)i + (3t -6)j ........(1)

要找 closet distance , 就表示 (1) 的 magnitude 要 minimum . 也就是说

sqrt[(2-3t)^2 + (3-6t)^2] 要 minimum . squareroot 里面是 quadratic ,要把它 minimize 的话可以用 completing the square 或 dy/dx

得到 time , t 后自然 得到 distance = ?

12) two particle A and B ,start to move at the same time from the point with position vectors 3i+4j and 8i+4j respectively .The particles move with constant velocity and the velocity of B is given by i+4j.

a)        If the velocity of A is 5i+2j, show that the closest distance between A and B is sqrt5 m.
b)        If the speed of A is 5 ms^-1 and the particle collide in the movement that follows ,find the direction of the movement of A.

a) 可以用类是 第8题的做法（也就是把他写成 time t 的 format 然后 completing the square).不然就用 relative velocity 的方法。

V_A = 5i + 2j , V_B = i + 4j
Relative velocity of A to B = V_A - V_B = 4i - 2j
Relative position of A to B = OA - OB = (3i+4j) - (8i+4j) = -5i

之后画vector diagram ,像一个 right angle triangle ,hypotenus = 5 , angle from x axis = Q , 那么

|AB| = 5 , shortest distance = |AB|sin Q = sqrt5
where tan Q = 1/2


b) 要 collide 的话，Relative velocity of A to B 必须 parallel to Relative position of A to B . 因为 A = 5ms-1 , 可以设 V_A = 5 cos Q i + 5 sin Q i
那么 V_A - V_B = (5 cos Q - 1)i + (5 sin Q - 4)j
OA - OB = -5i
所以 5 sin Q - 4 = 0 ==> sin Q = 4/5 , Q = ...

13)The engine of a ship ,A has broken down and the ship drift eastward at a speed of 3 knots At 11 am a rescue boat leaves a port which is situated 40 nautical miles south west of ship A.If the rescue boat sails at speed of 10 knots what is the bearing in which the ship should sail so that it can overtake ship A and when will the overtaking occur ?

overtake = collide
这体相等于找 collide 的时候的时间。

method 1 ) 画 triangle , 3 个 vertices 分别是 rescue boat（original position) , ship A(original position) , collide  point 的所在地 .

Rescue boat --> ship A = 40 nau
ship A --> collide point = 3t nau  (t = time)
rescue boat --> collide point = 10t nau (t = time)

然后用 cosine rule 来找 t （注意到 angle at ship A = 45 + 90 = 135 度)

method 2 : 因为 relative velocity of rescue boat to ship A 必须和 relative position of ship A & rescue boat parallel , 所以 他的 relative velocity 应该是 north east direction .也就是说它的 relative velocity 应该是 p i + p j 的 pattern .

又因为 relative velocity = (x-3)i + y j    (where xi + yj = velocity of rescue boat)

所以 x-3 = p , y = p ==> x-3 = y

那么 velocity of rescue boat = xi + (x-3)j
他的 magnitude = 10 ==> x^2 + (x-3)^2 = 100 ...... x = ...

[ 本帖最后由 dunwan2tellu 于 19-6-2007 12:04 PM 编辑 ]

dunwan2tellu

14)A ship is sailing on the baering of 210 degree at a speed 12 knots,A patrol boat is sailing on a bearing 315 degree .when patrol boat is 1 nautical mile south of the ship,the patrol boat and the ship are both sailing in the path of collision.what is the speed of the patrol ?If both the ship and the patrol boat do not change course how long does it take before the collision occurs.

画一个 115,30,45 度 的 triangle , 然后用 sine rule 分别找出它们的 velocity
第二 part 同样用 sine rule 分别找出它们的 distance ,然后用 time = distance/velocity 来继续

15) At 1200 hours ,a submatine is traveling northwards at a speed of 10 knots .At that moment the submarine is 15 nautical miles north –east of a destroyer ,that can travel with a maximum speed of 30 knots .Find the direction in which the destroyer should be steered so that it will collide with the submarine in the shortest time ,calculate
a) the time whne the destroyer collide with the submarine
b)the distance between the destroyer and the submarine at 1215 hours

先画一个 vector diagram , 用 sine /cosine rule 分别找出所有的 velocity 和 angle . 之后就可以找到他们的 distance 和 time

img3nius

bezakan terhadap x

tanh^-1 (kos x)
sinh^-1 (tan x)
tanh^-1 (sin x)

这一类的问题要怎样做?

dunwan2tellu

原帖由 img3nius 于 19-6-2007 12:17 PM 发表
bezakan terhadap x

tanh^-1 (kos x)
sinh^-1 (tan x)
tanh^-1 (sin x)

这一类的问题要怎样做?

比较“清楚”的做法是

let tanh^-1(kos x) = y , then tanh y = cos x

differentiate both side , sech^2 y dy/dx = -sin x

=> dy/dx = - sin x/sech^2 y = -sin x/(1 - cos^2 x)     因为 1 - tanh^2 y = sech^2 y

img3nius

可是答案是 - kosek x?
是答案有问题吗?

dunwan2tellu

原帖由 img3nius 于 19-6-2007 01:02 PM 发表
可是答案是 - kosek x?
是答案有问题吗?

当然没问题 因为  1 - cos^2 x = sin^2 x

所以 -sin x/(1- cos^2 x) = -1/sin x = -cosec x

img3nius

-(x^2 + y) = dy
  y^2 + x    dx

d^2y
dx^x = 什么?

img3nius

不好意思 我看懂了
原来把 tanh x  = cos x
substitube 回去
刚没注意到 哈哈

[ 本帖最后由 img3nius 于 19-6-2007 01:51 PM 编辑 ]

dunwan2tellu

原帖由 img3nius 于 19-6-2007 01:38 PM 发表
-(x^2 + y) = dy
  y^2 + x    dx

d^2y
dx^x = 什么?

我重写成

-(x^2 + y) = dy/dx * (y^2 + x)

differentiate both side

-(2x + dy/dx) = d^2y/dx^2 * (y^2 + x) + dy/dx * (2y dy/dx + 1)

d^2y/dx^2 = [-(2x^2 + dy/dx) - dy/dx * (2y dy/dx + 1)]/(y^2 + x)
          = [-2x^2 - 2dy/dx - 2y(dy/dx)^2]/(y^2 + x)

[ 本帖最后由 dunwan2tellu 于 19-6-2007 02:48 PM 编辑 ]

img3nius

书上给的答案是
2xy/(y^2 + x)^3
我把算式简化 好像拿不到一样的答案


sin x sin y / cos x cos y  可以写成  tan x tan y  对吗?

dunwan2tellu

原帖由 img3nius 于 19-6-2007 02:21 PM 发表
书上给的答案是
2xy/(y^2 + x)^3
我把算式简化 好像拿不到一样的答案


sin x sin y / cos x cos y  可以写成  tan x tan y  对吗?

有些小错误，不过我也没得到你给的答案。算出来的东西很复杂，不能 further simplified.

trigo 的问题：是的。

img3nius

算不到没关系
不过还是谢谢你的算式
至少我知道这一类的问题该怎么解了

img3nius

请问
1. buktikan bahawa pada titik persilangan lengkunglengkung  y = 1 + kosh x dan y = e^x , sinh x = 1


2 . dapatkan koordinatkoordinat titik penukaran pada lengkung y = 4x - 5sinh^-1 x dan lakarkan lengkung itu


这样的问题 在问什么?我看不明白 又应该怎样做???

hamilan911

原帖由 img3nius 于 20-6-2007 01:21 PM 发表
请问
1. buktikan bahawa pada titik persilangan lengkunglengkung  y = 1 + kosh x dan y = e^x , sinh x = 1


2 . dapatkan koordinatkoordinat titik penukaran pada lengkung y = 4x - 5sinh^-1 x  ...

1。
1 + （e^x + e^-x)/2 = e^x
简化得 e^x - 2 - e^-x = 0
sinhx = (e^x - e^-x)/2 = 2/2 = 1

2。先derive一下 differentiate arc sinhx 等于什么
   let arc sinh x = A
   sinhA = x
   coshA dA/dx = 1
   dA/dx = 1/coshA = 1/sqrt[1+x^2]

所以 dy/dx = 4 - 5/sqrt[1+x^2]
titik penukaran -> dy/dx = 0
所以 4 - 5/sqrt[1+x^2] = 0
sqrt[1+x^2] = 5/4
x^2 = 9/16
x = 3/4，-3/4
d2y/dx2 = 5x/[1+x^2]^(3/2)
分别把 3/4 和 -3/4放进去就可知道哪个是max and min，再找出它的koordinat

zxteh

1.)Define the congruences a=b(mod m)
Solve each of the congruences x^3=2(mod 3) and x^3=2(mod 5),
Deduce the set of positive integers which satisfy both the
congruences.

Hence, find the positive integers x and y which satisfy the
equation  x^3+15xy=12152

Leong13

1. solve the differential equation dy/dx = y(y+1)/x(x+1), given x=1 and y=2
2. solve the differential equation ( 1+cos2x) dy/dx - (1+e^y)sin2x = 0, given that y=0 and      
   x=180/4

hamilan911

原帖由 Leong13 于 10-7-2007 09:39 PM 发表
1. solve the differential equation dy/dx = y(y+1)/x(x+1), given x=1 and y=2
2. solve the differential equation ( 1+cos2x) dy/dx - (1+e^y)sin2x = 0, given that y=0 and      
   x=180/4

1.
int 1/y(y+1) dy = int 1/x(x+1) dx
int [1/y - 1/(y+1)] dy = int [1/x - 1/(x+1)] dx
lny -  ln(y+1) =  lnx - ln(x+1) + c
ln [y/(y+1)] = ln[x/(x+1)] +c
x=1,y=2
ln2/3 = ln1/2 + c
得 c = ln2/3 - ln1/2 = ln4/3

所以 ln[y/(y+1)] = ln[4x/3(x+1)]
y/(y+1) = 4x/3(x+1)
3xy+3y = 4xy+4x
3y-xy = 4x
y(3-x) = 4x
y = 4x/(3-x)

2.
int 1/(1+e^y) dy = int 2sinxcosx/(2cos^2 x) dx
int [1 -e^y/(1+e^y)] dy = int sinx/cosx dx
y - ln(1+e^y) = - lncosx + c
y=0,x=45
-ln2 = -ln[1/sqrt2] +c
c=ln[1/2sqrt2]
e^y/(1+e^y) = secx/2sqrt2
e^y(2sqrt2-secx) = secx
e^y = secx/(2sqrt2-secx)
y = ln [secx/(2sqrt2-secx)]